text stringlengths 45 10k |
|---|
ur answer. Developing Skills 3. When p ∧ q is true and q ∨ r is true, what is the truth value of r? 4. When p → q is false and q ∨ r is true, what is the truth value of r? 5. When p → q is false, what is the truth value of q → r? 6. When p → q and p ∧ q are both true, what are the truth values of p and of q? 14365C02.pgs 7/9/07 4:40 PM Page 84 84 Logic 7. When p → q and p ∧ q are both false, what are the truth values of p and of q? 8. When p → q is true and p ∧ q is false, what are the truth values of p and of q? 9. When p → q is true, p ∨ r is true and q is false, what is the truth value of r? Applying Skills 10. Laura, Marta, and Shanti are a lawyer, a doctor, and an investment manager. • The lawyer is Marta’s sister. • Laura is not a doctor. • Either Marta or Shanti is a lawyer. What is the profession of each woman? 11. Alex, Tony, and Kevin each have a different job: a plumber, a bookkeeper, and a teacher. • Alex is a plumber or a bookkeeper. • Tony is a bookkeeper or a teacher. • Kevin is a teacher. What is the profession of each person? 12. Victoria owns stock in three companies: Alpha, Beta, and Gamma. • Yesterday, Victoria sold her shares of Alpha or Gamma. • If she sold Alpha, then she bought more shares of Beta. • Victoria did not buy more shares of Beta. Which stock did Victoria sell yesterday? 13. Ren, Logan, and Kadoogan each had a different lunch. The possible lunches are: a ham sandwich, pizza, and chicken pot pie. • Ren or Logan had chicken pot pie. • Kadoogan did not have pizza. • If Logan did not have pizza, then Kadoogan had pizza. Which lunch did each person have? 14. Zach, Steve, and David each play a different sport: basketball, soccer, or baseball. Zach made each of the following true statements. • I do not play basketball. • If Steve does not play soccer, then David plays baseball. • David does not play baseball. What sport does each person play? 14365C02.pgs 7/9/07 4:40 PM Page 85 Chapter Summary 85 15. Taylor, Melissa, and Lauren each study one language: French, Spanish, and Latin. • If Melissa does not study French, then Lauren studies Latin. • If Lauren studies Latin, then Taylor studies Spanish. • Taylor does not study Spanish. What language does each person study? 16. Three friends, Augustus, Brutus, and Caesar, play a game in which each decides to be either a liar or a truthteller. A liar must always lie and a truthteller must always tell the truth. When you met these friends, you asked Augustus which he had chosen to be. You didn’t hear his answer but Brutus volunteered, “Augustus said that he is a liar.” Caesar added, “If one of us is a liar, then we are all liars.” Can you determine, for each person, whether he is a liar or a truthteller? CHAPTER SUMMARY Definitions to Know • Logic is the study of reasoning. • In logic, a mathematical sentence is a sentence that contains a complete thought and can be judged to be true or false. • A phrase is an expression that is only part of a sentence. • An open sentence is any sentence that contains a variable. • The domain or replacement set is the set of numbers that can replace a variable. • The solution set or truth set is the set of all replacements that will change an open sentence to true sentences. • A statement or a closed sentence is a sentence that can be judged to be true or false. • A closed sentence is said to have a truth value, either true (T) or false (F). • The negation of a statement has the opposite truth value of a given statement. • In logic, a compound sentence is a combination of two or more mathematical sentences formed by using the connectives not, and, or, if . . . then, or if and only if. • A conjunction is a compound statement formed by combining two simple statements, called conjuncts, with the word and. The conjunction p and q is written symbolically as p ∧ q. • A disjunction is a compound statement formed by combining two simple statements, called disjuncts, with or. The disjunction p or q is written symbolically as p ∨ q. • A truth table is a summary of all possible truth values of a logic statement. • A conditional is a compound statement formed by using the words if . . . then to combine two simple statements. The conditional if p then q is written symbolically as p → q. 14365C02.pgs 7/9/07 4:40 PM Page 86 86 Logic • A hypothesis, also called a premise or antecedent, is an assertion that begins an argument. The hypothesis usually follows the word if. • A conclusion, also called a consequent, is an ending or a sentence that closes an argument. The conclusion usually follows the word then. • Beginning with a statement (p → q), the inverse (p → q) is formed by negating the hypothesis and negating the conclusion. • Beginning with a statement (p → q), the converse (q → p) is formed by interchanging the hypothesis and the conclusion. • Beginning with a conditional (p → q), the contrapositive (q → p) is formed by negating both the hypothesis and the conclusion, and then interchanging the resulting negation. • Two statements are logically equivalent––or logical equivalents––if they always have the same truth value. • A biconditional (p ↔ q) is a compound statement formed by the conjunc- tion of the conditional p → q and its converse q → p. • A valid argument uses a series of statements called premises that have known truth values to arrive at a conclusion. Logic Statements Negation: p Conjunction: Disjunction: Conditional: Inverse: Converse: Contrapositive: Biconditional not p p and q p or q if p then q if p then q if q then p if q then p p if and only if q The truth values of the logic connectives can be summarized as follows: n tio tio c n nju o C p ∧ q n tio c n Disju al n ditio e s r e v In sitiv al n ditio n o Bic p ↔ q T F F T 14365C02.pgs 7/9/07 4:40 PM Page 87 Laws of Logic • The Law of Detachment states that when p → q is true and p is true, then q must be true. • The Law of Disjunctive Inference states that when p ∨ q is true and p is false, then q must be true. Review Exercises 87 VOCABULARY 2-1 Logic • Truth value • Mathematical sentence • Phrase • Open sentence • Domain • Replacement Set • Solution set • Truth set • Statement • Closed sentence • Truth value (T and F) • Negation 2-2 Compound sentence • Compound statement • Conjunction • Conjunct • p and q • Tree diagram • Truth table 2-3 Disjunction • Disjunct • p or q • Inclusive or • Exclusive or 2-4 Conditional • If p then q • Hypothesis • Premise • Antecedent • Conclusion • Consequent 2-5 Inverse • Converse • Contrapositive • Logical equivalents 2-6 Biconditional 2-7 Laws of logic • Valid argument • Premises • Law of Detachment • Law of Disjunctive Inference REVIEW EXERCISES 1. The statement “If I go to school, then I do not play basketball” is false. Using one or both of the statements “I go to school” and “I do not play basketball” or their negations, write five true statements. 2. The statement “If I go to school, then I do not play basketball” is false. Using one or both of the statements “I go to school” and “I do not play basketball” or their negations, write five false statements. 3. Mia said that the biconditional p ↔ q and the biconditional p ↔ q always have the opposite truth values. Do you agree with Mia? Explain why or why not. In 4 and 5: a. Identify the hypothesis p. b. Identify the conclusion q. 4. If at first you don’t succeed, then you should try again. 5. You will get a detention if you are late one more time. In 6–12, tell whether each given statement is true or false. 6. If July follows June, then August follows July. 7. July follows June and July is a winter month in the northern hemisphere. 14365C02.pgs 7/9/07 4:40 PM Page 88 88 Logic 8. July is a winter month in the northern hemisphere or July follows June. 9. If August follows July, then July does not follow June. 10. July is a winter month if August is a winter month. 11. August does not follow July and July is not a winter month. 12. July follows June if and only if August follows July. 13. Which whole number, when substituted for y, will make the following sen- tence true? (y 5 9) ∧ (y 6) In 14–17, supply the word, phrase, or symbol that can be placed in each blank to make the resulting statement true. 14. (p) has the same truth value as ________. 15. When p is true and q is false, then p ∧ q is _______. 16. When p ∨ q is false, then p is _______ and q is _______. 17. If the conclusion q is true, then p → q must be _______. In 18–22, find the truth value of each sentence when a, b, and c are all true. 18. a 20. b → c 21. a ∨ b 19. b ∧ c 22. a ↔ b In 23–32, let p represent “x 5,” and let q represent “x is prime.” Use the domain {1, 2, 3, 4, . . . , 10} to find the solution set for each of the following. 23. p 28. p ∧ q 24. p 29. p ∧ q 25. q 30. p → q 26. q 31. p → q 27. p ∨ q 32. p ↔ q 33. For the conditional “If I live in Oregon, then I live in the Northwest,” write: a. the inverse, b. the converse, c. the contrapositive, d. the biconditional. 34. Assume that the given sentences are true. Write a simple sentence that could be a conclusion. • If A is the vertex angle of isosceles ABC, then AB AC. • AB AC 35. Elmer Megabucks does not believe that girls should marry before the age of 21, and he disapproves of smoking. Therefore, he put the following provision in his will: I leave $100,000 to each of my nieces who, at the time of my death, is over 21 or unmarried, and does not smoke. Each of his nieces is described below at the time of Elmer’s death. Which nieces will inherit $100,000? • Judy is 24, married, and smokes. • Diane is 20, married, and does not smoke. 14365C02.pgs 7/9/07 4:40 PM Page 89 Review Exercises 89 • Janice is 26, unmarried, and does not smoke. • Peg is 19, unmarried, and smokes. • Sue is 30, unmarried, and smokes. • Sarah is 18, unmarried, and does not smoke. • Laurie is 28, married, and does not smoke. • Pam is 19, married, and smokes. 36. Some years after Elmer Megabucks prepared his will, he amended the conditions, by moving a comma, to read: I leave |
$100,000 to each of my nieces who, at the time of my death, is over 21, or unmarried and does not smoke. Which nieces described in Exercise 35 will now inherit $100,000? 37. At a swim meet, Janice, Kay, and Virginia were the first three finishers of a 200-meter backstroke competition. Virginia did not come in second. Kay did not come in third. Virginia came in ahead of Janice. In what order did they finish the competition? 38. Peter, Carlos, and Ralph play different musical instruments and different sports. The instruments that the boys play are violin, cello, and flute. The sports that the boys play are baseball, tennis, and soccer. From the clues given below, determine what instrument and what sport each boy plays. • The violinist plays tennis. • Peter does not play the cello. • The boy who plays the flute does not play soccer. • Ralph plays baseball. 39. Let p represent “x is divisible by 6.” Let q represent “x is divisible by 2.” a. If possible, find a value of x that will: (2) make p true and q false. (1) make p true and q true. (3) make p false and q true. (4) make p false and q false. b. What conclusion can be drawn about the truth value of p → q? 40. Each of the following statements is true. • Either Peter, Jim, or Tom is Maria’s brother. • If Jim is Maria’s brother, then Peter is Alice’s brother. • Alice has no brothers. • Tom has no sisters. Who is Maria’s brother? 14365C02.pgs 7/9/07 4:40 PM Page 90 90 Logic Exploration 1. A tautology is a statement that is always true. For instance, the disjunction p ∨ p is a tautology because if p is true, the disjunction is true, and if p is false, p is true and the disjunction is true. a. Which of the following statements are tautologies? (1) p → (p ∨ q) (2) Either it will rain or it will snow. (3) (p ∧ q) → p b. Construct two tautologies, one using symbols and the other using words. 2. A contradiction is a statement that is always false, that is, it cannot be true under any circumstances. For instance, the conjunction p ∧ p is a contradiction because if p is true, p is false and the conjunction is false, and if p is false, the conjunction is false. a. Which of the following statements are contradictions? (1) (p ∨ q) ∧ p (2) It is March or it is Tuesday, and it is not March and it is not Tuesday. (3) (p ∧ p) ∧ q b. Construct two contradictions, one using symbols and the other using words. 3. One way to construct a tautology is to use a contradiction as the hypothesis of a conditional. For instance, since we know that p ∧ p is a contradiction, the conditional (p ∧ p) → q is a tautology for any conclusion q. a. Construct two conditionals using any two contradictions as the premises. b. Show that the two conditionals from part a are tautologies. c. Explain why the method of using a contradiction as the hypothesis of a conditional always results in a tautology. CUMULATIVE REVIEW CHAPTERS 1–2 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following is an undefined term? (3) line (2) angle (1) ray (4) line segment 2. The statement “If x is a prime then x is odd” is false when x equals (1) 1 (2) 2 (3) 3 (4) 4 14365C02.pgs 7/9/07 4:40 PM Page 91 Cumulative Review 91 3. Points J, K, and L lie on a line. The coordinate of J is 17, the coordinate of K is 8, and the coordinate of L is 13. What is the coordinate of M, the midpoint of (1) 8 (3) 2 (2) 4 (4) 2 JL ? 4. When “Today is Saturday” is false, which of the following statements could be either true or false? (1) If today is Saturday, then I do not have to go to school. (2) Today is Saturday and I do not have to go to school. (3) Today is Saturday or I have to go to school. (4) Today is not Saturday. 5. Which of the following is not a requirement in order for point H to be (3) GH HI GI (4) G, H, and I are distinct points. between points G and I? (1) GH HI (2) G, H, and I are collinear. h UW mTUV? (1) 2.5° (2) 32.5° 6. bisects TUV. If mTUV 34x and mVUW 5x 30, what is (3) 42.5° (4) 85° 7. Which of the following must be true when AB BC AC? AB AC (1) B is the midpoint of (2) AB BC (3) C is a point on (4) B is a point on AC 8. Which of the following equalities is an example of the use of the commu- tative property? (1) 3(2 x) 6 3x (2) 3 (2 x) (3 2) x (3) 3 (0 x) 3 x (4) 3(2 x) 3(x 2) 9. Which of the following must be true when p is true? (1) p ∧ q (2) p → q (3) p ∨ q (4) ~p ∨ q 10. The solution set of x 1 x is (2) {0} (3) {–1} (4) {all real numbers} (1) Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 14365C02.pgs 7/9/07 4:40 PM Page 92 92 Logic 11. The first two sentences below are true. Determine whether the third sentence is true, is false, or cannot be found to be true or false. Justify your answer. I win the ring toss game. If I win the ring toss game, then I get a goldfish. I get a goldfish. 12. On the number line, the coordinate of R is 5 and the coordinate of S is 1. What is the coordinate of T if S is the midpoint of RT ? Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Let A and B be two points in a plane. Explain the meanings of the sym- bols g AB h , AB h BD 14. The ray perpendicular to AB , and AB. , is the bisector of ABC, a straight angle. Explain why . Use definitions to justify your explanation. g ABC h BD is Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The steps used to simplify an algebraic expression are shown below. Name the property that justifies each of the steps. a(2b 1) a(2b) a(1) (a 2) b a(1) (2 a) b a(1) 2ab a 16. The statement “If x is divisible by 12, then x is divisible by 4” is always true. a. Write the converse of the statement. b. Write the inverse of the statement. c. Are the converse and inverse true for all positive integers x? Justify your answer. d. Write another statement using “x is divisible by 12” and “x is divisible by 4” or the negations of these statements that is true for all positive integers x. 14365C03.pgs 7/12/07 2:52 PM Page 93 PROVING STATEMENTS IN GEOMETRY After proposing 23 definitions, Euclid listed five postulates and five “common notions.” These definitions, postulates, and common notions provided the foundation for the propositions or theorems for which Euclid presented proof. Modern mathematicians have recognized the need for additional postulates to establish a more rigorous foundation for these proofs. David Hilbert (1862–1943) believed that mathematics should have a logical foundation based on two principles: 1. All mathematics follows from a correctly chosen finite set of assumptions or axioms. 2. This set of axioms is not contradictory. Although mathematicians later discovered that it is not possible to formalize all of mathematics, Hilbert did succeed in putting geometry on a firm logical foundation. In 1899, Hilbert published a text, Foundations of Geometry, in which he presented a set of axioms that avoided the limitations of Euclid. CHAPTER 3 CHAPTER TABLE OF CONTENTS 3-1 Inductive Reasoning 3-2 Definitions as Biconditionals 3-3 Deductive Reasoning 3-4 Direct and Indirect Proofs 3-5 Postulates,Theorems, and Proof 3-6 The Substitution Postulate 3-7 The Addition and Subtraction Postulates 3-8 The Multiplication and Division Postulates Chapter Summary Vocabulary Review Exercises Cumulative Review 93 14365C03.pgs 7/12/07 2:52 PM Page 94 94 Proving Statements in Geometry 3-1 INDUCTIVE REASONING Gina was doing a homework assignment on factoring positive integers. She made a chart showing the number of factors for each of the numbers from 1 to 10. Her chart is shown below. Number Number of factors 10 3 4 She noticed that in her chart, only the perfect square numbers, 1, 4, and 9 had an odd number of factors. The other numbers had an even number of factors. Gina wanted to investigate this observation further so she continued her chart to 25. Number 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Number of factors Again her chart showed that only the perfect square numbers, 16 and 25, had an odd number of factors and the other numbers had an even number of factors. Gina concluded that this is true for all positive integers. Gina went from a few specific cases to a generalization. Was Gina’s conclusion valid? Can you find a counterexample to prove that a perfect square does not always have an odd number of factors? Can you find a non-perfect square that has an odd number of factors? Scientists perform experiments once, twice, or many times in order to come to a conclusion. A scientist who is searching for a vaccine to prevent a disease will test the vaccine repeatedly to see if it is effective. This method of reasoning, in which a series of particular examples leads to a conclusion, is called inductive reasoning. In geometry, we also perform experiments to discover properties of lines, angles, and polygons and to determine geometric relationships. Most of these experiments involve measurements. Because direct measurements depend on the type of instrument used to measure and the care with which the measurement is made, results can be only approximate. This is the first weakness in attempting to reach conclusions by inductive reasoning. Use a ruler to draw a pair of parallel line segments by drawing a line segme |
nt along opposite edges of the ruler. Turn the ruler and draw another pair of parallel line segments that intersect the first pair. Label the intersections of the line segments A, B, C, and D. The figure, ABCD, is a parallelogram. It appears that the opposite sides of the parallelogram have equal measures. Use the ruler to show that this is true. To convince yourself that this relationship is true in other parallelograms, draw other parallelograms and measure their opposite sides. In each experiment you will find that the opposite sides have the same measure. From these ex- C B D A 14365C03.pgs 7/12/07 2:52 PM Page 95 Inductive Reasoning 95 periments, you can arrive at a general conclusion: The opposite sides of a parallelogram have equal measures. This is an example of inductive reasoning in geometry. Suppose a student, after examining several parallelograms, made the generalization “All parallelograms have two acute angles and two obtuse angles.” Here, a single counterexample, such as a parallelogram which is a rectangle and has right angles, is sufficient to show that the general conclusion that the student reached is false. When we use inductive reasoning, we must use extreme care because we are arriving at a generalization before we have examined every possible example. This is the second weakness in attempting to reach conclusions by inductive reasoning. When we conduct an experiment we do not give explanations for why things are true. This is the third weakness of inductive reasoning. In light of these weaknesses, when a general conclusion is reached by inductive reasoning alone, it can at best be called probably true. Such statements, that are likely to be true but not yet been proven true by a deductive proof, are called conjectures. Then why study inductive reasoning? Simply because it mimics the way we naturally make new discoveries. Most new knowledge first starts with specific cases, then, through careful study, to a generalization. Only afterwards, is a proof or explanation usually sought. Inductive reasoning is therefore a powerful tool in discovering new mathematical facts. SUMMARY 1. Inductive reasoning is a powerful tool in discovering and making conjec- tures. 2. Generalizations arising from direct measurements of specific cases are only approximate. 3. Care must be taken when applying inductive reasoning to ensure that all relevant examples are examined (no counterexamples exist). 4. Inductive reasoning does not prove or explain conjectures. Exercises Writing About Mathematics 1. After examining several triangles, Mindy concluded that the angles of all triangles are acute. Is Mindy correct? If not, explain to Mindy why she is incorrect. 2. Use a counterexample to show that the whole numbers are not closed under subtraction. 14365C03.pgs 7/12/07 2:52 PM Page 96 96 Proving Statements in Geometry Developing Skills 3. a. Using geometry software or a pencil, ruler, and protractor, draw three right triangles that have different sizes and shapes. b. In each right triangle, measure the two acute angles and find the sum of their measures. c. Using inductive reasoning based on the experiments just done, make a conjecture about the sum of the measures of the acute angles of a right triangle. 4. a. Using geometry software or a pencil, ruler, and protractor, draw three quadrilaterals that have different sizes and shapes. b. For each quadrilateral, find the midpoint of each side of the quadrilateral, and draw a new quadrilateral using the midpoints as vertices. What appears to be true about the quadrilateral that is formed? c. Using inductive reasoning based on the experiments just done, make a conjecture about a quadrilateral with vertices at the midpoints of a quadrilateral. 5. a. Using geometry software or a pencil, ruler, and protractor, draw three equilateral trian- gles of different sizes. b. For each triangle, find the midpoint of each side of the triangle, and draw line segments joining each midpoint. What appears to be true about the four triangles that are formed? c. Using inductive reasoning based on the experiments just done, make a conjecture about the four triangles formed by joining the midpoints of an equilateral triangle. In 6–11, describe and perform a series of experiments to investigate whether each statement is probably true or false. 6. If two lines intersect, at least one pair of congruent angles is formed. 7. The sum of the degree measures of the angles of a quadrilateral is 360. 8. If a2 b2, then a b. 9. In any parallelogram ABCD, AC BD. 10. In any quadrilateral DEFG, DF bisects D and F. 11. The ray that bisects an angle of a triangle intersects a side of the triangle at its midpoint. 12. Adam made the following statement: “For any counting number n, the expression n2 n 41 will always be equal to some prime number.” He reasoned: • When n 1, then n2 n 41 1 1 41 43, a prime number. • When n 2, then n2 n 41 4 2 41 47, a prime number. Use inductive reasoning, by letting n be many different counting numbers, to show that Adam’s generalization is probably true, or find a counterexample to show that Adam’s generalization is false. 14365C03.pgs 7/12/07 2:52 PM Page 97 Definitions as Biconditionals 97 Applying Skills In 13–16, state in each case whether the conclusion drawn was justified. 13. One day, Joe drove home on Route 110 and found traffic very heavy. He decided never again to drive on this highway on his way home. 14. Julia compared the prices of twenty items listed in the advertising flyers of store A and store B. She found that the prices in store B were consistently higher than those of store A. Julia decided that she will save money by shopping in store A. 15. Tim read a book that was recommended by a friend and found it interesting. He decided that he would enjoy any book recommended by that friend. 16. Jill fished in Lake George one day and caught nothing. She decided that there are no fish in Lake George. 17. Nathan filled up his moped’s gas tank after driving 92 miles. He concluded that his moped could go at least 92 miles on one tank of gas. Hands-On Activity STEP 1. Out of a regular sheet of paper, construct ten cards numbered 1 to 10. STEP 2. Place the cards face down and in order. STEP 3. Go through the cards and turn over every card. STEP 4. Go through the cards and turn over every second card starting with the second card. STEP 5. Go through the cards and turn over every third card starting with the third card. STEP 6. Continue this process until you turn over the tenth (last) card. If you played this game correctly, the cards that are face up when you finish should be 1, 4, and 9. a. Play this same game with cards numbered 1 to 20. What cards are face up when you finish? What property do the numbers on the cards all have in common? b. Play this same game with cards numbered 1 to 30. What cards are face up when you finish? What property do the numbers on the cards all have in common? c. Make a conjecture regarding the numbers on the cards that remain facing up if you play this game with any number of cards. 3-2 DEFINITIONS AS BICONDITIONALS In mathematics, we often use inductive reasoning to make a conjecture, a statement that appears to be true. Then we use deductive reasoning to prove the conjecture. Deductive reasoning uses the laws of logic to combine definitions and general statements that we know to be true to reach a valid conclusion. 14365C03.pgs 7/12/07 2:52 PM Page 98 98 Proving Statements in Geometry Before we discuss this type of reasoning, it will be helpful to review the list of definitions in Chapter 1 given on page 29 that are used in the study of Euclidean geometry. Definitions and Logic Each of the definitions given in Chapter 1 can be written in the form of a conditional. For example: A scalene triangle is a triangle that has no congruent sides. Using the definition of a scalene triangle, we know that: 1. The definition contains a hidden conditional statement and can be rewrit- ten using the words If . . . then as follows: t: A triangle is scalene. p: A triangle has no congruent sides. t → p: If a triangle is scalene, then the triangle has no congruent sides. 2. In general, the converse of a true statement is not necessarily true. However, the converse of the conditional form of a definition is always true. For example, the following converse is a true statement: p → t: If a triangle has no congruent sides, then the triangle is scalene. 3. When a conditional and its converse are both true, the conjunction of these statements can be written as a true biconditional statement. Thus, (t → p) ∧ (p → t) is equivalent to the biconditional (t ↔ p). Therefore, since both the conditional statement and its converse are true, we can rewrite the above definition as a biconditional statement, using the words if and only if, as follows: A triangle is scalene if and only if the triangle has no congruent sides. Every good definition can be written as a true biconditional. Definitions will often be used to prove statements in geometry. EXAMPLE 1 A collinear set of points is a set of points all of which lie on the same straight line. a. Write the definition in conditional form. b. Write the converse of the statement given in part a. c. Write the biconditional form of the definition. 14365C03.pgs 7/12/07 2:52 PM Page 99 Definitions as Biconditionals 99 Solution a. Conditional: If a set of points is collinear, then all the points lie on the same straight line. b. Converse: If a set of points all lie on the same straight line, then the set of points is collinear. c. Biconditional: A set of points is collinear if and only if all the points lie on the same straight line. Exercises Writing About Mathematics 1. Doug said that “A container is a lunchbox if and only if it can be used to carry food” is not a definition because one part of the biconditional is false. Is Doug correct? If so, give a counterexample to show that Doug is correct. 2. a. Give a counterexample to show |
that the statement “If true. a b , 1 , then a b” is not always b. For what set of numbers is the statement “If a b , 1 , then a b” always true? Developing Skills In 3–8: a. Write each definition in conditional form. b. Write the converse of the conditional given in part a. c. Write the biconditional form of the definition. 3. An equiangular triangle is a triangle that has three congruent angles. 4. The bisector of a line segment is any line, or subset of a line that intersects the segment at its midpoint. 5. An acute angle is an angle whose degree measure is greater than 0 and less than 90. 6. An obtuse triangle is a triangle that has one obtuse angle. 7. A noncollinear set of points is a set of three or more points that do not all lie on the same straight line. 8. A ray is a part of a line that consists of a point on the line, called an endpoint, and all the points on one side of the endpoint. In 9–14, write the biconditional form of each given definition. 9. A point B is between A and C if A, B, and C are distinct collinear points and AB BC AC. 10. Congruent segments are segments that have the same length. 11. The midpoint of a line segment is the point of that line segment that divides the segment into two congruent segments. 14365C03.pgs 7/12/07 2:52 PM Page 100 100 Proving Statements in Geometry 12. A right triangle is a triangle that has a right angle. 13. A straight angle is an angle that is the union of opposite rays and whose degree measure is 180. 14. Opposite rays are two rays of the same line with a common endpoint and no other point in common. Applying Skills In 15–17, write the biconditional form of the definition of each given term. 15. equilateral triangle 16. congruent angles 17. perpendicular lines 3-3 DEDUCTIVE REASONING A proof in geometry is a valid argument that establishes the truth of a statement. Most proofs in geometry rely on logic. That is, they are based on a series of statements that are assumed to be true. Deductive reasoning uses the laws of logic to link together true statements to arrive at a true conclusion. Since definitions are true statements, they are used in a geometric proof. In the examples that follow, notice how the laws of logic are used in the proofs of geometric statements. Using Logic to Form a Geometry Proof Let ABC be a triangle in which angle. In this proof, use the following definition: AB ' BC . We can prove that ABC is a right • Perpendicular lines are two lines that intersect to form right angles. This definition contains a hidden conditional and can be rewritten as follows: • If two lines are perpendicular, then they intersect to form right angles. Recall that this definition is true for perpendicular line segments as well as for perpendicular lines. Using the specific information cited above, let p repre,” and let r represent “ABC is a right angle.” sent “ AB ' BC The proof is shown by the reasoning that follows: p: ⊥ BC AB p is true because it is given. p → r: If AB ⊥ BC , then ABC is a right angle. p → r is true because it is the definition of perpendicular lines. r: ABC is a right angle. r is true by the Law of Detachment. 14365C03.pgs 7/12/07 2:52 PM Page 101 A B A B Deductive Reasoning 101 In the logic-based proof above, notice that the Law of Detachment is cited as a reason for reaching our conclusion. In a typical geometry proof, however, the laws of logic are used to deduce the conclusion but the laws are not listed among the reasons. Let us restate this proof in the format used often in Euclidean geometry. We write the information known to be true as the “given” statements and the conclusion to be proved as the “prove”. Then we construct a two-column proof. In the left column, we write statements that we know to be true, and in the right column, we write the reasons why each statement is true. Given: In ABC, . Prove: ABC is a right angle. AB ' BC C Proof: Statements Reasons AB ' BC 1. 2. ABC is a right angle. 1. Given. 2. If two lines are perpendicular, then they intersect to form right angles. Notice how the Law of Detachment was used in this geometry proof. By combining statement 1 with reason 2, the conclusion is written as statement 2, just as p and p → r led us to the conclusion r using logic. In reason 2, we used the conditional form of the definition of perpendicular lines. Often in a proof, we find one conclusion in order to use that statement to find another conclusion. For instance, in the following proof, we will begin by proving that ABC is a right angle and then use that fact with the definition of a right triangle to prove that ABC is a right triangle. Given: In ABC, Prove: ABC is a right triangle. AB ' BC . C Proof: Statements Reasons AB ' BC 1. 2. ABC is a right angle. 1. Given. 2. If two lines are perpendicular, then they intersect to form right angles. 3. ABC is a right triangle. 3. If a triangle has a right angle, then it is a right triangle. Note that in this proof, we used the conditional form of the definition of perpendicular lines in reason 2 and the converse form of the definition of a right triangle in reason 3. The proof can also be written in paragraph form, also called a paragraph proof. Each statement must be justified by stating a definition or another 14365C03.pgs 7/12/07 2:52 PM Page 102 102 Proving Statements in Geometry statement that has been accepted or proved to be true. The proof given on page 101 can be written as follows: Proof: We are given that AB ' BC . If two lines are perpendicular, then they intersect to form right angles. Therefore, ABC is a right angle. A right triangle is a triangle that has a right angle. Since ABC is an angle of ABC, ABC is a right triangle. EXAMPLE 1 Given: DEF with DE DF Prove: DEF is an isosceles triangle. Proof We will need the following definitions: • An isosceles triangle is a triangle that has two congruent sides. D E F • Congruent segments are segments that have the same measure. We can use these two definitions to first prove that the two segments with equal measure are congruent and then to prove that since the two segments, the sides, are congruent, the triangle is isosceles. Statements Reasons 1. DE = DF 2. DE > DF 3. DEF is isosceles. 1. Given. 2. Congruent segments are segments that have the same measure. 3. An isosceles triangle is a triangle that has two congruent sides. The converse of the definition of congruent segments states that if two segments have the same measure, then they are congruent. The given statement, DE DF, means that have the same measure. Therefore, these two sides of DEF are congruent. The converse of the definition of an isosceles triangle states that if a triangle has two congruent sides, then it is isosceles. Therefore, since are congruent sides of DEF, DEF is isosceles. and and DE DF DE DF Alternative Proof EXAMPLE 2 h BD is the bisector of ABC. Given: Prove: mABD mDBC Proof We will need the following definitions: A D C B 14365C03.pgs 7/12/07 2:52 PM Page 103 Deductive Reasoning 103 • The bisector of an angle is a ray whose endpoint is the vertex of the angle and that divides the angle into two congruent angles. • Congruent angles are angles that have the same measure. First use the definition of an angle bisector to prove that the angles are congruent. Then use the definition of congruent angles to prove that the angles have equal measures. Statements is the bisector of ABC. h BD 1. 2. ABD DBC 3. mABD mDBC Exercises Writing About Mathematics Reasons 1. Given. 2. The bisector of an angle is a ray whose endpoint is the vertex of the angle and that divides the angle into two congruent angles. 3. Congruent angles are angles that have the same measure. 1. Is an equilateral triangle an isosceles triangle? Justify your answer. 2. Is it possible that the points A, B, and C are collinear but AB BC AC? Justify your answer. Developing Skills In 3–6, in each case: a. Draw a diagram to illustrate the given statement. b. Write a definition or definitions from geometry, in conditional form, that can be used with the given statement to justify the conclusion. 3. Given: h SP bisects RST. Conclusion: RSP PST 5. Given: g BCD g ' ACE Conclusion: mACD 90 4. Given: ABC is a scalene triangle. Conclusion: AB BC 6. Given: AB BC AC with ABC Conclusion: B is between A and C. In 7–12, in each case: a. Draw a diagram to illustrate the given statement. b. Write a definition from geometry, in conditional form, that can be used with the given statement to make a conclusion. c. From the given statement and the definition that you chose, draw a conclusion. 7. Given: LMN with 9. Given: PQ QR PR with LM ' MN PQR 11. Given: M is the midpoint of LN . 8. Given: g AB h 10. Given: ST 12. Given: 0 mA 90 bisects h SR and DE at F. are opposite rays. 14365C03.pgs 7/12/07 2:52 PM Page 104 104 Proving Statements in Geometry Applying Skills In 13–15: a. Give the reason for each statement of the proof. b. Write the proof in paragraph form. 13. Given: M is the midpoint of . AMB 14. Given: RST with RS ST. Prove: AM MB A M B Prove: RST is an isosceles triangle. S Statements Reasons Statements Reasons R T 1. M is the midpoint of . AMB AM 2. MB 3. AM MB 1. 2. 3. 1. RS ST RS > ST 2. 3. RST is isosceles. 1. 2. 3. 15. Given: In ABC, h CE bisects ACB. Prove: mACE mBCE Statements Reasons B E h CE bisects ACB. 1. 2. ACE ECB 3. mACE mECB 1. 2. 3. C A 16. Complete the following proof by writing the statement for each step. Given: DEF with DE EF. Prove: E is the midpoint of DEF . D E F Statements Reasons 1. 2. 3. 1. Given. 2. Congruent segments are segments that have the same measure. 3. The midpoint of a line segment is the point of that line segment that divides the segment into congruent segments. 14365C03.pgs 7/12/07 2:52 PM Page 105 17. In ABC, mA 90 and mB 90. If ABC is an obtuse triangle, why is mC 90? Direct and Indirect Proofs 105 Justify your answer with a definition. 18. Explain why the following proof is incorrect. Given: Isosceles ABC with A as the vertex angle. Prove: BC AC Statemen |
ts 1. ABC is isosceles. BC > AC 2. 3. BC AC Reasons 1. Given. 2. An isosceles triangle has two congruent sides. 3. Congruent segments are segments that have the same measure. 3-4 DIRECT AND INDIRECT PROOFS A proof that starts with the given statements and uses the laws of logic to arrive at the statement to be proved is called a direct proof. A proof that starts with the negation of the statement to be proved and uses the laws of logic to show that it is false is called an indirect proof or a proof by contradiction. An indirect proof works because when the negation of a statement is false, the statement must be true. Therefore, if we can show that the negation of the statement to be proved is false, then we can conclude that the statement is true. Direct Proof All of the proofs in Section 3-3 are direct proofs. In most direct proofs we use definitions together with the Law of Detachment to arrive at the desired conclusion. Example 1 uses direct proof. Given: ABC is an acute triangle. Prove: mA 90 In this proof, we will use the following definitions: • An acute triangle is a triangle that has three acute angles. • An acute angle is an angle whose degree measure is greater than 0 and less than 90. C A B In the proof, we will use the conditional form of these definitions. EXAMPLE 1 14365C03.pgs 7/12/07 2:52 PM Page 106 106 Proving Statements in Geometry Proof C Statements 1. ABC is an acute triangle. 2. A, B, and C are all acute. A 3. mA 90 B Reasons 1. Given. 2. If a triangle is acute, then the triangle has three acute angles. 3. If an angle is acute, then its degree measure is greater than 0 and less than 90. In this proof, the “and” statement is important for the conclusion. In Note: statement 2, the conjunction can be rewritten as “A is acute, and B is acute, and C is acute.” We know from logic that when a conjunction is true, each conjunct is true. Also, in Reason 3, the conclusion of the conditional is a conjunction: “The degree measure is greater than 0 and the degree measure is less than 90.” Again, since this conjunction is true, each conjunct is true. Indirect Proof In an indirect proof, let p be the given and q be the conclusion. Take the following steps to show that the conclusion is true: 1. Assume that the negation of the conclusion is true. 2. Use this assumption to arrive at a statement that contradicts the given statement or a true statement derived from the given statement. 3. Since the assumption leads to a contradiction, it must be false. The nega- tion of the assumption, the desired conclusion, must be true. Let us use an indirect proof to prove the following statement: If the mea- sures of two segments are unequal, then the segments are not congruent. EXAMPLE 2 Given: AB and CD such that AB CD. Prove: AB and CD are not congruent segments A C B D Proof Start with an assumption that is the negation of the conclusion. 14365C03.pgs 7/12/07 2:52 PM Page 107 Direct and Indirect Proofs 107 Statements Reasons CD 1. and AB segments. 2. AB CD are congruent 1. Assumption. 2. Congruent segments are segments that have the same measure. 3. AB CD 3. Given. 4. CD and AB segments. are not congruent 4. Contradiction in 2 and 3. Therefore, the assumption is false and its negation is true. In this proof, and most indirect proofs, our reasoning reflects the contra- positive of a definition. For example: Definition: Congruent segments are segments that have the same measure. Conditional: If segments are congruent, then they have the same measure. Contrapositive: If segments do not have the same measure, then they are not congruent. Note: To learn how the different methods of proof work, you will be asked to prove some simple statements both directly and indirectly in this section. Thereafter, you should use the method that seems more efficient, which is usually a direct proof. However, in some cases, only an indirect proof is possible. EXAMPLE 3 Write a direct and an indirect proof of the following: C Given: mCDE 90 Prove: CD is not perpendicular to DE . In this proof, we will use two definitions: D E • If an angle is a right angle, then its degree measure is 90. • If two intersecting lines are perpendicular, then they form right angles. Direct Proof We will use the contrapositive form of the two definitions: • If the degree measure of an angle is not 90, then it is not a right angle. • If two intersecting lines do not form right angles, then they are not per- pendicular. 14365C03.pgs 7/12/07 2:52 PM Page 108 108 Proving Statements in Geometry C Statements Reasons 1. mCDE 90 2. CDE is not a right angle. D E 1. Given. 2. If the degree measure of an angle is not 90, then it is not a right angle. 3. CD is not perpendicular to . DE 3. If two intersecting lines do not form right angles, then they are not perpendicular. Indirect Proof We will use the negation of the statement that is to be proved as an assumption, and then arrive at a contradiction using the conditional form of the two definitions. Statements Reasons CD 1. is perpendicular to 2. CDE is a right angle. . DE 3. mCDE 90 1. Assumption. 2. If two intersecting lines are perpendicular, then they form right angles. 3. If an angle is a right angle, then its degree measure is 90. 4. mCDE 90. 4. Given. 5. CD is not perpendicular to . DE 5. Contradiction in 3 and 4. Therefore, the assumption is false and its negation is true. Exercises Writing About Mathematics 1. If we are given ABC, is it true that intersects g BC ? Explain why or why not. g AB g ABC 2. Glen said that if we are given line , then we know that A, B, and C are collinear and B is between A and C. Do you agree with Glen? Justify your answer. Developing Skills In 3–8: a. Draw a figure that represents the statement to be proved. b. Write a direct proof in twocolumn form. c. Write an indirect proof in two-column form. 14365C03.pgs 7/12/07 2:52 PM Page 109 3. Given: LM MN Prove: LM > MN 5. Given: PQR is a straight angle. Prove: h QP and h QR are opposite rays. 7. Given: PQR is a straight angle. Postulates,Theorems, and Proof 109 4. Given: PQR is a straight angle. Prove: mPQR 180 6. Given: h QP and h QR are opposite rays. Prove: P, Q, and R are on the same line. bisects DEF. h EG 8. Given: Prove: P, Q, and R are on the same line. Prove: mDEG mGEF 9. Compare the direct proofs to the indirect proofs in problems 3–8. In these examples, which proofs were longer? Why do you think this is the case? 10. Draw a figure that represents the statement to be proved and write an indirect proof: h EG does not bisect DEF. Given: Prove: DEG is not congruent to GEF. Applying Skills 11. In order to prove a conditional statement, we let the hypothesis be the given and the con- clusion be the prove. h BD If is perpendicular to g ABC , then h BD is the bisector of ABC. a. Write the hypothesis of the conditional as the given. b. Write the conclusion of the conditional as the prove. c. Write a direct proof for the conditional. 12. If mEFG 180, then h FE and h FG are not opposite rays. a. Write the hypothesis of the conditional as the given. b. Write the conclusion of the conditional as the prove. c. Write an indirect proof for the conditional. 3-5 POSTULATES, THEOREMS, AND PROOF A valid argument that leads to a true conclusion begins with true statements. In Chapter 1, we listed undefined terms and definitions that we accept as being true. We have used the undefined terms and definitions to draw conclusions. At times, statements are made in geometry that are neither undefined terms nor definitions, and yet we know these are true statements. Some of these statements seem so “obvious” that we accept them without proof. Such a statement is called a postulate or an axiom. 14365C03.pgs 7/12/07 2:52 PM Page 110 110 Proving Statements in Geometry Some mathematicians use the term “axiom” for a general statement whose truth is assumed without proof, and the term “postulate” for a geometric statement whose truth is assumed without proof. We will use the term “postulate” for both types of assumptions. DEFINITION A postulate is a statement whose truth is accepted without proof. When we apply the laws of logic to definitions and postulates, we are able to prove other statements. These statements are called theorems. DEFINITION A theorem is a statement that is proved by deductive reasoning. The entire body of knowledge that we know as geometry consists of undefined terms, defined terms, postulates, and theorems which we use to prove other theorems and to justify applications of these theorems. The First Postulates Used in Proving Statements Geometry is often concerned with measurement. In Chapter 1 we listed the properties of the number system. These properties include closure, the commutative, associative, inverse, identity, and distributive properties of addition and multiplication and the multiplication property of zero. We will use these properties as postulates in geometric proof. Other properties that we will use as postulates are the properties of equality. When we state the relation “x is equal to y,” symbolized as “x y,” we mean that x and y are two different names for the same element of a set, usually a number. For example: 1. When we write AB CD, we mean that length of AB and the length of CD are the same number. 2. When we write mP mN, we mean that P and N contain the same number of degrees. Many of our definitions, for example, congruence, midpoint, and bisector, state that two measures are equal. There are three basic properties of equality. The Reflexive Property of Equality: a a The reflexive property of equality is stated in words as follows: Postulate 3.1 A quantity is equal to itself. 14365C03.pgs 7/12/07 2:52 PM Page 111 Postulates,Theorems, and Proof 111 For example, in CDE, observe that: C 1. The length of a segment is equal to itself: CD CD DE DE EC EC 2. The measure of an angle is equal to itself: D E mC mC mD mD mE mE The Symmetric Property of Equality: If a b, then b a. The symmetric property of equality is stated i |
n words as follows: Postulate 3.2 An equality may be expressed in either order. For example: 1. If LM NP, then NP LM. 2. If mA mB, then mB mA. L A M N P B The Transitive Property of Equality: If a b and b c, then a c. This property states that, if a and b have the same value, and b and c have the same value, it follows that a and c have the same value. The transitive property of equality is stated in words as follows: Postulate 3.3 Quantities equal to the same quantity are equal to each other. The lengths or measures of segments and angles are numbers. In the set of real numbers, the relation “is equal to” is said to be reflexive, symmetric, and transitive. A relation for which these postulates are true is said to be an equivalence relation. Congruent segments are segments with equal measures and congruent angles are angles with equal measures. This suggests that “is congruent to” is also an equivalence relation for the set of line segments. For example: 1. AB > AB A line segment is congruent to itself. 2. If AB > CD , then CD > AB . Congruence can be stated in either order. 3. If AB > CD and CD > EF , then AB > EF . Segments congruent to the same segment are congruent to each other. 14365C03.pgs 7/12/07 2:52 PM Page 112 112 Proving Statements in Geometry Therefore, we can say that “is congruent to” is an equivalence relation on the set of line segments. We will use these postulates of equality in deductive reasoning. In con- structing a valid proof, we follow these steps: 1. A diagram is used to visualize what is known and how it relates to what is to be proved. 2. State the hypothesis or premise as the given, in terms of the points and lines in the diagram. The premises are the given facts. 3. The conclusion contains what is to be proved. State the conclusion as the prove, in terms of the points and lines in the diagram. 4. We present the proof, the deductive reasoning, as a series of statements. Each statement in the proof should be justified by the given, a definition, a postulate, or a previously proven theorem. EXAMPLE 1 If AB > BC and BC > CD , then AB = CD. Given: AB > BC and BC > CD Prove: AB = CD A B C D Proof Statements Reasons 1. AB > BC 2. AB = BC 3. BC > CD 4. BC = CD 5. AB = CD 1. Given. 2. Congruent segments are segments that have the same measure. 3. Given. 4. Congruent segments are segments that have the same measure. 5. Transitive property of equality (steps 2 and 4). Alternative Proof Statements Reasons 1. AB > BC 2. BC > CD 3. AB > CD 4. AB = CD 1. Given. 2. Given. 3. Transitive property of congruence. 4. Congruent segments are segments that have the same measure. 14365C03.pgs 7/12/07 2:52 PM Page 113 EXAMPLE 2 AB ' BC If mABC mLMN. and LM ' MN , then Given: AB ' BC and LM ' MN Prove: mABC = mLMN Proof Statements AB ' BC 1. 2. ABC is a right angle. 3. mABC 90 LM ' MN 4. 5. LMN is a right angle. 6. mLMN 90 7. 90 mLMN 8. mABC mLMN Postulates,Theorems, and Proof 113 A B M L C N Reasons 1. Given. 2. Perpendicular lines are two lines that intersect to form right angles. 3. A right angle is an angle whose degree measure is 90. 4. Given. 5. Perpendicular lines are two lines that intersect to form right angles. 6. A right angle is an angle whose degree measure is 90. 7. Symmetric property of equality. 8. Transitive property of equality (steps 3 and 7). Exercises Writing About Mathematics 1. Is “is congruent to” an equivalence relation for the set of angles? Justify your answer. 2. Is “is perpendicular to” an equivalence relation for the set of lines? Justify your answer. Developing Skills In 3–6, in each case: state the postulate that can be used to show that each conclusion is valid. 3. CD CD 4. 2 3 5 and 5 1 4. Therefore, 2 3 4 1. 5. 10 a 7. Therefore a 7 10. 6. mA 30 and mB 30. Therefore, mA mB. 14365C03.pgs 7/12/07 2:52 PM Page 114 114 Proving Statements in Geometry Applying Skills In 7–10, write the reason of each step of the proof. 7. Given: y x 4 and y 7 8. Given: AB BC AC and AB BC 12 Reasons Prove: x 4 7 Statements 1. y x 4 2. x 4 y 3. y 7 4. x 4 7 1. 2. 3. 4. Reasons Prove: AC 12 Statements 1. AB BC AC 2. AC AB BC 3. AB BC 12 4. AC 12 1. 2. 3. 4. 9. Given: M is the midpoint of LN and N is the midpoint of MP . Prove: LM NP Statements Reasons 1. M is the midpoint of LN . LM > MN 2. 3. LM MN 4. N is the midpoint of MP . MN > NP 5. 6. MN NP 7. LM NP 1. 2. 3. 4. 5. 6. 7. 10. Given: mFGH mJGK and mHGJ mJGK is the bisector of FGJ. Prove: h GH Reasons G Statements 1. mFGH mJGK 2. mHGJ mJGK 3. mFGH mHGJ 4. FGH HGJ h GH 5. is the bisector of FGJ. 1. 2. 3. 4. 5. F H J K 14365C03.pgs 7/12/07 2:52 PM Page 115 11. Explain why the following proof is incorrect. B Given: ABC with D a point on Prove: ADB ADC . BC The Substitution Postulate 115 A D C Statements 1. ADB and ADC are right angles. 2. mADB 90 and mADC 90. 3. mADB mADC 4. ADB ADC Reasons 1. Given (from the diagram). 2. A right angle is an angle whose degree measure is 90. 3. Transitive property of equality. 4. If two angles have the same measure, then they are congruent. Hands-On Activity Working with a partner: a. Determine the definitions and postulates that can be used with the given statement to write a proof. b. Write a proof in two-column form. Given: ABC and BCD are equilateral. C Prove: AB 5 CD A D B 3-6 THE SUBSTITUTION POSTULATE The substitution postulate allows us to replace one quantity, number, or measure with its equal. The substitution postulate is stated in words as follows: Postulate 3.4 A quantity may be substituted for its equal in any statement of equality. From x y and y 8, we can conclude that x 8. This is an example of the transitive property of equality, but we can also say that we have used the substitution property by substituting 8 for y in the equation x y. 14365C03.pgs 7/12/07 2:52 PM Page 116 116 Proving Statements in Geometry From y x 7 and x 3, we can conclude that y 3 7. This is not an example of the transitive property of equality. We have used the substitution property to replace x with its equal, 3, in the equation y x 7. We use this postulate frequently in algebra. For example, if we find the solution of an equation, we can substitute that solution to show that we have a true statement. 4x 1 3x 7 x 8 Check 4x 1 3x 7 4(8) 1 3(8) 7 5? 31 31 ✔ In a system of two equations in two variables, x and y, we can solve one equation for y and substitute in the other equation. For example, we are using the substitution postulate in the third line of this solution: 3x 2y 13 y x 1 3x 2(x 1) 13 3x 2x 2 13 5x 15 x 3 y 3 1 2 EXAMPLE 1 Given: CE 2CD and CD DE Prove: CE 2DE Proof Statements 1. CE 2CD 2. CD DE 3. CE 2DE Reasons 1. Given. 2. Given. 3. Substitution postulate. (Or: A quantity may be substituted for its equal in any expression of equality.) 14365C03.pgs 7/12/07 2:52 PM Page 117 The Substitution Postulate 117 EXAMPLE 2 Given: mABD mDBC 90 and mABD mCBE Prove: mCBE mDBC 90 Proof Statements 1. mABD mDBC 90 2. mABD mCBE 3. mCBE mDBC 90 Reasons 1. Given. 2. Given. 3. Substitution postulate. A B D C E Exercises Writing About Mathematics 1. If we know that PQ > RS and that RS > ST , can we conclude that PQ > ST ? Justify your answer. 2. If we know that clude that PQ ' ST PQ > RS , and that ? Justify your answer. RS ' ST , can we use the substitution postulate to con- Developing Skills In 3–10, in each case write a proof giving the reason for each statement in your proof. and RM MT Prove: RM 3. Given: MT 1 2RT 1 2RT 5. Given: ma mb 180 and ma mc Prove: mc mb 180 7. Given: 12 x y and x 8 Prove: 12 8 y 9. Given: AB CD, 1 2GH EF, and " CD 5 EF " Prove: AB 1 2GH 4. Given: AD DE AE and AD EB Prove: EB DE AE 6. Given: y x 5 and y 7 2x Prove: x 5 7 2x 8. Given: BC 2 AB2 AC 2 and AB DE Prove: BC 2 DE 2 AC 2 10. Given: mQ mR mS 75, mQ mS mT, and mR mT mU Prove: mU 75 14365C03.pgs 7/12/07 2:52 PM Page 118 118 Proving Statements in Geometry 3-7 THE ADDITION AND SUBTRACTION POSTULATES The Partition Postulate When three points, A, B, and C, lie on the same line, the symbol of indicating the following equivalent facts about these points: g ABC is a way • B is on the line segment . AC • B is between A and C. • AB 1 BC 5 AC Since A, B, and C lie on the same line, we can also conclude that into two segments whose sum is . This fact is expressed in the following postulate called the partition postulate. AB BC AC. In other words, B separates AC AC Postulate 3.5 A whole is equal to the sum of all its parts. This postulate applies to any number of segments or to their lengths. This postulate also applies to any number of angles or their measures. A B C D • If B is between A and C, then A, B, and C are collinear. AB BC AC AB 1 BC 5 AC • If C is between B and D, then B, C, and D are collinear. BC CD BD BC 1 CD 5 BD • If A, B, and C are collinear and B, C, and D are collinear, then A, B, C, and D are collinear, that is, ABCD may conclude: . We AB BC CD AD AB 1 BC 1 CD 5 AD A C D E B • If is a ray in the interior of h BC ABD: mABC mCBD mABD ABC CBD ABD h BD CBE: is a ray in the interior of • If mCBD mDBE mCBE CBD DBE CBE • We can also conclude that: mABC mCBD mDBE mABE ABC CBD DBE ABE 14365C03.pgs 7/12/07 2:52 PM Page 119 The Addition and Subtraction Postulates 119 Note: We write x y to represent the sum of the angles, x and y, only if x and y are adjacent angles. Since AB DE indicates that and mABC mDEF indicates that ABC DEF, that is, since equality implies congruency, we can restate the partition postulate in terms of congruent segments and angles. AB > DE Postulate 3.5.1 A segment is congruent to the sum of all its parts. Postulate 3.5.2 An angle is congruent to the sum of all its parts. The Addition Postulate The addition postulate may be stated in symbols or in words as follows: Postulate 3.6 If a b and c d, then a c b d. If equal quantities are added to equal quantities, the sums are equal. The following example proof uses the addition postulate. EXAMPLE 1 Given: g ABC and g DEF with AB DE and BC EF. Prove: AC DF A D B C E F Proof Statements |
Reasons 1. AB DE 2. BC EF 3. AB BC DE EF 4. AB BC AC DE EF DF 5. AC DF 1. Given. 2. Given. 3. Addition postulate. (Or: If equal quantities are added to equal quantities, the sums are equal.) 4. Partition postulate. (Or: A whole is equal to the sum of all its parts.) 5. Substitution postulate. 14365C03.pgs 7/12/07 2:52 PM Page 120 120 Proving Statements in Geometry Just as the partition postulate was restated for congruent segments and congruent angles, so too can the addition postulate be restated in terms of congruent segments and congruent angles. Recall that we can add line segments AB and if and only if B is between A and C. BC Postulate 3.6.1 If congruent segments are added to congruent segments, the sums are congruent. The example proof just demonstrated can be rewritten in terms of the con- gruence of segments. EXAMPLE 2 Given: g ABC g DEF and with AB > DE and BC > EF . Prove: AC > DF Proof Statements Reasons A D B E C F 1. AB > DE 2. BC > EF 3. AB 1 BC > DE 1 EF 4. AB 1 BC 5 AC 5. DE 1 EF 5 DF 6. AC > DF 1. Given. 2. Given. 3. Addition postulate. (Or: If congruent segments are added to congruent segments, the sums are congruent segments.) 4. Partition postulate. 5. Partition postulate. 6. Substitution postulate (steps 3, 4, 5). When the addition postulate is stated for congruent angles, it is called the angle addition postulate: Postulate 3.6.2 If congruent angles are added to congruent angles, the sums are congruent. Recall that to add angles, the angles must have a common endpoint, a com- mon side between them, and no common interior points. In the diagram, ABC CBD ABD. A B C D 14365C03.pgs 7/12/07 2:52 PM Page 121 The Addition and Subtraction Postulates 121 The Subtraction Postulate The subtraction postulate may also be stated in symbols or in words. Postulate 3.7 If a b, and c d, then a c b d. If equal quantities are subtracted from equal quantities, the differences are equal. Just as the addition postulate was restated for congruent segments and congruent angles, so too may we restate the subtraction postulate in terms of congruent segments and congruent angles. Postulate 3.7.1 If congruent segments are subtracted from congruent segments, the differences are congruent. Postulate 3.7.2 If congruent angles are subtracted from congruent angles, the differences are congruent. In Example 3, equal numbers are subtracted. In Example 4, congruent lengths are subtracted. EXAMPLE 3 Given: x 6 14 Prove: x 8 Proof Statements Reasons 1. x 6 14 2. 6 6 3. x 8 1. Given. 2. Reflexive property. 3. Subtraction postulate. 14541C03.pgs 1/25/08 3:51 PM Page 122 122 Proving Statements in Geometry EXAMPLE 4 Given: DEF , E is between D and F D E F Prove: DE > DF 2 EF Proof Statements Reasons 1. E is between D and F. 1. Given. 2. DE 1 EF > DF 3. EF > EF 2. Partition postulate. 3. Reflexive property. 4. DE > DF 2 EF 4. Subtraction postulate. Exercises Writing About Mathematics 1. Cassie said that we do not need the subtraction postulate because subtraction can be expressed in terms of addition. Do you agree with Cassie? Explain why or why not. 2. In the diagram, mABC 30, mCBD 45, and mDBE 15. a. Does mCBD mABC mDBE? Justify your E D answer. b. Is /CBD > /ABC 1 /DBE ? Justify your answer. C Developing Skills In 3 and 4, in each case fill in the missing statement or reason in the proof to show that the conclusion is valid. B A 3. Given: and AED Prove: AD BC BFC , AE BF, and ED FC Statements Reasons BFC AED 1. and 2. AE ED AD BF FC BC 1. Given. 2. 3. AE BF and ED FC 3. Given. 4. 5. AD BC 4. Substitution postulate. 5. Transitive property (steps 2, 4). D E A C F B 14365C03.pgs 7/12/07 2:52 PM Page 123 The Addition and Subtraction Postulates 123 4. Given: SPR QRP and RPQ PRS Prove: SPQ QRS R Q Statements Reasons 1. 2. 1. Given. 2. If congruent angles are added to congruent angles, the sums are congruent. S P 3. SPR RPQ SPQ 3. 4. 5. SPQ QRS 4. Partition postulate. 5. In 5–8, in each case write a proof to show that the conclusion is valid. 5. Given: AC > BC and MC > NC 6. Given: ABCD with AB > CD Prove: AM > BN Prove: AC > BD . Given: LMN PMQ Prove: LMQ NMP 8. Given: mAEB 180 and mCED 180 Prove: mAEC mBED 14365C03.pgs 7/12/07 2:52 PM Page 124 124 Proving Statements in Geometry 3-8 THE MULTIPLICATION AND DIVISION POSTULATES The postulates of multiplication and division are similar to the postulates of addition and subtraction. The postulates in this section are stated in symbols and in words. The Multiplication Postulate Postulate 3.8 If a b, and c d, then ac bd. If equals are multiplied by equals, the products are equal. When each of two equal quantities is multiplied by 2, we have a special case of this postulate, which is stated as follows: Postulate 3.9 Doubles of equal quantities are equal. The Division Postulate Postulate 3.10 a If a b, and c d, then c b d (c 0 and d 0). If equals are divided by nonzero equals, the quotients are equal. When each of two equal quantities is divided by 2, we have a special case of this postulate, which is stated as follows: Postulate 3.11 Halves of equal quantities are equal. Note: Doubles and halves of congruent segments and angles will be handled in Exercise 10. Powers Postulate Postulate 3.12 If a b, then a2 b2. The squares of equal quantities are equal. If AB 7, then (AB)2 (7)2, or (AB)2 49. 14365C03.pgs 7/12/07 2:52 PM Page 125 The Multiplication and Division Postulates 125 Roots Postulate Postulate 3.13 If a b and a 0, then ! a b . ! Recall that a and b are the positive square roots of a and of b, and so ! the postulate can be rewritten as: ! Postulate 3.13 Positive square roots of positive equal quantities are equal. If (AB)2 49, then (AB)2 5 " EXAMPLE 1 49 , or AB 7. " Given: AB CD, RS 3AB, LM 3CD Prove: RS LM Proof Statements Reasons 1. AB CD 2. 3AB 3CD 3. RS 3AB 4. RS 3CD 5. LM 3CD 6. RS LM EXAMPLE 2 Given: 5x 3 38 Prove: x 7 1. Given. 2. Multiplication postulate. 3. Given. 4. Transitive property of equality (steps 2 and 3). 5. Given. 6. Substitution postulate (steps 4 and 5). Proof Statements Reasons 1. 5x 3 38 2. 3 3 3. 5x 35 4. x 7 1. Given. 2. Reflexive property of equality. 3. Subtraction postulate. 4. Division postulate. 14365C03.pgs 7/12/07 2:52 PM Page 126 126 Proving Statements in Geometry EXAMPLE 3 Given: mABM 1 2m/ABC , mABC 2mMBC Prove: h BM bisects ABC. C M B A Proof Statements Reasons 1 1. mABM 2m/ABC 2. mABC 2mMBC 3. mABC mMBC 1 2 4. mABM mMBC 5. ABM MBC h BM 6. bisects ABC 1. Given. 2. Given. 3. Division postulate (or halves of equal quantities are equal). 4. Transitive property of equality. 5. Congruent angles are angles that have the same measure. 6. The bisector of an angle is a ray whose endpoint is the vertex of the angle and that divides the angle into two congruent angles. Exercises Writing About Mathematics 1. Explain why the word “positive” is needed in the postulate “Positive square roots of posi- tive equal quantities are equal.” 2. Barry said that “c 0” or “d 0” but not both could be eliminated from the division postu- late. Do you agree with Barry? Explain why or why not. 14365C03.pgs 7/12/07 2:52 PM Page 127 The Multiplication and Division Postulates 127 Developing Skills In 3 and 4, in each case fill in the missing statement or reason in the proof to show that the conclusion is valid. 3. Given: AB 1 4BC Prove: CD 4AB and BC CD 4. Given: ma 3mb and mb 20 Prove: ma 60 Statements Reasons Statements Reasons 1 4BC 1. AB 2. 4 4 3. 4AB BC 4. BC CD 5. 4AB CD 6. CD 4AB 1. 2. 3. 4. 5. 6. 1. 2. 3. 3mb 60 1. Given. 2. Given. 3. 4. 4. Transitive property of equality. In 5–7, in each case write a proof to show that the conclusion is valid. 5. Given: LM 2MN and MN 1 2NP Prove: LM > NP 6. Given: 2(3a 4) 16 Prove: a 4 7. Given: PQRS , PQ 3QR, and QR 1 3RS Prove: PQ > RS Applying Skills 8. On Monday, Melanie walked twice as far as on Tuesday. On Wednesday, she walked one- third as far as on Monday and two-thirds as far as on Friday. Prove that Melanie walked as far on Friday as she did on Tuesday. 9. The library, the post office, the grocery store, and the bank are located in that order on the same side of Main Street. The distance from the library to the post office is four times the distance from the post office to the grocery store. The distance from the grocery store to the bank is three times the distance from the post office to the grocery store. Prove that the distance from the library to the post office is equal to the distance from the post office to the bank. (Think of Main Street as the line segment LPGB .) 10. Explain why the following versions of Postulates 3.9 and 3.11 are valid: Doubles of congruent segments are congruent. Halves of congruent segments are congruent. Doubles of congruent angles are congruent. Halves of congruent angles are congruent. 14365C03.pgs 8/2/07 5:39 PM Page 128 128 Proving Statements in Geometry CHAPTER SUMMARY Postulates Geometric statements can be proved by using deductive reasoning. Deductive reasoning applies the laws of logic to a series of true statements to arrive at a conclusion. The true statements used in deductive reasoning may be the given, definitions, postulates, or theorems that have been previously proved. Inductive reasoning uses a series of particular examples to lead to a general conclusion. Inductive reasoning is a powerful tool in discovering and making conjectures. However, inductive reasoning does not prove or explain conjectures; generalizations arising from direct measurements of specific cases are only approximate; and care must be taken to ensure that all relevant examples are examined. A proof using deductive reasoning may be either direct or indirect. In direct reasoning, a series of statements that include the given statement lead to the desired conclusion. In indirect reasoning, the negation of the desired conclusion leads to a statement that contradicts a given statement. The Reflexive Property of Equality: a a 3.1 The Symmetric Property of Equality: If a b, then b a. 3.2 The Transitive Property of Eq |
uality: If a b and b c, then a c. 3.3 3.4 A quantity may be substituted for its equal in any statement of equality. 3.5 A whole is equal to the sum of all its parts. 3.5.1 A segment is congruent to the sum of all its parts. 3.5.2 An angle is congruent to the sum of all its parts. If equal quantities are added to equal quantities, the sums are equal. 3.6 3.6.1 If congruent segments are added to congruent segments, the sums are congruent. 3.6.2 If congruent angles are added to congruent angles, the sums are 3.7 congruent. If equal quantities are subtracted from equal quantities, the differences are equal. 3.7.1 If congruent segments are subtracted from congruent segments, the differences are congruent. 3.7.2 If congruent angles are subtracted from congruent angles, the differences are congruent. If equals are multiplied by equals, the products are equal. 3.8 3.9 Doubles of equal quantities are equal. 3.10 If equals are divided by nonzero equals, the quotients are equal. 3.11 Halves of equal quantities are equal. 3.12 The squares of equal quantities are equal. 3.13 Positive square roots of equal quantities are equal. VOCABULARY 3-1 Generalization • Inductive reasoning • Counterexample • Conjecture 14365C03.pgs 7/12/07 2:52 PM Page 129 Review Exercises 129 3-3 Proof • Deductive reasoning • Given • Prove • Two-column proof • Paragraph proof 3-4 Direct proof • Indirect proof • Proof by contradiction 3-5 Postulate • Axiom • Theorem • Reflexive property of equality • Symmetric property of equality • Transitive property of equality • Equivalence relation 3-6 Substitution postulate 3-7 Partition postulate • Addition postulate • Angle addition postulate • Subtraction postulate 3-8 Multiplication postulate • Division postulate • Powers postulate • Roots postulate REVIEW EXERCISES In 1–3, in each case: a. Write the given definition in a conditional form. b. Write the converse of the statement given as an answer to part a. c. Write the biconditional form of the definition. 1. An obtuse triangle is a triangle that has one obtuse angle. 2. Congruent angles are angles that have the same measure. 3. Perpendicular lines are two lines that intersect to form right angles. 4. Explain the difference between a postulate and a theorem. 5. Name the property illustrated by the following statement: If AB CD, then CD AB. 6. Is the relation “is greater than” an equivalence relation for the set of real numbers? Explain your answer by demonstrating which (if any) of the properties of an equivalence relation are true and which are false. In 7–12, in each case draw a figure that illustrates the given information and write a proof to show that the conclusion is valid. 7. Given: g bisects AB Prove: CM MD CD at M. 8. Given: Prove: is a line segment, RMST RM > ST RM > MS , and MS > ST . 9. Given: Prove ABCD AB > CD is a line segment and AC > BD . 10. Given: SQRP is a line segment and SQ RP. Prove: SR QP 14541C03.pgs 1/25/08 3:51 PM Page 130 130 Proving Statements in Geometry 11. Given: h BC bisects ABD and mCBD mPQR. Prove: mABC mPQR 12. Given: CD Prove: CD AB and AB bisect each other at E and CE BE. 13. A student wrote the following proof: Given: Prove: AB > CD CD ' BC and AB ' BC Statements Reasons 1. 2. 3. AB > CD AB ' BC CD ' BC 1. Given. 2. Given. 3. Substitution postulate. What is the error in this proof? 14. A palindrome is a sequence of numbers or letters that reads the same from left to right as from right to left. a. Write the definition of a palindrome as a conditional statement. b. Write the converse of the conditional statement in a. c. Write the definition as a biconditional. Exploration The following “proof” leads to the statement that twice a number is equal to the number. This would mean, for example, that if b 1, then 2(1) 1, which is obviously incorrect. What is the error in the proof? Given: a b Prove: b 2b Statements 1. a b 2. a2 ab 3. a2 b2 ab b2 4. (a b)(a b) b(a b) 5. a b b 6. b b b 7. 2b b Reasons 1. Given. 2. Multiplication postulate. 3. Subtraction postulate. 4. Substitution postulate. 5. Division postulate. 6. Substitution postulate. 7. Substitution postulate. 14365C03.pgs 7/12/07 2:52 PM Page 131 CUMULATIVE REVIEW Part I Cumulative Review 131 CHAPTERS 1–3 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. If y 2x 7 and y 3, then x is equal to (1) 1 (2) 1 (3) 5 (4) 5 2. The property illustrated in the equality 2(a 4) 2(4 a) is (1) the distributive property. (2) the associative property. (3) the identity property. (4) the commutative property. 3. In biconditional form, the definition of the midpoint of a line segment can be written as (1) A point on a line segment is the midpoint of that segment if it divides the segment into two congruent segments. (2) A point on a line segment is the midpoint of that segment if it divides the segment into two congruent segments. (3) A point on a line segment is the midpoint of that segment only if it divides the segment into two congruent segments. (4) A point on a line segment is the midpoint of that segment if and only if it divides the segment into two congruent segments. 4. The multiplicative identity element is (2) 1 (3) 0 (1) 1 5. The angle formed by two opposite rays is (4) not a real number (1) an acute angle. (2) a right angle. (3) an obtuse angle. (4) a straight angle. 6. The inverse of the statement “If two angles are right angles, then they are congruent” is (1) If two angles are not congruent, then they are not right angles. (2) If two angles are not right angles, then they are not congruent. (3) If two angles are congruent, then they are right angles. (4) Two angles are not right angles if they are not congruent. 7. The statements “Today is Saturday or I go to school” and “Today is not Saturday” are both true statements. Which of the following statements is also true? (1) Today is Saturday. (2) I do not go to school. (3) I go to school. (4) Today is Saturday and I go to school. 14365C03.pgs 7/12/07 2:52 PM Page 132 132 Proving Statements in Geometry 8. is a line segment and B is the midpoint of AC . Which of the fol- ABCD lowing must be true? (1) C is the midpoint of (2) AB BC BD (3) AC CD (4) AC BD AD 9. The statements “AB BC” and “DC BC” are true statements. Which of the following must also be true? (1) AB BC AC (2) A, B, and C are collinear (3) B, C, and D are collinear (4) AB DC 10. Triangle LMN has exactly two congruent sides. Triangle LMN is (3) an isosceles triangle. (4) an equilateral triangle. (1) a right triangle. (2) a scalene triangle. Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Give a reason for each step used in the solution of the equation. 3x 7 13 7 7 6 2 __________ __________ __________ Given 3x x 12. Given: DE ' EF Prove: DEF is a right triangle. Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Given: ABC with D a point on and AC AD DB. AB Prove: ABC is isosceles. is a line segment. PQ 4a 3, QR 3a 2 and PR 8a 6. Is Q 14. PQR the midpoint of PQR ? Justify your answer. 14365C03.pgs 7/12/07 2:52 PM Page 133 Cumulative Review 133 Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. h bisects ABC, mABD 3x 18 and mDBC 5x 30. If BD mABC 7x 12, is ABC a straight angle? Justify your answer. 16. For each statement, the hypothesis is true. Write the postulate that justifies the conclusion. a. If x 5, then x 7 5 7. b. If 2y 3 represents a real number, then 2y 3 2y 3. RST is a line segment, then c. If d. If y 2x 1 and y 15, then 2x 1 15. e. If a 3, then RS 1 ST 5 RT . 5 5 3 a 5 . 14365C04.pgs 7/12/07 3:04 PM Page 134 CHAPTER 4 CHAPTER TABLE OF CONTENTS 4-1 Postulates of Lines, Line Segments, and Angles 4-2 Using Postulates and Definitions in Proofs 4-3 Proving Theorems About Angles 4-4 Congruent Polygons and Corresponding Parts 4-5 Proving Triangles Congruent Using Side, Angle, Side 4-6 Proving Triangles Congruent Using Angle, Side, Angle 4-7 Proving Triangles Congruent Using Side, Side, Side Chapter Summary Vocabulary Review Exercises Cumulative Review 134 CONGRUENCE OF LINE SEGMENTS, ANGLES, AND TRIANGLES One of the common notions stated by Euclid was the following:“Things which coincide with one another are equal to one another.” Euclid used this common notion to prove the congruence of triangles. For example, Euclid’s Proposition 4 states, “If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.” In other words, Euclid showed that the equal sides and angle of the first triangle can be made to coincide with the sides and angle of the second triangle so that the two triangles will coincide.We will expand on Euclid’s approach in our development of congruent triangles presented in this chapter. 14365C04.pgs 7/31/07 1:22 PM Page 135 4-1 POSTULATES OF LINES, LINE SEGMENTS, AND ANGLES Postulates of Lines, Line Segments, and Angles 135 Melissa planted a new azalea bush in the fall and wants to protect it from the cold and snow this winter. She dro |
ve four parallel stakes into the ground around the bush and covered the structure with burlap fabric. During the first winter storm, this protective barrier was pushed out of shape. Her neighbor suggested that she make a tripod of three stakes fastened together at the top, forming three triangles. Melissa found that this arrangement was able to stand up to the storms. Why was this change an improvement? What geometric figure occurs most frequently in weight-bearing structures? In this chapter we will study the properties of triangles to discover why triangles keep their shape. g AB Recall that a line, , is an infinite set of points that extends endlessly in both directions, but a line g segment, and has a finite length. AB to form a line We can choose some point of segment of any length. When we do this, we say that we are extending the line segment. that is not a point of , is a part of g AB AB AB Postulate 4.1 A line segment can be extended to any length in either direction. D B A Postulate 4.2 When we choose point D on g AB so that B is the midpoint of AD , we say that we have extended we have chosen D so that AB BD and AD 2AB. We will also accept the following postulates: but AD AB is not the original segment, AB . In this case, Through two given points, one and only one line can be drawn. Two points determine a line. Through given points C and D, one and only one line can be drawn. C D Postulate 4.3 Two lines cannot intersect in more than one point. g AEB and g CED any other point. intersect at E and cannot intersect at B D C A E 14365C04.pgs 7/12/07 3:04 PM Page 136 136 Congruence of Line Segments, Angles, and Triangles Postulate 4.4 One and only one circle can be drawn with any given point as center and the length of any given line segment as a radius. Only one circle can be drawn that has point O as its center and a radius r equal in length to segment r. O We make use of this postulate in constructions when we use a compass to locate points at a given distance from a given point. Postulate 4.5 At a given point on a given line, one and only one perpendicular can be drawn to the line. At point P on g APB drawn perpendicular to g APB is perpendicular to , exactly one line, g APB and no other line through P g PD , can be D . P A B Postulate 4.6 From a given point not on a given line, one and only one perpendicular can be drawn to the line. From point P not on drawn perpendicular to perpendicular to g CD . g CD g CD , exactly one line, g PE , can be and no other line from P is P E C D Postulate 4.7 For any two distinct points, there is only one positive real number that is the length of the line segment joining the two points. A B Postulate 4.8 E C B A D resented by AB, which is the length of For the distinct points A and B, there is only one positive real number, repAB Since AB is also called the distance from A to B, we refer to Postulate 4.7 . as the distance postulate. The shortest distance between two points is the length of the line segment joining these two points. The figure shows three paths that can be taken in going from A to B. 14365C04.pgs 7/12/07 3:04 PM Page 137 Postulates of Lines, Line Segments, and Angles 137 The length of (the path through C, a point collinear with A and B) is less than the length of the path through D or the path through E. The measure of the shortest path from A to B is the distance AB. AB Postulate 4.9 A line segment has one and only one midpoint. has a midpoint, point M, and no other point is a AB midpoint of AB . Postulate 4.10 An angle has one and only one bisector. Angle ABC has one bisector, h BD , and no other ray bisects ABC. A M B C D A B EXAMPLE 1 Use the figure to answer the following questions: a. What is the intersection of m and n? Answers point B b. Do points A and B determine line m line m, n, or l? EXAMPLE 2 Lines p and n are two distinct lines that intersect line m at A. If line n is perpendicular to line m, can line p be perpendicular to line m? Explain. Solution No. Only one perpendicular can be drawn to a line at a given point on the line. Since line n is perpendicular to m and lines n and p are distinct, line p cannot be perpendicular to m. Answer 14365C04.pgs 7/12/07 3:04 PM Page 138 138 Congruence of Line Segments, Angles, and Triangles EXAMPLE 3 h BD bisects ABC and point E is not a point on h BD , can If h BE be the bisector of ABC? Solution No. An angle has one and only bisector. Since point E is not a point on h BE Therefore, h BD h , is not the same ray as BD cannot be the bisector of ABC. Answer h BE . A D B E C Conditional Statements as They Relate to Proof To prove a statement in geometry, we start with what is known to be true and use definitions, postulates, and previously proven theorems to arrive at the truth of what is to be proved. As we have seen in the text so far, the information that is known to be true is often stated as given and what is to be proved as prove. When the information needed for a proof is presented in a conditional statement, we use the information in the hypothesis to form a given statement, and the information in the conclusion to form a prove statement. Numerical and Algebraic Applications EXAMPLE 4 Rewrite the conditional statement in the given and prove format: If a ray bisects a straight angle, it is perpendicular to the line determined by the straight angle. Solution Draw and label a diagram. D Use the hypothesis, “a ray bisects a straight angle,” as the given. Name a straight angle using the three letters from the diagram and state in the given that this angle is a straight angle. Name the ray that bisects the angle, using the vertex of the angle as the endpoint of the ray that is the bisector. State in the given that the ray bisects the angle: A B C Given: ABC is a straight angle and h BD bisects ABC. Use the conclusion, “if (the bisector) is perpendicular to the line determined by the straight angle,” to write the prove. We have called the bisector the line determined by the straight angle is h Prove: BD g ' AC g ABC . h BD , and 14365C04.pgs 7/12/07 3:04 PM Page 139 Postulates of Lines, Line Segments, and Angles 139 Answer Given: ABC is a straight angle and g ' AC Prove: h BD h BD bisects ABC. In geometry, we are interested in proving that statements are true. These true statements can then be used to help solve numerical and algebraic problems, as in Example 5. EXAMPLE 5 h bisects RST, mRSQ 4x, and SQ mQST 3x 20. Find the measures of RSQ and QST. R Q Solution The bisector of an angle separates the angle into two congruent angles. Therefore, RSQ QST. Then since congruent angles have equal measures, we may write an equation that states that mRSQ mQST. 4x S 3x 20 T 4x 3x 20 x 20 mRSQ 4x mQST 3x 20 4(20) 80 3(20) 20 60 20 80 Answer mRSQ mQST 80 Exercises Writing About Mathematics 1. If two distinct lines g AEB and g CED intersect at a point F, what must be true about points E and F? Use a postulate to justify your answer. 2. If LM 10, can LM be extended so that LM 15? Explain why or why not. 14365C04.pgs 7/12/07 3:04 PM Page 140 140 Congruence of Line Segments, Angles, and Triangles Developing Skills In 3–12, in each case: a. Rewrite the conditional statement in the Given/Prove format. b. Write a formal proof. 3. If AB AD and DC AD, then AB DC. 5. If m1 m2 90 and mA m2, then m1 mA 90. 4. If then BD > CD AD > CD AD > BD and . 6. If mA mB, m1 mB, and m2 mA, then m1 m2. , 7. If AB > CD EF > CD , then AB > EF , and . 9. If CE CF, CD 2CE, and CB 2CF, then CD CB. 11. If AD BE and BC CD, then AC CE. E D C B A In 13–15, Given: g RST , h SQ , and h SP . 8. If 2EF DB and GH 1 2DB , then EF GH. 10. If RT RS, RD then RD RE. 1 2RT , and RE 1 2RS , 12. If CDB CBD and ADB ABD, then CDA CBA. D B C A 13. If QSR PST and mQSP 96, find mQSR. h 14. If mQSR 40 and SQ 15. If mPSQ is twice mQSR and , find mPST. h ' SP h ' SP , find h SQ Q R S P T mPST. Applying Skills In 16–19, use the given conditional to a. draw a diagram with geometry software or pencil and paper, b. write a given and a prove, c. write a proof. 16. If a triangle is equilateral, then the measures of the sides are equal. 17. If E and F are distinct points and two lines intersect at E, then they do not intersect at F. 18. If a line through a vertex of a triangle is perpendicular to the opposite side, then it separates the triangle into two right triangles. 19. If two points on a circle are the endpoints of a line segment, then the length of the line segment is less than the length of the portion of the circle (the arc) with the same endpoints. 20. Points F and G are both on line l and on line m. If F and G are distinct points, do l and m name the same line? Justify your answer. 14365C04.pgs 7/12/07 3:04 PM Page 141 4-2 USING POSTULATES AND DEFINITIONS IN PROOFS Using Postulates and Definitions in Proofs 141 A theorem was defined in Chapter 3 as a statement proved by deductive reasoning. We use the laws of logic to combine definitions and postulates to prove a theorem. A carefully drawn figure is helpful in deciding upon the steps to use in the proof. However, recall from Chapter 1 that we cannot use statements that appear to be true in the figure drawn for a particular problem. For example, we may not assume that two line segments are congruent or are perpendicular because they appear to be so in the figure. On the other hand, unless otherwise stated, we will assume that lines that appear to be straight lines in a figure actually are straight lines and that points that appear to be on a given line actually are on that line in the order shown. EXAMPLE 1 Given: g ABCD with AB > CD Prove: AC > BD A B C D Proof Statements Reasons 1. A, B, C, and D are collinear 1. Given. with B between A and C and C between B and D. 2. AB 1 BC BC 1 CD AC BD 3. AB > CD 4. BC > BC 5. AB 1 BC CD 1 BC 6. AC > BD 2. Partition postulate. 3. Given. 4. Reflexive property. 5. Addition postulate. 6. Substitution postulate. EXAMPLE 2 Given: M is the midpoint of AB . Prove: AM 1 2AB and MB 1 2 |
AB A M B 14365C04.pgs 7/12/07 3:04 PM Page 142 142 Congruence of Line Segments, Angles, and Triangles M A Proof B Statements Reasons 1. M is the midpoint of AB . 1. Given. AM > MB 2. 3. AM MB 4. AM MB AB 5. AM AM AB or 2AM AB 6. AM 1 2AB 7. MB MB AB or 2MB AB 8. MB 1 2AB 2. Definition of midpoint. 3. Definition of congruent segments. 4. Partition postulate. 5. Substitution postulate. 6. Halves of equal quantities are equal. 7. Substitution postulate. 8. Halves of equal quantities are equal. Note: We used definitions and postulates to prove statements about length. We could not use information from the diagram that appears to be true. Exercises Writing About Mathematics 1. Explain the difference between the symbols ABC and ACB . Could both of these describe segments of the same line? g ABC g DBE CBE, and EBA also 90? Justify your answer. and 2. Two lines, , intersect and mABD is 90. Are the measures of DBC, Developing Skills In 3–12, in each case: a. Rewrite the conditional statement in the Given/Prove format. b. Write a proof that demonstrates that the conclusion is valid. 3. If AB , CB bisects CB FD bisects AD AB , and . CE , then FE 4. If h CA bisects DCB, DAB, and DCB DAB, then CAB DCA. bisects h AC 14365C04.pgs 7/12/07 3:04 PM Page 143 5. If AD BE , then AE . BD 6. If ABCD is a segment and AB CD, then AC BD. Using Postulates and Definitions in Proofs 143 C A D E B ABCD AC 7. If of AB BC CD. is a segment, B is the midpoint , then , and C is the midpoint of BD A B C D A B C D 8. If P and T are distinct points and P is , then T is not the RS the midpoint of . midpoint of RS R T P S 9. If DF BE , then DE . BF 10. If AD BC , E is the midpoint of , then BC , AD and F is the midpoint of AE . FC D C E F B A 11. If AC DB AC each other at E, then and and AE DB . EB bisect D A C B E Applying Skills D E A B C F 12. If h DR bisects CDA, 3 1, and 4 2, then 3 4 13. The rays h DF and h DG separate CDE into three congruent angles, CDF, FDG, and GDE. If mCDF 7a 10 and mGDE 10a 2, find: a. mCDG d. Is CDE acute, right, or obtuse? c. mCDE b. mFDE 14365C04.pgs 7/12/07 3:04 PM Page 144 144 Congruence of Line Segments, Angles, and Triangles 14. Segment is a bisector of BD BC x 10. a. Draw a diagram that shows ABC and BD . ABC and is perpendicular to ABC . AB 2x 30 and b. Find AB and BC. c. Find the distance from A to BD . Justify your answer. 15. Two line segments, and and RN LN. RS 3x 9, and LM 5x 17. a. Draw a diagram that shows and LM LM RS RS . , bisect each other and are perpendicular to each other at N, b. Write the given using the information in the first sentence. c. Prove that RS LM. d. Find RS and LM. e. Find the distance from L to RS . Justify your answer. 4-3 PROVING THEOREMS ABOUT ANGLES In this section we will use definitions and postulates to prove some simple theorems about angles. Once a theorem is proved, we can use it as a reason in later proofs. Like the postulates, we will number the theorems for easy reference. Theorem 4.1 If two angles are right angles, then they are congruent. Given ABC and DEF are right angles. C Prove ABC DEF Proof Statements 1. ABC and DEF are right angles. 2. mABC 90 and mDEF 90 3. mABC mDEF 4. ABC DEF F B A E D Reasons 1. Given. 2. Definition of right angle. 3. Transitive property of equality. 4. Definition of congruent angles. We can write this proof in paragraph form as follows: Proof: A right angle is an angle whose degree measure is 90. Therefore, mABC is 90 and mDEF is 90. Since mABC and mDEF are both equal to the same quantity, they are equal to each other. Since ABC and DEF have equal measures, they are congruent. 14365C04.pgs 7/12/07 3:04 PM Page 145 Theorem 4.2 If two angles are straight angles, then they are congruent. Proving Theorems About Angles 145 Given ABC and DEF are straight angles. A B C Prove ABC DEF D E F The proof of this theorem, which is similar to the proof of Theorem 4.1, is left to the student. (See exercise 17.) Definitions Involving Pairs of Angles DEFINITION Adjacent angles are two angles in the same plane that have a common vertex and a common side but do not have any interior points in common Angle ABC and CBD are adjacent angles because as their common they have B as their common vertex, h BC side, and no interior points in common. However, XWY and XWZ are not adjacent angles. B D C A Although XWY and XWZ have W as their common vertex and as their common side, they have interior points in common. For example, point P is in the interior of both XWY and XWZ. h WX DEFINITION Complementary angles are two angles, the sum of whose degree measures is 90. Each angle is called the complement of the other. If mc 40 and md 50, then c and d are complementary angles. If ma 35 and mb 55, then a and b are complementary angles. Complementary angles may be adjacent, as in the case of c and d, or they may be nonadjacent, as in the case of a and b. Note that if the two complementary angles are adjacent, their sum is a right angle: c d WZY, a right angle. Since mc md 90, we say that c is the complement of d, and that d is the complement of c. When the degree measure of an angle is k, the degree measure of the complement of the angle is (90 k) because k (90 k) 90. DEFINITION Supplementary angles are two angles, the sum of whose degree measures is 180. 14365C04.pgs 7/12/07 3:04 PM Page 146 146 Congruence of Line Segments, Angles, and Triangles When two angles are supplementary, each angle is called the supplement of the other. If mc 40 and md 140, then c and d are supplementary angles. If ma 35 and mb 145, then a and b are supplementary angles Supplementary angles may be adjacent, as in the case of c and d, or they may be nonadjacent, as in the case of a and b. Note that if the two supplementary angles are adjacent, their sum is a straight angle. Here c d TQR, a straight angle. Since mc md 180, we say that c is the supplement of d and that d is the supplement of c. When the degree measure of an angle is k, the degree measure of the supplement of the angle is (180 k) because k (180 k) 180. EXAMPLE 1 Find the measure of an angle if its measure is 24 degrees more than the measure of its complement. Solution Let x measure of complement of angle. Then x 24 measure of angle. The sum of the degree measures of an angle and its complement is 90. x x 24 90 2x 24 90 2x 66 x 33 x 24 57 Answer The measure of the angle is 57 degrees. Theorems Involving Pairs of Angles Theorem 4.3 If two angles are complements of the same angle, then they are congruent. Given 1 is the complement of 2 and 3 is the comple- ment of 2. Prove 1 3 C D E 23 1 B A 14365C04.pgs 7/12/07 3:04 PM Page 147 Proof Statements 1. 1 is the complement of 2. 2. m1 m2 90 3. 3 is the complement of 2. 4. m3 m2 90 Proving Theorems About Angles 147 Reasons 1. Given. 2. Complementary angles are two angles the sum of whose degree measures is 90. 3. Given. 4. Definition of complementary angles. 5. m1 m2 m3 m2 5. Transitive property of equality 6. m2 m2 7. m1 m3 8. 1 3 (steps 2 and 4). 6. Reflexive property of equality. 7. Subtraction postulate. 8. Congruent angles are angles that have the same measure. Note: In a proof, there are two acceptable ways to indicate a definition as a reason. In reason 2 of the proof above, the definition of complementary angles is stated in its complete form. It is also acceptable to indicate this reason by the phrase “Definition of complementary angles,” as in reason 4. We can also give an algebraic proof for the theorem just proved. Proof: In the figure, mCBD x. Both ABD and CBE are complements to CBD. Thus, mABD 90 x and mCBE 90 x, and we conclude ABD and CBE have the same measure. Since angles that have the same measure are congruent, ABD CBE. C D x E (90 x) B ( 9 0 x ) A Theorem 4.4 If two angles are congruent, then their complements are congruent. Given ABD EFH CBD is the complement of ABD. GFH is the complement of EFH. C G D H Prove CBD GHF B A F E This theorem can be proved in a manner similar to Theorem 4.3 but with the use of the substitution postulate. We can also use an algebraic proof. 14365C04.pgs 7/12/07 3:05 PM Page 148 148 Congruence of Line Segments, Angles, and Triangles C D B A F G Proof Congruent angles have the same measure. If ABD EFH, we can represent the measure of each angle by the same variable: mABD mEFH x. Since CBD is the complement of ABD, and GFH is the complement of EFH, then mCBD 90 x and mGFH 90 x. Therefore, mCBD mGFH and CBD GHF. H E Theorem 4.5 If two angles are supplements of the same angle, then they are congruent. Given ABD is the supplement of DBC, and EBC is the supplement of DBC. Prove ABD EBC B C E Theorem 4.6 If two angles are congruent, then their supplements are congruent. Given ABD EFH, CBD is the supplement of ABD, and GFH is the supplement of EFH. Prove CBD GFH The proofs of Theorems 4.5 and 4.6 are similar to the proofs of Theorems 4.3 and 4.4 and will be left to the student. (See exercises 18 and 19.) More Definitions and Theorems Involving Pairs of Angles DEFINITION A linear pair of angles are two adjacent angles whose sum is a straight angle. C In the figure, ABD is a straight angle and C is not on . Therefore, ABC + CBD ABD. Note that ABC and CBD are adjacent angles and whose remaining sides are opposite rays that whose common side is h BC g ABD D B A together form a straight line, g . AD Theorem 4.7 If two angles form a linear pair, then they are supplementary. Given ABC and CBD form a linear pair. Prove ABC and CBD are supplementary. C D A B 14365C04.pgs 7/12/07 3:05 PM Page 149 Proving Theorems About Angles 149 Proof , and their remaining sides, In the figure, ABC and CBD form a linear pair. They share a common side, h , are opposite rays. The sum of and BC a linear pair of angles is a straight angle, and the degree measure of a straight angle is 180. Therefore, mABC mCBD 180. Then, ABC and CBD are supplementary because supplementary angles are two angles the sum of whose degree measure is 180. h BD h BA Theorem 4.8 If two lines intersect to for |
m congruent adjacent angles, then they are perpendicular. Given Prove g ABC g ABC g DBE and g ' DBE with ABD DBC Proof The union of the opposite rays, , and h is the straight angle, ABC. The BC measure of straight angle is 180. By the partition postulate, ABC is the sum of ABD and DBC. Thus, h BA D B E A C mABD mDBC mABC 180 Since ABD DBC, they have equal measures. Therefore, mABD = mDBC 1 2(180) 90 The angles, ABD and DBC, are right angles. Therefore, because perpendicular lines intersect to form right angles. g ABC g ' DBE DEFINITION Vertical angles are two angles in which the sides of one angle are opposite rays to the sides of the second angle. A C E D In the figure, AEC and DEB are a pair of vertical angles because h EB are opposite rays and h EA are opposite rays. Also, AED and h are opposite rays and EA h and ED CEB are a pair of vertical angles because h ED are opposite rays. In each pair of vertical angles, the opposite rays, h EB h EC h EC and and and B which are the sides of the angles, form straight lines, g AB and g CD . When two straight lines intersect, two pairs of vertical angles are formed. 14365C04.pgs 7/12/07 3:05 PM Page 150 150 Congruence of Line Segments, Angles, and Triangles Theorem 4.9 If two lines intersect, then the vertical angles are congruent. Given g AEB and g CED intersect at E. Prove BEC AED Proof 1. Statements g CED and intersect at E. g AEB h EA h ED 2. and h EB h EC 3. BEC and AED are and are opposite rays. are opposite rays. vertical angles. 4. BEC and AEC are a linear pair. AEC and AED are a linear pair. 5. BEC and AEC are supplementary. AEC and AED are supplementary. 6. BEC AED B D C A E Reasons 1. Given. 2. Definition of opposite rays. 3. Definition of vertical angles. 4. Definition of a linear pair. 5. If two angles form a linear pair, they are supplementary. (Theorem 4.7) 6. If two angles are supplements of the same angle, they are congruent. (Theorem 4.5) In the proof above, reasons 5 and 6 demonstrate how previously proved theorems can be used as reasons in deducing statements in a proof. In this text, we have assigned numbers to theorems that we will use frequently in proving exercises as well as in proving other theorems. You do not need to remember the numbers of the theorems but you should memorize the statements of the theorems in order to use them as reasons when writing a proof. You may find it useful to keep a list of definitions, postulates, and theorems in a special section in your notebook or on index cards for easy reference and as a study aid. In this chapter, we have seen the steps to be taken in presenting a proof in geometry using deductive reasoning: 1. As an aid, draw a figure that pictures the data of the theorem or the prob- lem. Use letters to label points in the figure. 2. State the given, which is the hypothesis of the theorem, in terms of the figure. 3. State the prove, which is the conclusion of the theorem, in terms of the figure. 14365C04.pgs 7/12/07 3:05 PM Page 151 Proving Theorems About Angles 151 4. Present the proof, which is a series of logical arguments used in the demonstration. Each step in the proof should consist of a statement about the figure. Each statement should be justified by the given, a definition, a postulate, or a previously proved theorem. The proof may be presented in a two-column format or in paragraph form. Proofs that involve the measures of angles or of line segments can often be presented as an algebraic proof. EXAMPLE 2 If g ABC h BC g DBE and intersect at B bisects EBF, prove that and CBF ABD. A D Solution Given: g and ABC at B and EBF. g hDBE BC bisects intersect B F Prove: CBF ABD Proof Statements E C Reasons h BC bisects EBF. 1. 2. EBC CBF g ABC g intersect at B. 3. DBE 4. EBC and ABD are vertical and angles. 5. EBC ABD 6. CBF ABD 1. Given. 2. Definition of a bisector of an angle. 3. Given. 4. Definition of vertical angles. 5. If two lines intersect, then the vertical angles are congruent. 6. Transitive property of congruence (steps 2 and 5). Alternative Proof An algebraic proof can be given: Let mEBF 2x. It is given that h BC bisects EBF. The bisector of an angle separates the angle into two congruent angles: EBC and CBF. Congruent angles have equal measures. Therefore, mEBC mCBF x. It is also given that intersect at B. If two lines intersect, g DBE g ABC and the vertical angles are congruent and therefore have equal measures: mABD mEBC x. Then since mCBF = x and mABD x, mCBF = mABD and CBF ABD. 14365C04.pgs 7/12/07 3:05 PM Page 152 152 Congruence of Line Segments, Angles, and Triangles Exercises Writing About Mathematics 1. Josh said that Theorem 4.9 could also have been proved by showing that AEC BED. Do you agree with Josh? Explain. 2. The statement of Theorem 4.7 is “If two angles form a linear pair then they are supplemen- tary.” Is the converse of this theorem true? Justify your answer. Developing Skills In 3–11, in each case write a proof, using the hypothesis as the given and the conclusion as the statement to be proved. 3. If mACD mDCB 90, B DCA, and A DCB, then A and B are complements. C A D B h ACFG h BCDE 4. If and intersect at C and ADC BFC, then ADE BFG. 5. If ADB is a right angle and CE ' DBE , then ADB CEB. If ABC and BCD are right angles, and EBC ECB, then EBA ECD. 7. If ABC is a right angle and DBF is a right angle, then ABD CBF 14365C04.pgs 7/12/07 3:05 PM Page 153 Proving Theorems About Angles 153 8. If g EF intersects g DC and mBHG mCGH, then BHG DGE. at H and g AB at G, 9. If h ABD and h ACE intersect at A, and ABC ACB, then DBC ECB AEB g CED 10. If and AEC CEB, then intersect at E, and g ' CED g AEB . C B A E D 11. If ABC is a right angle, and BAC is complementary to DBA, then BAC CBD. B A D C In 12–15, g AEB and g CED intersect at E. 12. If mBEC 70, find mAED, mDEB, and mAEC. 13. If mDEB 2x 20 and mAEC 3x – 30, find mDEB, mAEC, mAED, and mCEB. 14. If mBEC 5x – 25 and mDEA 7x – 65, find mBEC, mDEA, mDEB, and mAEC. 15. If mBEC y, mDEB 3x, and mDEA 2x – y, find mCEB, mBED, mDEA, and mAEC. D A E B C 14365C04.pgs 7/12/07 3:05 PM Page 154 154 Congruence of Line Segments, Angles, and Triangles g RS 16. intersects g LM at P, mRPL x y, mLPS 3x 2y, mMPS 3x – 2y. a. Solve for x and y. b. Find mRPL, mLPS, and mMPS. 17. Prove Theorem 4.2, “If two angles are straight angles, then they are congruent.” 18. Prove Theorem 4.5, “If two angles are supplements of the same angle, then they are congruent.” 19. Prove Theorem 4.6, “If two angles are congruent, then their supplements are congruent.” Applying Skills 20. Two angles form a linear pair. The measure of the smaller angle is one-half the measure of the larger angle. Find the degree measure of the larger angle. 21. The measure of the supplement of an angle is 60 degrees more than twice the measure of the angle. Find the degree measure of the angle. 22. The difference between the degree measures of two supplementary angles is 80. Find the degree measure of the larger angle. 23. Two angles are complementary. The measure of the larger angle is 5 times the measure of the smaller angle. Find the degree measure of the larger angle. 24. Two angles are complementary. The degree measure of the smaller angle is 50 less than the degree measure of the larger. Find the degree measure of the larger angle. 25. The measure of the complement of an angle exceeds the measure of the angle by 24 degrees. Find the degree measure of the angle. 4-4 CONGRUENT POLYGONS AND CORRESPONDING PARTS Fold a rectangular sheet of paper in half by placing the opposite edges together. If you tear the paper along the fold, you will have two rectangles that fit exactly on one another. We call these rectangles congruent polygons. Congruent polygons are polygons that have the same size and shape. Each angle of one polygon is congruent to an angle of the other and each edge of one polygon is congruent to an edge of the other. In the diagram at the top of page 155, polygon ABCD is congruent to polygon EFGH. Note that the congruent polygons are named in such a way that each vertex of ABCD corresponds to exactly one vertex of EFGH and each vertex of EFGH corresponds to exactly one vertex of ABCD. This relationship is called a one-to-one correspondence. The order in which the vertices are named shows this one-to-one correspondence of points. 14365C04.pgs 7/12/07 3:05 PM Page 155 Congruent Polygons and Corresponding Parts 155 ABCD EFGH indicates that: • A corresponds to E; E corresponds to A. • B corresponds to F; F corresponds to B. • C corresponds to G; G corresponds to C. • D corresponds to H; H corresponds to D. A D B E F H C G Congruent polygons should always be named so as to indicate the corre- spondences between the vertices of the polygons. Corresponding Parts of Congruent Polygons In congruent polygons ABCD and EFGH shown above, vertex A corresponds to vertex E. Angles A and E are called corresponding angles, and A E. In this example, there are four pairs of such corresponding angles: A E B F C G D H In congruent polygons, corresponding angles are congruent. In congruent polygons ABCD and EFGH, since A corresponds to E and B corresponds to F, EF In this example, there are four pairs of such corresponding sides: are corresponding sides, and and AB AB EF . AB EF BC > FG CD > GH DA > HE In congruent polygons, corresponding sides are congruent. The pairs of congruent angles and the pairs of congruent sides are called the corresponding parts of congruent polygons. We can now present the formal definition for congruent polygons. DEFINITION Two polygons are congruent if and only if there is a one-to-one correspondence between their vertices such that corresponding angles are congruent and corresponding sides are congruent. This definition can be stated more simply as follows: Corresponding parts of congruent polygons are congruent. Congruent Triangles The smallest number of sides that a polygon can have is three. A triangle is a polygon with exactly three sides. In the figure, ABC and DEF ar |
e congruent triangles. C F A B D E 14365C04.pgs 7/12/07 3:05 PM Page 156 156 Congruence of Line Segments, Angles, and Triangles C F B E A D The correspondence establishes six facts about these triangles: three facts about corresponding sides and three facts about corresponding angles. In the table at the right, these six facts are stated as equalities. Since each congruence statement is equivalent to an equality statement, we will use whichever notation serves our purpose better in a particular situation. Congruences Equalities AB > DE BC > EF AC > DF A D B E C F AB DE BC EF AC DF mA mD mB mE mC mF For example, in one proof, we may and in another AC > DF prefer to write proof to write AC = DF. In the same way, we might write C F or we might write mC = mF. From the definition, we may now say: Corresponding parts of congruent triangles are equal in measure. In two congruent triangles, pairs of corresponding sides are always opposite pairs of corresponding angles. In the preceding figure, ABC DEF. The order in which we write the names of the vertices of the triangles indicates the one-to-one correspondence. 1. A and D are corresponding congruent angles. 2. BC is opposite A, and EF is opposite D. 3. BC and EF are corresponding congruent sides. Equivalence Relation of Congruence In Section 3-5 we saw that the relation “is congruent to” is an equivalence relation for the set of line segments and the set of angles. Therefore, “is congruent to” must be an equivalence relation for the set of triangles or the set of polygons with a given number of sides. 1. Reflexive property: ABC ABC. 2. Symmetric property: If ABC DEF, then DEF ABC. 3. Transitive property: If ABC DEF and DEF RST, then ABC RST. Therefore, we state these properties of congruence as three postulates: Postulate 4.11 Any geometric figure is congruent to itself. (Reflexive Property) Postulate 4.12 A congruence may be expressed in either order. (Symmetric Property) 14365C04.pgs 7/12/07 3:05 PM Page 157 Congruent Polygons and Corresponding Parts 157 Postulate 4.13 Two geometric figures congruent to the same geometric figure are congruent to each other. (Transitive Property) Exercises Writing About Mathematics 1. If ABC DEF, then AB > DE . Is the converse of this statement true? Justify your answer. 2. Jesse said that since RST and STR name the same triangle, it is correct to say RST STR. Do you agree with Jesse? Justify your answer. Developing Skills In 3–5, in each case name three pairs of corresponding angles and three pairs of corresponding sides in the given congruent triangles. Use the symbol to indicate that the angles named and also the sides named in your answers are congruent. 3. ABD CBD 4. ADB CBD 5. ABD EBC In 6–10, LMNP is a square and each statement. LSN and PSM bisect each other. Name the property that justifies 6. LSP LSP 7. If LSP NSM, then NSM LSP. 8. If LSP NSM and NSM NSP, then LSP NSP. 9. If LS SN, then SN LS. 10. If PLM PNM, then PNM PLM. P N S L M 14365C04.pgs 7/12/07 3:05 PM Page 158 158 Congruence of Line Segments, Angles, and Triangles 4-5 PROVING TRIANGLES CONGRUENT USING SIDE, ANGLE, SIDE The definition of congruent polygons states that two polygons are congruent if and only if each pair of corresponding sides and each pair of corresponding angles are congruent. However, it is possible to prove two triangles congruent by proving that fewer than three pairs of sides and three pairs of angles are congruent. Hands-On Activity In this activity, we will use a protractor and ruler, or geometry software. Use the procedure below to draw a triangle given the measures of two sides and of the included angle. STEP 1. Use the protractor or geometry software to draw an angle with the given measure. STEP 2. Draw two segments along the rays of the angle with the given lengths. The two segments should share the vertex of the angle as a common endpoint. STEP 3. Join the endpoints of the two segments to form a triangle. STEP 4. Repeat steps 1 through 3 to draw a second triangle using the same angle measure and segment lengths. a. Follow the steps to draw two different triangles with each of the given side-angle-side measures. (1) 3 in., 90°, 4 in. (2) 5 in., 40°, 5 in. (3) 5 cm, 115°, 8 cm (4) 10 cm, 30°, 8 cm b. For each pair of triangles, measure the side and angles that were not given. Do they have equal measures? c. Are the triangles of each pair congruent? Does it appear that when two sides and the included angle of one triangle are congruent to the corresponding sides and angle of another, that the triangles are congruent? This activity leads to the following statement of side-angle-side or SAS triangle congruence, whose truth will be assumed without proof: Postulate 4.14 Two triangles are congruent if two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of the other. (SAS) In ABC and DEF, BA > ED , ABC DEF and . It follows that ABC DEF. The postulate used here is abbreviated SAS. BC > EF B E A C D F 14365C04.pgs 7/12/07 3:05 PM Page 159 Proving Triangles Congruent Using Side, Angle, Side 159 Note: When congruent sides and angles are listed, a correspondence is established. Since the vertices of congruent angles, ABC and DEF, are B and E, B corresponds to E. Since BA ED and B corresponds to E, then A corresponds to D, and since BC EF and B corresponds to E, then C corresponds to F. We can write ABC DEF. But, when naming the triangle, if we change the order of the vertices in one triangle we must change the order in the other. For example, ABC, BAC, and CBA name the same triangle. If ABC DEF, we may write BAC EDF or CBA FED, but we may not write ABC EFD. EXAMPLE 1 Given: ABC, CD CD ' AB . is the bisector of AB , and C Prove: ACD BCD Prove the triangles congruent by SAS. Proof Statements 1. CD bisects AB . 1. Given. A D B Reasons 2. D is the midpoint of AB . S 3. AD > DB CD ' AB 4. 5. ADC and BDC are right angles. A 6. ADC BDC S 7. CD > CD 8. ACD BCD 2. The bisector of a line segment intersects the segment at its midpoint. 3. The midpoint of a line segment divides the segment into two congruent segments. 4. Given. 5. Perpendicular lines intersect to form right angles. 6. If two angles are right angles, then they are congruent. 7. Reflexive property of congruence. 8. SAS (steps 3, 6, 7). Note that it is often helpful to mark segments and angles that are congruent with the same number of strokes or arcs. For example, in the diagram for this are marked with a single stroke, ADC and BDC are proof, is marked with an “” to indimarked with the symbol for right angles, and cate a side common to both triangles. and AD DB CD 14365C04.pgs 7/12/07 3:05 PM Page 160 160 Congruence of Line Segments, Angles, and Triangles Exercises Writing About Mathematics 1. Each of two telephone poles is perpendicular to the ground and braced by a wire that extends from the top of the pole to a point on the level ground 5 feet from the foot of the pole. The wires used to brace the poles are of unequal lengths. Is it possible for the telephone poles to be of equal height? Explain your answer. 2. Is the following statement true? If two triangles are not congruent, then each pair of corre- sponding sides and each pair of corresponding angles are not congruent. Justify your answer. Developing Skills In 3–8, pairs of line segments marked with the same number of strokes are congruent. Pairs of angles marked with the same number of arcs are congruent. A line segment or an angle marked with an “” is congruent to itself by the reflexive property of congruence. In each case, is the given information sufficient to prove congruent triangles using SAS? 3. B F 4. A C D E 6. A C D B B D A 7. C C F A B D E 5. D C A B 8. A B E C D In 9–11, two sides or a side and an angle are marked to indicate that they are congruent. Name the pair of corresponding sides or corresponding angles that would have to be proved congruent in order to prove the triangles congruent by SAS. 9. D E A B C 10. C 11. A B A D B E D C 14365C04.pgs 7/12/07 3:05 PM Page 161 Proving Triangles Congruent Using Angle, Side, Angle 161 Applying Skills In 12–14: a. Draw a diagram with geometry software or pencil and paper and write a given state- ment using the information in the first sentence. b. Use the information in the second sentence to write a prove statement. c. Write a proof. and ABC DBE bisect each other. Prove that ABE CBD. 12. 13. ABCD is a quadrilateral; AB = CD; BC = DA; and DAB, ABC, BCD, and CDA are separates the quadrilateral into two congruent tri- AC right angles. Prove that the diagonal angles. 14. PQR and RQS are a linear pair of angles that are congruent and PQ QS. Prove that PQR RQS. 4-6 PROVING TRIANGLES CONGRUENT USING ANGLE, SIDE, ANGLE In the last section we saw that it is possible to prove two triangles congruent by proving that fewer than three pairs of sides and three pairs of angles are congruent, that is, by proving that two sides and the included angle of one triangle are congruent to the corresponding parts of another. There are also other ways of proving two triangles congruent. Hands-On Activity In this activity, we will use a protractor and ruler, or geometry software. Use the procedure below to draw a triangle given the measures of two angles and of the included side. STEP 1. Use the protractor or geometry software to draw an angle with the first given angle measure. Call the vertex of that angle A. STEP 2. Draw a segment with the given length along one of the rays of A. One of the endpoints of the segment should be A. Call the other endpoint B. A A B 14365C04.pgs 7/12/07 3:05 PM Page 162 162 Congruence of Line Segments, Angles, and Triangles STEP 3. Draw a second angle of the triangle using the other given angle measure. Let B be the ver- tex of this second angle and let the sides of this angle. h BA be one of STEP 4. Let C be the intersection of the rays of A and B that are not on g AB . STEP 5. Repeat steps 1 through 4. Let D the v |
ertex of the first angle, E the vertex of the second angle, and F the intersection of the rays of D and E. C A B F D E a. Follow the steps to draw two different triangles with each of the given angle-side-angle measures. (1) 65°, 4 in., 35° (2) 60°, 4 in., 60° (3) 120°, 9 cm, 30° (4) 30°, 9 cm, 30° b. For each pair of triangles, measure the angle and sides that were not given. Do they have equal measures? c. For each pair of triangles, you can conclude that the triangles are congruent. The triangles formed, ABC and DEF, can be placed on top of one another so that A and D coincide, B and E coincide, and C and F coincide. Therefore, ABC DEF. This activity leads to the following statement of angle-side-angle or ASA triangle congruence, whose truth will be assumed without proof: Postulate 4.15 Two triangles are congruent if two angles and the included side of one triangle are congruent, respectively, to two angles and the included side of the other. (ASA) Thus, in ABC and DEF, if B E, , and A D, it follows that ABC DEF. The postulate used here is abbreviated ASA. We will now use this postulate to prove two triangles congruent. BA > ED 14365C04.pgs 7/12/07 3:05 PM Page 163 Proving Triangles Congruent Using Angle, Side, Angle 163 EXAMPLE 1 Given: CED and AEB midpoint of BD ' BE . intersect at E, E is the AC ' AE , and AEB , Prove: AEC BDE Prove the triangles congruent by ASA. D B E A C Proof Statements Reasons g AEB 1. and g CED intersect at E. 1. Given. A 2. AEC BED 2. If two lines intersect, the vertical angles are congruent. 3. E is the midpoint of . AEB 3. Given. S 4. AE > BE AC ' AE and 5. 6. A and B are right angles. BD ' BE A 7. A B 8. AEC BED Exercises Writing About Mathematics 4. The midpoint of a line segment divides the segment into two congruent segments. 5. Given. 6. Perpendicular lines intersect to form right angles. 7. If two angles are right angles, they are congruent. 8. ASA (steps 2, 4, 7). 1. Marty said that if two triangles are congruent and one of the triangles is a right triangle, then the other triangle must be a right triangle. Do you agree with Marty? Explain why or why not. 2. In ABC and DEF, AB > DE , and /B > /E . C F a. Dora said that if , then it follows that ABC DEF. Do you agree with Dora? Justify your answer. is congruent to DF AC A B D E b. Using only the SAS and ASA postulates, if ABC is not congruent to DEF, what sides and what angles cannot be congruent? 14365C04.pgs 7/12/07 3:05 PM Page 164 164 Congruence of Line Segments, Angles, and Triangles Developing Skills In 3–5, tell whether the triangles in each pair can be proved congruent by ASA, using only the marked congruent parts in establishing the congruence. Give the reason for your answer. 3. C F 4. C 5 In 6–8, in each case name the pair of corresponding sides or the pair of corresponding angles that would have to be proved congruent (in addition to those pairs marked congruent) in order to prove that the triangles are congruent by ASA. 6. D C 7. D B 8. D A B Applying Skills E A C A C B 9. Given: E C, EDA CDB, . EC and D is the midpoint of Prove: DAE DBC A B 10. Given: h DB h BD bisects ADC and bisects ABC. Prove: ABD CBD D A C B and h AD bisects 12. Given: DBC GFD and AE E D C 11. Given: AD ' BC BAC. Prove: ADC ADB A C D B at D. bisects Prove: DFE DBA FB G F E D A B C 14365C04.pgs 7/12/07 3:05 PM Page 165 4-7 PROVING TRIANGLES CONGRUENT USING SIDE, SIDE, SIDE Proving Triangles Congruent Using Side, Side, Side 165 Cut three straws to any lengths and put their ends together to form a triangle. Now cut a second set of straws to the same lengths and try to form a different triangle. Repeated experiments lead to the conclusion that it cannot be done. As shown in ABC and DEF, when all three pairs of corresponding sides of a triangle are congruent, the triangles must be congruent. The truth of this statement of side-side-side or SSS triangle congruence is assumed without proof. C F A B D E Postulate 4.16 Two triangles are congruent if the three sides of one triangle are congruent, respectively, to the three sides of the other. (SSS) Thus, in ABC and DEF above, AB > DE , AC > DF , and BC > EF . It follows that ABC DEF. The postulate used here is abbreviated SSS. EXAMPLE 1 Given: Isosceles JKL with midpoint of . JL Prove: JKM LKM JK > KL and M the K Proof Prove the triangles congruent by using SSS. J M L Statements S 1. JK > KL 2. M is the midpoint of . JL S 3. JM > LM S 4. KM > KM 5. JKM LKM Reasons 1. Given. 2. Given. 3. Definition of midpoint. 4. Reflexive property of congruence. 5. SSS (steps 1, 3, 4). 14365C04.pgs 7/12/07 3:05 PM Page 166 166 Congruence of Line Segments, Angles, and Triangles Exercises Writing About Mathematics 1. Josh said that if two triangles are congruent and one of the triangles is isosceles, then the other must be isosceles. Do you agree with Josh? Explain why or why not. 2. Alvan said that if two triangles are not congruent, then at least one of the three sides of one triangle is not congruent to the corresponding side of the other triangle. Do you agree with Alvan? Justify your answer. Developing Skills In 3–5, pairs of line segments marked with the same number of strokes are congruent. A line segment marked with “” is congruent to itself by the reflexive property of congruence. In each case, is the given information sufficient to prove congruent triangles? 3. C A B D 4. C D F E B A 5. W Z R S T In 6–8, two sides are marked to indicate that they are congruent. Name the pair of corresponding sides that would have to be proved congruent in order to prove the triangles congruent by SSS. 6. B 7. D C 8 In 9–14, pairs of line segments marked with the same number of strokes are congruent. Pairs of angles marked with the same number of arcs are congruent. A line segment or an angle marked with “” is congruent to itself by the reflexive property of congruence. In each case, is the given information sufficient to prove congruent triangles? If so, write the abbreviation for the postulate that proves the triangles congruent. 9. B C D A 10. P S 11. D C T Q R A B 14365C04.pgs 8/2/07 5:40 PM Page 167 12. C 13. A M B A D B Chapter Summary 167 C 14. S R P Q 15. If two sides and the angle opposite one of those sides in a triangle are congruent to the corresponding sides and angle of another triangle, are the two triangles congruent? Justify your answer or draw a counterexample proving that they are not. (Hint: Could one triangle be an acute triangle and the other an obtuse triangle?) Applying Skills 16. Given: AEB bisects , CED AC ' CED , and BD ' CED . Prove: EAC EBD 17. Given: ABC is equilateral, D is the midpoint of AB . Prove: ACD BCD 18. Given: Triangle PQR with S on PQ and RS ' PQ ; PSR is not congruent to QSR. Prove: PS QS 19. Gina is drawing a pattern for a kite. She wants it to consist of two congruent triangles that share a common side. She draws an angle with its vertex at A and marks two points, B and C, one on each of the rays of the angle. Each point, B and C, is 15 inches from the vertex of the angle. Then she draws the bisector of BAC, marks a point D on the angle bisector and . Prove that the triangles that she drew are congruent. draws and BD CD CHAPTER SUMMARY Definitions to Know • Adjacent angles are two angles in the same plane that have a common vertex and a common side but do not have any interior points in common. • Complementary angles are two angles the sum of whose degree measures is 90. • Supplementary angles are two angles the sum of whose degree measures is 180. • A linear pair of angles are two adjacent angles whose sum is a straight angle. • Vertical angles are two angles in which the sides of one angle are opposite rays to the sides of the second angle. 14365C04.pgs 8/2/07 5:41 PM Page 168 168 Congruence of Line Segments, Angles, and Triangles • Two polygons are congruent if and only if there is a one-to-one correspondence between their vertices such that corresponding angles are congruent and corresponding sides are congruent. Corresponding parts of congruent polygons are congruent. Corresponding parts of congruent polygons are equal in measure. Postulates 4.1 A line segment can be extended to any length in either direction. 4.2 Through two given points, one and only one line can be drawn. (Two points determine a line.) 4.3 Two lines cannot intersect in more than one point. 4.4 One and only one circle can be drawn with any given point as center and the length of any given line segment as a radius. 4.5 At a given point on a given line, one and only one perpendicular can be drawn to the line. 4.6 From a given point not on a given line, one and only one perpendicular can be drawn to the line. 4.7 For any two distinct points, there is only one positive real number that is the length of the line segment joining the two points. (Distance Postulate) 4.8 The shortest distance between two points is the length of the line segment joining these two points. 4.9 A line segment has one and only one midpoint. 4.10 An angle has one and only one bisector. 4.11 Any geometric figure is congruent to itself. (Reflexive Property) 4.12 A congruence may be expressed in either order. (Symmetric Property) 4.13 Two geometric figures congruent to the same geometric figure are congru- ent to each other. (Transitive Property) 4.14 Two triangles are congruent if two sides and the included angle of one triangle are congruent, respectively, to two sides and the included angle of the other. (SAS) 4.15 Two triangles are congruent if two angles and the included side of one triangle are congruent, respectively, to two angles and the included side of the other. (ASA) 4.16 Two triangles are congruent if the three sides of one triangle are congru- ent, respectively, to the three sides of the other. (SSS) 4.1 If two angles are right angles, then they are congruent. 4.2 If two angles are straight angles, then they are congruent. 4.3 If two angles are complements of the same angle, then they are congruent. 4.4 If two angles are co |
ngruent, then their complements are congruent. 4.5 If two angles are supplements of the same angle, then they are congruent. 4.6 If two angles are congruent, then their supplements are congruent. 4.7 If two angles form a linear pair, then they are supplementary. 4.8 If two lines intersect to form congruent adjacent angles, then they are perpendicular. 4.9 If two lines intersect, then the vertical angles are congruent. Theorems 14365C04.pgs 7/12/07 3:05 PM Page 169 Review Exercises 169 VOCABULARY 4-1 Distance postulate 4-3 Adjacent angles • Complementary angles • Complement • Supplementary angles • Supplement • Linear pair of angles • Vertical angles 4-4 One-to-one correspondence • Corresponding angles • Corresponding sides • Congruent polygons 4-5 SAS triangle congruence 4-6 ASA triangle congruence 4-7 SSS triangle congruence REVIEW EXERCISES 1. The degree measure of an angle is 15 more than twice the measure of its complement. Find the measure of the angle and its complement. 2. Two angles, LMP and PMN, are a linear pair of angles. If the degree measure of LMP is 12 less than three times that of PMN, find the measure of each angle. 3. Triangle JKL is congruent to triangle PQR and mK 3a 18 and mQ 5a 12. Find the measure of K and of Q. 4. If and ABC DBE tulate that justifies your answer. intersect at F, what is true about B and F? State a pos- 5. If g LM ' MN and g KM ' MN , what is true about g LM and g KM ? State a postulate that justifies your answer. 6. Point R is not on LMN . Is LM MN less than, equal to, or greater than LR RN? State a postulate that justifies your answer. 7. If h BD and h BE are bisectors of ABC, does E lie on g BD ? State a postulate that justifies your answer. 8. The midpoint of AB is M. If g MN and g PM are bisectors of AB , does P lie g MN on ? Justify your answer. 9. The midpoint of AB is M. If g MN and g PM are perpendicular to AB , does P lie on ? Justify your answer. g MN 14365C04.pgs 8/2/07 5:41 PM Page 170 170 Congruence of Line Segments, Angles, and Triangles 10. Given: mA 50, mB 45, AB = 10 cm, mD 50, mE 45, and DE 10 cm. Prove: ABC DEF E 11. Given: bisects GEH mD mF. DEF and Prove: GFE HDE D H C A B D F G 12. Given: AB > DE not congruent to DEF. BC > EF , , ABC is E F C F Prove: B is not congruent to E. Exploration A B D E 1. If three angles of one triangle are congruent to the corresponding angles of another triangle, the triangles may or may not be congruent. Draw diagrams to show that this is true. 2. STUVWXYZ is a cube. Write a paragraph proof that would convince someone that STX, UTX, and STU are all congruent to one another. Z V W S X T Y U CUMULATIVE REVIEW CHAPTERS 1–4 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following is an illustration of the associative property of addition? (1) 3(4 7) 3(7 4) (2) 3(4 7) 3(4) 3(7) (3) 3 (4 7) 3 (7 4) (4) 3 (4 7) (3 4) 7 14365C04.pgs 7/12/07 3:05 PM Page 171 Cumulative Review 171 2. If the sum of the measures of two angles is 90, the angles are (1) supplementary. (2) complementary. (3) a linear pair. (4) adjacent angles. 3. If AB BC AC, which of the following may be false? . AC (1) B is the midpoint of (2) B is a point of AC . (3) B is between A and C. h BC h BA and (4) are opposite rays. 4. If b is a real number, then b has a multiplicative inverse only if (1) b 1 (2) b 0 (3) b 0 (4) b 0 5. The contrapositive of “Two angles are congruent if they have the same measures” is (1) Two angles are not congruent if they do not have the same measures. (2) If two angles have the same measures, then they are congruent. (3) If two angles are not congruent, then they do not have the same measures. (4) If two angles do not have the same measures, then they are not congruent. 6. The statement “Today is Saturday and I am going to the movies” is true. Which of the following statements is false? (1) Today is Saturday or I am not going to the movies. (2) Today is not Saturday or I am not going to the movies. (3) If today is not Saturday, then I am not going to the movies. (4) If today is not Saturday, then I am going to the movies. 7. If ABC BCD, then ABC and BCD must be (2) scalene. (3) isosceles. (4) equilateral. (1) obtuse. g ABC 8. If and g DBE (1) congruent vertical angles. (2) supplementary vertical angles. intersect at B, ABD and CBE are (3) congruent adjacent angles. (4) supplementary adjacent angles. 9. LMN and NMP form a linear pair of angles. Which of the following statements is false? (1) mLMN mNMP 180 (2) LMN and NMP are supplementary angles. h ML h ML (3) (4) and and h MP h MN are opposite rays. are opposite rays. 10. The solution set of the equation 3(x 2) 5x is (1) {x x 3} (2) {x x 3} (3) {x x 1} (4) {x x 1} 14365C04.pgs 7/12/07 3:05 PM Page 172 172 Congruence of Line Segments, Angles, and Triangles Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Given: bisects PQ RS and R S. at M 12. Given: Quadrilateral DEFG with DE DG and EF GF. Prove: RMQ SMP Prove: DEF DGF P E R M S G D F Q Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. The following statements are true: If our team does not win, we will not celebrate. We will celebrate or we will practice. We do not practice. Did our team win? Justify your answer. 14. The two angles of a linear pair of angles are congruent. If the measure of one angle is represented by 2x y and the measure of the other angle by x 4y, find the values of x and of y. 14365C04.pgs 7/12/07 3:05 PM Page 173 Cumulative Review 173 Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Josie is making a pattern for quilt pieces. One pattern is a right triangle with two acute angles that are complementary. The measure of one of the acute angles is to be 12 degrees more than half the measure of the other acute angle. Find the measure of each angle of the triangle. 16. Triangle DEF is equilateral and equiangular. The midpoint of is L. Line segments is N, and of DE are drawn. , and MN ML FD NL EF , is M, of a. Name three congruent triangles. b. Prove that the triangles named in a are congruent. c. Prove that NLM is equilateral. d. Prove that NLM is equiangular. 14365C05.pgs 7/10/07 8:41 AM Page 174 CHAPTER 5 CHAPTER TABLE OF CONTENTS 5-1 Line Segments Associated with Triangles 5-2 Using Congruent Triangles to Prove Line Segments Congruent and Angles Congruent 5-3 Isosceles and Equilateral Triangles 5-4 Using Two Pairs of Congruent Triangles 5-5 Proving Overlapping Triangles Congruent 5-6 Perpendicular Bisector of a Line Segment 5-7 Basic Constructions Chapter Summary Vocabulary Review Exercises Cumulative Review 174 CONGRUENCE BASED ON TRIANGLES The SSS postulate tells us that a triangle with sides of given lengths can have only one size and shape. Therefore, the area of the triangle is determined. We know that the area of a triangle is one-half the product of the lengths of one side and the altitude to that side. But can the area of a triangle be found using only the lengths of the sides? A formula to do this was known by mathematicians of India about 3200 B.C. In the Western world, Heron of Alexandria, who lived around 75 B.C., provided in his book Metrica a formula that we now call Heron’s formula: If A is the area of the triangle with sides of length a, b, and c, and the semiperimeter, s, is one-half the perimeter, that is, s (a b c), then 1 2 A 5 " s(s 2 a)(s 2 b)(s 2 c) In Metrica, Heron also provided a method of finding the approximate value of the square root of a number. This method, often called the divide and average method, continued to be used until calculators made the pencil and paper computation of a square root unnecessary. 14365C05.pgs 7/10/07 8:41 AM Page 175 5-1 LINE SEGMENTS ASSOCIATED WITH TRIANGLES Line Segments Associated with Triangles 175 Natalie is planting a small tree. Before filling in the soil around the tree, she places stakes on opposite sides of the tree at equal distances from the base of the tree. Then she fastens cords from the same point on the trunk of the tree to the stakes. The cords are not of equal length. Natalie reasons that the tree is not perpendicular to the ground and straightens the tree until the cords are of equal lengths. Natalie used her knowledge of geometry to help her plant the tree. What was the reasoning that helped Natalie to plant the tree? Geometric shapes are all around us. Frequently we use our knowledge of geometry to make decisions in our daily life. In this chapter you will write formal and informal proofs that will enable you to form the habit of looking for logical relationships before making a decision. Altitude of a Triangle DEFINITION An altitude of a triangle is a line segment drawn from any vertex of the triangle, perpendicular to and ending in the line that contains the opposite side In ABC, if K AB CD CD GH g EF , then , then is perpendicular to is the is perpendicular to altitude from vertex C to the opposite side. In EFG, if , the line that contains the side GH is the altitude from vertex G to the opposite EF side. In an obtuse triangle such as EFG above, the altitude from each of the acute angles lies outside the triangle. In right TSR, if is the altitude from vertex R to the opposite |
side . RS In a right triangle such as TSR above, the altitude from each vertex of an acute angle is a leg of the triangle. Every triangle has three altitudes as shown in JKL. is perpendicular to TS TS is the altitude from T to the opposite side , then and RS RS TS L J Median of a Triangle DEFINITION A median of a triangle is a line segment that joins any vertex of the triangle to the midpoint of the opposite side. 14365C05.pgs 7/10/07 8:41 AM Page 176 176 Congruence Based on Triangles angle bisector P D Q In ABC, if M is the midpoint of is the median drawn from CM . We may also draw a median from vertex A to the midpoint . Thus, every , and a median from vertex B to the midpoint of side , then AB AC AB vertex C to side of side BC triangle has three medians. Angle Bisector of a Triangle DEFINITION An angle bisector of a triangle is a line segment that bisects any angle of the triangle and terminates in the side opposite that angle. In PQR, if D is a point on is the angle bisector from R in PQR. We may also draw an angle bisector from the , and an angle bisector from the vertex Q to some vertex P to some point on point on . Thus, every triangle has three angle bisectors. such that PRD QRD, then RD QR PQ PR In a scalene triangle, the altitude, the median, and the angle bisector drawn from any common vertex are three distinct line segments. In ABC, from the common vertex B, three line segments are drawn: B 1. 2. 3. is the altitude from B because BD BD ' AC . is the angle bisector from B BE because ABE EBC. BF is the midpoint of is the median from B because F AC . A F E D C median angle bisector altitude In some special triangles, such as an isosceles triangle and an equilateral triangle, some of these segments coincide, that is, are the same line. We will consider these examples later. EXAMPLE 1 Given: KM and is the angle bisector from K in JKL, LK > JK . K Prove: JKM LKM L M J 14365C05.pgs 7/10/07 8:41 AM Page 177 Line Segments Associated with Triangles 177 Proof Statements Reasons 1. LK > JK 2. is the angle bisector from KM K in JKL. 1. Given. 2. Given. 3. KM bisects JKL. 3. Definition of an angle bisector of a triangle. 4. JKM LKM 4. Definition of the bisector of an angle. KM > KM 5. 6. JKM LKM 5. Reflexive property of congruence. 6. SAS (steps 1, 4, 5). Exercises Writing About Mathematics 1. Explain why the three altitudes of a right triangle intersect at the vertex of the right angle. 2. Triangle ABC is a triangle with C an obtuse angle. Where do the lines containing the three altitudes of the triangle intersect? Developing Skills 3. Use a pencil, ruler, and protractor, or use geometry software, to draw ABC, an acute, sca- lene triangle with altitude CD , angle bisector CE , and median . CF a. Name two congruent angles that have their vertices at C. b. Name two congruent line segments. c. Name two perpendicular line segments. d. Name two right angles. 4. Use a pencil, ruler, and protractor, or use geometry software, to draw several triangles. Include acute, obtuse, and right triangles. a. Draw three altitudes for each triangle. b. Make a conjecture regarding the intersection of the lines containing the three altitudes. 5. Use a pencil, ruler, and protractor, or use geometry software, to draw several triangles. Include acute, obtuse, and right triangles. a. Draw three angle bisectors for each triangle. b. Make a conjecture regarding the intersection of these three angle bisectors. 14365C05.pgs 7/10/07 8:41 AM Page 178 178 Congruence Based on Triangles 6. Use a pencil, ruler, and protractor, or use geometry software, to draw several triangles. Include acute, obtuse, and right triangles. a. Draw three medians to each triangle. b. Make a conjecture regarding the intersection of these three medians. In 7–9, draw and label each triangle described. Complete each required proof in two-column format. 7. Given: In PQR, PR > QR , P Q, and RS Prove: PSR QSR is a median. 8. Given: In DEF, EG bisector and an altitude. is both an angle Prove: DEG FEG 9. Given: CD Prove: CD is a median of ABC but ADC is not congruent to BDC. is not an altitude of ABC. (Hint: Use an indirect proof.) Applying Skills In 10–13, complete each required proof in paragraph format. 10. In a scalene triangle, LNM, show that an altitude, NO , cannot be an angle bisector. (Hint: Use an indirect proof.) 11. A telephone pole is braced by two wires that are fastened to the pole at point C and to the . If the ground at points A and B. The base of the pole is at point D, the midpoint of pole is perpendicular to the ground, are the wires of equal length? Justify your answer. AB 12. The formula for the area of a triangle is A and h the length of the altitude to that side. In ABC, side and M is the midpoint of angles of equal area, AMC and BMC. AB AB 1 2bh with b the length of one side of a triangle is the altitude from vertex C to . Show that the median separates ABC into two tri- CD 13. A farmer has a triangular piece of land that he wants to separate into two sections of equal area. How can the land be divided? 5-2 USING CONGRUENT TRIANGLES TO PROVE LINE SEGMENTS CONGRUENT AND ANGLES CONGRUENT The definition of congruent triangles tells us that when two triangles are congruent, each pair of corresponding sides are congruent and each pair of corresponding angles are congruent. We use three pairs of corresponding parts, SAS, ASA, or SSS, to prove that two triangles are congruent. We can then conclude that each of the other three pairs of corresponding parts are also congruent. In this section we will prove triangles congruent in order to prove that two line segments or two angles are congruent. 14365C05.pgs 7/10/07 8:41 AM Page 179 Using Congruent Triangles to Prove Line Segments Congruent and Angles Congruent 179 EXAMPLE 1 Given: , A is a right angle, AE > DF , ABCD D is a right angle, AB > CD . Prove: EC > FB E F Proof The line segments that we want to prove congruent are corresponding sides of EAC and FDB. Therefore we will first prove that EAC FDB. Then use that corresponding parts of congruent triangles are congruent. A C B D Statements Reasons 1. A is a right angle, D is a 1. Given. right angle. 2. A D 3. AE > DF 4. AB > CD 2. If two angles are right angles, then they are congruent. (Theorem 4.1) 3. Given. 4. Given. 5. AB 1 BC > BC 1 CD 5. Addition postulate. 6. ABCD 7. AB 1 BC 5 AC BC 1 CD 5 BD AC > BD 8. 9. EAC FDB 10. EC > FB 6. Given. 7. Partition postulate. 8. Substitution postulate (steps 5, 7). 9. SAS (steps 3, 2, 8). 10. Corresponding parts of congruent triangles are congruent. Exercises Writing About Mathematics 1. Triangles ABC and DEF are congruent triangles. If A and B are complementary angles, are D and E also complementary angles? Justify your answer. 2. A leg and the vertex angle of one isosceles triangle are congruent respectively to a leg and the vertex angle of another isosceles triangle. Is this sufficient information to conclude that the triangles must be congruent? Justify your answer. 14365C05.pgs 7/10/07 8:41 AM Page 180 180 Congruence Based on Triangles Developing Skills In 3–8, the figures have been marked to indicate pairs of congruent angles and pairs of congruent segments. a. In each figure, name two triangles that are congruent. b. State the reason why the triangles are congruent. c. For each pair of triangles, name three additional pairs of parts that are congruent because they are corresponding parts of congruent triangles. 3. M R P S 6. D C E A B 9. Given: CA > CB of . AB Prove: A B C 4. D A C B 7. X S R P Q Y 5. D C E A B 8. D C E F A B and D is the midpoint 10. Given: Prove: and AB > CD CAB ACD AD > CB D C A D B A B CED 11. Given: and AEB other. Prove: C D B C E D A bisect each 12. Given: KLM and NML are right angles and KL = NM. Prove: K N K L N M 14365C05.pgs 7/10/07 8:41 AM Page 181 Isosceles and Equilateral Triangles 181 13. Triangle ABC is congruent to triangle DEF, AB 3x 7, DE 5x 9, and BC 4x. Find: a. x b. AB c. BC d. EF 14. Triangle PQR is congruent to triangle LMN, mP 7a, mL 4a 15, and P and Q are complementary. Find: a. a b. mP c. mQ d. mM Applying Skills In 15 and 16, complete each required proof in paragraph format. 15. a. Prove that the median from the vertex angle of an isosceles triangle separates the triangle into two congruent triangles. b. Prove that the two congruent triangles in a are right triangles. 16. a. Prove that if each pair of opposite sides of a quadrilateral are congruent, then a diagonal of the quadrilateral separates it into two congruent triangles. b. Prove that a pair of opposite angles of the quadrilateral in a are congruent. In 17 and 18, complete each required proof in two-column format. 17. a. Points B and C separate into three congruent segments. P is a point not on such that PA > PD and . Draw a diagram that shows these line segments and ABCD PB > PC g AD write the information in a given statement. b. Prove that APB DPC. c. Prove that APC DPB. 18. The line segment PM is both the altitude and the median from P to LN in LNP. a. Prove that LNP is isosceles. b. Prove that PM is also the angle bisector from P in LNP. 5-3 ISOSCELES AND EQUILATERAL TRIANGLES When working with triangles, we observed that when two sides of a triangle are congruent, the median, the altitude, and the bisector of the vertex angle separate the triangle into two congruent triangles. These congruent triangles can be used to prove that the base angles of an isosceles triangle are congruent. This observation can be proved as a theorem called the Isosceles Triangle Theorem. 14365C05.pgs 7/10/07 8:41 AM Page 182 182 Congruence Based on Triangles Theorem 5.1 If two sides of a triangle are congruent, the angles opposite these sides are congruent. Given ABC with AC > BC Prove A B C Proof In order to prove this theorem, we will use the median to the base to separate the triangle into two congruent triangles. A D B Statements Reasons 1. Draw D, the midpoint of . AB 1. A line segment has one and only one midpoint. 2. is |
the median from CD vertex C. 2. Definition of a median of a triangle. 3. CD > CD 4. AD > DB AC > BC 5. 6. ACD BCD 7. A B 3. Reflexive property of congruence. 4. Definition of a midpoint. 5. Given. 6. SSS (steps 3, 4, 5). 7. Corresponding parts of congruent triangles are congruent. A corollary is a theorem that can easily be deduced from another theorem. We can prove two other statements that are corollaries of the isosceles triangle theorem because their proofs follow directly from the proof of the theorem. Corollary 5.1a The median from the vertex angle of an isosceles triangle bisects the vertex angle. Proof: From the preceding proof that ACD BCD, we can also conclude that ACD BCD since they, too, are corresponding parts of congruent triangles. Corollary 5.1b The median from the vertex angle of an isosceles triangle is perpendicular to the base. 14365C05.pgs 7/10/07 8:41 AM Page 183 Isosceles and Equilateral Triangles 183 Proof: Again, from ACD BCD, we can say that CDA CDB because they are corresponding parts of congruent triangles. If two lines intersect to form congruent adjacent angles, then they are perpendicular. Therefore, CD ' AB . Properties of an Equilateral Triangle The isosceles triangle theorem has shown that in an isosceles triangle with two congruent sides, the angles opposite these sides are congruent. We may prove another corollary to this theorem for any equilateral triangle, where three sides are congruent. Corollary 5.1c Every equilateral triangle is equiangular. Proof: If ABC is equilateral, then By the isosceles triangle theorem, since A C, and since fore, A B C. AB > BC > CA . AB > BC , , B A. There- BC > CA Given: E not on ABCD , AB > CD , and EB > EC . Prove: AE > DE C B A E DE and are corresponding sides of AE ABE and DCE, and we will prove these triangles congruent by SAS. We are given two pairs of congruent corresponding sides and must prove that the included angles are congruent. A B C D EXAMPLE 1 Proof E Statements EB > EC 1. 2. EBC ECB A B C D 3. ABCD Reasons 1. Given. 2. Isosceles triangle theorem (Or: If two sides of a triangle are congruent, the angles opposite these sides are congruent.). 3. Given. Continued 14365C05.pgs 7/10/07 8:41 AM Page 184 184 Congruence Based on Triangles (Continued) E Statements 4. ABE and EBC are supplementary. DCE and ECB are supplementary. 5. ABE DCE A B C D AB > CD 6. 7. ABE DCE 8. AE > DE Reasons 4. If two angles form a linear pair, they are supplementary. 5. The supplements of congruent angles are congruent. 6. Given. 7. SAS (steps 1, 5, 6). 8. Corresponding parts of congruent triangles are congruent. Exercises Writing About Mathematics 1. Joel said that the proof given in Example 1 could have been done by proving that ACE DBE. Do you agree with Joel? Justify your answer. 2. Abel said that he could prove that equiangular triangle ABC is equilateral by drawing and showing that ABD CBD. What is wrong with Abel’s reasoning? median BD Developing Skills 3. In ABC, if 4. Triangle RST is an isosceles right triangle with RS ST and R and T complementary , mB 3x 15 and mC 7x 5, find mB and mC. AB > AC angles. What is the measure of each angle of the triangle? 5. In equilateral DEF, mD 3x y, mE 2x 40, and mF 2y. Find x, y, mD, mE, and mF. DABE 6. Given: C not on CA > CB Prove: CAD CBE and D A B E C 7. Given: Quadrilateral ABCD with AD > CD AB > CB Prove: BAD BCD and A B D C 14365C05.pgs 7/10/07 8:41 AM Page 185 Isosceles and Equilateral Triangles 185 8. Given: AC > CB Prove: CDE CED and DA > EB 9. Given: AC > BC Prove: CAD CBD and DAB DBA C D E C D A B A B 10. Given: Isosceles ABC with AC > BC and F is the midpoint of , D is the midpoint of AB . BC E is the midpoint of a. Prove: ADF BEF b. Prove: DEF is isosceles. AC , C D E A F B In 11 and 12, complete each given proof with a partner or in a small group. 11. Given: ABC with AB AC, BG EC, , and 12. Given: E not on BE ' DE CG ' GF and EB . AB > CD , , ABCD is not congruent to . EC Prove: BD > CF Prove: AE is not congruent to DE . A D F B E G C Applying Skills E A B C D (Hint: Use an indirect proof.) 13. Prove the isosceles triangle theorem by drawing the bisector of the vertex angle instead of the median. 14. Prove that the line segments joining the midpoints of the sides of an equilateral triangle are congruent. › ‹ 15. C is a point not on FBDG 16. In PQR, mR mQ. Prove that PQ PR. and BC DC. Prove that FBC GDC. 14365C05.pgs 7/10/07 8:41 AM Page 186 186 Congruence Based on Triangles 5-4 USING TWO PAIRS OF CONGRUENT TRIANGLES Often the line segments or angles that we want to prove congruent are not corresponding parts of triangles that can be proved congruent by using the given information. However, it may be possible to use the given information to prove a different pair of triangles congruent. Then the congruent corresponding parts of this second pair of triangles can be used to prove the required triangles congruent. The following is an example of this method of proof. EXAMPLE 1 Given: , AEB AC > AD , and CB > DB Prove: CE > DE C A B E Proof Since CE DE and are corresponding parts of ACE and ADE, we can prove these two line segments congruent if we can prove ACE and ADE congruent. From the given, we cannot prove immediately that ACE and ADE congruent. However, we can prove that CAB DAB. Using corresponding parts of these larger congruent triangles, we can then prove that the smaller triangles are congruent. D Statements Reasons A A C D C D 1. AC > AD 2. CB > DB 3. AB > AB B E 4. CAB DAB 5. CAB DAB B 6. AE > AE E 7. CAE DAE 8. CE > DE 1. Given. 2. Given. 3. Reflexive property of congruence. 4. SSS (steps 1, 2, 3). 5. Corresponding parts of congruent triangles are congruent. 6. Reflexive property of congruence. 7. SAS (steps 1, 5, 6). 8. Corresponding parts of congruent triangles are congruent. 14365C05.pgs 7/10/07 8:41 AM Page 187 Using Two Pairs of Congruent Triangles 187 Exercises Writing About Mathematics 1. Can Example 1 be proved by proving that BCE BDE? Justify your answer. 2. Greg said that if it can be proved that two triangles are congruent, then it can be proved that the medians to corresponding sides of these triangles are congruent. Do you agree with Greg? Explain why or why not. Developing Skills 3. Given: ABC DEF, M is the , and N is . DE midpoint of AB the midpoint of Prove: AMC DNF B M A C F 4. Given: ABC DEF, bisects ACB, and bisects DFE. CG FH Prove: CG > FH . Given: AEC FEG and DEB intersects bisect each other, at G and AB CD at F. 6. Given: AME BMF and DE > CF Prove: E is the midpoint of . FEG Prove: AD > BC 14365C05.pgs 7/10/07 8:41 AM Page 188 188 Congruence Based on Triangles 7. Given: Prove: BC > BA h DB bisects CDA. h BD and bisects CBA. 8. Given: RP RQ and SP SQ Prove: RT ' PQ R S P T Q B A C D Applying Skills 9. In quadrilateral ABCD, AB = CD, BC = DA, and M is the midpoint of through M intersects AB at E and CD at F. Prove that BMD bisects 10. Complete the following exercise with a partner or in a small group: BD EMF . A line segment at M. Line l intersects AB SA = SB, then M is the midpoint of AB and l is perpendicular to AB . at M, and P and S are any two points on l. Prove that if PA = PB and a. Let half the group treat the case in which P and S are on the same side of AB . b. Let half the group treat the case in which P and S are on opposite sides of AB . c. Compare and contrast the methods used to prove the cases. 5-5 PROVING OVERLAPPING TRIANGLES CONGRUENT AD > BC DB > CA , can we prove If we know that and that DBA CAB? These two triangles overlap and share a common side. To make it easier to visualize the overlapping triangles that we want to prove congruent, it may be helpful to outline each with a different color as shown in the figure. AB Or the triangles can be redrawn as separate triangles. is a side of The segment each of the triangles DBA and CAB. Therefore, to the given inforDB > CA mation, , we and prove that can add DBA CAB by SSS. AB > AB AD > BC and 14365C05.pgs 7/10/07 8:41 AM Page 189 EXAMPLE 1 Given: BE and are medians CD to the legs of isosceles ABC. Proving Overlapping Triangles Congruent 189 A A A D E DE Prove: CD > BE B BC C Proof Separate the triangles to see more clearly the triangles to be proved congruent. We know that the legs of an isosceles triangle are congruent. Therefore, AB > AC . We also know that the median is a line segment from a vertex to the midpoint of the opposite side. Therefore, D and E are midpoints of the congruent legs. The midpoint divides the line segment into two congruent segments, that is, in half, and halves of congruent segments are congruent: AE > AD AE > AD is A, and A is congruent to itself by the reflexive property of congruence. Therefore, ABE ACD by SAS and because corresponding parts of congruent triangles are congruent. . Now we have two pair of congruent sides: . The included angle between each of these pairs of congruent sides AB > AC CD > BE and EXAMPLE 2 Solution Using the results of Example 1, find the length of CD x 15. BE if BE 5x 9 and BE CD 5x 9 x 15 4x 24 x 6 BE 5x 9 5(6) 9 30 9 21 CD x 15 6 15 21 Answer 21 Exercises Writing About Mathematics 1. In Example 1, the medians to the legs of isosceles ABC were proved to be congruent by proving ABE ACD. Could the proof have been done by proving DBC ECB? Justify your answer. 14365C05.pgs 7/10/07 8:41 AM Page 190 190 Congruence Based on Triangles 2. In Corollary 5.1b, we proved that the median to the base of an isosceles triangle is also the altitude to the base. If the median to a leg of an isosceles triangle is also the altitude to the leg of the triangle, what other type of triangle must this triangle be? Developing Skills 3. Given: AE > FB , DA > CB , , AEFB and A and B are right angles. DF > CE Prove: DAF CBE and 4. Given: SPR > SQT , Prove: SRQ STP and R T PR > QT . Given: DA > CB CB ' AB , DA ' AB , and 6. Given: AC > BD Prove: , BAE CBF, ABCD BCE CDF, AE > BF and E F AB > CD Prove: DAB CBA and TR 8. Given: AD > CE Prove: ADC CEA and DB > EB 7. Given: TM |
> TN and N is the midpoint of RN > SM Prove: , M is the midpoint of . TS Applying Skills In 9–11, complete each required proof in paragraph format. 9. Prove that the angle bisectors of the base angles of an isosceles triangle are congruent. 10. Prove that the triangle whose vertices are the midpoints of the sides of an isosceles triangle is an isosceles triangle. 11. Prove that the median to any side of a scalene triangle is not the altitude to that side. 14365C05.pgs 7/10/07 8:41 AM Page 191 Perpendicular Bisector of a Line Segment 191 5-6 PERPENDICULAR BISECTOR OF A LINE SEGMENT Perpendicular lines were defined as lines that intersect to form right angles. We also proved that if two lines intersect to form congruent adjacent angles, then they are perpendicular. (Theorem 4.8) The bisector of a line segment was defined as any line or subset of a line that intersects a line segment at its midpoint In the diagrams, g , PM g , NM QM , and h MR are all bisectors of AB since they each intersect AB at its midpoint, M. Only is both perpendicular to AB and g NM g NM is the perpendicular bisector of . AB the bisector of AB . DEFINITION The perpendicular bisector of a line segment is any line or subset of a line that is perpendicular to the line segment at its midpoint. In Section 3 of this chapter, we proved as a corollary to the isosceles triangle theorem that the median from the vertex angle of an isosceles triangle is perpendicular to the base. In the diagram below, since g CM CM ' AB . Therefore, ABC, CM is the median to the base of isosceles is the perpendicular bisector of . AB (1) M is the midpoint of AB : AM = MB. C M is equidistant, or is at an equal distance, from the endpoints of AB . (2) AB is the base of isosceles ABC: AC = BC. C is equidistant from the endpoints of AB . These two points, M and C, determine the perpendicular bisector of . This suggests the following theorem. AB A M B 14365C05.pgs 7/10/07 8:41 AM Page 192 192 Congruence Based on Triangles Theorem 5.2 If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment. Given and points P and T such that PA = PB and AB TA = TB. Prove g PT is the perpendicular bisector of . AB A Strategy Let g PT AB intersect at M. Prove APT BPT by SSS. Then, using the congruent corresponding angles, prove APM BPM by SAS. Consequently, is a bisector. Also, AMP BMP. AM > MB Since two lines that intersect to form congruent adjacent g PT , so angles are perpendicular, the perpendicular bisector of g AB ' PT AB . . Therefore, g PT is B P T A B P M T The details of the proof of Theorem 5.2 will be left to the student. (See exercise 7.) Theorem 5.3a If a point is equidistant from the endpoints of a line segment, then it is on the perpendicular bisector of the line segment. Given Point P such that PA = PB. Prove P lies on the perpendicular bisector of . AB Proof Choose any other point that is equidistant from AB . Then AB by Theorem 5.2. (If two points , for example, M, the midg is the perpendicular PM the endpoints of point of AB bisector of are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment.) P lies on g PM . The converse of this theorem is also true. P P B A A M B 14365C05.pgs 7/10/07 8:41 AM Page 193 Theorem 5.3b If a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the line segment. Perpendicular Bisector of a Line Segment 193 Given Point P on the perpendicular bisector of . AB Prove PA = PB AB Proof Let M be the midpoint of PM > PM AM > BM . Then . Perpendicular lines intersect to and form right angles, so PMA PMB. By SAS, PMA PMB. Since corresponding parts of congruent triangles are congruent, PA PB. PA > PB and C P M D B A Theorems 5.3a and 5.3b can be written as a biconditional. Theorem 5.3 A point is on the perpendicular bisector of a line segment if and only if it is equidistant from the endpoints of the line segment. Methods of Proving Lines or Line Segments Perpendicular To prove that two intersecting lines or line segments are perpendicular, prove that one of the following statements is true: 1. The two lines form right angles at their point of intersection. 2. The two lines form congruent adjacent angles at their point of intersection. 3. Each of two points on one line is equidistant from the endpoints of a seg- ment of the other. Intersection of the Perpendicular Bisectors of the Sides of a Triangle When we draw the three perpendicular bisectors of the sides of a triangle, it appears that the three lines are concurrent, that is, they intersect in one point. Theorems 5.3a and 5.3b allow us to prove the following perpendicular bisector concurrence theorem. C N P L A M B 14365C05.pgs 7/10/07 8:41 AM Page 194 194 Congruence Based on Triangles Theorem 5.4 The perpendicular bisectors of the sides of a triangle are concurrent. Given g MQ , the perpendicular bisector of AB C C Proof EXAMPLE 1 g NR , the perpendicular bisector of AC g LS , the perpendicular bisector of BC Prove g , MQ g NR , and g LS intersect in P MQ A (1) We can assume from the diagram that and g NR intersect. Let us call the point of intersection P. (2) By theorem 5.3b, since P is a point on g MQ , the perpendicular bisector of AB , P is equidistant from A and B. (3) Similarly, since P is a point on g NR , the perpendicular bisector of AC , P is equidistant from A and C. (4) In other words, P is equidistant from A, B, and C. By theorem 5.3a, since P , P is on the perpendicular bisector BC is equidistant from the endpoints of of BC . (5) Therefore, g MQ , g NR , and g LS , the three perpendicular bisectors of ABC, intersect in a point, P. The point where the three perpendicular bisectors of the sides of a triangle intersect is called the circumcenter. Prove that if a point lies on the perpendicular bisector of a line segment, then the point and the endpoints of the line segment are the vertices of an isosceles triangle. Given: P lies on the perpendicular bisector of . RS Prove: RPS is isosceles. P R S 14365C05.pgs 7/10/07 8:41 AM Page 195 Perpendicular Bisector of a Line Segment 195 Proof Statements Reasons 1. P lies on the perpendicular RS bisector of . 2. PR = PS 3. PR > PS 4. RPS is isosceles. 1. Given. 2. If a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the line segment. (Theorem 5.3b) 3. Segments that have the same mea- sure are congruent. 4. An isosceles triangle is a triangle that has two congruent sides. Exercises Writing About Mathematics 1. Justify the three methods of proving that two lines are perpendicular given in this section. 2. Compare and contrast Example 1 with Corollary 5.1b, “The median from the vertex angle of an isosceles triangle is perpendicular to the base.” Developing Skills 3. If RS ASB is the perpendicular bisector of , prove that ARS BRS. R S B A 5. Polygon ABCD is equilateral (AB = BC = CD = DA). Prove that AC BD other and are perpendicular to each other. and bisect each D C E B A 4. If PR PS and QR QS, prove PQ ' RS and RT ST. that R P T Q S 6. Given CED and ADB with ACE BCE and AED BED, prove that is the perpendicular bisector of CED . ADB C E A D B 14365C05.pgs 7/10/07 8:41 AM Page 196 196 Congruence Based on Triangles Applying Skills 7. Prove Theorem 5.2. 8. Prove that if the bisector of an angle of a triangle is perpendicular to the opposite side of the triangle, the triangle is isosceles. 9. A line through one vertex of a triangle intersects the opposite side of the triangle in adjacent angles whose measures are represented by 27 and 15. Is the line perpendicular to the side of the triangle? Justify your answer. 3 2a 1 2a 5-7 BASIC CONSTRUCTIONS A geometric construction is a drawing of a geometric figure done using only a pencil, a compass, and a straightedge, or their equivalents. A straightedge is used to draw a line segment but is not used to measure distance or to determine equal distances. A compass is used to draw circles or arcs of circles to locate points at a fixed distance from given point. The six constructions presented in this section are the basic procedures used for all other constructions. The following postulate allows us to perform these basic constructions: Postulate 5.1 Radii of congruent circles are congruent. Construction 1 Construct a Line Segment Congruent to a Given Line Segment. Given AB Construct CD , a line segment congruent to .AB . With a straightedge, draw a ray, h .CX 2. Open the compass so that the point is on A and the point of the pencil is on B. 3. Using the same compass radius, place the point on C and, with the pencil, draw an arc that h CX intersects of intersection D. . Label this point 14365C05.pgs 8/2/07 5:43 PM Page 197 Conclusion CD > AB Proof Since AB and CD are radii of congruent circles, they are congruent. Basic Constructions 197 Construction 2 Construct an Angle Congruent to a Given Angle. Given A Construct EDF BAC A D 1. Draw a ray with endpoint D. C B F E A D 4. Draw h .DF . With A as center, draw an arc that intersects each ray of A. Label the points of intersection B and C. Using the same radius, draw an arc with D as the center that intersects the ray from D at E. 3. With E as the center, draw an arc with radius equal to BC that intersects the arc drawn in step 3. Label the intersection F. Conclusion EDF BAC Proof We used congruent radii to draw , and BC > EF AB > DE DEF ABC by SSS and EDF BAC because they are corresponding parts of congruent triangles. . Therefore, AC > DF , C B F E A D 14365C05.pgs 8/2/07 5:43 PM Page 198 198 Congruence Based on Triangles Construction 3 Given Construct AB g CD Construct the Perpendicular Bisector of a Given Line Segment and the Midpoint of a Given Line Segment. ' AB at M, the midpoint of AB . . Open the compass to a radius that is greater than one-half of AB. 2. Place the point of the compass at A a |
nd draw an arc above AB arc below and an .AB D 3. Using the same radius, place the point of the compass at B and draw an arc above and an arc AB interbelow AB secting the arcs drawn in step 2. Label the intersections C and D. Conclusion g CD ' AB at M, the midpoint of AB . Proof Since they are congruent radii, AD > BD from A and B. If two points are each equidistant . Therefore, C and D are both equidistant AC > BC and from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment (Theorem 5.2). Thus, is the perpen- g CD dicular bisector of AB . Finally, M is the point on AB where the perpendicular bisector intersects AB , so AM = BM. By definition, M is the midpoint of AB . M D B 4. Use a straight- edge to draw g CD intersecting AB at M. C A M D B 14365C05.pgs 8/2/07 5:43 PM Page 199 Basic Constructions 199 Construction 4 Bisect a Given Angle. Given ABC Construct h BF , the bisector of ABC . With B as center and any convenient radius, draw an arc that 2. With D as center, draw an arc in the interior of ABC. intersects h BC D and h BA at at E. 3. Using the same 4. Draw h .BF radius, and with E as center, draw an arc that intersects the arc drawn in step 2. Label this intersection F. Conclusion h BF bisects ABC; ABF FBC. Proof We used congruent radii to draw BD > BE and DF > EF . By the reflexive property of congruence, so by SSS, FBD FBE. Therefore, BF > BF ABF FBC because they are corresponding parts of congruent triangles. Then bisects ABC h BF because an angle bisector separates an angle into two congruent angles. B D E A F C Construction 5 will be similar to Construction 4. In Construction 4, any given angle is bisected. In Construction 5, APB is a straight angle that is bisected by the construction. Therefore, APE and BPE are right angles and g PE g .AB ⊥ 14365C05.pgs 8/2/07 5:43 PM Page 200 200 Congruence Based on Triangles Construction 5 Construct a Line Perpendicular to a Given Line Through a Given Point on the Line. Given Point P on g . AB Construct g PE g ' AB . With P as center and any convenient radius, draw arcs that interh PB at C and h PA sect at D. 3. Draw g .EP 2. With C and D as centers and a radius greater than that used in step 1, draw arcs intersecting at E. Conclusion g PE g ' AB Proof Since points C and D were constructed using congruent radii, CP = PD and P is equidistant to C and D. Similarly, since E was constructed using congruent radii, CE ED, and E is equidistant to C and D. If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendic- ular bisector of the line segment (Theorem 5.2). Therefore, g PE CD . Since CD is the perpendicular bisector of g ' AB is a subset of line g AB g PE , . E P C D A B 14365C05.pgs 8/2/07 5:43 PM Page 201 Basic Constructions 201 Construction 6 Construct a Line Perpendicular to a Given Line Through a Point Not on the Given Line. Given Point P not on g . AB Construct g PE g ' AB . With P as center and any 2. Open the compass to a convenient radius, draw an arc that intersects two points, C and D. g AB in radius greater than onehalf of CD. With C and D as centers, draw intersecting arcs. Label the point of intersection E. Conclusion g PE g ' AB 3. Draw g PE intersecting g AB at F. P F E A C D B Proof Statements Reasons 1. CP > PD , CE > DE 1. Radii of congruent circles are congruent. 2. CP PD, CE DE g PE 3. ' CD g PE g ' AB 4. 2. Segments that are congruent have the same measure. 3. If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment. (Theorem 5.2) 4. CD is a subset of line g . AB 14365C05.pgs 7/10/07 8:42 AM Page 202 202 Congruence Based on Triangles EXAMPLE 1 Construct the median to AB in ABC. Construction A median of a triangle is a line segment that joins any vertex of the triangle to the midpoint of the opposite side. To construct the median to first find the midpoint of , we must AB AB . 1. Construct the perpendicular bisector of to locate the midpoint. Call the mid- AB point M. 2. Draw . CM Conclusion CM is the median to AB in ABC. A C R M S B Exercises Writing About Mathematics 1. Explain the difference between the altitude of a triangle and the perpendicular bisector of a side of a triangle. 2. Explain how Construction 3 (Construct the perpendicular bisector of a given segment) and Construction 6 (Construct a line perpendicular to a given line through a point not on the given line) are alike and how they are different. Developing Skills 3. Given: AB Construct: a. A line segment congruent to . AB b. A line segment whose measure is 2AB. A B c. The perpendicular bisector of d. A line segment whose measure is . AB 11 . 2AB 4. Given: A Construct: a. An angle congruent to A. b. An angle whose measure is 2mA. c. The bisector of A. d. An angle whose measure is 21 2m/A . A 14365C05.pgs 7/10/07 8:42 AM Page 203 5. Given: Line segment . ABCD Construct: a. A line segment congruent to . BC Basic Constructions 203 A B C D b. A triangle with sides congruent to , AB BC , and . CD c. An isosceles triangle with the base congruent to AB and with legs congruent to . BC d. An equilateral triangle with sides congruent to . CD 6. Given: A with mA 60. Construct: a. An angle whose measure is 30. b. An angle whose measure is 15. c. An angle whose measure is 45. 7. Given: ABC Construct: a. The median from vertex C. b. The altitude to AB . c. The altitude to BC . A C d. The angle bisector of the triangle from vertex A. A B 8. a. Draw ABC. Construct the three perpendicular bisectors of the sides of ABC. Let P be the point at which the three perpendicular bisectors intersect. b. Is it possible to draw a circle that passes through each of the vertices of the triangle? Explain your answer. Hands-On Activity Compass and straightedge constructions can also be done on the computer by using only the point, line segment, line, and circle creation tools of your geometry software and no other software tools. Working with a partner, use either a compass and straightedge, or geometry software to complete the following constructions: a. A square with side AB . b. An equilateral triangle with side . AB c. 45° angle ABD. d. 30° angle ABD. e. A circle passing through points A, B and C. (Hint: See the proof of Theorem 5.4 or use Theorem 5.3.) A B 14365C05.pgs 8/2/07 5:43 PM Page 204 204 Congruence Based on Triangles CHAPTER SUMMARY Definitions to Know • An altitude of a triangle is a line segment drawn from any vertex of the triangle, perpendicular to and ending in the line that contains the opposite side. • A median of a triangle is a line segment that joins any vertex of the trian- gle to the midpoint of the opposite side. • An angle bisector of a triangle is a line segment that bisects any angle of the triangle and terminates in the side opposite that angle. • The perpendicular bisector of a line segment is a line, a line segment, or a ray that is perpendicular to the line segment at its midpoint. Postulates 5.1 Radii of congruent circles are congruent. Theorems and Corollaries 5.1 If two sides of a triangle are congruent, the angles opposite these sides are congruent. 5.1a The median from the vertex angle of an isosceles triangle bisects the ver- tex angle. 5.1b The median from the vertex angle of an isosceles triangle is perpendicu- lar to the base. 5.1c Every equilateral triangle is equiangular. 5.2 If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment. 5.3 A point is on the perpendicular bisector of a line segment if and only if it is equidistant from the endpoints of the line segment. 5.4 The perpendicular bisectors of the sides of a triangle are concurrent. VOCABULARY 5-1 Altitude of a triangle • Median of a triangle • Angle bisector of a triangle 5-3 Isosceles triangle theorem • Corollary 5-6 Perpendicular bisector of a line segment • Equidistant • Concurrent • Perpendicular bisector concurrence theorem • Circumcenter 5-7 Geometric construction • Straightedge • Compass REVIEW EXERCISES 1. If g LMN ' KM , mLMK x y and mKMN 2x y, find the value of x and of y. 2. The bisector of PQR in PQR is QS . If mPQS x 20 and mSQR 5x, find mPQR. 14365C05.pgs 8/2/07 5:44 PM Page 205 Review Exercises 205 3. In ABC, CD is both the median and the altitude. If AB 5x 3, AC 2x 8, and BC 3x 5, what is the perimeter of ABC? 4. Angle PQS and angle SQR are a linear pair of angles. If mPQS 5a 15 and mSQR 8a 35, find mPQS and mSQR. 5. Let D be the point at which the three perpendicular bisectors of the sides of equilateral ABC intersect. Prove that ADB, BDC, and CDA are congruent isosceles triangles. side 6. Prove that if the median, , then , to side CD is not congruent to AB is the base of isosceles ABC and g CD AB ABD. Prove that AC 7. AB BC of ABC is not the altitude to . AB is also the base of isosceles is the perpendicular bisector of AB . 8. In ABC, CD is the median to AB mA mB mACB. (Hint: Use Theorem 5.1, “If two sides of a triangle are congruent, the angles opposite these sides are congruent.”) . Prove that CD > DB and 9. a. Draw a line, g ADB . Construct g CD ' ADB . b. Use ADC to construct ADE such that mADE 45. c. What is the measure of EDC? d. What is the measure of EDB? 10. a. Draw obtuse PQR with the obtuse angle at vertex Q. b. Construct the altitude from vertex P. Exploration As you have noticed, proofs may be completed using a variety of methods. In this activity, you will explore the reasoning of others. 1. Complete a two-column proof of the following: Points L, M, and N separate gruent segments. Point C is not on is an altitude of CLM. Prove that into four conand AB CM CA > CB . AB 2. Cut out each statement and each reason from your proof, omitting the step numbers. 3. Trade proofs with a partner. 4. Attempt to reassemble your partner’s proof. 5. Answer the following questions: C A L M N B a. Were you able to |
reassemble the proof? b. Did the reassembled proof match your partner’s original proof? c. Did you find any components missing or have any components left- over? Why? 14365C05.pgs 7/10/07 8:42 AM Page 206 206 Congruence Based on Triangles CUMULATIVE REVIEW Chapters 1–5 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The symbol g ABC represents (1) a line segment with B between A and C. (2) a line with B the midpoint of (3) a line with B between A and C. (4) a ray with endpoint A. AB . 2. A triangle with no two sides congruent is (1) a right triangle. (2) an equilateral triangle. 3. Opposite rays have (1) no points in common. (2) one point in common. (3) an isosceles triangle. (4) a scalene triangle. (3) two points in common. (4) all points in common. 4. The equality a 1 1 a is an illustration of (1) the commutative property of addition. (2) the additive inverse property. (3) the multiplicative identity property. (4) the closure property of addition. 5. The solution set of the equation 1.5x 7 0.25x 8 is (1) 120 (2) 12 (3) 11 (4) 1.2 6. What is the inverse of the statement “When spiders weave their webs by noon, fine weather is coming soon”? (1) When spiders do not weave their webs by noon, fine weather is not coming soon. (2) When fine weather is coming soon, then spiders weave their webs by noon. (3) When fine weather is not coming soon, spiders do not weave their webs by noon. (4) When spiders weave their webs by noon, fine weather is not coming soon. 7. If ABC and CBD are a linear pair of angles, then they must be (1) congruent angles. (2) complementary angles. (3) supplementary angles. (4) vertical angles. 14365C05.pgs 7/10/07 8:42 AM Page 207 Chapter Summary 207 8. Which of the following is not an abbreviation for a postulate that is used to prove triangles congruent? (1) SSS (2) SAS (3) ASA (4) SSA 9. If the statement “If two angles are right angles, then they are congruent” is true, which of the following statements must also be true? (1) If two angles are not right angles, then they are not congruent. (2) If two angles are congruent, then they are right angles. (3) If two angles are not congruent, then they are not right angles. (4) Two angles are congruent only if they are right angles. 10. If is a line, which of the following may be false? g ABC (1) B is on (2) . AC AB 1 BC 5 AC (3) B is between A and C. (4) B is the midpoint of AC . Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Angle PQS and angle SQR are a linear pair of angles. If mPQS 3a 18 and mSQR = 7a 2, find the measure of each angle of the linear pair. 12. Give a reason for each step in the solution of the given equation. 5(4 x) 32 x 20 5x 32 x 20 5x x 32 x x 20 6x 32 0 20 6x 32 6x 12 x 2 Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 14365C05.pgs 7/10/07 8:42 AM Page 208 208 Congruence Based on Triangles 13. Given: , ABF is the AE > BF supplement of A, and AB > CD . Prove: AEC BFD E F 14. Given: AB > CB point on of ABC. h BD and E is any , the bisector Prove: AE > CE C E D A B C D B A Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Given: Point P is not on ABCD and P PB PC. Prove: ABP DCP A B C D 16. Prove that if AC and BD are perpendicular bisectors of each other, quadrilateral ABCD is equilateral (AB BC CD DA). 14365C06.pgs 7/12/07 2:57 PM Page 209 TRANSFORMATIONS AND THE COORDINATE PLANE In our study of mathematics, we are accustomed to representing an equation as a line or a curve. This blending of geometry and algebra was not always familiar to mathematicians. In the seventeenth century, René Descartes (1596–1650), a French philosopher and mathematician, applied algebraic principles and methods to geometry. This blending of algebra and geometry is known as coordinate geometry or analytic geometry. Pierre de Fermat (1601–1665) independently developed analytic geometry before Descartes but Descartes was the first to publish his work. Descartes showed that a curve in a plane could be completely determined if the distances of its points from two fixed perpendicular lines were known. We call these distances the Cartesian coordinates of the points; thus giving Descartes’ name to the system that he developed. CHAPTER 6 CHAPTER TABLE OF CONTENTS 6-1 The Coordinates of a Point in a Plane 6-2 Line Reflections 6-3 Line Reflections in the Coordinate Plane 6-4 Point Reflections in the Coordinate Plane 6-5 Translations in the Coordinate Plane 6-6 Rotations in the Coordinate Plane 6-7 Glide Reflections 6-8 Dilations in the Coordinate Plane 6-9 Transformations as Functions Chapter Summary Vocabulary Review Exercises Cumulative Review 209 14365C06.pgs 7/12/07 2:57 PM Page 210 210 Transformations and the Coordinate Plane 6-1 THE COORDINATES OF A POINT IN A PLANE y 3 2 1 –3 –2 –1 –1 –2 –3 O 1 2 3 x In this chapter, we will review what you know about the coordinate plane and you will study transformations and symmetry, which are common elements in nature, art, and architecture. Two intersecting lines determine a plane. The coordinate plane is determined by a horizontal line, the x-axis, and a vertical line, the y-axis, which are perpendicular and intersect at a point called the origin. Every point on a plane can be described by two numbers, called the coordinates of the point, usually written as an ordered pair. The first number in the pair, called the x-coordinate or the abscissa, is the distance from the point to the y-axis. The second number, the y-coordinate or the ordinate is the distance from the point to the x-axis. In general, the coordinates of a point are represented as (x, y). Point O, the origin, has the coordinates (0, 0). We will accept the following postulates of the coordinate plane. Postulate 6.1 Two points are on the same horizontal line if and only if they have the same y-coordinates. Postulate 6.2 The length of a horizontal line segment is the absolute value of the difference of the x-coordinates. Postulate 6.3 Two points are on the same vertical line if and only if they have the same x-coordinate. Postulate 6.4 The length of a vertical line segment is the absolute value of the difference of the y-coordinates. Postulate 6.5 Each vertical line is perpendicular to each horizontal line. Locating a Point in the Coordinate Plane An ordered pair of signed numbers uniquely determines the location of a point in the plane. 14365C06.pgs 7/12/07 2:57 PM Page 211 The Coordinates of a Point in a Plane 211 Procedure To locate a point in the coordinate plane: 1. From the origin, move along the x-axis the number of units given by the xcoordinate. Move to the right if the number is positive or to the left if the number is negative. If the x-coordinate is 0, there is no movement along the x-axis. 2. Then, from the point on the x-axis, move parallel to the y-axis the number of units given by the y-coordinate. Move up if the number is positive or down if the number is negative. If the y-coordinate is 0, there is no movement in the y direction. For example, to locate the point A(3, 4), from O, move 3 units to the left along the x-axis, then 4 units down, parallel to the y-axis. y O x A Finding the Coordinates of a Point in a Plane The location of a point in the coordinate plane uniquely determines the coordinates of the point. Procedure To find the coordinates of a point: 1. From the point, move along a vertical line to the x-axis.The number on the x-axis is the x-coordinate of the point. 2. From the point, move along a horizontal line to the y-axis.The number on the y-axis is the y-coordinate of the point. For example, from point R, move in the vertical direction to 5 on the x-axis and in the horizontal direction to 6 on the y-axis. The coordinates of R are (5, 6). Note: The coordinates of a point are often referred to as rectangular coordinates(5, –6) 14365C06.pgs 7/12/07 2:57 PM Page 212 212 Transformations and the Coordinate Plane Graphing Polygons A quadrilateral (a four-sided polygon) can be represented in the coordinate plane by locating its vertices and then drawing the sides connecting the vertices in order. y 1 A(3, 2) O 1 x The graph shows the quadrilateral ABCD. The vertices are A(3, 2), B(3, 2), C(3, 2) and D(3, 2). 1. Points A and B have the same y-coordinate and are on the same horizontal line. 2. Points C and D have the same y-coordinate and are on the same horizontal line. 3. Points B and C have the same x-coordinate and are on the same vertical line. 4. Points A and D have the same x-coordinate and are on the same vertical line. 5. Every vertical line is perpendicular to every horizontal line. 6. Perpendicular lines are lines that intersect to form right angles. Each angle of the quadrilateral is a right angle: mA mB mC mD 90 From the graph, we can find the dimensions of this quadrilateral. To find AB and CD, we can count the number of units from A to B or from C to D. AB CD 6 Points on the same horizontal line have the same y-coordinate. Therefore, we can also find AB and CD by subtracting their x-coordinates. AB CD 3 (3) 3 3 6 To find BC and DA, we can count the number of units from B to C or from D to A. BC DA 4 Points on the same vertical line have the same x-coordinate. Therefore, we |
can find BC and DA by subtracting their y-coordinates. BC DA 2 (2) 2 2 4 14365C06.pgs 7/12/07 2:57 PM Page 213 The Coordinates of a Point in a Plane 213 EXAMPLE 1 Graph the following points: A(4, 1), B(1, 5), C(2,1). Then draw ABC and find its area. Solution The graph shows ABC. To find the area of the triangle, we need to know the lengths of the base and of the altitude drawn to that base. The base of ABC is AC , a horizontal line segment. AC 4 (2) 4 2 6 y B(1, 5) O D(1, 1) A(4, 1) 1 x The vertical line segment drawn from B perpendicular to AC is the altitude . BD BD 5 1 4 Area 1 2(AC)(BD) 1 2(6)(4) 12 Answer The area of ABC is 12 square units. Exercises Writing About Mathematics 1. Mark is drawing isosceles right triangle ABC on the coordinate plane. He locates points A(2, 4) and C(5, 4). He wants the right angle to be C. What must be the coordinates of point B? Explain how you found your answer. / 2. Phyllis graphed the points D(3, 0), E(0, 5), F(2, 0), and G(0, 4) on the coordinate plane and joined the points in order. Explain how Phyllis can find the area of this polygon. Developing Skills In 3–12: a. Graph the points and connect them with straight lines in order, forming a polygon. b. Find the area of the polygon. 3. A(1, 1), B(8, 1), C(1, 5) 5. C(8, 1), A(9, 3), L(4, 3), F(3, 1) 4. P(0, 0), Q(5, 0), R(5, 4), S(0, 4) 6. H(4, 0), O(0, 0), M(0, 4), E(4, 4) 14365C06.pgs 7/12/07 2:57 PM Page 214 214 Transformations and the Coordinate Plane 7. H(5, 3), E(5, 3), N(2, 0) 9. B(3, 2), A(2, 2), R(2, 2), N(3, 2) 11. R(4, 2), A(0, 2), M(0, 7) 8. F(5, 1), A(5, 5), R(0, 5), M(2, 1) 10. P(3, 0), O(0, 0), N(2, 2), D(1, 2) 12. M(1, 1), I(3, 1), L(3, 3), K(1, 3) 13. Graph points A(1, 1), B(5, 1), and C(5, 4). What must be the coordinates of point D if ABCD is a quadrilateral with four right angles? 14. Graph points P(1, 4) and Q(2, 4). What are the coordinates of R and S if PQRS is a quadrilateral with four right angles and four congruent sides? (Two answers are possible.) 15. a. Graph points S(3, 0), T(0, 4), A(3, 0), and R(0, 4), and draw the polygon STAR. b. Find the area of STAR by adding the areas of the triangles into which the axes sepa- rate the polygon. 16. a. Graph points P(2, 0), L(1, 1), A(1, 1), N(2, 0). E(1, 1), and T(1, 1), and draw the hexagon PLANET. b. Find the area of PLANET. (Hint: Use two vertical lines to separate the hexagon into parts.) 6-2 LINE REFLECTIONS It is often possible to see the objects along the shore of a body of water reflected in the water. If a picture of such a scene is folded, the objects can be made to coincide with their images. Each point of the reflection is an image point of the corresponding point of the object. The line along which the picture is folded is the line of reflection, and the correspondence between the object points and the image points is called a line reflection. This common experience is used in mathematics to study congruent figures. If the figure at the left were folded along line k, ABC would coincide with ABC. Line k is the line of reflection, point A corresponds to point A (in symbols, A → A) and point B corresponds to point B (B → B). Point C is a fixed point because it is a point on the line of reflection. In other words, C corresponds to itself (C → C). Under a reflection in line k, then, ABC corresponds to ABC (ABC → ABC). Each of the points A, B, and C is called a preimage and each of the points A, B, and C is called an image. C B B A A k 14365C06.pgs 7/12/07 2:57 PM Page 215 DEFINITION A transformation is a one-to-one correspondence between two sets of points, S and S, such that every point in set S corresponds to one and only one point in set S, called its image, and every point in S is the image of one and only one point in S, called its preimage. S S Line Reflections 215 The sets S and S can be the same set and that set of points is frequently the set of points in a plane. For example, let S be the set of points in the coordinate plane. Let the image of (x, y), a point in S, be (2 x, y), a point in S. Under this transformation: (0, 1) → (2 0, 1) (2, 1) (4, 3) → (2 (4), 3) (2, 3) (5, 1) → (2 5, 1) (3, 1) Every image (2 x, y) in S is a point of the coordinate plane, that is, a point of S. Therefore, S S. P P P k DEFINITION A reflection in line k is a transformation in a plane such that: 1. If point P is not on k, then the image of P is P where k is the perpendicular bisector of PPr . 2. If point P is on k, the image of P is P. From the examples that we have seen, it appears that the size and shape of the image is the same as the size and shape of the preimage. We want to prove that this is true. Theorem 6.1 Under a line reflection, distance is preserved. Given Under a reflection in line k, the image of A is A and the image of B is B. Prove AB AB B D A C Proof We will prove this theorem by using the definition of a reflection and SAS to show that BDC BDC and that using the fact that if two angles are congruent, their complements are congruent (Theorem 4.4), we can show that ACB ACB. From this last statement, we can conclude that AB AB. BC > BrC . Then, k 14365C06.pgs 7/12/07 2:57 PM Page 216 216 Transformations and the Coordinate Plane (1) Let the image of A be A and the image of B be B under reflection in k. Let C be the and D be the midpoint of midpoint of BBr . Points C and D are on k, since k is the perpendicular bisector of . and of AAr AAr BBr (2) since D is the midpoint of BD > BrD , BDC BDC since perpendicular lines intersect to form right angles, and CD > CD BDC BDC by SAS. by the reflexive property. Thus, BBr (3) From step 2, we can conclude that and BCD BCD since BC > BrC corresponding parts of congruent triangles are congruent. (4) Since k is the perpendicular bisector of , ACD and ACD are right angles. AAr Thus, ACB and BCD are complementary angles. Similarly, ACB and BCD are complementary angles. If two angles are congruent, their complements are congruent. Since BCD and BCD were shown to be congruent in step 3, their complements are congruent: ACB ACB. (5) By SAS, ACB ACB. Since corresponding parts of congruent trian. Therefore, AB = AB by the definition of AB > ArBr gles are congruent, congruent segments. A similar proof can be given when A and B are on opposite sides of the line k. The proof is left to the student. (See exercise 11.) Since distance is preserved under a line reflection, we can prove that the image of a triangle is a congruent triangle and that angle measure, collinearity, and midpoint are also preserved. For each of the following corollaries, the image of A is A, the image of B is B, and the image of C is C. Therefore, in the diagram, ABC ABC by SSS. We will use this fact to prove the following corollaries: B C D A M k M 14365C06.pgs 7/12/07 2:57 PM Page 217 Corollary 6.1a Under a line reflection, angle measure is preserved. Line Reflections 217 Proof: We found that ABC ABC. Therefore, ABC ABC because the angles are corresponding parts of congruent triangles. Corollary 6.1b Under a line reflection, collinearity is preserved. D A B C k Proof: Let D be a point on AB whose image is D. Since distance is preserved: AD AD DB DB A, D, and B are collinear, D is between A and B, and AD DB AB. By substitution, AD DB AB. AB = AB If D were not on , AD DB AB because a straight line is the shortest distance between two points. But by substitution, AD DB AB. Therefore, A, D and B are collinear and D is between A and B. ArBr Corollary 6.1c Under a line reflection, midpoint is preserved. A M M B C k Proof: Let M be the midpoint of preserved under a line reflection, AM AM, and MC MC. AC Since M is the midpoint of AC , AM MC and, by the substitution postu- and M the image of M. Since distance is late, AM MC. Therefore, M is the midpoint of ArCr , that is, midpoint is preserved under a line reflection. We can summarize Theorem 6.1 and its corollaries in the following statement: Under a line reflection, distance, angle measure, collinearity, and midpoint are preserved. We use rk as a symbol for the image under a reflection in line k. For example, rk(A) B means rk(ABC) ABC means “The image of A under a reflection in line k is B.” “The image of ABC under a reflection in line k is ABC.” 14365C06.pgs 7/12/07 2:57 PM Page 218 218 Transformations and the Coordinate Plane EXAMPLE 1 If rk(CD) CrDr , construct CrDr . Construction 1. Construct the perpendicular line from C to k. Let the point of intersection be M. 2. Construct the perpendicular line from D to k. Let the point of intersection be N. 3. Construct point C on such that CM MC and point D on g such that DN ND. DN g CM 4. Draw CrDr . Line Symmetry We know that the altitude to the base of an isosceles triangle is also the median and the perpendicular bisector of the base. If we imagine that isosceles triangle ABC, shown at the left, is folded along the perpendicular bisector of the base so that A falls on C, the line along which it folds, k, is a reflection line. Every point of the triangle has as its image a point of the triangle. Points B and D are fixed points because they are points of the line of reflection. Thus, under the line reflection in k: 1. All points of ABC are reflected so that C → A 2. The sides of ABC are reflected; that is, A → C E → F that the legs of an isosceles triangle are congruent. Also, ing that the base is its own image AB S CB , a statement verifying AC S CA , show- B E F A D C k 14365C06.pgs 7/12/07 2:57 PM Page 219 Line Reflections 219 3. The angles of ABC are reflected; that is, BAD → BCD, a statement verifying that the base angles of an isosceles triangle are congruent. Also, ABC → CBA, showing that the vertex angle is its own image. We can note some properties of a line reflection by considering the reflec- tion of isosceles triangle ABC in line k: 1. Distance is preserved (unchanged). AB S CB and AB CB AD S CD and AD CD 2. Angle measure is preserved. BAD → BCD and mBAD mBCD BDA → BDC and mBDA mBDC 3. The line of reflection is the perpendicular bisector o |
f every segment join- ing a point to its image. The line of reflection, bisector of AC . g BD , is the perpendicular 4. A figure is always congruent to its image: ABC CBA. In nature, in art, and in industry, many forms have a pleasing, attractive appearance because of a balanced arrangement of their parts. We say that such forms have symmetry. In each of the figures above, there is a line on which the figure could be folded so that the parts of the figure on opposite sides of the line would coincide. If we think of that line as a line of reflection, each point of the figure has as its image a point of the figure. This line of reflection is a line of symmetry, or an axis of symmetry, and the figure has line symmetry. DEFINITION A figure has line symmetry when the figure is its own image under a line reflection. 14365C06.pgs 7/12/07 2:57 PM Page 220 220 Transformations and the Coordinate Plane It is possible for a figure to have more than one axis of symmetry. In the square g XY is an axis of symme- PQRS at the right, g VW try and is a second axis of symmetry. The diagonals, PR and QS , are also segments of axes of symmetry. Lines of symmetry may be found for some letters and for some words, as shown at the left. BD Not every figure, however, has line symmetry. If the parallelogram ABCD at the right is reflected in the line that contains the , the image of A is A and the diagonal image of C is C. Points A and C, however, are not points of the original parallelogram. The image of parallelogram ABCD under a is ABCD. Therefore, g . BD ABCD is not symmetric with respect to reflection in g BD We have used the line that contains the diagonal as a line of reflection, but note that it is not a line of symmetry. There is no line along which the parallelogram can be folded so that the image of every point of the parallelogram will be a point of the parallelogram under a reflection in that line. BD EXAMPLE 2 How many lines of symmetry does the letter H have? Solution The horizontal line through the crossbar is a line of symmetry. The vertical line midway between the vertical segments is also a line of symmetry. Answer The letter H has two lines of symmetry. Exercises Writing About Mathematics 1. Explain how a number line illustrates a one-to-one correspondence between the set of points on a line and the set of real numbers. 2. Is the correspondence (x, y) → (2, y) a transformation? Explain why or why not. 14365C06.pgs 7/12/07 2:57 PM Page 221 Line Reflections 221 Developing Skills 3. If rk(PQR) PQR and PQR is isosceles with PQ QR, prove that PQR is isosceles. 4. If rk(LMN) LMN and LMN is a right triangle with mN 90, prove that LMN is a right triangle. In 5–9, under a reflection in line k, the image of A is A, the image of B is B, the image of C is C, and the image of D is D. 5. Is ABC ABC? Justify your answer. 6. If AC ' BC , is ArCr ' BrCr ? Justify your answer? 7. The midpoint of AB is M. If the image of M is M, is M the midpoint of ArBr ? Justify your answer. 8. If A, M, and B lie on a line, do A, M, and B lie on a line? Justify your answer. 9. Which point lies on the line of reflection? Justify your answer. 10. A triangle, RST, has line symmetry with respect to the altitude from S but does not have line symmetry with respect to the altitudes from R and T. What kind of a triangle is RST? Justify your answer. Applying Skills 11. Write the proof of Theorem 6.1 for the case when points A and B are on opposite sides of line k. A k B 12. A baseball diamond is in the shape of a rhombus with sides 90 feet in length. Describe the lines of symmetry of a baseball diamond in terms of home plate and the three bases. 13. Print three letters that have line symmetry with respect to only one line and draw the line of symmetry. 14. Print three letters that have line symmetry with respect to exactly two lines and draw the lines of symmetry. 14365C06.pgs 7/12/07 2:57 PM Page 222 222 Transformations and the Coordinate Plane 6-3 LINE REFLECTIONS IN THE COORDINATE PLANE We can apply the definition of a line reflection to points in the coordinate plane. Reflection in the y-axis In the figure, ABC is reflected in the y-axis. Its image under the reflection is ABC. From the figure, we see that: A(1, 2) → A(1, 2) B(3, 4) → B(3, 4) C(1, 5) → C(1, 5) C(1, 5) B(3, 4) A(1, 2) y 1 C(1, 5) B(3, 4) A(1, 2) For each point and its image under a reflection in the y-axis, the y-coordinate of the image is the same as the y-coordinate of the point; the x-coordinate of the image is the opposite of the x-coordinate of the point. Note that for a reflection in the y-axis, the image of (1, 2) is (1, 2) and the image of (1, 2) is (1, 2). O x 1 A reflection in the y-axis can be designated as ry-axis. For example, if the image of (1, 2) is (1, 2) under a reflection in the y-axis, we can write: ry-axis(1, 2) (1, 2) From these examples, we form a general rule that can be proven as a theorem. Theorem 6.2 Under a reflection in the y-axis, the image of P(a, b) is P(a, b). Given A reflection in the y-axis. Prove The image of P(a, b) under a reflection in the y-axis is P(a, b). P(a, b) y O Q(0, b) P(a, b) x Proof By the definition of a reflection in a line, a point P is the image of P under a reflection in a given line if and only if the line is the perpendicular bisector of . Therefore, we can prove that P is the image of P under a reflection in PPr the y-axis by showing that the y-axis is the perpendicular bisector of PPr . (1) The y-axis is perpendicular to . The line of reflection, the y-axis, is a PPr vertical line. P(a, b) and P(a, b) have the same y-coordinates. is a segment of a horizontal line because two points are on the same horizontal line if and only if they have the same y-coordinates. Every vertical line is perpendicular to every horizontal line. Therefore, the y-axis is perpendicular to .PPr PPr 14365C06.pgs 7/12/07 2:57 PM Page 223 Line Reflections in the Coordinate Plane 223 PPr (2) The y-axis bisects . Let Q be the point at which y-axis. The x-coordinate of every point on the y-axis is 0. The length of a horizontal line segment is the absolute value of the difference of the x-coordinates of the endpoints. intersects the PPr PQ a 0 a and PQ a 0 = a Since PQ PQ, Q is the midpoint of Steps 1 and 2 prove that if P has the coordinates (a, b) and P has the coor, and therefore, dinates (a, b), the y-axis is the perpendicular bisector of the image of P(a, b) is P(a, b). or the y-axis bisects PPr PPr PPr . Reflection in the x-axis In the figure, ABC is reflected in the x-axis. Its image under the reflection is ABC. From the figure, we see that: A(1, 2) → A(1, 2) B(3, 4) → B(3, 4) C(1, 5) → C(1, 5) For each point and its image under a reflection in the x-axis, the x-coordinate of the image is the same as the x-coordinate of the point; the y-coordinate of the image is the opposite of the y-coordinate of the point. Note that for a reflection in the x-axis, the image of (1, 2) is (1, 2) and the image of (1, 2) is (1, 2). y C(1, 5) B(3, 4) A(1, 2) 1 x 1 O A reflection in the x-axis can be designated as rx-axis. For example, if the image of (1, 2) is (1, 2) under a reflection in the x-axis, we can write: rx-axis(1, 2) (1, 2) From these examples, we form a general rule that can be proven as a theorem. Theorem 6.3 Under a reflection in the x-axis, the image of P(a, b) is P(a, b). The proof follows the same general pattern as that for a reflection in the y-axis. Prove that the x-axis is the perpendicular bisector of . The proof is left to the student. (See exercise 18.) PPr y O P(a, b) x Q(a, 0) P(a, b) 14365C06.pgs 7/12/07 2:57 PM Page 224 224 Transformations and the Coordinate Plane EXAMPLE 1 On graph paper: a. Locate A(3, 1). b. Locate A, the image of A under a reflection in the y-axis, and write its coordinates. c. Locate A, the image of A under a reflection in the x-axis, and write its coordinates. Answers y 1 1O x Reflection in the Line y = x In the figure, ABC is reflected in the line y x. Its image under the reflection is ABC. From the figure, we see that: y C(1, 5) B(3, 4) A(1, 2) → A(2, 1) B(3, 4) → B(4, 3) C(1, 5) → C(5, 1) For each point and its image under a reflection in the line y = x, the x-coordinate of the image is the y-coordinate of the point; the y-coordinate of the image is the x-coordinate of the point. Note that for a reflection in the line y x, the image of (1, 2) is (2, 1) and the image of (2, 1) is (1, 2). B(4, 3) C(5, 1) x A(1, 2) A(2, 1) O 11 A reflection in the line y x can be designated as ry = x. For example, if the image of (1, 2) is (2, 1) under a reflection in y x, we can write: From these examples, we form a general rule that can be proven as a theorem. ry = x(1, 2) (2, 1) Theorem 6.4 Under a reflection in the line y = x, the image of P(a, b) is P(b, a). Given A reflection in the line whose equation is y x. Prove The image of P(a, b) is P(b, a). 14365C06.pgs 7/12/07 2:57 PM Page 225 Line Reflections in the Coordinate Plane 225 Proof For every point on the line y x, the xcoordinate and the y-coordinate are equal. Locate the points Q(a, a) and R(b, b) on the line y x. If two points have the same x-coordinate, then the distance between them is the absolute value of their y-coordinates and if two points have the same y-coordinate, then the distance between them is the absolute value of their x-coordinates. PQ b a and PQ b a Therefore, Q is equidistant from P and P. y P(a, b) R(b, b) Q(a, a) P(b, a) x PR b a and PR = b a Therefore, R is equidistant from P and P. If two points are each equidistant from the endpoints of a line segment, they lie on the perpendicular bisector of the line segment (Theorem 5.3). Therefore, . By RQ the definition of a line reflection across the line y x, the image of P(a, b) is P(b, a). , which is a subset of the line y x, is the perpendicular bisector of PPr EXAMPLE 2 The vertices of DEF are D(2, 2), E(0, 3), and F(2, 0). a. Draw DEF on graph paper. b. Draw the image of DEF under a reflection in the line whose eq |
uation is y = x. Answers y E(3, 0) D(2, 2) 1 O 1 F(2, 0) x Exercises Writing About Mathematics 1. When finding the distance from P(a, b) to Q(a, a), Allison wrote PQ b a and Jacob wrote PQ a b. Explain why both are correct. 2. The image of point A is in the third quadrant under a reflection in the y-axis and in the first quadrant under a reflection in the x-axis. In what quadrant is the image of A under a reflection in the line y x? 14365C06.pgs 7/12/07 2:57 PM Page 226 226 Transformations and the Coordinate Plane Developing Skills In 3–7: a. On graph paper, locate each point and its image under rx-axis. b. Write the coordinates of the image point. 3. (2, 5) 4. (1, 3) 5. (2, 3) 6. (2, 4) 7. (0, 2) In 8–12: a. On graph paper, locate each point and its image under ry-axis. b. Write the coordinates of the image point. 8. (3, 5) 9. (1, 4) 10. (2, 3) 11. (2, 3) 12. (1, 0) In 13–17: a. On graph paper, locate each point and its image under ryx. b. Write the coordinates of the image point. 13. (3, 5) 14. (3, 5) 15. (4, 2) 16. (1, 5) 17. (2, 2) Applying Skills 18. Prove Theorem 6.3, “Under a reflection in the x-axis, the image of P(a, b) is P(a, b).” 19. When the points A(4, 0), B(0, 4), C(4, 0) and D(0, 4) are connected in order, square ABCD is drawn. a. Show that the line y = x is a line of symmetry for the square. b. Show that the y-axis is a line of symmetry for the square. 20. Show that the y-axis is not a line of symmetry for the rectangle whose vertices are E(0, 3), F(5, 3), G(5, 3), and H(0, 3). 21. Write the equation of two lines that are lines of symmetry for the rectangle whose vertices are E(0, 3), F(6, 3), G(6, 3), and H(0, 3). Hands-On Activity 1 In this activity, you will learn how to construct a reflection in a line using a compass and a straightedge, or geometry software. (Note: Compass and straightedge constructions can also be done on the computer by using only the point, line segment, line, and circle creation tools of your geometry software and no other software tools.) STEP 1. Draw a line segment. Label the endpoints A and B. Draw a reflection line k. STEP 2. Construct line l perpendicular to line k through point A. Let M be the point where lines l and k intersect. STEP 3. Construct line segment along line l. Using point M as the center, AM draw a circle with radius equal to AM. Let A be the point where the circle intersects line l on the ray that is the opposite ray of congruent to h . MA ArM STEP 4. Repeat steps 2 and 3 for point B in order to construct B. STEP 5. Draw ArBr . Result: ArBr is the image of AB under a reflection in line k. 14365C06.pgs 7/12/07 2:57 PM Page 227 For each figure, construct the reflection in the given line. Point Reflections in the Coordinate Plane 227 AB a. Segment with vertices A(4, 2) and B(2, 4), and line k through points C(2, 1) and D(0, 5). b. Angle EFG with vertices E(4, 3), F(1, 6), and G(6, 6), and line l through points H(1, 3) and I(3, 1). c. Triangle JKL with vertices J(2, 2), K(5, 1), and L(1, 4), and line p through points M(7, 3) and N(2, 0). PQ d. Segments and RS points T(0, 3) and U(3, 0). with vertices P(3, 4), Q(3, 2), R(1, 0), and S(1, 6), and line q through Hands-On Activity 2 In each figure, one is the image of the other under a reflection in a line. For each figure, construct the reflection line using a compass and a straightedge. a. A A b. B A c. Triangle ABC with vertices A(1, 4), B(4, 3), and (2, 8), and triangle ABC with vertices A(3, 2), B(2, 3), C(7, 3). d. Triangle ABC with vertices A(0, 4), B(7, 1), and C(6, 6), and triangle ABC with vertices A(7, 3), B(4, 4), C(9, 3). 6-4 POINT REFLECTIONS IN THE COORDINATE PLANE The figure at the right illustrates another type of reflection, a reflection in a point. In the figure, ABC is the image of ABC under a reflection in point P. If a line segment is drawn connecting any point to its image, then the point of reflection is the midpoint of that segment. In the figure: • Point A is on midpoint of , AP PA, and P is the g AP AAr . g BP g CP • Point B is on , BP PB, and P is the midpoint of • Point C is on , CP PC, and P is the midpoint of .CCr A C P B BBr . DEFINITION A point reflection in P is a transformation of the plane such that: 1. If point A is not point P, then the image of A is A and P the midpoint of 2. The point P is its own image. AAr . 14365C06.pgs 7/12/07 2:57 PM Page 228 228 Transformations and the Coordinate Plane Properties of Point Reflections Looking at triangles ABC and ABC and point of reflection P, we observe some properties of a point reflection: A C P B 1. Distance is preserved: AB AB. 2. Angle measure is preserved: mACB mACB. 3. A figure is always congruent to its image. 4. Collinearity is preserved. The image of any point on AB is a point on ArBr . 5. Midpoint is preserved. The image of the midpoint of AB is the midpoint of ArBr . We can state the first property as a theorem and the remaining properties as corollaries to this theorem. Theorem 6.5 Under a point reflection, distance is preserved. Given Under a reflection in point P, the image of A is A and the image of B is B. Prove AB AB Proof Since APB and APB are vertical angles, APB APB. In a point reflection, if a point X is not point P, then the image of X is X XXr and P the midpoint of . Therefore, P is the AAr BBr BP > BPr and midpoint of both . By SAS, APB APB and AP > PAr and AB AB. , and A B P B A The case when either point A or B is point P is left to the student. (See exercise 9.) Corollary 6.5a Under a point reflection, angle measure is preserved. Corollary 6.5b Under a point reflection, collinearity is preserved. Corollary 6.5c Under a point reflection, midpoint is preserved. 14365C06.pgs 7/12/07 2:57 PM Page 229 Point Reflections in the Coordinate Plane 229 We can prove that angle measure, collinearity, and midpoint are preserved using the same proofs that we used to prove the corollaries of Theorem 6.1. (See exercises 10–12.) Theorem 6.5 and its corollaries can be summarized in the following statement. Under a point reflection, distance, angle measure, collinearity, and midpoint are preserved. We use RP as a symbol for the image under a reflection in point P. For example, RP(A) B means “The image of A under a reflection in point P is B.” R(1, 2)(A) A means “The image of A under a reflection in point (1, 2) is A.” Point Symmetry DEFINITION A A figure has point symmetry if the figure is its own image under a reflection in a point. P B A circle is the most common example of a figure with point symmetry. Let P be the center of a circle, A be any point on the circle, and B be the other point intersect the circle. Since every point on a circle is equidistant from g AP at which the center, PA PB, P is the midpoint of image of A under a reflection in P. AB and B, a point on the circle, is the Other examples of figures that have point symmetry are letters such as S and N and numbers such as 8. Point Reflection in the Coordinate Plane In the coordinate plane, the origin is the most common point that is used to used to define a point reflection. In the diagram, points A(3, 5) and B(2, 4) are reflected in the origin. The coordinates of A, the image of A, are (3, 5) and the coordinates of B, the image of B, are (2, 4). These examples suggest the following theorem. A(3, 5) B(–2, 4) y 1 O 1 x B(2, 4) 14365C06.pgs 7/12/07 2:57 PM Page 230 230 Transformations and the Coordinate Plane Theorem 6.6 Under a reflection in the origin, the image of P(a, b) is P(a, b). Given A reflection in the origin. Prove Under a reflection in the origin, the image of P(a, b) is P(a, b). y P(a, b) x O B(a, 0) Proof Let P be the image of P(a, b) under a reflection in the origin, O. Then: OP OP and OP > OPr Let B(a, 0) be the point of intersection of the x-axis and a vertical line through P. Then: OB 0 a a Let B be the point (a, 0). Then: OB 0 (a) a OB OB and OB > OBr POB POB because vertical angles are congruent. Therefore, POB POB by SAS. In particular, P PB PB 0 b b Since OBP is a right angle, OBP is a right angle and is a vertical line. Therefore, P has the same x-coordinate as B and is b units in the opposite direction from the x-axis as P. The coordinates of P are (a, b). PrBr Note: a and b are the opposites of a and b. The image of (4, 2) is (4, 2) and the image of (5, 1) is (5, 1). We can symbolize the reflection in the origin as RO. Therefore, we can write: RO(a, b) (a, b) EXAMPLE 1 a. What are the coordinates of B, the image of A(3, 2) under a reflection in the origin? b. What are the coordinates of C, the image of A(3, 2) under a reflection in the x-axis? 14365C06.pgs 7/12/07 2:57 PM Page 231 Point Reflections in the Coordinate Plane 231 c. What are the coordinates of D, the image of C under a reflection in the y-axis? d. Does a reflection in the origin give the same result as a reflection in the x-axis followed by a reflection in the y-axis? Justify your answer. Solution a. Since RO(a, b) (a, b), RO(3, 2) (3, 2). The coordinates of B are (3, 2). Answer b. Since rx-axis (a, b) (a, b), rx-axis (3, 2) (3, 2). The coordinates of C are (3, 2). Answer c. Since ry-axis (a, b) (a, b), ry-axis (3, 2) (3, 2). The coordinates of D are (3, 2). Answer d. For the point A(3, 2), a reflection in the origin gives the same result as a reflection in the x-axis followed by a reflection in the y-axis. In general, RO(a, b) (a, b) while rx-axis(a, b) (a, b) ry-axis(a, b) (a, b) Therefore, a reflection in the origin gives the same result as a reflection in the x-axis followed by a reflection in the y-axis. Answer Exercises Writing About Mathematics 1. Ada said if the image of AB under a reflection in point P is ArBr , then the image of ArBr under a reflection in point P is AB . Do you agree with Ada? Justify your answer. 2. Lines l and m intersect at P. ArBr is the image of AB under a reflection in line l and is the image of P? Justify your answer. ArBr under a reflection in line m. Is the result the same if AB ArrBrr is reflected in Developing Skills In 3–8, give t |
he coordinates of the image of each point under RO. 3. (1, 5) 4. (2, 4) 5. (1, 0) 6. (0, 3) 7. (6, 6) 8. (1, 5) 14365C06.pgs 7/12/07 2:57 PM Page 232 232 Transformations and the Coordinate Plane Applying Skills 9. Write a proof of Theorem 6.5 for the case when either point A or B is point P, that is, when the reflection is in one of the points. 10. Prove Corollary 6.5a, “Under a point reflection, angle measure is preserved.” 11. Prove Corollary 6.5b, “Under a point reflection, collinearity is preserved.” 12. Prove Corollary 6.5c, “Under a point reflection, midpoint is preserved.” 13. The letters S and N have point symmetry. Print 5 other letters that have point symmetry. 14. Show that the quadrilateral with vertices P(5, 0), Q(0, 5), R(5, 0), and S(0, 5) has point symmetry with respect to the origin. 15. a. What is the image of A(2, 6) under a reflection in P(2, 0)? b. What is the image of B(3, 6) under a reflection in P(2, 0)? c. Is the reflection in the point P(2, 0) the same as the reflection in the x-axis? Justify your answer. 16. a. What is the image of A(4, 4) under a reflection in the point P(4, 2)? b. What is the image of C(1, 2) under a reflection in the point P(4, 2)? c. The point D(2, 0) lies on the segment AB . Does the image of D lie on the image of AB ? Justify your answer. 6-5 TRANSLATIONS IN THE COORDINATE PLANE It is often useful or necessary to move objects from one place to another. If we move a table from one place in the room to another, the bottom of each leg moves the same distance in the same direction. DEFINITION A translation is a transformation of the plane that moves every point in the plane the same distance in the same direction. C If ABC is the image of ABC under a translation, AA BB CC. It appears that the size and shape of the figure are unchanged, so that ABC ABC. Thus, under a translation, as with a reflection, a figure is congruent to its image. The following example in the coordinate plane shows that this is true. In the coordinate plane, the distance is given in terms of horizontal distance (change in the x-coordinates) and vertical distance (change in the y-coordinates). A B 14365C06.pgs 7/12/07 2:57 PM Page 233 Translations in the Coordinate Plane 233 In the figure, DEF is translated by moving every point 4 units to the right and 5 units down. From the figure, we see that: y F(1, 6) E(4, 2) D(1, 2) F(5, 1) 1 x 1 O D(5, 3) E(8, 3) D(1, 2) → D(5, 3) E(4, 2) → E(8, 3) F(1, 6) → F(5, 1) • DE DrEr and DE 1 4 3 are horizontal segments. DE 5 8 3 • DF and DrFr are vertical segments. DF 2 6 DF 3 1 4 4 • FDE and FDE are right angles. Therefore, DEF DEF by SAS. This translation moves every point 4 units to the right (4) and 5 units down (5). 1. The x-coordinate of the image is 4 more than the x-coordinate of the point: x → x 4 2. The y-coordinate of the image is 5 less than the y-coordinate of the point: From this example, we form a general rule: y → y 5 DEFINITION A translation of a units in the horizontal direction and b units in the vertical direction is a transformation of the plane such that the image of P(x, y) is P(x a, y b). Note: If the translation moves a point to the right, a is positive; if it moves a point to the left, a is negative; if the translation moves a point up, b is positive; if it moves a point down, b is negative. 14365C06.pgs 7/12/07 2:57 PM Page 234 234 Transformations and the Coordinate Plane Theorem 6.7 Under a translation, distance is preserved. Given A translation in which the image b) of A(x1, y1) is A(x1 a, y1 and the image of B(x2, y2) is B(x2 a, y2 b). Prove AB = AB y B(x2 a, y2 b) B(x2, y2) A(x1 a, y1 b) C(x2 a, y1 b) C(x2, y1) A(x1, y1) Proof Locate C(x2, y1), a point on the same horizontal line as A and the same vertical line as B. Then by definition, the image of C is O x Thus: C(x2 a, y1 b). BC 5 y2 2 y1 BrCr 5 (y2 1 b) 2 (y1 1 b) 5 y2 2 y1 1 b 2 b 5 y2 2 y1 AC 5 x1 2 x2 ArCr 5 (x1 1 a) 2 (x2 1 a) 5 x1 2 x2 1 a 2 a 5 x1 2 x2 Since horizontal and vertical lines are perpendicular, BCA and BCA are both right angles. By SAS, ABC ABC. Then, AB and AB are the equal measures of the corresponding sides of congruent triangles. When we have proved that distance is preserved, we can prove that angle measure, collinearity, and midpoint are preserved. The proofs are similar to the proofs of the corollaries of Theorem 6.1 and are left to the student. (See exercise 10.) Corollary 6.7a Under a translation, angle measure is preserved. Corollary 6.7b Under a translation, collinearity is preserved. Corollary 6.7c Under a translation, midpoint is preserved. 14365C06.pgs 7/31/07 1:21 PM Page 235 Translations in the Coordinate Plane 235 We can write Theorem 6.7 and its corollaries as a single statement. Under a translation, distance, angle measure, collinearity, and midpoint are preserved. Let Ta,b be the symbol for a translation of a units in the horizontal direction and b units in the vertical direction. We can write: Ta,b(x, y) (x a, y b) EXAMPLE 1 The coordinates of the vertices of ABC are A(3, 3), B(3, 2), and C(6, 1). a. Find the coordinates of the vertices of ABC, the image of ABC under T5,3. b. Sketch ABC and ABC on graph paper. Solution a. Under the given translation, every point moves 5 units to the left and 3 units up. b. y B(2, 5) A(3, 3) → A(3 5, 3 3) A(2, 0) B(3, 2) → B(3 5, 2 3) B(2, 5) C(6, 1) → C(6 5, 1 3) C(1, 4) C(1, 4) B(3, 2) C(6, 1) 1 A(2, 0) O 1 x A(3, 3) Translational Symmetry DEFINITION A figure has translational symmetry if the image of every point of the figure is a point of the figure. Patterns used for decorative purposes such as wallpaper or borders on clothing often appear to have translational symmetry. True translational symmetry would be possible, however, only if the pattern could repeat without end. 14365C06.pgs 7/12/07 2:57 PM Page 236 236 Transformations and the Coordinate Plane Exercises Writing About Mathematics 1. Explain why there can be no fixed points under a translation other than T0,0. 2. Hunter said that if A is the image of A under a reflection in the y-axis and A is the image of A under a reflection in the line x 3, then A is the image of A under the translation (x, y) → (x 6, y). a. Do you agree with Hunter when A is a point with an x-coordinate greater than or equal to 3? Justify your answer. b. Do you agree with Hunter when A is a point with an x-coordinate greater than 0 but less than 3? Justify your answer. c. Do you agree with Hunter when A is a point with a negative x-coordinate? Justify your answer. Developing Skills 3. The diagram consists of nine congruent rectangles. Under a A B C D translation, the image of A is G. Find the image of each of the given points under the same translation. a. J b. B c. I d. F e. a. On graph paper, draw and label ABC, whose vertices have the coordinates A(1, 2), B(6, 3), and C(4, 6). b. Under the translation P(x, y) → P(x 5, y 3), the image of ABC is ABC. Find the coordinates of A, B, and C. c. On the same graph drawn in part a, draw and label ABC. 5. a. On graph paper, draw and label ABC if the coordinates of A are (2, 2), the coordi- nates of B are (2, 0), and the coordinates of C are (3, 3). b. On the same graph, draw and label ABC, the image of ABC under the transla- tion T4,7. c. Give the coordinates of the vertices of ABC. 6. If the rule of a translation is written (x, y) → (x a, y b), what are the values of a and b for a translation where every point moves 6 units to the right on a graph? 7. In a translation, every point moves 4 units down. Write a rule for this translation in the form (x, y) → (x a, y b). 14365C06.pgs 7/12/07 2:57 PM Page 237 Translations in the Coordinate Plane 237 8. The coordinates of ABC are A(2, 1), B(4, 1), and C(5, 5). a. On graph paper, draw and label ABC. b. Write a rule for the translation in which the image of A is C(5, 5). c. Use the rule from part b to find the coordinates of B, the image of B, and C, the image of C, under this translation. d. On the graph drawn in part a, draw and label CBC, the image of ABC. 9. The coordinates of the vertices of ABC are A(0, 1), B(2, 1), and C(3, 3). a. On graph paper, draw and label ABC. b. Under a translation, the image of C is B(2, 1). Find the coordinates of A, the image of A, and of B, the image of B, under this same translation. c. On the graph drawn in part a, draw and label ABB, the image of ABC. d. How many points, if any, are fixed points under this translation? Applying Skills 10. Prove the corollaries of Theorem 6.7. a. Corollary 6.7a, “Under a translation, angle measure is preserved.” b. Corollary 6.7b, “Under a translation, collinearity is preserved.” c. Corollary 6.7c, “Under a translation, midpoint is preserved.” (Hint: See the proofs of the corollaries of Theorem 6.1 on page 217.) 11. The coordinates of the vertices of LMN are L(6, 0), M(2, 0), and N(2, 2). a. Draw LMN on graph paper. b. Find the coordinates of the vertices of LMN, the image of LMN under a reflection in the line x 1, and draw LMN on the graph drawn in a. c. Find the coordinates of the vertices of LMN, the image of LMN under a reflec- tion in the line x = 4, and draw LMN on the graph drawn in a. d. Find the coordinates of the vertices of PQR, the image of LMN under the translation T10,0. e. What is the relationship between LMN and PQR? 12. The coordinates of the vertices of DEF are D(2, 3), E(1, 3), and F(0, 0). a. Draw DEF on graph paper. b. Find the coordinates of the vertices of DEF, the image of DEF under a reflection in the line y 0 (the x-axis), and draw DEF on the graph drawn in a. c. Find the coordinates of the vertices of DEF, the image of DEF under a reflec- tion in the line y 3, and draw DEF on the graph drawn in a. d. Find the coordinates of the vertices of RST, the image of DEF under the translation T0,6. e. What is the relationship between DEF and RST? 14365C06.pgs 7/12/07 2:57 PM Page 238 238 Transformations and the Coordinate Plane 13. The coordinates of the vertices of ABC are A(1, 2), B(5, 2), and C(4, 5 |
). a. Draw ABC on graph paper. b. Find the coordinates of the vertices of ABC, the image of ABC under a reflection in the line y 0 (the x-axis), and draw ABC on the graph drawn in a. c. Find the coordinates of the vertices of ABC, the image of ABC under a reflec- tion in the line y 3, and draw ABC on the graph drawn in a. d. Is there a translation, Ta,b, such that ABC is the image of ABC? If so, what are the values of a and b? 14. In exercises, 11, 12, and 13, is there a relationship between the distance between the two lines of reflection and the x-values and y-values in the translation? 15. The coordinates of the vertices of ABC are A(1, 2), B(5, 2), and C(4, 5). a. Draw ABC on graph paper. b. Find the coordinates of the vertices of ABC, the image of ABC under a reflection in any horizontal line, and draw ABC on the graph drawn in a. c. Find the coordinates of the vertices of ABC, the image of ABC under a reflection in the line that is 3 units below the line of reflection that you used in b, and draw ABC on the graph drawn in a. d. What is the translation Ta,b such that ABC is the image of ABC? Is this the same translation that you found in exercise 13? 6-6 ROTATIONS IN THE COORDINATE PLANE Think of what happens to all of the points of a wheel as the wheel is turned. Except for the fixed point in the center, every point moves through a part of a circle, or arc, so that the position of each point is changed by a rotation of the same number of degrees. DEFINITION A rotation is a transformation of a plane about a fixed point P through an angle of d degrees such that: 1. For A, a point that is not the fixed point P, if the image of A is A, then PA = PA and mAPA d. 2. The image of the center of rotation P is P. In the figure, P is the center of rotation. If A is rotated about P to A, and B is rotated the same number of degrees to B, then mAPA mBPB. Since P is the center of rotation, PA PA and PB PB, and mAPA mBPA mAPB mBPB mBPA mAPB B A P 14365C06.pgs 7/12/07 2:57 PM Page 239 Rotations in the Coordinate Plane 239 Therefore, mAPB = mAPB and APB APB by SAS. Because corresponding parts of congruent triangles are congruent and equal in measure, AB AB, that is, that distance is preserved under a rotation. We have just proved the following theorem: Theorem 6.8 Distance is preserved under a rotation about a fixed point. As we have seen when studying other transformations, when distance is pre- served, angle measure, collinearity, and midpoint are also preserved. Under a rotation about a fixed point, distance, angle measure, collinearity, and midpoint are preserved. We use RP,d as a symbol for the image under a rotation of d degrees about point P. For example, the statement “RO,30°(A) B” can be read as “the image of A under a rotation of 30° degrees about the origin is B.” A rotation in the counterclockwise direction is called a positive rotation. For instance, B is the image of A under a rotation of 30° about P. A rotation in the clockwise direction is called a negative rotation. For instance, C is the image of A under a rotation of 45° about P. B A P C Rotational Symmetry DEFINITION A figure is said to have rotational symmetry if the figure is its own image under a rotation and the center of rotation is the only fixed point. Many letters, as well as designs in the shapes of wheels, stars, and polygons, have rotational symmetry. When a figure has rotational symmetry under a rotation of do, we can rotate the figure by do to an image that is an identical figure. Each figure shown below has rotational symmetry. 14365C06.pgs 7/12/07 2:57 PM Page 240 240 Transformations and the Coordinate Plane Any regular polygon (a polygon with all sides congruent and all angles congruent) has rotational sym3608 , metry. When regular pentagon ABCDE is rotated 5 or 72°, about its center, the image of every point of the figure is a point of the figure. Under this rotation The figure would also have rotational symmetry if it were rotated through a multiple of 72° (144°, 216°, or 288°). If it were rotated through 360°, every point would be its own image. Since this is true for every figure, we do not usually consider a 360° rotation as rotational symmetry. Rotations in the Coordinate Plane The most common rotation in the coordinate plane is a quarter turn about the origin, that is, a counterclockwise rotation of 90° about the origin. In the diagram, the vertices of right triangle ABC are A(0,0), B(3, 4) and C(3, 0). When rotated 90° about the origin, A remains fixed because it is the center of rotation. The image of C, which is on the x-axis and 3 units from the origin, is C(0, 3) on the y-axis and 3 units from the origin. Since is a vertical line 4 units long, its image is a horizontal line 4 units long and to the left of the y-axis. Therefore, the image of B is B(4, 3). Notice that the x-coordinate of B is the negative of the y-coordinate of B and the y-coordinate of B is the x-coordinate of B. CB y B(3, 4) A(0, 0) C(3, 0) x The point B(3, 4) and its image B(4, 3) in the above example suggest a rule for the coordinates of the image of any point P(x, y) under a counterclockwise rotation of 90° about the origin. Theorem 6.9 Under a counterclockwise rotation of 90° about the origin, the image of P(a, b) is P(–b, a). Proof: We will prove this theorem by using a rectangle with opposite vertices at the origin and at P. Note that in quadrants I and II, when b is positive, b is negative, and in quadrants III and IV, when b is negative, b is positive. 14365C06.pgs 7/12/07 2:57 PM Page 241 S(0, b) y P(a, b) x O R(a, 0) Rotations in the Coordinate Plane 241 y P(a, b) S(0, b) x R(a, 0) O y y R(a, 0) O x P(a, b) S(0, b) R(a, 0) x P(a, b) O S(0, b) Let P(a, b) be any point not on an axis and O(0, 0) be the origin. Let R(a, 0) be the point on the x-axis on the same vertical line as P and S(0, b) be the point on the y-axis on the same horizontal line as P. Therefore, PR b 0 b and PS a 0 a. P(b, a) y R(0, a) S(0, b) P(a, b) c b x S(b, 0) O a R(a, 0) c OR Under a rotation of 90° about the origin, the image of ORPS is ORPS. Then, OR OR and mROR 90. This means that since is a horizontal ORr is a vertical segment with segment, the same length. Therefore, since R is on the x-axis, R is on the y-axis and the coordinates of R are (0, a). Similarly, since S is on the y-axis, S is on the x-axis and the coordinates of S are (b, 0). Point P is on the same horizontal line as R and therefore has the same y-coordinate as R, and P is on the same vertical line as S and has the same x-coordinate as S. Therefore, the coordinates of P are (b, a). The statement of Theorem 6.9 may be written as: RO,90°(x, y) (y, x) or R90°(x, y) (y, x) When the point that is the center of rotation is not named, it is understood to be O, the origin. Note: The symbol R is used to designate both a point reflection and a rotation. 1. When the symbol R is followed by a letter that designates a point, it repre- sents a reflection in that point. 2. When the symbol R is followed by both a letter that designates a point and the number of degrees, it represents a rotation of the given number of degrees about the given point. 3. When the symbol R is followed by the number of degrees, it represents a rotation of the given number of degrees about the origin. 14365C06.pgs 7/12/07 2:57 PM Page 242 242 Transformations and the Coordinate Plane EXAMPLE 1 Point P is at the center of equilateral triangle ABC (the point at which the perpendicular bisectors of the sides intersect) so that: PA PB PC and mAPB mBPC mCPA A P Under a rotation about P for which the image of A is B, find: B C a. The number of degrees in the rotation. b. The image of B. c. The image of d. The image of CAB. . CA Answers a. 360o 3 5 120o b. C c. AB d. ABC EXAMPLE 2 What are the coordinates of the image of P(2, 3) under R90°? Solution The image of (x, y) is (y, x). Therefore, the image of (2, 3) is (3, 2). Answer (3, 2) Exercises Writing About Mathematics 1. A point in the coordinate plane is rotated 180° about the origin by rotating the point counterclockwise 90° and then rotating the image counterclockwise 90°. a. Choose several points in the coordinate plane and find their images under a rotation of 180°. What is the image of P(x, y) under this rotation? b. For what other transformation does P(x, y) have the same image? c. Is a 180° rotation about the origin equivalent to the transformation found in part b? 2. Let Q be the image of P(x, y) under a clockwise rotation of 90° about the origin and R be the image of Q under a clockwise rotation of 90° about the origin. For what two different transformations is R the image of P? 14365C06.pgs 7/12/07 2:57 PM Page 243 Developing Skills For 3 and 4, refer to the figure at the right. 3. What is the image of each of the given points under R90°? a. A e. H b. B f. J c. C g. K d. G h. L 4. What is the image of each of the given points under R–90°? a. A b. B c. C d. G Glide Reflections 243 y C B A O x LK J D EF G H I e. H h. L 5. The vertices of rectangle ABCD are A(2, 1), B(5, 1), C(5, 4), and D(2, 4). g. K f. J a. What are the coordinates of the vertices of ABCD, the image of ABCD under a counterclockwise rotation of 90° about the origin? b. Is ABCD a rectangle? c. Find the coordinates of the midpoint of AB and of ArBr . Is the midpoint of ArBr the image of the midpoint of AB under this rotation? 6. a. What are the coordinates of Q, the image of P(1, 3) under a counterclockwise rotation of 90° about the origin? b. What are the coordinates of R, the image of Q under a counterclockwise rotation of 90° about the origin? c. What are the coordinates of S, the image of R under a counterclockwise rotation of 90° about the origin? d. What are the coordinates of P, the image of P under a clockwise rotation of 90° about the origin? e. Explain why S and P are the same point. 6-7 GLIDE REFLECTIONS When two transformations are performed, one following the other, we have a composition of transformations. The first tr |
ansformation produces an image and the second transformation is performed on that image. One such composition that occurs frequently is the composition of a line reflection and a translation. DEFINITION A glide reflection is a composition of transformations of the plane that consists of a line reflection and a translation in the direction of the line of reflection performed in either order. 14365C06.pgs 7/12/07 2:57 PM Page 244 244 Transformations and the Coordinate Plane y C The vertices of ABC are A(1, 2), B(5, 3), and C(3, 4). Under a reflection in the y-axis, the image of ABC is ABC whose vertices are A(1, 2), B(5, 3), and C(3, 4). Under the translation T0,–4, the image of ABC is ABC whose vertices are A(1, 2), B(5, 1), and C(3, 0). Under a glide reflection, the image of ABC is ABC. Note that the line of reflection, the y-axis, is a vertical line. The translation is in the vertical direction because the x-coordinate of the translation is 0. Distance is preserved under a line reflection and under a translation: ABC ABC ABC. Distance is preserved under a glide reflection. We have just proved the following theorem: O A B 1 1 x Theorem 6.10 Under a glide reflection, distance is preserved. As we have seen with other transformations, when distance is preserved, angle measure, collinearity, and midpoint are also preserved. Therefore, we may make the following statement: Under a glide reflection, distance, angle measure, collinearity, and midpoint are preserved. In this chapter, we have studied five transformations: line reflection, point reflection, translation, rotation, and glide reflection. Each of these transformations is called an isometry because it preserves distance. DEFINITION An isometry is a transformation that preserves distance. EXAMPLE 1 The vertices of PQR are P(2, 1), Q(4, 1), and R(4, 3). a. Find PQR, the image of PQR under ryx followed by T3,3. b. Find PQR, the image of PQR under T3,3 followed by ryx. c. Are PQR and PQR the same triangle? d. Are ryx followed by T3,3 and T3,3 followed by ryx the same glide reflection? Explain. e. Write a rule for this glide reflection. 14365C06.pgs 7/12/07 2:57 PM Page 245 Solution T3,3(y, x) (y 3, x 3) a. ryx(x, y) (y, x) T3,3(1, 2) (2, 1) ryx(2, 1) (1, 2) T3,3(1, 4) (2, 1) ryx(4, 1) (1, 4) ryx(4, 3) (3, 4) T3,3(3, 4) (0, 1) The vertices of PQR are P(2, 1), Q(2, 1), and R(0, 1). Answer Glide Reflections 245 y R Q x P b. T3,3(x, y) (x 3, y 3) T3,3(2, 1) (1, 2) T3,3(4, 1) (1, 2) T3,3(4, 3) (1, 0) The vertices of PQR are P (2, 1), Q(2, 1), and R(0, 1). Answer ryx(x 3, y 3) (y – 3, x 3) ryx(1, 2) (2, 1) ryx(1, 2) (2, 1) ryx(1, 0) (0, 1. PQR and PQR are the same triangle. Answer d. The image of (x, y) under ryx followed by T3,3 is (x, y) → (y, x) → (y 3, x 3) The image of (x, y) under T3,3 followed by ryx is (x, y) → (x 3, y 3) → (y 3, x 3) ryx followed by T3,3 and T3,3 followed by ryx are the same glide reflection. Answer e. (x, y) → (y 3, x 3) Answer EXAMPLE 2 Is a reflection in the y-axis followed by the translation T5,5 a glide reflection? Solution The y-axis is a vertical line. The translation T5,5 is not a translation in the verti- cal direction. Therefore, a reflection in the y-axis followed by the translation T5,5 is not a glide reflection. 14365C06.pgs 7/12/07 2:57 PM Page 246 246 Transformations and the Coordinate Plane Exercises Writing About Mathematics 1. Does a glide reflection have any fixed points? Justify your answer. 2. In a glide reflection, the line reflection and the translation can be done in either order. Is the composition of any line reflection and any translation always the same in either order? Justify your answer. Developing Skills In 3–8, a. find the coordinates of the image of the triangle whose vertices are A(1, 1), B(5, 4), and C(3, 5) under the given composition of transformations. b. Sketch ABC and its image. c. Explain why the given composition of transformations is or is not a glide reflection. d. Write the coordinates of the image of (a, b) under the given composition of transformations. 3. A reflection in the x-axis followed by a translation of 4 units to the right. 4. A reflection in the y-axis followed by a translation of 6 units down. 5. T5,0 followed by rx-axis 7. T3,3 followed by ry=x Applying Skills 6. T3,3 followed by ry-axis 8. R90 followed by T1,2 9. The point A(4, 5) is the image of A under a glide reflection that consists of a reflection in the x-axis followed by the translation T2,0. What are the coordinates of A? 10. Under a glide reflection, the image of A(2, 4) is A(2, 7) and the image of B(3, 7) is B(3, 4). a. If the line of reflection is a vertical line, write an equation for the line reflection and a rule for the translation. b. What are the coordinates of C, the image of C(0, 7), under the same glide reflection? 11. Under a glide reflection, the image of A(3, 4) is A(1, 4) and the image of B(5, 5) is B(1, 5). a. If the line of reflection is a horizontal line, write a rule for the line reflection and a rule for the translation. b. What are the coordinates of C, the image of C(4, 2), under the same glide reflection? 12. a. For what transformation or transformations are there no fixed points? b. For what transformation or transformations is there exactly one fixed point? c. For what transformation or transformations are there infinitely many fixed points? 14365C06.pgs 7/12/07 2:57 PM Page 247 Dilations in the Coordinate Plane 247 6-8 DILATIONS IN THE COORDINATE PLANE In this chapter, we have learned about transformations in the plane that are isometries, that is, transformations that preserve distance. There is another transformation in the plane that preserves angle measure but not distance. This transformation is a dilation. For example, in the coordinate plane, a dilation of 2 with center at the origin will stretch each ray by a factor of 2. If the image of A is A, h then A is a point on OA and OA 2OA . y O A A x DEFINITION A dilation of k is a transformation of the plane such that: 1. The image of point O, the center of dilation, is O. 2. When k is positive and the image of P is P, then h OP and h OPr and OP kOP. 3. When k is negative and the image of P is P, then rays and OP kOP. are the same ray h OP and h OPr are opposite Note: In step 3, when k is negative, k is positive. In the coordinate plane, the center of dilation is usually the origin. If the center of dilation is not the origin, the coordinates of the center will be given. In the coordinate plane, under a dilation of k with the center at the origin: P(x, y) → P(kx, ky) or Dk(x, y) (kx, ky) For example, the image of ABC is ABC under a dilation of . The vertices of ABC are A(2, 6), B(6, 4), and C(4, 0). Under a dilation of , the rule is 1 2 1 2 A (x, y) 5 2x, 1 1 D1 2y B 2 A(2, 6) → A(1, 3) B(6, 4) → B(3, 2) C(4, 0) → C(2, 0 14365C06.pgs 7/12/07 2:57 PM Page 248 248 Transformations and the Coordinate Plane By the definition of a dilation, distance is not preserved. We will prove in Chapter 12 that angle measure, collinearity, and midpoint are preserved. Under a dilation about a fixed point, distance is not preserved. EXAMPLE 1 The coordinates of parallelogram EFGH are E(0, 0), F(3, 0), G(4, 2), and H(1, 2). Under D3, the image of EFGH is EFGH. a. Find the coordinates of the vertices of EFGH. b. Let M be the midpoint of EF . Find the coordinates of M and of M, the image of M. Verify that the midpoint is preserved. Solution a. D3(x, y) (3x, 3y). Therefore, E(0, 0), F(9, 0), G(12, 6), and H(3, 6). Answer b. Since E and F lie on the x-axis, EF is the absolute value of the difference of their x-coordinates. EF 3 0 3. 11 Therefore, the midpoint, M, is 2 from E or from F on the x-axis. The 5 coordinates of M are units , 0 0 1 11 11 2, 0 A Since E and F lie on the x-axis, EF is the absolute value of the difference of their x-coordinates. EF 9 0 9. Therefore, the midpoint, 41 units from E or from F on the x-axis. The coordinates of M are M, is 2 0 1 41 2, 0 B 11 D3 2, 0 A preserved. Answer . Therefore, the image of M is M and midpoint is 41 . 2, 0 B 41 2, 0 A 5 5 A B B A EXAMPLE 2 Find the coordinates of P, the image of P(4, 5) under the composition of transformations: D2 followed by rx-axis. Solution The dilation is to be performed first: D2(4, 5) (8, 10). Then perform the reflection, using the result of the dilation: rx-axis(8, 10) (8, 10). Answer P (8, 10) 14365C06.pgs 7/12/07 2:57 PM Page 249 Dilations in the Coordinate Plane 249 Exercises Writing About Mathematics 1. Let ABC be the image of ABC under a dilation of k. When k 1, how do the lengths of the sides of ABC compare with the lengths of the corresponding sides of ABC? Is a dilation an isometry? 2. Let ABC be the image of ABC under a dilation of k. When 0 k 1, how do the lengths of the sides of ABC compare with the lengths of the corresponding sides of ABC? Developing Skills In 3–6, use the rule (x, y) → (4x, 4y) to find the coordinates of the image of each given point. 3. (3, 7) 5. (2, 0) 4. (4, 2) 6. (1, 9) In 7–10, find the coordinates of the image of each given point under D5. 7. (2, 2) 9. (3, 5) 8. (1, 10) 10. (0, 4) In 11–14, each given point is the image under D2. Find the coordinates of each preimage. 11. (6, 2) 13. (6, 5) 12. (4, 0) 14. (10, 7) In 15–20, find the coordinates of the image of each given point under the given composition of transformations. 15. D4(1, 5) followed by ry-axis 17. T2,1(4, 2) followed by D1 2 19. T1,1(4, 2) followed by D2 16. D2(3, 2) followed by R90° 18. D3(5, 1) followed by rx-axis 20. D2(6, 3) followed by ryx In 21–24, each transformation is the composition of a dilation and a reflection in either the x-axis or the y-axis. In each case, write a rule for composition of transformations for which the image of A is A. 21. A(2, 5) → A(4, 10) 23. A(10, 4) → A(5, 2) Applying Skills 22. A(3, 1) → A(21, 7) 24. A(20, 8) → A(5, 2) 25. The vertices of rectangle ABCD are A(1, 1), B(3, 1), C(3, 3), and D(1, 3). a. Find the coordinates of the vertices of ABCD, the image of ABC |
D under D4. b. Show that ABCD is a rectangle. 14365C06.pgs 7/12/07 2:57 PM Page 250 250 Transformations and the Coordinate Plane 26. Show that when k 0, a dilation of k with center at the origin followed by a reflection in the origin is the same as a dilation of k with center at the origin. 27. a. Draw ABC, whose vertices are A(2, 3), B(4, 3), and C(4, 6). b. Using the same set of axes, graph ABC, the image of ABC under a dilation of 3. c. Using ABC and its image ABC, show that distance is not preserved under the dilation. Hands-On Activity In this activity, we will verify that angle measure, collinearity, and midpoint are preserved under a dilation. Using geometry software, or a pencil, ruler and a protractor, draw the pentagon A(1, 2), B(4, 2), C(6, 4), D(2, 8), E(1, 6), and point P(4, 6) on . For each given dilation: a. Measure the corresponding angles. Do they appear to be congruent? b. Does the image of P appear ? c. Does the image of the midpoint of to be on the image of appear to be the midCD point of the image of ? AE (1) D3 (2) D–3 D1 2 D21 AE CD (4) (3) 2 6-9 TRANSFORMATIONS AS FUNCTIONS A function is a set of ordered pairs in which no two pairs have the same first element. For example, the equation y x 5 describes the set of ordered pairs of real numbers in which the second element, y, is 5 more than the first element, x. The set of first elements is the domain of the function and the set of second ele3, 17 2 ments is the range. Some pairs of this function are (3, 2), (2.7, 2.3), , 3 B (0, 5), and (1, 6). Since the set of real numbers is infinite, the set of pairs of this function cannot be listed completely and are more correctly defined when they are described. Each of the following notations describes the function in which the second element is 5 more than the first. A f {(x, y) y x 5} f: x → x 5 y x 5 f(x) x 5 Note that f(x) and y each name the second element of the ordered pair. Each of the transformations defined in this chapter is a one-to-one function. It is a set of ordered pairs in which the first element of each pair is a point of the plane and the second element is the image of that point under a transformation. For each point in the plane, there is one and only one image. For example, compare the function f(x) x 5 and a line reflection. 14365C06.pgs 7/12/07 2:57 PM Page 251 f: A one-to-one algebraic function f(x) x 5 y f(x) x 5 1 O 1 x Transformations as Functions 251 rm: A reflection in line m A B D C E m 1. For every x in the domain there is one and only one y in the range. For example: f(3) 8 f(0) 5 f(2) 3 2. Every f(x) or y in the range corresponds to one and only one value of x in the range. For example: If f(x) 4, then x 1. If f(x) 9, then x 4. 3. For the function f: domain {real numbers} range {real numbers} 1. For every point in the plane, there is one and only one image of that point in the plane. For example: rm(A) A rm(C) C rm(D) D 2. Every point is the image of one and only one preimage. For example: The point B has the preimage B. The point E has itself as the preimage. The point C has the preimage C. 3. For the function rm: domain {points in the plane} range {points in the plane} Composition of Transformations We defined the composition of transformations as a combination of two transformations in which the first transformation produces an image and the second transformation is performed on that image. For example, to indicate that A is the image of A under a reflection in the line y = x followed by the translation T2,0, we can write T2,0(ryx(A)) A. 14365C06.pgs 7/12/07 2:57 PM Page 252 252 Transformations and the Coordinate Plane If the coordinates of A are (2, 5), we start with A and perform the transformations from right to left as indicated by the parentheses, evaluating what is in parentheses first. ryx(2, 5) (5, 2) followed by T2,0(5, 2) (7, 2) or T2,0(ry 5 x(2, 5)) 5 T2,0(5, 2) 5 (7, 2) A small raised circle is another way of indicating a composition of transfor- mations. T2,0(ryx (A)) A can also be written as + ry 5 x(A) 5 Ar . Again, we start with the coordinates of A and move from right to left. Find the image of A under the reflection and then, using the coordinates of that image, find the coordinates under the translation. T2,0 Orientation The figures below show the images of ABC under a point reflection, a translation, and a rotation. In each figure, the vertices, when traced from A to B to C are in the clockwise direction, called the orientation of the points. The vertices of the images, when traced in the same order, from A to B to C are also in the clockwise direction. Therefore, we say that under a point reflection, a translation, or a rotation, orientation is unchanged Point Reflection Translation P Rotation DEFINITION A direct isometry is a transformation that preserves distance and orientation. Point reflection, rotation, and translation are direct isometries. 14365C06.pgs 7/12/07 2:57 PM Page 253 The figure at the right shows ABC and its image under a line reflection. In this figure, the vertices, when traced from A to B to C, are again in the clockwise direction. The vertices of the images, when traced in the same order, from A to B to C, are in the counterclockwise direction. Therefore, under a line reflection, orientation is changed or reversed. Transformations as Functions 253 B A C Line Reflection DEFINITION An opposite isometry is a transformation that preserves distance but changes the order or orientation from clockwise to counterclockwise or from counterclockwise to clockwise. A line reflection is an opposite isometry. EXAMPLE 1 The vertices of RST are R(1,1), S(6, 3), and T(2, 5). a. Sketch RST. b. Find the coordinates of the vertices of RST under the composition rx-axis + ry-axis and sketch RST. c. Is the composition a direct isometry? d. For what single transformation is the image the same as that of rx-axis Solution a. + ry-axis ? y 1 R O1 T b. rx-axis + ry-axis( S x rx-axis + ry-axis( rx-axis + ry-axis( 1, 1) rx-axis(1, 1) (1, 1) 6, 3) rx-axis(6, 3) (6, 3) 2, 5) rx-axis(2, 5) (2, 5) c. A line reflection is an opposite d. isometry. The composition of two opposite isometries is a direct isometry. This composition is a direct isometry. + ry-axis( x, y) (x, y) rx-axis and RO(x, y) (–x, y). The composition of a reflection in the y-axis followed by a reflection in the x-axis is a reflection in the origin. 14365C06.pgs 7/12/07 2:57 PM Page 254 254 Transformations and the Coordinate Plane EXAMPLE 2 The transformation (x, y) → (y 3, x 2) is the composition of what two transformations? Solution Under a rotation of 90° about the origin, (x, y) → (y, x). When this rotation is followed by the translation T3,–2(y, x) (y 3, x 2). Answer T3,22 + R908(x,y) T3,2(y, x) (y 3, x 2) This solution is not unique. Alternative Solution Under the translation T2,3(x, y) (x 2, y 3). When this translation is followed by the rotation R90°(x 2, y 3) ((y 3), x 2) (y 3, x 2). Answer R908 + T22,23(x, y) R90°(x 2, y 3) (y 3, x 2) Exercises Writing About Mathematics 1. Owen said that since a reflection in the x-axis followed by a reflection in the y-axis has the same result as a reflection in the y-axis followed by a reflection in the x-axis, the composition of line reflections is a commutative operation. Do you agree with Owen? Justify your answer. 2. Tyler said that the composition of an even number of opposite isometries is a direct isometry and that the composition of an odd number of opposite isometries is an opposite isometry. Do you agree with Tyler? Justify your answer. Developing Skills In 3–11, find the image of A(3, 2) under the given composition of transformations. 3. 6. rx-axis R908 ry-axis + ry-axis + RO + ry 5 x 4. 7. ry-axis T21,4 RO + rx-axis + ry 5 x + T3,22 9. 12. A reflection in the line y x followed by a rotation of 90° about the origin is equivalent to 10. 11. what single transformation? 13. A rotation of 90° about the origin followed by another rotation of 90° about the origin is equivalent to what single transformation? 5. 8. RO R908 T22,4 + rx-axis + R908 + D22 14365C06.pgs 8/2/07 5:45 PM Page 255 14. The vertices of DEF are D(3, 2), E(5, 5), and F(4, 1). Chapter Summary 255 a. If T23,0 + rx-axis(nDEF) T23,0 DEF, find the coordinates of the vertices of DEF. + rx-axis is an example of what type of transformation? b. The composition c. Is the composition T23,0 + rx-axis a direct or an opposite isometry? Applying Skills 15. Prove that under a reflection in the line y x, A(a, b) → A(b, a). Suggested plan: Let B(a, a) and C(b, b) be points on the line y x. Show that the line y x is the perpendicular bisector of AAr . 16. Prove that the composition of a reflection in the line y x followed by a reflection in the line y x is a reflection in the origin. 17. Prove that the composition of a reflection in the line y x followed by a reflection in the line y x is a reflection in the origin. CHAPTER SUMMARY Definitions to Know • A transformation is a one-to-one correspondence between two sets of points, S and S, when every point in S corresponds to one and only one point in S called its image, and every point in S is the image of one and only one point in S called its preimage. • A reflection in line k is a transformation in a plane such that: 1. If point A is not on k, then the image of A is A where k is the perpen- dicular bisector of AAr . 2. If point A is on k, the image of A is A. • A figure has line symmetry when the figure is its own image under a line reflection. • A point reflection in P is a transformation plane such that: 1. If point A is not point P, then the image of A is A where P the mid- point of AAr . 2. The point P is its own image. • A translation is a transformation in a plane that moves every point in the plane the same distance in the same direction. • A translation of a units in the horizontal direction and b units in the vertical direction is a transformation in a plane such that the image of P(x, y) is P(x a, y b). 14365C06.pgs 8/2/07 5:45 |
PM Page 256 256 Transformations and the Coordinate Plane • A rotation is a transformation of a plane about a fixed point P through an angle of d degrees such that: 1. For A, a point that is not the fixed point, if the image of A is A, then PA PA and mAPA d. 2. The image of the center of rotation P is P. • A quarter turn is a counterclockwise rotation of 90°. • A composition of transformations is a combination of two transformations in which the first transformation produces an image and the second transformation is performed on that image. • A glide reflection is a composition of transformations of the plane that consists of a line reflection and a translation in the direction of the line of reflection in either order. • An isometry is a transformation that preserves distance. • A dilation of k is a transformation of the plane such that: 1. The image of point O, the center of dilation, is O. 2. When k is positive and the image of P is P, then h OP and h OPr are the same ray and OP = kOP. 3. When k is negative and the image of P is P, then opposite rays and OP = kOP. h OP and h OPr are • A function is a set of ordered pairs in which no two pairs have the same first element. • The set of first elements is the domain of the function. • The set of second elements is the range of the function. • A direct isometry is a transformation that preserves distance and orienta- tion. • An opposite isometry is a transformation that preserves distance and reverses orientation. Postulates 6.1 Two points are on the same horizontal line if and only if they have the same y-coordinates. 6.2 The length of a horizontal line segment is the absolute value of the differ- ence of the x-coordinates. 6.3 Two points are on the same vertical line if and only if they have the same x-coordinate. 6.4 The length of a vertical line segment is the absolute value of the difference of the y-coordinates. 6.5 Each vertical line is perpendicular to each horizontal line. Theorems and Corollaries 6.1 Under a line reflection, distance is preserved. 6.1a Under a reflection, angle measure is preserved. 6.1b Under a reflection, collinearity is preserved. 6.1c Under a reflection, midpoint is preserved. 14365C06.pgs 7/12/07 2:57 PM Page 257 Review Exercises 257 6.2 Under a reflection in the y-axis, the image of P(a, b) is P(a, b). 6.3 Under a reflection in the x-axis, the image of P(a, b) is P(a, b). 6.4 Under a reflection in the line y x, the image of P(a, b) is P(b, a). 6.5 Under a point reflection, distance is preserved. 6.5a Under a point reflection, angle measure is preserved. 6.5b Under a point reflection, collinearity is preserved. 6.5c Under a point reflection, midpoint is preserved. 6.6 Under a reflection in the origin, the image of P(a, b) is P(a, b). 6.7 Under a translation, distance is preserved. 6.7a Under a translation, angle measure is preserved. 6.7b Under a translation, collinearity is preserved. 6.7c Under a translation, midpoint is preserved. 6.8 Distance is preserved under a rotation about a fixed point. 6.9 Under a counterclockwise rotation of 90° about the origin, the image of P(a, b) is P(b, a). 6.10 Under a glide reflection, distance is preserved. VOCABULARY 6-1 Coordinate plane • x-axis • y-axis • Origin • Coordinates • Ordered pair • x-coordinate (abscissa) • y-coordinate (ordinate) • (x, y) 6-2 Line of reflection • Line reflection • Fixed point • Transformation • Preimage • Image • Reflection in line k • rk • Axis of symmetry • Line symmetry 6-3 ry-axis • rx-axis • ryx 6-4 Point reflection in P • RP • Point symmetry • RO 6-5 Translation • Translation of a units in the horizontal direction and b units in the vertical direction • Ta,b • Translational symmetry 6-6 Rotation • RP,d • Positive rotation • Negative rotation • Rotational symmetry • Quarter turn 6-7 Composition of transformations • Glide reflection • Isometry 6-8 Dilation • Dk 6-9 Function • Domain • Range • Orientation • Direct isometry • Opposite Isometry REVIEW EXERCISES 1. Write the equation of the vertical line that is 2 units to the left of the origin. 2. Write the equation of the horizontal line that is 4 units below the origin. 14365C06.pgs 7/12/07 2:57 PM Page 258 258 Transformations and the Coordinate Plane 3. a. On graph paper, draw the polygon ABCD whose vertices are A(4, 0), B(0, 0), C(3, 3), and D(4, 3). b. Find the area of polygon ABCD. In 4–11: a. Find the image of P(5, 3) under each of the given transformations. b. Name a fixed point of the transformation if one exists. 4. rx-axis 8. rx-axis 7. T0,3 11. R(2,2) 12. Draw a quadrilateral that has exactly one line of symmetry. 6. R90° 10. ry 5 x 5. RO 9. + rx-axis + ry 5 x + T2,2 rx-axis + T5,23 13. Draw a quadrilateral that has exactly four lines of symmetry. 14. Print a letter that has rotational symmetry. 15. The letters S and N have point symmetry. Print another letter that has point symmetry. 16. What transformations are opposite isometries? 17. What transformation is not an isometry? 18. a. On graph paper, locate the points A(3, 2), B(3, 7), and C(2, 7). Draw ABC. b. Draw ABC, the image of ABC under a reflection in the origin, and write the coordinates of its vertices. c. Draw ABC, the image of ABC under a reflection in the y-axis, and write the coordinates of its vertices. d. Under what single transformation is ABC the image of ABC? 19. a. On graph paper, locate the points R(4, 1), S(1, 1), and T(1, 2). Draw RST. b. Draw RST, the image of RST under a reflection in the origin, and write the coordinates of its vertices. c. Draw RST , the image of RST under a reflection in the line whose equation is y 4, and write the coordinates of its vertices. 20. a. Write the coordinates of the image, under the correspondence (x, y) → (x, 0), of each of the following ordered pairs: (3, 5), (3, 3), (1, 1), (1, 5). b. Explain why (x, y) → ( x, 0) is not a transformation. 21. The vertices of MAT have coordinates M(1, 3), A(2, 2), and T(2, 2). + D3 . + rx-axis equivalent transformations? Justify your a. Find MAT, the image of MAT under the composition b. Find MAT , the image of MAT under the composition c. Are rx-axis D3 and . D3 + rx-axis + D3 rx-axis answer. 14365C06.pgs 8/2/07 5:45 PM Page 259 Cumulative Review 259 Exploration Designs for wallpaper, wrapping paper, or fabric often repeat the same pattern in different arrangements. Such designs are called tessellations. Find examples of the use of line reflections, point reflections, translations and glide reflections in these designs. CUMULATIVE REVIEW Chapters 1–6 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The product 7a(3a 1) can be written as (1) –21a 7a (2) –21a 7a (3) –21a2 7a (4) –21a2 1 2. In the coordinate plane, the point whose coordinates are (2, 0) is (1) on the x-axis (2) on the y-axis 3. If D is not on ABC (1) AD DC AC (2) AD DC AC , then (3) in the first quadrant (4) in the fourth quadrant (3) AD DC AC (4) AB BC AD DC 4. If is the perpendicular bisector of DBE , which of the following could g ABC be false? (1) AD AE (4) AB BC 5. DEF is not congruent to LMN, DE LM, and EF MN. Which of (3) CD CE (2) DB BE the following must be true? (1) DEF and LMN are not both right triangles. (2) mD mL (3) mF mN (4) mE mM 6. Under a reflection in the origin, the image of (2, 4) is (1) (2, 4) (2) (2, 4) (3) (2, 4) (4) (4, 2) 7. If PQ RQ, then which of the following must be true? . PR (1) Q is the midpoint of (2) Q is on the perpendicular bisector of (3) PQ QR PR (4) Q is between P and R. PR . 14365C06.pgs 7/12/07 2:57 PM Page 260 260 Transformations and the Coordinate Plane 8. Which of the following always has a line of symmetry? (1) a right triangle (2) a scalene triangle (3) an isosceles triangle (4) an acute triangle 9. In the coordinate plane, two points lie on the same vertical line. Which of the following must be true? (1) The points have the same x-coordinates. (2) The points have the same y-coordinates. (3) The points lie on the y-axis. (4) The points lie on the x-axis. 10. Angle A and angle B are complementary angles. If mA x 42 and mB 2x 12, the measure of the smaller angle is (1) 20 (2) 28 (3) 62 (4) 88 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The image of ABC under a reflection in the line y x is ADE. If the coordinates of A are (2, b), what is the value of b? Explain your answer. 12. What are the coordinates of the midpoint of a line segment whose end- points are A(2, 5) and B(2, 3)? Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. In quadrilateral ABCD, prove that ABC ADC if . BED pendicular bisector of is the per- AEC 14. The measure of R is 12 degrees more than three times the measure of S. If S and R are supplementary angles, find the measure of each angle. 14365C06.pgs 7/12/07 2:57 PM Page 261 Cumulative Review 261 Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The measures of the angles of ABC are unequal (mA mB, mB mC, and mA mC). Prove that ABC is a scalene triangle. 16. Given: T3,2(ABC) ABC and XYZ with XY AB, YZ BC, and Y B. Prove: ABC XYZ 14365C07.pgs 7/10/07 8:45 AM Page 262 CHAPTER 7 CHAPTER TABLE OF CONTENTS 7-1 Basic Inequality Postulates 7-2 Inequality Postulates Involving Addition and |
Subtraction 7-3 Inequality Postulates Involving Multiplication and Division 7-4 An Inequality Involving the Lengths of the Sides of a Triangle 7-5 An Inequality Involving an Exterior Angle of a Triangle 7-6 Inequalities Involving Sides and Angles of a Triangle Chapter Summary Vocabulary Review Exercises Cumulative Review 262 GEOMETRIC INEQUALITIES Euclid’s Proposition 20 of Book 1 of the Elements states,“In any triangle, two sides taken together in any manner are greater than the remaining one.” The Epicureans, a group of early Greek philosophers, ridiculed this theorem, stating that it is evident even to a donkey since if food is placed at one vertex of a triangle and the donkey at another, the donkey will make his way along one side of the triangle rather than traverse the other two, to get to the food. But no matter how evident the truth of a statement may be, it is important that it be logically established in order that it may be used in the proof of theorems that follow. Many of the inequality theorems of this chapter depend on this statement for their proof. 14365C07.pgs 7/10/07 8:45 AM Page 263 7-1 BASIC INEQUALITY POSTULATES Basic Inequality Postulates 263 Each time the athletes of the world assemble for the Olympic Games, they attempt to not only perform better than their competitors at the games but also to surpass previous records in their sport. News commentators are constantly comparing the winning time of a bobsled run or a 500-meter skate with the world records and with individual competitors’ records. In previous chapters, we have studied pairs of congruent lines and pairs of congruent angles that have equal measures. But pairs of lines and pairs of angles are often not congruent and have unequal measures. In this chapter, we will apply the basic inequality principles that we used in algebra to the lengths of line segments and the measures of angles. These inequalities will enable us to formulate many important geometric relationships. Postulate Relating a Whole Quantity and Its Parts In Chapter 3 we stated postulates of equality. Many of these postulates suggest related postulates of inequality. Consider the partition postulate: A whole is equal to the sum of all its parts. This corresponds to the following postulate of inequality: Postulate 7.1 A whole is greater than any of its parts. In arithmetic: Since 14 9 5, then 14 9 and 14 5. In algebra: If a, b, and c represent positive numbers and a b c, then a b and a c. In geometry: The lengths of line segments and the measures of angles are positive numbers. Consider these two applications: A E C B D F G ACB is a line segment, then AB AC CB, AB AC, and AB CB. • If • If DEF and FEG are adjacent angles, mDEG mDEF mFEG, mDEG mDEF, and mDEG mFEG. Transitive Property Consider this statement of the transitive property of equality: If a, b, and c are real numbers such that a b and b c, then a c. 14365C07.pgs 7/10/07 8:45 AM Page 264 264 Geometric Inequalities This corresponds to the following transitive property of inequality: Postulate 7.2 If a, b, and c are real numbers such that a b and b c, then a c. In arithmetic: If 12 7 and 7 3, then 12 3. In algebra: If 5x 1 2x and 2x 16, then 5x 1 16. In geometry: If BA BD and BD BC, then BA BC. Also, if mBCA mBCD and mBCD mBAC, then mBCA mBAC. C B D A Substitution Postulate Consider the substitution postulate as it relates to equality: A quantity may be substituted for its equal in any statement of equality. Substitution also holds for inequality, as demonstrated in the following postulate: Postulate 7.3 A quantity may be substituted for its equal in any statement of inequality. In arithmetic: If 10 2 5 and 2 5 7, then 10 7. In algebra: If 5x 1 2y and y 4, then 5x 1 2(4). In geometry: If AB BC and BC AC, then AB AC. Also, if mC mA and mA = mB, then mC mB. C A B The Trichotomy Postulate We know that if x represents the coordinate of a point on the number line, then x can be a point to the left of 3 when x 3, x can be the point whose coordinate is 3 if x 3, or x can be a point to the right of 3 if x 3. We can state this as a postulate that we call the trichotomy postulate, meaning that it is divided into three cases. Postulate 7.4 Given any two quantities, a and b, one and only one of the following is true: a b a b. a b or or 14365C07.pgs 7/10/07 8:45 AM Page 265 Basic Inequality Postulates 265 EXAMPLE 1 Given: mDAC mDAB mBAC and mDAB mABC A D Prove: mDAC mABC C B Proof Statements Reasons 1. mDAC mDAB mBAC 2. mDAC mDAB 1. Given. 2. A whole is greater than any of its 3. mDAB mABC 4. mDAC mABC parts. 3. Given. 4. Transitive property of inequality. EXAMPLE 2 Given: Q is the midpoint of PS and RS QS. Prove: RS PQ P Q R S Proof Statements Reasons 1. Q is the midpoint of . PS 1. Given. 2. PQ > QS 3. PQ = QS. 4. RS QS 5. RS PQ 2. The midpoint of a line segment is the point that divides the segment into two congruent segments. 3. Congruent segments have equal measures. 4. Given. 5. Substitution postulate. 14365C07.pgs 7/10/07 8:45 AM Page 266 266 Geometric Inequalities Exercises Writing About Mathematics 1. Is inequality an equivalence relation? Explain why or why not. 2. Monica said that when AB BC is false, AB BC must be true. Do you agree with Monica? Explain your answer. Developing Skills In 3–12: a. Draw a diagram to illustrate the hypothesis and tell whether each conclusion is true or false. b. State a postulate or a definition that justifies your answer. 3. If ADB is a line segment, then DB AB. , then CD DA CA. AC 4. If D is not on 5. If BCD DCA = BCA, then mBCD mBCA. h DB are opposite rays with point C not on h DA h DB 6. If and h DA or , then mBDC mCDA 180. h DA h DB and 7. If are opposite rays and mBDC 90, then mCDA 90. is a line segment, then DA BD, or DA BD, or DA BD. ADB 8. If 9. If AT AS and AS AR, then AT AR. 10. If m1 m2 and m2 m3, then m1 m3. 11. If SR KR and SR TR, then TR KR. 12. If m3 m2 and m2 m1, then m3 m1. Applying Skills 13. Given: ABC is isosceles, AC BC , mCBD mCBA Prove: mCBD mA C A B D 14. Given: PQRS Prove: a. PR PQ and PQ RS b. PR RS P Q R S 14365C07.pgs 7/10/07 8:45 AM Page 267 In 15 and 16, use the figure to the right. N Inequality Postulates Involving Addition and Subtraction 267 KLM 15. If 16. If KM KN, KN NM, and NM NL, prove that KM NL. , prove that KM NM. NM and LM LK M 7-2 INEQUALITY POSTULATES INVOLVING ADDITION AND SUBTRACTION Postulates of equality and examples of inequalities involving the numbers of arithmetic can help us to understand the inequality postulates presented here. Consider the addition postulate: If equal quantities are added to equal quantities, then the sums are equal. Addition of inequalities requires two cases: Postulate 7.5 If equal quantities are added to unequal quantities, then the sums are unequal in the same order. Postulate 7.6 If unequal quantities are added to unequal quantities in the same order, then the sums are unequal in the same order. E D C In arithmetic: Since 12 5, then 12 3 5 3 or 15 8. Since 12 5 and 3 2, then 12 3 5 2 or 15 7. In algebra: If x 5 10, then x 5 5 10 5 or x 15. If x 5 10 and 5 3, then x 5 5 10 3 or x 13. In geometry: If and AB CD, then AB BC BC CD or ABCD AC BD. If ABCDE AB BC CD DE or AC CE. , AB CD, and BC DE, then B A We can subtract equal quantities from unequal quantities without changing the order of the inequality, but the result is uncertain when we subtract unequal quantities from unequal quantities. Consider the subtraction postulate: If equal quantities are subtracted from equal quantities, then the differences are equal. 14365C07.pgs 7/10/07 8:45 AM Page 268 268 Geometric Inequalities Subtraction of inequalities is restricted to a single case: Postulate 7.7 If equal quantities are subtracted from unequal quantities, then the differences are unequal in the same order. However, when unequal quantities are subtracted from unequal quantities, the results may or may not be unequal and the order of the inequality may or may not be the same. For example: • 5 2 and 4 1, but it is not true that 5 4 2 1 since 1 1. • 12 10 and 7 1, but it is not true that 12 7 10 1 since 5 9. • 12 10 and 2 1, and it is true that 12 2 10 1 since 10 9. EXAMPLE 1 Given: mBDE mCDA Prove: mBDC mEDA E C Proof Statements 1. mBDE mCDA 2. mBDE mEDC mEDC mCDA B D A Reasons 1. Given. 2. If equal quantities are added to unequal quantities, then the sums are unequal in the same order. 3. mBDC mBDE mEDC 3. The whole is equal to the sum of its parts. 4. mEDA mEDC mCDA 4. The whole is equal to the sum of 5. mBDC mEDA its parts. 5. Substitution postulate for inequalities. 14365C07.pgs 7/10/07 8:45 AM Page 269 Inequality Postulates Involving Addition and Subtraction 269 Exercises Writing About Mathematics 1. Dana said that 13 11 and 8 3. Therefore, 13 8 11 3 tells us that if unequal quantities are subtracted from unequal quantities, the difference is unequal in the opposite order. Do you agree with Dana? Explain why or why not. 2. Ella said that if unequal quantities are subtracted from equal quantities, then the differences are unequal in the opposite order. Do you agree with Ella? Explain why or why not. Developing Skills In 3–10, in each case use an inequality postulate to prove the conclusion. 3. If 10 7, then 18 15. 5. If x 3 12, then x 9. 7. If 8 6 and 5 3, then 13 9. 9. If y 8, then y 1 7. 4. If 4 14, then 15 25. 6. If y 5 5, then y 10. 8. If 7 12, then 5 10. 10. If a b, then 180 a 90 b. Applying Skills 11. Given: AB AD, BC DE 12. Given: AE BD, AF BF Prove: AC AE A D B C E 13. Given: mDAC mDBC and AE BE Prove: a. mEAB mEBA b. mDAB mCBA Prove: FE FD 14. In August, Blake weighed more than Caleb. In the next two months, Blake and Caleb had each gained the same number of pounds. Does Blake still weigh more than Caleb? Justify your answer. 15. In December, Blake weighed more than Andre. In the next two months, Blake lost more than Andre lost. Does Blake still weigh more than Andre? Justify your answer. 14541C07.pgs 1/25/08 3:50 PM Page 270 270 Geometric I |
nequalities 7-3 INEQUALITY POSTULATES INVOLVING MULTIPLICATION AND DIVISION Since there are equality postulates for multiplication and division similar to those of addition and subtraction, we would expect that there are inequality postulates for multiplication and division similar to those of addition and subtraction. Consider these examples that use both positive and negative numbers. If 9 3, then 9(4) 3(4) or 36 12. If 1 5, then 1(3) 5(3) or 3 15. If 9 3, then 9(4) 3(4) or 36 12. If 1 5, then 1(3) 5(3) or 3 15. If 9 3, then 9(4) 3(4) or 36 12. If 1 5, then 1(3) 5(3) or 1 15. If 9 3, then 9(4) 3(4) or 36 12. If 1 5, then 1(3) 5(3) or 3 15. Notice that in the top four examples, we are multiplying by positive numbers and the order of the inequality does not change. In the bottom four examples, we are multiplying by negative numbers and the order of the inequality does change. These examples suggest the following postulates of inequality: Postulate 7.8 If unequal quantities are multiplied by positive equal quantities, then the products are unequal in the same order. Postulate 7.9 If unequal quantities are multiplied by negative equal quantities, then the products are unequal in the opposite order. Since we know that division by a 0 is the same as multiplication by and that a and are always either both positive or both negative, we can write similar postulates for division of inequalities. 1 a 1 a Postulate 7.10 If unequal quantities are divided by positive equal quantities, then the quotients are unequal in the same order. 14365C07.pgs 7/10/07 8:45 AM Page 271 Inequality Postulates Involving Multiplication and Division 271 Postulate 7.11 If unequal quantities are divided by negative equal quantities, then the quotients are unequal in the opposite order. Care must be taken when using inequality postulates involving multiplication and division because multiplying or dividing by a negative number will reverse the order of the inequality. EXAMPLE 1 Given: BA 3BD, BC 3BE, and BE BD Prove: BC BA B D E A C Proof Statements Reasons 1. BE BD 2. 3BE 3BD 3. BC 3BE, BA 3BD 4. BC BA 1. Given. 2. If unequal quantities are multi- plied by positive equal quantities, then the products are unequal in the same order. 3. Given. 4. Substitution postulate for inequalities. EXAMPLE 2 Given: mABC mDEF, ABC, h EH h BG bisects DEF. bisects C G Prove: mABG mDEH B A E F H D Proof An angle bisector separates the angle into two congruent parts. Therefore, the measure of each part is one-half the measure of the angle that was bisected, so 1 mABG 2m/DEF and mDEH 1 2m/ABC . Since we are given that mABC mDEF, 1 2m/DEF because if unequal quantities are multiplied by positive equal quantities, the products are unequal in the same order. Therefore, by the substitution postulate for inequality, mABG mDEH. 1 2m/ABC 14365C07.pgs 7/10/07 8:45 AM Page 272 272 Geometric Inequalities Exercises Writing About Mathematics 1. Since 1 2, is it always true that a 2a? Explain why or why not. 2. Is it always true that if a b and c d, then ac bd? Justify your answer. Developing Skills In 3–8, in each case state an inequality postulate to prove the conclusion. 3. If 8 7, then 24 21. 5. If 8 6, then 4 3. 7. If 4, then x 8. x 2 4. If 30 35, then 6 7. 6. If 3x 15, then x 5. 8. If 3, then y 18. y 6 In 9–17: If a, b, and c are positive real numbers such that a b and b c, tell whether each relationship is always true, sometimes true, or never true. If the statement is always true, state the postulate illustrated. If the statement is sometimes true, give one example for which it is true and one for which it is false. If the statement is never true, give one example for which it is false. 9. ac bc 12. a c b c 15. c . b a c Applying Skills 10. a c b c 13. a b b c 16. ac bc 11. c a c b c a . c 14. b 17. a c 18. Given: BD BE, D is the midpoint BA of Prove: BA BC , E is the midpoint of . BC B D A E C 20. Given: AB AD, AE AF 1 2AD Prove: AE AF 1 2AB , D F C A E B 19. Given: mDBA mCAB, mCBA 2mDBA, mDAB 2mCAB Prove: mCBA mDAB D C E A B 21. Given: mCAB mCBA, bisects CAB, BE AD bisects CBA. Prove: mDAB mEBA C D B E A 14365C07.pgs 7/10/07 8:45 AM Page 273 An Inequality Involving the Lengths of the Sides of a Triangle 273 7-4 AN INEQUALITY INVOLVING THE LENGTHS OF THE SIDES OF A TRIANGLE The two quantities to be compared are often the lengths of line segments or the distances between two points. The following postulate was stated in Chapter 4. The shortest distance between two points is the length of the line segment joining these two points. The vertices of a triangle are three is noncollinear points. The length of AB, the shortest distance from A to B. Therefore, AB AC CB. Similarly, BC BA AC and AC AB BC. We have just proved the following theorem, called the triangle inequality theorem: AB C A B Theorem 7.1 The length of one side of a triangle is less than the sum of the lengths of the other two sides. In the triangle shown above, AB AC BC. To show that the lengths of three line segments can be the measures of the sides of a triangle, we must show that the length of any side is less than the sum of the other two lengths of the other two sides. EXAMPLE 1 Which of the following may be the lengths of the sides of a triangle? (1) 4, 6, 10 (2) 8, 8, 16 (3) 6, 8, 16 (4) 10, 12, 14 Solution The length of a side of a triangle must be less than the sum of the lengths of the other two sides. If the lengths of the sides are a b c, then a b means that a b c and b c means that b c a. Therefore, we need only test the longest side. (1) Is 10 4 6? No (2) Is 16 8 8? No (3) Is 16 6 8? No (4) Is 14 10 12? Yes Answer 14365C07.pgs 7/10/07 8:45 AM Page 274 274 Geometric Inequalities EXAMPLE 2 Two sides of a triangle have lengths 3 and 7. Find the range of possible lengths of the third side. Solution (1) Let s length of third side of triangle. (2) Of the lengths 3, 7, and s, the longest side is either 7 or s. (3) If the length of the longest side is s, then s 3 7 or s 10. (4) If the length of the longest side is 7, then 7 s 3 or 4 s. Answer 4 s 10 EXAMPLE 3 Given: Isosceles triangle ABC with AB CB and M . AC the midpoint of B Prove: AM AB A M C Proof Statements Reasons 1. AC AB CB 1. The length of one side of a triangle is less than the sum of the lengths of the other two sides. 2. AB CB 3. AC AB AB or AC 2AB 2. Given. 3. Substitution postulate for . AC 4. M is the midpoint of 5. AM MC 6. AC AM MC 7. AC AM AM 2AM 8. 2AM 2AB inequalities. 4. Given. 5. Definition of a midpoint. 6. Partition postulate. 7. Substitution postulate. 8. Substitution postulate for inequality. 9. AM AB 9. Division postulate for inequality. 14365C07.pgs 7/10/07 8:45 AM Page 275 An Inequality Involving the Lengths of the Sides of a Triangle 275 Exercises Writing About Mathematics 1. If 7, 12, and s are the lengths of three sides of a triangle, and s is not the longest side, what are the possible values of s? 2. a. If a b c are any real numbers, is a b c always true? Justify your answer. b. If a b c are the lengths of the sides of a triangle, is a b c always true? Justify your answer. Developing Skills In 3–10, tell in each case whether the given lengths can be the measures of the sides of a triangle. 3. 3, 4, 5 7. 2, 2, 3 4. 5, 8, 13 8. 1, 1, 2 5. 6, 7, 10 9. 3, 4, 4 6. 3, 9, 15 10. 5, 8, 11 In 11–14, find values for r and t such that the inequality r s t best describes s, the length of the third sides of a triangle for which the lengths of the other two sides are given. 11. 2 and 4 12. 12 and 31 13. 13 2 and 13 2 14. 9.6 and 12.5 15. Explain why x, 2x, and 3x cannot represent the lengths of the sides of a triangle. 16. For what values of a can a, a 2, a 2 represent the lengths of the sides of a triangle? Justify your answer. Applying Skills 17. Given: ABCD is a quadrilateral. Prove: AD AB BC CD D C A B 19. Given: Point P in the interior of XYZ, YPQ Prove: PY PZ XY XZ 18. Given: ABC with D a point on and AD DC. BC Prove: AB BC B D C Q A X P Y Z 14365C07.pgs 7/10/07 8:45 AM Page 276 276 Geometric Inequalities Hands-On Activity One side of a triangle has a length of 6. The lengths of the other two sides are integers that are less than or equal to 6. a. Cut one straw 6 inches long and two sets of straws to integral lengths of 1 inch to 6 inches. Determine which lengths can represent the sides of a triangle. Use geometry software to determine which lengths can represent the sides of a triangle. b. List all sets of three integers that can be the lengths of the sides of the triangle. Or For example, is one set of lengths. {6, 3, 5} c. List all sets of three integers less than or equal to 6 that cannot be the lengths of the sides of the triangle. d. What patterns emerge in the results of parts b and c? 7-5 AN INEQUALITY INVOLVING AN EXTERIOR ANGLE OF A TRIANGLE Exterior Angles of a Polygon At each vertex of a polygon, an angle is formed that is the union of two sides of the polygon. Thus, for polygon ABCD, DAB is an angle of the polygon, often called an interior angle. If, at h vertex A, we draw , AD we form BAE, an exterior angle of the polygon at vertex A. , the opposite ray of h AE C D A E B DEFINITION An exterior angle of a polygon is an angle that forms a linear pair with one of the interior angles of the polygon. 14365C07.pgs 7/10/07 8:45 AM Page 277 An Inequality Involving an Exterior Angle of a Triangle 277 h AF h AB At vertex A, we can also draw , the , to form DAF, another opposite ray of exterior angle of the polygon at vertex A. At each vertex of a polygon, two exterior angles can be drawn. Each of these exterior angles forms a linear pair with the interior angle at A, and the angles in a linear pair are supplementary. The two exterior angles at A are congruent angles because they are vertical angles. Either can be drawn as the exterior angle at A. C D F A E B Exterior Angles of a Triangle An exterior angle of a triangle is formed outside the triangle by extending a side of the triangle. The figure to the left shows ABC whos |
e three interior angles are CAB, ABC, and BCA. By extending each side of ABC, three exterior angles are formed, namely, DAC, EBA, and FCB. For each exterior angle, there is an adjacent interior angle and two remote or nonadjacent interior angles. For ABC, these angles are as follows: FC B E D A Vertex Exterior Angle Adjacent Interior Angle Nonadjacent Interior Angles A B C DAC EBA FCB CAB ABC BCA ABC and BCA CAB and BCA CAB and ABC With these facts in mind, we are now ready to prove another theorem about inequalities in geometry called the exterior angle inequality theorem. Theorem 7.2 The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle. Given ABC with exterior BCD at vertex C; A and B are nonadjacent interior angles with respect to BCD. Prove mBCD mB B E M A C D 14541C07.pgs 1/25/08 3:50 PM Page 278 278 Geometric Inequalities Proof E B M Statements Reasons 1. Let M be the midpoint of BC . 1. Every line segment has one and 2. Draw h AM , extending the ray A C D through M to point E so only one midpoint. 2. Two points determine a line. A line segment can be extended that AM . EM to any length. . EC 3. Draw 4. mBCD mBCE mECD 3. Two points determine a line. 4. A whole is equal to the sum of its 5. BM CM EM AM 6. 7. AMB EMC 8. AMB EMC 9. B MCE parts. 5. Definition of midpoint. 6. Construction (step 2). 7. Vertical angles are congruent. 8. SAS (steps 5, 7, 6). 9. Corresponding parts of congruent triangles are congruent. 10. mBCD mMCE 10. A whole is greater than any of its 11. mBCD mB parts. 11. Substitution postulate for inequalities. These steps prove that the measure of an exterior angle is greater than the measure of one of the nonadjacent interior angles, B. A similar proof can be used to prove that the measure of an exterior angle is greater than the measure of the other nonadjacent interior angle, A. This second proof uses N, the midpoint of , and a point F extending ray h BC AC through C. The details of this proof will be left to the student. (See exercise 14.) , a line segment BN > NG BNG with EXAMPLE 1 The point D is on AB of ABC. a. Name the exterior angle at D of ADC. A D b. Name two nonadjacent interior angles of the exterior angle at D of ADC. C B c. Why is mCDB mDCA? d. Why is AB AD? 14365C07.pgs 7/10/07 8:45 AM Page 279 An Inequality Involving an Exterior Angle of a Triangle 279 Solution a. CDB b. DCA and A c. The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle. d. The whole is greater than any of its parts. EXAMPLE 2 Given: Right triangle ABC, mC 90, BAD is an exterior angle at A. C Prove: BAD is obtuse. A D B Proof Statements Reasons 1. BAD is an exterior angle. 2. mBAD mC 1. Given. 2. Exterior angle inequality 3. mC 90 4. mBAD 90 5. mBAD mBAC 180 6. 180 mBAD 7. 180 mBAD 90 8. BAD is obtuse. theorem. 3. Given. 4. Substitution postulate for inequalities. 5. If two angles form a linear pair, then they are supplementary. 6. The whole is greater than any of its parts. 7. Steps 4 and 6. 8. An obtuse angle is an angle whose degree measure is greater than 90 and less than 180. 14541C07.pgs 1/25/08 3:50 PM Page 280 280 Geometric Inequalities Exercises Writing About Mathematics 1. Evan said that every right triangle has at least one exterior angle that is obtuse. Do you agree with Evan? Justify your answer. 2. Connor said that every right triangle has at least one exterior angle that is a right angle. Do you agree with Connor? Justify your answer. Developing Skills T 3. a. Name the exterior angle at R. b. Name two nonadjacent interior angles of the exterior angle at R. S R P In 4–13, ABC is scalene and side AB . CM is a median to a. Tell whether each given statement is True or False. b. If the statement is true, state the definition, postulate, or A M B C theorem that justifies your answer. 4. AM MB 6. mAMC mABC 8. mCMB mACM 10. BA MB 12. mBCA mMCA Applying Skills 14. Given: ABC with exterior BCD at vertex C; A and B are nonadjacent interior angles with respect to BCD. Prove: mBCD mA 5. mACB mACM 7. AB AM 9. mCMB mCAB 11. mACM mBCM 13. mBMC mAMC 15. Given: ABD DBE ABE and ABE EBC ABC Prove: mABD mABC (Complete the proof of Theorem 7.2). A B A C D B D E C 14365C07.pgs 7/10/07 8:45 AM Page 281 Inequalities Involving Sides and Angles of a Triangle 281 16. Given: Isosceles DEF with DE FE and exterior EFG Prove: mEFG mEFD E 17. Given: Right ABC with mC 90 Prove: A is acute. B D F G 18. Given: SMR with STM extended through M to P Prove: mRMP mSRT R A C 19. Given: Point F not on › ‹ ABCDE and FC FD Prove: mABF mEDF F S T M P A CB D E 7-6 INEQUALITIES INVOLVING SIDES AND ANGLES OF A TRIANGLE A 22° We know that if the lengths of two sides of a triangle are equal, then the measures of the angles opposite these sides are equal. Now we want to compare the measures of two angles opposite sides of unequal length. Let the measures of the sides of ABC be AB 12, BC 5, and CA 9. 12 9 115° C 43° 5 B Theorem 7.3 Write the lengths in order: 12 9 5 Name the sides in order: AB CA BC Name the angles opposite these sides in order: mC mB mA Notice how the vertex of the angle opposite a side of the triangle is always the point that is not an endpoint of that side. If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal and the larger angle lies opposite the longer side. 14365C07.pgs 7/10/07 8:45 AM Page 282 282 Geometric Inequalities To prove this theorem, we will extend the shorter side of a triangle to a length equal to that of the longer side, forming an isosceles triangle. We can then use the isosceles triangle theorem and the exterior angle inequality theorem to compare angle measures. Given ABC with AB BC Prove mACB mBAC A B C D Proof Statements 1. ABC with AB BC. Reasons 1. Given. 2. Extend through C to point D so that BD BA. BC . 3. Draw AD 4. ABD is isosceles. 5. mBAD mBDA 6. For ACD, mBCA mBDA. 7. mBCA mBAD 2. A line segment may be extended to any length. 3. Two points determine a line. 4. Definition of isosceles triangle. 5. Base angles of an isosceles triangle are equal in measure. 6. Exterior angle inequality theorem. 7. Substitution postulate for inequalities. 8. mBAD mBAC 8. A whole is greater than any of its parts. 9. mBCA mBAC 9. Transitive property of inequality. The converse of this theorem is also true, as can be seen in this example: Let the measures of the angles of ABC be mA 40, mB 80, and mC = 60. Write the angle measures in order: 80 60 40 Name the angles in order: mB mC mA Name the sides opposite these angles in order: AC AB BC If the measures of two angles of a triangle are unequal, then the lengths of the sides opposite these angles are unequal and the longer side lies opposite the larger angle. A 3.7 40° 4.2 B 2.8 80° 60° C Theorem 7.4 14365C07.pgs 7/10/07 8:45 AM Page 283 Inequalities Involving Sides and Angles of a Triangle 283 We will write an indirect proof of this theorem. Recall that in an indirect proof, we assume the opposite of what is to be proved and show that the assumption leads to a contradiction. Given DEF with mD mE F Prove FE FD D E Proof By the trichotomy postulate: FE FD or FE FD or FE FD. We assume the negation of the conclusion, that is, we assume FE FD. Therefore, either FE FD or FE FD. If FE FD, then mD mE because base angles of an isosceles triangle are equal in measure. This contradicts the given premise, mD mE. Thus, FE FD is a false assumption. If FE FD, then, by Theorem 7.3, we must conclude that mD mE. This also contradicts the given premise that mD mE. Thus, FE FD is also a false assumption. Since FE FD and FE FD are both false, FE FD must be true and the theorem has been proved. EXAMPLE 1 One side of ABC is extended to D. If mA 45, mB 50, and mBCD 95, which is the longest side of ABC? B Solution The exterior angle and the interior angle at vertex C form a linear pair and are supplementary. Therefore: mBCA 180 mBCD 180 95 85 A C D Since 85 50 45, the longest side of the triangle is BCA. Answer BA , the side opposite EXAMPLE 2 In ADC, CB that CD CA. is drawn to ABD and CA > CB . Prove C Proof Consider CBD. The measure of an exterior angle is greater than the measure of a nonadjacent interior angle, so mCBA mCDA. Since angles of an isosceles triangle have equal measures, so mA mCBA. A quantity may be substituted for its equal in an inequality, so mA mCDA. A , ABC is isosceles. The base CA > CB D B 14365C07.pgs 7/10/07 8:45 AM Page 284 284 Geometric Inequalities C If the measures of two angles of a triangle are unequal, then the lengths of the sides opposite these angles are unequal and the longer side is opposite the larger angle. Therefore, CD AC. A B D Exercises Writing About Mathematics 1. a. Write the contrapositive of the statement “If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal.” b. Is this contrapositive statement true? 2. The Isosceles Triangle Theorem states that if two sides of a triangle are congruent, then the angles opposite these sides are congruent. a. Write the converse of the Isosceles Triangle Theorem. b. How is this converse statement related to the contrapositive statement written in exer- cise 1? Developing Skills 3. If AB 10, BC 9, and CA 11, name the largest angle of ABC. 4. If mD 60, mE 70, and mF 50, name the longest side of DEF. In 5 and 6, name the shortest side of ABC, using the given information. 5. In ABC, mC 90, mB 35, and mA 55. 6. In ABC, mA 74, mB 58, and mC 48. In 7 and 8, name the smallest angle of ABC, using the given information. 7. In ABC, AB 7, BC 9, and AC 5. 8. In ABC, AB 5, BC 12, and AC 13. 9. In RST, an exterior angle at R measures 80 degrees. If mS mT, name the shortest side of the triangle. 10. If ABD is an exterior angle of BCD, mABD 118, mD 60, and mC 58, list the sides of BCD in order starting with the longest. 11. If EFH is an exterior angle of FGH, mEFH 125, mG 65, mH 60, l |
ist the sides of FGH in order starting with the shortest. 12. In RST, S is obtuse and mR mT. List the lengths of the sides of the triangle in order starting with the largest. 14365C07.pgs 8/2/07 5:46 PM Page 285 Applying Skills 13. Given: C is a point that is not on , ABD mABC mCBD. Prove: AC BC Chapter Summary 285 C A B D 14. Let ABC be any right triangle with the right angle at C and hypotenuse . AB a. Prove that A and B are acute angles. b. Prove that the hypotenuse is the longest side of the right triangle. 15. Prove that every obtuse triangle has two acute angles. CHAPTER SUMMARY Definitions to Know • An exterior angle of a polygon is an angle that forms a linear pair with one of the interior angles of the polygon. • Each exterior angle of a triangle has an adjacent interior angle and two remote or nonadjacent interior angles. Postulates 7.1 A whole is greater than any of its parts. If a, b, and c are real numbers such that a b and b c, then a c. 7.2 7.3 A quantity may be substituted for its equal in any statement of inequality. 7.4 Given any two quantities, a and b, one and only one of the following is 7.5 7.6 7.7 7.8 7.9 true: a b, or a b, or a b. If equal quantities are added to unequal quantities, then the sums are unequal in the same order. If unequal quantities are added to unequal quantities in the same order, then the sums are unequal in the same order. If equal quantities are subtracted from unequal quantities, then the differences are unequal in the same order. If unequal quantities are multiplied by positive equal quantities, then the products are unequal in the same order. If unequal quantities are multiplied by negative equal quantities, then the products are unequal in the opposite order. 7.10 If unequal quantities are divided by positive equal quantities, then the quotients are unequal in the same order. 7.11 If unequal quantities are divided by negative equal quantities, then the quotients are unequal in the opposite order. Theorems 7.1 The length of one side of a triangle is less than the sum of the lengths of the other two sides. 7.2 The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle. 14365C07.pgs 7/10/07 8:45 AM Page 286 286 Geometric Inequalities 7.3 If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal and the larger angle lies opposite the longer side. 7.4 If the measures of two angles of a triangle are unequal, then the lengths of the sides opposite these angles are unequal and the longer side lies opposite the larger angle. VOCABULARY 7-1 Transitive property of inequality • Trichotomy postulate 7-4 Triangle inequality theorem 7-5 Exterior angle of a polygon • Adjacent interior angle • Nonadjacent interior angle • Remote interior angle • Exterior angle inequality theorem REVIEW EXERCISES In 1–8, state a definition, postulate, or theorem that justifies each of the following statements about the triangles in the figure. D 1. AC BC 2. If DA DB and DB DC, then DA DC. A 3. mDBC mA 4. If mC mCDB, then DB BC. 5. If DA DB, then DA AC DB AC. 6. DA AC DC 7. If mA mC, then DC DA. 8. mADC mADB B C , AE BD, 9. Given: , BDC AEC and EC DC Prove: mB mA 10. Given: ABC CDA, AD DC Prove: a. mACD mCAD does not bisect A. B D b. AC B C A E C A D 14365C07.pgs 8/2/07 5:47 PM Page 287 Review Exercises 287 11. In isosceles triangle ABC, C, prove that DB DA. CA CB . If D is a point on AC between A and 12. In isosceles triangle RST, RS ST. Prove that SRP, the exterior angle at R, is congruent to STQ, the exterior angle at T. 13. Point B is 4 blocks north and 3 blocks east of A. All streets run north and south or east and west except a street that slants from C to B. Of the three paths from A and B that are marked: a. Which path is shortest? Justify your answer. b. Which path is longest? Justify your answer Exploration The Hinge Theorem states: If two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first triangle is greater than the third side of the second triangle. 1. a. With a partner or in a small group, prove the Hinge Theorem. b. Compare your proof with the proofs of the other groups. Were different diagrams used? Were different approaches used? Were these approaches valid? The converse of the Hinge Theorem states: If the two sides of one triangle are congruent to two sides of another triangle, and the third side of the first triangle is greater than the third side of the second triangle, then the included angle of the first triangle is larger than the included angle of the second triangle. 2. a. With a partner or in a small group, prove the converse of the Hinge Theorem. b. Compare your proof with the proofs of the other groups. Were different diagrams used? Were different approaches used? Were these approaches valid? 14365C07.pgs 7/10/07 8:45 AM Page 288 288 Geometric Inequalities CUMULATIVE REVIEW CHAPTERS 1–7 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The solution set of the equation 2x 3.5 5x 18.2 is (1) 49 (2) 49 (3) 4.9 (4) 4.9 2. Which of the following is an example of the transitive property of inequal- ity? (1) If a b, then b a. (2) If a b, then a c b c. 3. Point M is the midpoint of (1) AM MC ABMC (2) AB MC (3) If a b and c 0, then ac bc. (4) If a b and b c, then a c. . Which of the following is not true? (3) AM BC (4) BM MC 4. The degree measure of the larger of two complementary angles is 30 more than one-half the measure of the smaller. The degree measure of the smaller is (1) 40 (4) 100 (2) 50 (3) 80 5. Which of the following could be the measures of the sides of a triangle? (1) 2, 2, 4 (2) 1, 3, 5 (3) 7, 12, 20 (4) 6, 7, 12 6. Which of the following statements is true for all values of x? (1) x 5 and x 5 (2) x 5 or x 5 7. In ABC and DEF, (3) If x 5, then x 3. (4) If x 3, then x 5. , and A D. In order to prove AB > DE ABC DEF using ASA, we need to prove that (1) B E BC > EF (2) C F AC > DF (3) (4) 8. Under a reflection in the y-axis, the image of (2, 5) is (2) (2, 5) (3) (2, 5) (1) (2, 5) (4) (5, 2) 9. Under an opposite isometry, the property that is changed is (1) distance (2) angle measure (3) collinearity (4) orientation 10. Points P and Q lie on the perpendicular bisector of AB . Which of the fol- is the perpendicular bisector of PQ . AB lowing statements must be true? (1) (2) PA PB and QA QB. (3) PA QA and PB QB. (4) P is the midpoint of AB or Q is the midpoint of .AB 14365C07.pgs 7/10/07 8:45 AM Page 289 Cumulative Review 289 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Each of the following statements is true. If the snow continues to fall, our meeting will be cancelled. Our meeting is not cancelled. Can you conclude that snow does not continue to fall? List the logic principles needed to justify your answer. 12. The vertices of ABC are A(0, 3), B(4, 3), and C(3, 5). Find the coordi- nates of the vertices of ABC, the image of ABC under the composition ryx T4,5. + Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Given: PR Prove: AP BP bisects ARB but PR is not perpendicular to ARB . 14. Given: In quadrilateral ABCD, h AC bisects DAB and h CA bisects DCB. Prove: B D Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The intersection of g PQ and g RS is T. If mPTR x, mQTS y, and mRTQ 2x y, find the measures of PTR, QTS, RTQ, and PTS. 16. In ABC, mA mB and DCB is an exterior angle at C. The mea- sure of BCA 6x 8, and the measure of DCB 4x 12. a. Find mBCA and mDCB. b. List the interior angles of the triangle in order, starting with the smallest. c. List the sides of the triangles in order starting with the smallest. 14365C08.pgs 7/10/07 8:46 AM Page 290 CHAPTER 8 CHAPTER TABLE OF CONTENTS 8-1 The Slope of a Line 8-2 The Equation of a Line 8-3 Midpoint of a Line Segment 8-4 The Slopes of Perpendicular Lines 8-5 Coordinate Proof 8-6 Concurrence of the Altitudes of a Triangle Chapter Summary Vocabulary Review Exercises Cumulative Review 290 SLOPES AND EQUATIONS OF LINES In coordinate geometry, a straight line can be characterized as a line whose slope is a constant. Do curves have slopes and if so, can they be determined? In the late 17th and early 18th centuries, two men independently developed methods to answer these and other questions about curves and the areas that they determine. The slope of a curve at a point is defined to be the slope of the tangent to that curve at that point. Descartes worked on the problem of finding the slope of a tangent to a curve by considering the slope of a tangent to a circle that intersected the curve at a given point and had its center on an axis. Gottfried Leibniz (1646–1716) in Germany and Isaac Newton (1642–1727) in England each developed methods for determining the slope of a tangent to a curve at any point as well as methods for determining the area bounded by a curve or curves. Newton acknowledged the influence of the work of mathematicians and scientists who preceded him in his statement, “If I have seen further, it is by standing on th |
e shoulders of giants.” The work of Leibniz and Newton was the basis of differential and integral calculus. 14365C08.pgs 7/10/07 8:46 AM Page 291 The Slope of a Line 291 8-1 THE SLOPE OF A LINE In the coordinate plane, horizontal and vertical lines are used as reference lines. Slant lines intersect horizontal lines at acute and obtuse angles. The ratio that measures the slant of a line in the coordinate plane is the slope of the line. Finding the Slope of a Line Through two points, one and only one line can be drawn. In the coordinate plane, if the coordinates of two points are given, it is possible to use a ratio to determine the measure of the slant of the line. This ratio is the slope of the line. Through the points A(1, 2) and B(2, 7), g is drawn. Let C(2, 2) be the point at which AB the vertical line through B intersects the horiis the zontal line through A. The slope of ratio of the change in vertical distance, BC, to the change in horizontal distance, AC. Since B and C are on the same vertical line, BC is the difference in the y-coordinates of B and C. Since A and C are on the same horizontal line, AC is the difference in the x-coordinates of A and C. AB y B(2, 7) 1 1 O 1 A(1, 2) x C(2, 2) slope of AB 5 change in vertical distance change in horizontal distance 5 BC AC 7 2 (22) 2 2 (21) 5 5 9 3 5 3 This ratio is the same for any segment of the line . Suppose we change the order of the points (1, 2) and (2, 7) in performing the computation. We then have: g AB slope of BA 5 change in vertical distance change in horizontal distance 5 CB CA (22) 2 7 (21) 2 2 5 5 29 23 5 3 The result of both computations is the same. When we compute the slope of a line that is determined by two points, it does not matter which point is considered as the first point and which the second. 14365C08.pgs 7/10/07 8:46 AM Page 292 292 Slopes and Equations of Lines Also, when we find the slope of a line using two points on the line, it does not matter which two points on the line we use because all segments of a line have the same slope as the line. Procedure To find the slope of a line: 1. Select any two points on the line. 2. Find the vertical change, that is, the change in y-values by subtracting the y-coordinates in any order. 3. Find the horizontal change, that is, the change in x-values, by subtracting the x-coordinates in the same order as the y-coordinates. 4. Write the ratio of the vertical change to the horizontal change. In general, the slope, m, of a line that passes through any two points A( ) and 2 x2 B( , is the ratio of the x1 difference of the y-values of these points to the difference of the corresponding x-values. ), where x1, y1 x2, y2 y B(x2, y2) A(x1, y1) x x2 y1 y y2 x1 g AB m y2 2 y1 x2 2 x1 slope of x The difference in x-values, x2 – x1, can be represented by x, read as y1, can be represented by y, “delta x.” Similarly, the difference in y-values, y2 read as “delta y.” Therefore, we write: O slope of a line m Dy Dx The slope of a line is positive if the line slants upward from left to right, negative if the line slants downward from left to right, or zero if the line is horizontal. If the line is vertical, it has no slope. Positive Slope g AB . Let The points C and D are two points on the coordinates of C be (1, 2) and the coordinates of D be (3, 3). As the values of x increase, g the values of y also increase. The graph of AB slants upward. slope of g AB The slope of is positive. g AB (3, 3) C(1, 2) 1 x 14365C08.pgs 7/10/07 8:46 AM Page 293 Negative Slope The Slope of a Line 293 . Let the Points C and D are two points on coordinates of C be (2, 3) and the coordinates of D be (5, 1). As the values of x increase, the val- g EF ues of y decrease. The graph of ward. g EF slants down- g EF 5 1 2 3 5 2 2 5 22 3 slope of g EF The slope of is negative(2, 3) 3 D(5, 1) F x Zero Slope Points C and D are two points on . Let the coordinates of C be (2, 2) and the coordinates of D be (1, 2). As the values of x increase, the values of y is a horizontal line. remain the same. The graph of g GH g GH slope of g GH 5 22 2 (22) 1 2 (22) 5 0 3 5 0 The slope of The slope of any horizontal line is 0. is 0. g GH No Slope . Let the coordiPoints C and D are two points on nates of C be (2, 2) and the coordinates of D be (2, 1). The values of x remain the same for all points as the values of y increase. The graph of is a vertical line. The slope of g ML is 2 2 2 5 23 22 2 1 0 , which is g ML g ML undefined. g ML A vertical line has no slope. has no slope. y 1 11 G C(2, 2) O 1 x H D(1, 2) y 1 1 1 O 1 M D(2, 1) x C(2, 2) L A fundamental property of a straight line is that its slope is constant. Therefore, any two points on a line may be used to compute the slope of the line. 14365C08.pgs 7/10/07 8:46 AM Page 294 294 Slopes and Equations of Lines EXAMPLE 1 Find the slope of the line that is determined by points (2, 4) and (4, 2). Solution Plot points (2, 4) and (4, 2). Let point (2, 4) be P1(x1, y1) and let point (4, 2) be P2(x2, y2). 4, x2 2, y1 Then, x1 Dy Dx 5 g slope of P1P2 5 2. 4, and y2 y2 2 y1 x2 2 x1 5 2 2 4 4 2 (22) 5 22 6 5 21 3 Answer y P1(2, 4) 2 6 O 1 1 1 1 P2(4, 2) x EXAMPLE 2 Through point (1, 4), draw the line whose slope is 23 2 . Solution How to Proceed (1) Graph point A(1, 4). (2) Note that, since slope Dy 3 Dx 5 2 2 5 23 2 , when y decreases by 3, x increases by 2. Start at point A(1, 4) and move 3 units downward and 2 units to the right to locate point B. (3) Start at B and repeat these movements to locate point C. (4) Draw a line that passes through points A, B, and C. y 3 A(1, 4 Exercises Writing About Mathematics 1. How is the symbol y read and what is its meaning? 2. Brad said that since 0 is the symbol for “nothing,” no slope is the same as zero slope. Do you agree with Brad? Explain why or why not. 14365C08.pgs 7/10/07 8:46 AM Page 295 The Equation of a Line 295 Developing Skills In 3–11, in each case: a. Plot both points and draw the line that they determine. b. Find the slope of this line if the line has a defined slope. c. State whether the line through these points would slant upward, slant downward, be horizontal, or be vertical. 3. (0, 1) and (4, 5) 6. (1, 5) and (3, 9) 9. (1, 2) and (7, 8) 4. (1, 0) and (4, 9) 7. (5, 3) and (1, 1) 10. (4, 2) and (8, 2) 5. (0, 0) and (3, 6) 8. (2, 4) and (2, 2) 11. (1, 3) and (2, 3) In 12–23, in each case, draw a line with the given slope, m, through the given point. 12. (0, 1); m 2 15. (4, 5); 18. (1, 3); m 1 m 5 2 3 21. (1, 0); m 5 4 Applying Skills 13. (1, 3); m 3 16. (3, 2); m 0 m 5 23 19. (2, 3); 2 22. (0, 2); m 2 3 14. (2, 5); m 1 17. (4, 7); m 2 21 20. (1, 5); m 3 23. (2, 0); m 1 2 24. a. Graph the points A(2, 4) and B(8, 4). b. From point A, draw a line that has a slope of 2. c. From point B, draw a line that has a slope of 2. d. Let the intersection of the lines drawn in b and c be C. What are the coordinates of C? of ABC. Prove that ABC is an isosce- e. Draw the altitude from vertex C to base AB les triangle. 25. Points A(3, 2) and B(9, 2) are two vertices of rectangle ABCD whose area is 24 square units. Find the coordinates of C and D. (Two answers are possible.) 26. A path to the top of a hill rises 75 feet vertically in a horizontal distance of 100 feet. Find the slope of the path up the hill. 27. The doorway of a building under construction is 3 feet above the ground. A ramp to reach the doorway is to have a slope of . How far from the base of the building should the ramp begin? 2 5 8-2 THE EQUATION OF A LINE We have learned two facts that we can use to write the equation of a line. Two points determine a line. The slope of a straight line is constant. 14365C08.pgs 7/10/07 8:46 AM Page 296 296 Slopes and Equations of Lines The second statement on the bottom of page 295 can be written as a biconditional: Postulate 8.1 A, B, and C lie on the same line if and only if the slope of slope of .BC AB is equal to the y 1 O 1 A(3, 1) B(6, 5) x Let A(3, 1) and B(6, 5) be two points g . Let P(x, y) be any other point on AB on g AB using the following fact: . We can write the equation of g AB by slope of AB 5 slope of BP 21 2 5 23 2 6 5 26 29 5 2 3 5 3(y 2 5) 5 2(x 2 6) 3y 2 15 5 2x 2 12 3y 5 2x 1 3 y 5 2 3x 1 1 Recall that when the equation is solved for y in terms of x, the coefficient of x is the slope of the line and the constant term is the y-intercept, the y-coordinate of the point where the line intersects the y-axis. The x-intercept is the x-coordinate of the point where the line intersects the x-axis. Procedure To find the equation of a line given two points on the line: 1. Find the slope of the line using the coordinates of the two given points. 2. Let P(x, y) be any point on the line.Write a ratio that expresses the slope of the line in terms of the coordinates of P and the coordinates of one of the given points. 3. Let the slope found in step 2 be equal to the slope found in step 1. 4. Solve the equation written in step 3 for y. 14365C08.pgs 7/10/07 8:46 AM Page 297 The Equation of a Line 297 When we are given the coordinates of one point and the slope of the line, the equation of the line can be determined. For example, if (a, b) is a point on the line whose slope is m, then the equation is: y 2 b x 2 a 5 m This equation is called the point-slope form of the equation of a line. EXAMPLE 1 The slope of a line through the point A(3, 0) is 2. a. Use the point-slope form to write an equation of the line. b. What is the y-intercept of the line? c. What is the x-intercept of the line? Solution a. Let P(x, y) be any point on the line. The slope of AP 2. y 2 0 x 2 3 5 22 y 5 22(x 2 3) y 5 22x 1 6 y P(x, y) b. The y-intercept is the y-coordinate of the point at which the line intersects the y-axis, that is, the value of y when x is 0. When x 0, 1 O 1 A(3, 0) x y 5 22(0) 1 6 5 0 1 6 5 6 The y-intercept is 6. When the equation is solved for y, the y-intercept is the constant term. c. The x-intercept is the x-coordinate of the point at which the line intersects the x-ax |
is, that is, the value of x when y is 0. Since (3, 0) is a given point on the line, the x-intercept is 3. Answers a. y 2x 6 b. 6 c. 3 EXAMPLE 2 a. Show that the three points A(2, 3), B(0, 1), and C(3, 7) lie on a line. b. Write an equation of the line through A, B, and C. 14365C08.pgs 7/10/07 8:46 AM Page 298 298 Slopes and Equations of Lines Solution a. The points A(2, 3), B(0, 1), and C(3, 7) lie on the same line if and only if the slope of AB slope of AB is equal to the slope of 5 23 2 1 22 2 0 5 24 22 5 2 BC . slope of BC 5 1 2 7 0 2 3 5 26 23 5 2 The slopes are equal. Therefore, the three points lie on the same line. b. Use the point-slope form of the equation of a line. Let (x, y) be any other point on the line. You can use any of the points, A, B, or C, with (x, y) and the slope of the line, to write an equation. We will use A(2, 3). y 2 (23) x 2 (22) 5 2 y 1 3 5 2(x 1 2) y 1 3 5 2x 1 4 y 5 2x 1 1 Answers a. Since the slope of b. y 2x 1 AB is equal to the slope of BC , A, B, and C lie on a line. Alternative Solution (1) Write the slope-intercept form of an equation of a line: y mx b (2) Substitute the coordinates of A in that 3 m(2) b equation: (3) Substitute the coordinates of C in that equation: (4) Write the system of two equations from (2) and (3): 7 m(3) b 3 2m b 7 3m b (5) Solve the equation 3 2m b for b in b 2m 3 terms of m: (6) Substitute the value of b found in (5) for b in the second equation and solve for m: (7) Substitute this value of m in either equation to find the value of b: The equation is y 2x 1. 7 3m b 7 3m (2m 3) 7 5m 3 10 5m 2 m b 2m 3 b 2(2) 3 b 1 14365C08.pgs 7/10/07 8:46 AM Page 299 The Equation of a Line 299 We can show that each of the given points lies on the line whose equation is y 2x 1 by showing that each pair of values makes the equation true. (2, 3) y 2x 1 2(2) 1 5? 3 3 3 ✔ (0, 1) y 2x 1 2(0) 1 5? 1 1 1 ✔ (3, 7) y 2x 1 2(3) 1 5? 7 7 7 ✔ Answers a, b: Since the coordinates of each point make the equation y 2x 1 true, the three points lie on a line whose equation is y 2x 1. Exercises Writing About Mathematics 1. Jonah said that A, B, C, and D lie on the same line if the slope of . Do you agree with Jonah? Explain why or why not. CD of AB is equal to the slope 2. Sandi said that the point-slope form cannot be used to find the equation of a line with no slope. a. Do you agree with Sandi? Justify your answer. b. Explain how you can find the equation of a line with no slope. Developing Skills In 3–14, write the equation of each line. 3. Through (1, 2) and (5, 10) 5. Through (2, 2) and (0, 6) 7. Slope 4 through (1, 1) 9. Slope 3 and y-intercept 5 11. Through (1, 5) and (4, 5) 4. Through (0, 1) and (1, 0) 6. Slope 2 and through (2, 4) 8. Slope 1 2 through (5, 4) 10. Slope 1 and x-intercept 4 12. Through (1, 5) and (1, 2) 13. x-intercept 2 and y-intercept 4 15. a. Do the points P(3, 3), Q(5, 4), and R(1, 1) lie on the same line? 14. No slope and x-intercept 2 b. If P, Q, and R lie on the same line, find the equation of the line. If P, Q, and R do not lie on the same line, find the equations of the lines g PQ , g QR , and g .PR 14365C08.pgs 7/10/07 8:46 AM Page 300 300 Slopes and Equations of Lines 16. a. Do the points L(1, 3), M(5, 6), and N(4, 0) lie on the same line? b. If L, M, and N lie on the same line, find the equation of the line. If L, M, and N do not lie on the same line, find the equations of the lines Applying Skills g LM , g MN , and g LN . 17. At a TV repair shop, there is a uniform charge for any TV brought in for repair plus an hourly fee for the work done. For a TV that needed two hours of repair, the cost was $100. For a TV that needed one and a half hours of repair, the cost was $80. a. Write an equation that can be used to find the cost of repair, y, when x hours of work are required. Write the given information as ordered pairs, (2, 100) and (1.5, 80). b. What would be the cost of repairing a TV that requires 3 hours of work? c. What does the coefficient of x in the equation that you wrote in a represent? d. What does the constant term in the equation that you wrote in a represent? 18. An office manager buys printer cartridges from a mail order firm. The bill for each order includes the cost of the cartridges plus a shipping cost that is the same for each order. The bill for 5 cartridges was $98 and a later bill for 3 cartridges, at the same rate, was $62. a. Write an equation that can be used to find y, the amount of the bill for an order, when x cartridges are ordered. Write the given information as ordered pairs, (5, 98) and (3, 62). b. What would be the amount of the bill for 8 cartridges? c. What does the coefficient of x in the equation that you wrote in a represent? d. What does the constant term in the equation that you wrote in a represent? 19. Show that if the equation of the line can be written as x-axis at (a, 0) and the y-axis at (0, b). x a 1 y b 5 1 , then the line intersects the 8-3 MIDPOINT OF A LINE SEGMENT The midpoint of a line segment is the point of that line segment that divides the segment into two congruent segments. In the figure, A(1, 4) and B(7, 4) determine a horizontal segment, , whose midpoint, M, can be found by first finding the distance from A to B. Since AB 7 (1) 8 units, AM 4 units, and MB 4 units. A(1, 4) B(7, 4) AB M y We can find the x-coordinate of M by adding AM to the x-coordinate of A or by subtracting MB from the x-coordinate of B. The x-coordinate of M is 1 4 3 or 7 4 3. Since A, B, and M are all on the same hori- 1 1 O 1 1 x 14365C08.pgs 7/10/07 8:46 AM Page 301 Midpoint of a Line Segment 301 zontal line, they all have the same y-coordinate, 4. The coordinates of the midpoint M are (3, 4). The x-coordinate of M is the average of the x-coordinates of A and B. x-coordinate of M 5 21 1 7 2 y CD 5 6 2 5 3 Similarly, C(3, 3) and D(3, 1) determine a vertical segment, , whose midpoint, N, can be found by first finding the distance from C to D. Since CD 1 (3) 4 units, CN 2 units, and ND 2 units. We can find the y-coordinate of N by adding 2 to the y-coordinate of C or by subtracting 2 from the y-coordinate of D. The ycoordinate of N is 3 2 1 or 1 2 1. Since C, D, and N are all on the same vertical line, they all have the same x-coordinate, 3. The coordinates of the midpoint N are (3, 1). The y-coordinate of N is the average of the y-coordinates of C and D. C(3, 3) D(3, 1) 1 1 O N x 1 1 y-coordinate of N 5 1 1 (23) 2 5 22 2 5 21 These examples suggest the following relationships: If the endpoints of a horizontal segment are (a, c) and (b, c), then the coordinates of the midpoint are: a 1 b 2 If the endpoints of a vertical segment are (d, e) and (d, f ), then the , c A B coordinates of the midpoint are: d, e 1 f 2 A In the figure, P(2, 1) and Q(8, 5) are the endpoints of . A horizontal line PQ through P and a vertical line through Q intersect at R(8, 1). • The coordinates of S, the midpoint of PR , are 2 1 8 2 , 1 (5, 1). • The coordinates of T, the midpoint 8, 5 1 1 2 (8, 3). , are QR of A A B B B y Q(8, 5) T(8, 3) R(8, 1) x 1 O 1 P(2, 1) S(5, 1) 14365C08.pgs 7/10/07 8:46 AM Page 302 302 Slopes and Equations of Lines Now draw a vertical line through S and a horizontal line through T. These lines appear to intersect at a point on that we will call M. This point has PQ the coordinates (5, 3). We need to show that this point is a point on and is the midpoint of PQ The point M is on PQ . PQ if and only is equal to the slope if the slope of . of MQ PM y M(5, 3) 1 O 1 P(2, 1) S(5, 1) Q(8, 5) T(8, 3) R(8, 1) x slope of PM slope of MQ Since these slopes are equal, P, M, and Q lie on a line. The point M is the midpoint of if PM MQ. We can show that PQ PM MQ by showing that they are corresponding parts of congruent triangles. Therefore, • PS 5 2 3 and MT 8 5 3. PS > MT • SM 3 1 2 and TQ 5 3 2. SM > TQ Therefore, . . y • Since vertical lines are perpendicular to horizontal lines, PSM and MTQ are right angles and therefore congruent. 1 O 1 M(5, 3) P(2, 1) S(5, 1) Q(8, 5) T(8, 3) R(8, 1) x • Therefore, PSM MTQ by SAS and PM > MQ because correspond- ing parts of congruent triangles are congruent. We can conclude that the coordinates of the midpoint of a line segment whose endpoints are (2, 1) and (8, 5) are This example suggests the following theorem5, 3). B Theorem 8.1 If the endpoints of a line segment are (x1, y1) and nates of the midpoint of the segment are x1 1 x2 2 , A y Given The endpoints of B(x2, y2). AB are A(x1, y1) and (x2, y2) y1 1 y2 2 B , then the coordi- . Prove The coordinates of the midpoint of are A x1 1 x2 2 y1 1 y2 2 , . B AB O A(x1, y1) B(x2, y2) y1 y2 M( , ) 2 x1 x2 2 x 14365C08.pgs 7/10/07 8:46 AM Page 303 Midpoint of a Line Segment 303 Proof In this proof we will use the following facts from previous chapters that we have shown to be true: • Three points lie on the same line if the slope of the segment joining two of the points is equal to the slope of the segment joining one of these points to the third. • If two points lie on the same horizontal line, they have the same y-coordinate and the length of the segment joining them is the absolute value of the difference of their x-coordinates. • If two points lie on the same vertical line, they have the same x-coordinate and the length of the segment joining them is the absolute value of the difference of their y-coordinates. We will follow a strategy similar to the one used in the previous example. First, we will prove that the point M with coordinates A then we will use congruent triangles to show that tion of a midpoint of a segment, this will prove that M is the midpoint of B . From the defini- AB . is on AB , and , x1 1 x2 2 y1 1 y2 2 AM > MB (1) Show that M x1 1 x2 2 y1 1 y2 2 , A B lies on AB : slope of AM 5 y1 1 y2 2 2 y1 2 2 x1 x1 1 x2 slope of MB 5 y2 2 x2 2 y1 1 y2 2 x1 1 x2 2 5 y1 1 y2 2 2y1 x1 1 x2 2 2x1 y2 2 y1 x2 2 x1 2y2 2 (y1 1 y2) 2x2 2 (x1 1 x2) y2 2 y1 x2 2 x1 Points A, M, and B lie on the same line because the slope of to the slope of AM MB 5 |
5 5 . is equal (2) Let C be the point on the same vertical line as B and the same horizontal line as A. The coordinates of C are (x2, y1). y A(x1, y1) O 2 y 2 1 y 2 M( , ) 1 x 2 x B(x2, y2) y1 y2 E(x2, ) 2 x1 x2 D( , y1) 2 C(x2, y1) x The midpoint of AC is D The midpoint of BC is E . x1 1 x2 , y1 B 2 y1 1 y2 . 2 B A x2, A 14541C08.pgs 1/25/08 3:46 PM Page 304 304 Slopes and Equations of Lines y A(x1, y1) O 2 y 2 1 y M( , ) 1 x x 2 2 B(x2, y2) y1 y2 E(x2, ) 2 x1 x2 D( , y1) 2 C(x2, y1) x AD 5 5 P P 5 x1 1 x2 2 2 x1 P x1 1 x2 2 2x1 2 x2 2 x1 2 P P Therefore, AD = ME and P AD > ME . MD 5 5 5 P P P y1 1 y2 2 y1 2 2y1 2 y1 2 y2 2 y1 2 y2 2 P P P Therefore, MD = BE and MD > BE . ME 5 5 5 BE 5 5 5 P P P P P P x1 1 x2 2 x2 2 2x2 2 x1 2 x2 2 x2 2 x1 2 P y1 1 y2 2 2 y2 P y1 1 y2 2 2y2 2 y1 2 y2 2 P P P P . Therefore, ADM and MEB are right angles and are Vertical lines are perpendicular to horizontal lines. ME ' BE congruent. ADM is the midpoint of MEB by SAS and . AM > MB AB > . Therefore, M x1 1 x2 2 y1 1 y2 , 2 A B AD ' MD and We generally refer to the formula given in this theorem, that is, x1 1 x2 2 y1 1 y2 2 , , as the midpoint formula. B A EXAMPLE 1 Find the coordinates of the midpoint of the segment C(1, 5) and D(4, 1). CD whose endpoints are Solution Let (x1, y1) (1, 5) and (x2, y2) (4, 1). x1 1 x2 2 The coordinates of the midpoint are A y1 1 y2 2 , B 5 5 5 A A A 21 1 4 2 3 2, 4 2 B 3 2, 2 B 5 1 (21) 2 , B Answer 14365C08.pgs 7/10/07 8:46 AM Page 305 Midpoint of a Line Segment 305 EXAMPLE 2 M(1, 2) is the midpoint of coordinates of B. AB and the coordinates of A are (3, 2). Find the Solution Let the coordinates of A (x1, y1) (3, 2) and the coordinates of B (x2, y2). The coordinates of the midpoint are 23 1 x2 2 5 1 23 1 x2 5 2 x2 5 5 x1 1 x2 2 , A (1, 2). y1 1 y2 2 B 2 1 y2 2 5 22 2 1 y2 5 24 y2 5 26 Answer The coordinates of B are (5, 6). EXAMPLE 3 The vertices of ABC are A(1, 1), B(7, 3), and C(2, 6). Write an equation of the line that contains the median from C to . AB Solution A median of a triangle is a line segment that joins any vertex to the midpoint of the opposite side. Let M be the midpoint of AB . (1) Find the coordinates of M. Let (x1, y1) be (1, 1) and (x2, y2) be (7, 3). The coordinates of M are: y C(2, 6) B(7, 3) M(4, 2) x P(x, y) 1 1 O 1 A(1, 1) 1 x1 1 x2 2 y1 1 y2 4, 2) B , 1 1 3 2 B (2) Write the equation of the line through C(2, 6) and M(4, 2). Let P(x, y) be any other point on the line. y 2 6 slope of PC 5 slope of CM 22 y 2 6 5 22(x 2 2) 2 2 4 y 2 6 5 22x 1 4 y 5 22x 1 10 Answer y 2x 10 14365C08.pgs 7/10/07 8:46 AM Page 306 306 Slopes and Equations of Lines Exercises Writing About Mathematics 1. If P(a, c) and Q(b, c) are two points in the coordinate plane, show that the coordinates of the midpoint are a 1 b 2 a 2 , c A . Hint: Show that . If P(a, c) and Q(b, c) are two points in the coordinate plane, show that the coordinates of the midpoint are b 2 b 2 a 2 , c A . Hint: Show that Developing Skills In 3–14, find the midpoint of the each segment with the given endpoints. 3. (1, 7), (5, 1) 6. (0, 2), (4, 6) 9. (1, 0), (0, 8) 12. (7, 2), (1, 9) In 15–20, M is the midpoint of two of the points are given. 15. A(2, 7), M(1, 6) 17. B(4, 7), M(5, 5) 19. A(3, 3), B(1, 10) Applying Skills 4. (2, 5), (8, 7) 7. (5, 1), (5, 1) 10. (3, 8), (5, 8) 13. 1 2, 3 , B 1, 21 2 B A A 5. (0, 8), (10, 0) 8. (6, 6), (2, 5) 11. (3, 5), (1, 1) 14. 1 3, 9 , B A 2 3, 3 B A AB . Find the coordinates of the third point when the coordinates of 16. A(3, 3), M(3, 9) 18. B(4, 2), M 20. A(0, 7), M A B 3 2, 0 A 0, 7 2 B 21. The points A(1, 1) and C(9, 7) are the vertices of rectangle ABCD and B is a point on the same horizontal line as A. a. What are the coordinates of vertices B and D? b. Show that the midpoint of diagonal AC is also the midpoint of diagonal . BD 22. The points P(x1, y1) and R(x2, y2) are vertices of rectangle PQRS and Q is a point on the same horizontal line as P. a. What are the coordinates of vertices Q and S? b. Show that the midpoint of diagonal PR is also the midpoint of diagonal QS . 23. The vertices of ABC are A(1, 4), B(5, 2), and C(5, 6). a. What are the coordinates of M, the midpoint of g b. Write an equation of CM c. What are the coordinates of N, the midpoint of g BN d. Write an equation of AB ? AC ? that contains the median from B. that contains the median from C. 14365C08.pgs 7/10/07 8:46 AM Page 307 e. What are the coordinates of the intersection of g CM and g ? BN The Slopes of Perpendicular Lines 307 f. What are the coordinates of P, the midpoint of g AP g. Write an equation of BC ? that contains the median from A. g CM lie on g BN g AP and ? h. Does the intersection of i. Do the medians of this triangle intersect in one point? 8-4 THE SLOPES OF PERPENDICULAR LINES Let l1 be a line whose equation is y m1x where m1 is not equal to 0. Then O(0, 0) and A(1, m1) are two points on the line. m1 2 0 slope of l1 5 1 2 0 m1 5 1 5 m1 Under a counterclockwise rotation of 90° about the origin, the image of A(1, m1) is A(m1, 1). Since AOA is a g ' OAr right angle, g OA Let l2 be the line through A(m1, 1) and O(0, 0), and let the slope of l2 be m2. Then: g OAr . y 1 l2 A(m1, 1) l1 A(1, m1 m2 5 0 2 1 5 21 0 2 (2m1) 0 1 m1 5 2 1 m1 We have shown that when two lines through the origin are perpendicular, the slope of one is the negative reciprocal of the slope of the other. Is the rule that we found for the slopes of perpendicular lines through the origin true for perpendicular lines that do not intersect at the origin? We will show this by first establishing that translations preserve slope: Theorem 8.2 Under a translation, slope is preserved, that is, if line l has slope m, then under a translation, the image of l also has slope m. 14541C08.pgs 1/25/08 3:46 PM Page 308 308 Slopes and Equations of Lines Proof: Let P(x1, y1) and Q(x2, y2) be two points on line l. Then: slope of l 5 y2 2 y1 x2 2 x1 Under a translation Ta,b, the images of P and Q have coordinates . Therefore, the slope of l, the image Pr(x1 1 a, y1 1 b) of l, is Qr(x2 1 a, y2 1 b) and slope of lr 5 5 5 (y2 1 b) 2 (y1 1 b) (x2 1 a) 2 (x1 1 a) y2 2 y1 1 b 2 b x2 2 x1 1 a 2 a y2 2 y1 x2 2 x1 As a simple application of Theorem 8.2, we can show that the slopes of any two perpendicular lines are negative reciprocals of each other: Theorem 8.3a If two non-vertical lines are perpendicular, then the slope of one is the negative reciprocal of the other. y (x1, y1) l 2 (x1 a, y1 b) O (x2 a, y2 b) (x2, y2) l 1 l2 (a, b) x l1 Proof: Let l1 and l2 be two perpendicular lines that intersect at (a, b). Under the translation (x, y) → (x a, y b), the image of (a, b) is (0, 0). l2r l1r l1r and Theorem 8.2 tells us that if the slope of l1 is m, then the slope of its , is m. Since l1 and l2 are perimage, pendicular, their images, , are also perpendicular because translations preserve angle measure. Using what we established at the beginning of the section, since the slope of is l2r m, the slope of . Slope is preis served under a translation. Therefore, 21 . the slope of l2 is m 21 m l1r The proof of Theorem 8.3a is called a transformational proof since it uses transformations to prove the desired result. Is the converse of Theorem 8.3a also true? To demonstrate that it is, we need to show that if the slope of one line is the negative reciprocal of the slope of the other, then the lines are perpendicular. We will use an indirect proof. 14365C08.pgs 7/10/07 8:46 AM Page 309 Theorem 8.3b If the slopes of two lines are negative reciprocals, then the two lines are perpendicular. The Slopes of Perpendicular Lines 309 Given Two lines, g AB , that intersect at A. The slope of is m and the slope g AC of is 21 m , the negative reciprocal g AC and g AB of m. Prove g AB g ' AC C D B A y O x Proof Statements Reasons g AC 1. is not perpendicular g . AB to 1. Assumption. g AD perpendicular 2. At a point on a given line, one and 2. Construct g AB to at A. 3. Slope of g AB 4. The slope of is m. g AD is 21 m . only one perpendicular can be drawn. 3. Given. 4. If two lines are perpendicular, the slope of one is the negative reciprocal of the slope of the other. 5. The slope of g AC is 21 m . 5. Given. 6. A, C, and D are on the same g AC g AD and are line, that is, the same line. 6. Three points lie on the same line if and only if the slope of the segment joining two of the points is equal to the slope of a segment joining another pair of these points. g AC g ' AB 7. 7. Contradiction (steps 1, 6). 14365C08.pgs 7/10/07 8:46 AM Page 310 310 Slopes and Equations of Lines We can restate Theorems 8.3a and 8.3b as a biconditional. Theorem 8.3 Two non-vertical lines are perpendicular if and only if the slope of one is the negative reciprocal of the slope of the other. EXAMPLE 1 What is the slope of l2, the line perpendicular to l1, if the equation of l1 is x 2y 4? Solution (1) Solve the equation of l1 for y: x 1 2y 5 4 2y 5 2x 1 4 y 5 21 y 5 21 2x 1 2 2x 1 2 slope slope of l2 5 2 (2) Find the slope of l1: (3) Find the slope of l2 , the negative reciprocal of the slope of l1: Answer The slope of l2 is 2. EXAMPLE 2 Show that ABC is a right triangle if its vertices are A(1, 1), B(4, 3), and C(2, 6). Solution The slope of AB is 23 2 5 23 2 1 2 6 1 2 2 5 25 21 5 5 . . The slope of BC is The slope of AC The slope of BC is is the negative reciprocal of the slope of . Therefore, , B is a right angle, and ABC is a right triangle. g AB AB g BC is perpendicular to Exercises Writing About Mathematics 1. Explain why the slope of a line perpendicular to the line whose equation is x 5 cannot be found by using the negative reciprocal of the slope of that line. 2. The slope of a line l is 21 3 1 . Kim said that the slope of a line perpendicular to l is . 1 3 Santos said that the slope of a line perpendicular to l is 3. Who is correct? Explain your answer. 14365C08.pgs 7/10/07 8:46 AM Page 311 The Slopes of Perpendicular Lines 311 Developing Skills In 3–12: a. Find the slope of the given line. b. Find the slo |
pe of the line perpendicular to the given line. 3. y 4x 7 5. x y 8 7. 3x 5 2y 9. through (0, 4) and (2, 0) 11. through (4, 4) and (4, 2) 4. y x 2 6. 2x y 3 8. through (1, 1) and (5, 3) 10. y-intercept 2 and x-intercept 4 12. parallel to the x-axis through (5, 1) In 13–16, find the equation of the line through the given point and perpendicular to the given line. 13. 21 2, 22 ; 2x 7y 15 14. (0, 0); 2x 4y 12 B y 5 21 A 15. (7, 3); 17. Is the line whose equation is y 3x 5 perpendicular to the line whose equation is 16. (2, 2); y 1 3x 2 3 3x y 6? 18. Two perpendicular lines have the same y-intercept. If the equation of one of these lines is y x 1, what is the equation of the other line? 1 2 19. Two perpendicular lines intersect at (2, 1). If x y 3 is the equation of one of these lines, what is the equation of the other line? 20. Write an equation of the line that intersects the y-axis at (0, 1) and is perpendicular to the line whose equation is x + 2y 6. In 21–24, the coordinates of the endpoints of a line segment are given. For each segment, find the equation of the line that is the perpendicular bisector of the segment. 21. A(2, 2) , B(21, 1) 23. A(3, 9), B(3, 9) Applying Skills 22. A 21 2, 3 , B 3 2, 1 A B 24. A(4, 1), B(3, 3) A B 25. The vertices of DEF triangle are D(3, 4), E(1, 2), and F(3, 2). a. Find an equation of the altitude from vertex D of DEF. b. Is the altitude from D also the median from D? Explain your answer. c. Prove that DEF is isosceles. 26. If a four-sided polygon has four right angles, then it is a rectangle. Prove that if the vertices of a polygon are A(3, 2), B(5, 1), C(1, 5), and D(3, 2), then ABCD is a rectangle. 14365C08.pgs 7/10/07 8:46 AM Page 312 312 Slopes and Equations of Lines 27. The vertices of ABC are A(2, 2), B(6, 6) and C(6, 0). a. What is the slope of AB ? b. Write an equation for the perpendicular bisector of AB . c. What is the slope of BC ? d. Write an equation for the perpendicular bisector of BC . e. What is the slope of AC ? f. Write an equation for the perpendicular bisector of g. Verify that the perpendicular bisectors of ABC intersect in one point. 28. The coordinates of the vertices of ABC are A(2, 0), B(4, 0), and C(0, 4). AC . a. Write the equation of each altitude of the triangle. b. Find the coordinates of the point of intersection of these altitudes. 29. The coordinates of LMN are L(2, 5), M(2, 3), and N(7, 3). a. Using the theorems of Section 7-6, prove that L and N are acute angles. b. List the sides of the triangles in order, starting with the shortest. c. List the angles of the triangles in order, starting with the smallest. Hands-On Activity The following activity may be completed using graph paper, pencil, compass, and straightedge, or geometry software. In the exercises of Section 6-5, we saw how a translation in the horizontal direction can be achieved by a composition of two line reflections in vertical lines and a translation in the vertical direction can be achieved by a composition of two line reflections in horizontal lines. In this activity, we will see how any translation is a composition of two line reflections. . STEP 1. Draw any point A on a coordinate plane. STEP 2. Translate the point A to its image A under the given translation, Ta,b. g AAr STEP 3. Draw line STEP 4. Draw any line l1 perpendicular to STEP 5. Reflect the point A in line l1. Let its image be A. STEP 6. Let l2 be the line that is the perpendicular bisector of Result: The given translation, Ta,b + rl2 Ta,b 5 rl1 order, that is, , is the composition of the two line reflections, . (Recall that is performed first.) g AAr ArAs rl1 . . rl1 and rl2 , in that For a–c, using the procedure above write the equations of two lines under which reflections in the two lines are equal to the given translation. Check your answers using the given coordinates. a. T4,4 c. T1,3 b. T3,2 D(0, 0), E(5, 3), F(2, 2) D(1, 2), E(5, 3), F(5, 6) D(6, 6), E(2, 5), F(1, 6) 14365C08.pgs 7/10/07 8:46 AM Page 313 8-5 COORDINATE PROOF Coordinate Proof 313 Many of the proofs that we did in the preceding chapters can also be done using coordinates. In particular, we can use the formula for slope, the slopes of perpendicular lines, and the coordinates of the midpoint of a line segment presented in this chapter to prove theorems about triangles. In later chapters we will use coordinates to prove theorems about polygons, parallel lines, and distances. There are two types of proofs in coordinate geometry: 1. Proofs Involving Special Cases. When the coordinates of the endpoints of a segment or the vertices of a polygon are given as ordered pairs of numbers, we are proving something about a specific segment or polygon. (See Example 1.) 2. Proofs of General Theorems. When the given information is a figure that represents a particular type of polygon, we must state the coordinates of its vertices in general terms using variables. Those coordinates can be any convenient variables. Since it is possible to use a transformation that is an isometry to move a triangle without changing its size and shape, a geometric figure can be placed so that one of its sides is a segment of the x-axis. If two line segments or adjacent sides of a polygon are perpendicular, they can be represented as segments of the x-axis and the y-axis. (See Example 2.) To prove that line segments bisect each other, show that the coordinates of the midpoints of each segment are the same ordered pair, that is, are the same point. To prove that two lines are perpendicular to each other, show that the slope of one line is the negative reciprocal of the slope of the other. The vertices shown in the diagrams below can be used when working with triangles. y (0, b) y (0, b) y (0, b) y (0, b) (a, 0) O (c, 0) x O (a, 0) (c, 0) x (0, 0) (a, 0) x (a, 0) O (a, 0) x The triangle with vertices (a, 0), (0, b), (c, 0) can be any triangle. It is convenient to place one side of the triangle on the x-axis and the vertex opposite that side on the y-axis. The triangle can be acute if a and c have opposite signs or obtuse if a and c have the same sign. A triangle with vertices at (a, 0), (0, 0), (0, b) is a right triangle because it has a right angle at the origin. 14365C08.pgs 7/10/07 8:46 AM Page 314 314 Slopes and Equations of Lines A triangle with vertices at (a, 0), (0, b), (a, 0) is isosceles because the alti- tude and the median are the same line segment. When a general proof involves the midpoint of a segment, it is helpful to express the coordinates of the endpoints of the segment as variables divisible by 2. For example, if we had written the coordinates of the vertices of a right triangle as (d, 0), (0, e) and ( f, 0), we could simply let d 2a, e 2b, and f 2c so that the coordinates would be (2a, 0), (0, 2b) and (2c, 0). The coordinates of midpoints of the sides of this triangle would be simpler using these coordinates. EXAMPLE 1 AB and bisect each other and are perpendicular to each other Prove that if the coordinates of the endpoints of these segments are A(3, 5), B(5, 1), C(2, 3), and D(4, 9). CD Solution This is a proof involving a special case. B 5 1 1 2 23 1 5 2 The midpoint of , A The midpoint of , 23 1 9 2 22 1 4 2 A The slope of B AB The slope of CD is AB 2 5 2, 6 2 B 5 (1, 3) . 5 (1, 3) . A is CD 6 2 5 2, 2 B A 5 2 1 23 2 5 5 4 9 2 (23) 4 2 (22) 5 12 is is 28 5 21 2 . 6 5 2 . y A(3, 5) D(4, 9) (1, 3) 1 O 1 C(2, 3) B(5, 1) x CD CD and and AB AB rocal of the slope of the other. bisect each other because they have a common midpoint, (1, 3). are perpendicular because the slope of one is the negative recip- EXAMPLE 2 Prove that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices. Given: Right triangle ABC whose vertices are A(2a, 0), B(0, 2b), and C(0, 0). Let M be the midpoint of the hypotenuse . AB Prove: AM BM CM y B(0, 2b) M(a, b) C(0, 0) N(a, 0) A(2a, 0) x 14365C08.pgs 7/10/07 8:46 AM Page 315 Coordinate Proof 315 Proof This is a proof of a general theorem. Since it is a right triangle, we can place one vertex at the origin, one side of the triangle on the x-axis, and a second side on the y-axis so that these two sides form the right angle. We will use coordinates that are divisible by 2 to simplify computation of midpoints. (1) The midpoint of a line segment is a point of that line segment that sepa- rates the line segment into two congruent segments. Therefore, AM > BM . (2) The coordinates of M are 2a 1 0 2 , 0 1 2b 2 (a, b). B A (3) From M, draw a vertical segment that intersects AC at N. The x-coordi- nate of N is a because it is on the same vertical line as M. The y-coordinate of N is 0 because it is on the same horizontal line as A and C. The coordinates of N are (a, 0). (4) The midpoint of AC is and AN > NC . 0 1 2a a, 0) . N is the midpoint of AC (5) The vertical segment MN is perpendicular to the horizontal segment AC . Perpendicular lines are two lines that intersect to form right angles. Therefore, ANM and CMN are right angles. All right angles are congruent, so ANM CMN. Also, MN > MN . (6) Then, AMN CMN by SAS (steps 4 and 5). (7) AM > CM because they are corresponding parts of congruent triangles. (8) (step 1) and AM > BM AM > BM > CM or AM BM = CM. The midpoint of the hypotenuse of a right triangle is equidistant from the vertices. (step 7). Therefore, AM > CM Exercises Writing About Mathematics 1. Ryan said that (a, 0), (0, b), (c, 0) can be the vertices of any triangle but if b 5 2b a c , then the triangle is a right triangle. Do you agree with Ryan? Explain why or why not. 2. Ken said that (a, 0), (0, b), (c, 0) can be the vertices of any triangle but if a and c have the same sign, then the triangle is obtuse. Do you agree with Ken? Explain why or why not. Developing Skills 3. The coordinates of the endpoints of AB are A(0, 2) and B(4, 6). The coordinates of the endpoints of segments bisect each other. CD are C(4, 5) and D(8, 1). Using the midpoint formula, show that the line 14365C08.pgs 7/10/07 8:46 AM Page 3 |
16 316 Slopes and Equations of Lines 4. The vertices of polygon ABCD are A(2, 2), B(5, 2), C(9, 1), and D(6, 5). Prove that the diagonals AC and BD are perpendicular and bisect each other using the midpoint formula. 5. The vertices of a triangle are L(0, 1), M(2, 5), and N(6, 3). a. Find the coordinates K, the midpoint of the base, LN . MK is an altitude from M to b. Show that c. Using parts a and b, prove that LMN is isosceles. 6. The vertices of ABC are A(1, 7), B (9, 3), and C(3, 1). LN . a. Prove that ABC is a right triangle. b. Which angle is the right angle? c. Which side is the hypotenuse? d. What are the coordinates of the midpoint of the hypotenuse? e. What is the equation of the median from the vertex of the right angle to the hypotenuse? f. What is the equation of the altitude from the vertex of the right angle to the hypotenuse? g. Is the triangle an isosceles right triangle? Justify your answer using parts e and f. 7. The coordinates of the vertices of ABC are A(4, 0), B(0, 8), and C(12, 0). a. Draw the triangle on graph paper. b. Find the coordinates of the midpoints of each side of the triangle. c. Find the slope of each side of the triangle. d. Write the equation of the perpendicular bisector of each side of the triangle. e. Find the coordinates of the circumcenter of the triangle. 8. A rhombus is a quadrilateral with four congruent sides. The vertices of rhombus ABCD are A(2, 3), B(5, 1), C(10, 1), and D(7, 3). a. Prove that the diagonals AC and BD bisect each other. b. Prove that the diagonals AC and BD are perpendicular to each other. Applying Skills 9. The vertices of rectangle PQRS are P(0, 0), Q(a, 0), R(a, b), and S(0, b). Use congruent tri- angles to prove that the diagonals of a rectangle are congruent, that is, PR > QS . 10. The vertices of square EFGH are E(0, 0), F(a, 0), G(a, a), and H(0, a). Prove that the diago- nals of a square, point formula. EG and FH , are the perpendicular bisectors of each other using the mid- 11. Use congruent triangles to prove that (0, 0), (2a, 0), and (a, b) are the vertices of an isosce- les triangle. (Suggestion: Draw the altitude from (a, b).) 14365C08.pgs 7/10/07 8:46 AM Page 317 Concurrence of the Altitudes of a Triangle 317 12. Use a translation to prove that (a, 0), (0, b), and (a, 0) are the vertices of an isosceles tri- angle. (Hint: A translation will let you use the results of Exercise 11.) 13. The coordinates of the vertices of ABC are A(0, 0), B(2a, 2b), and C(2c, 2d). a. Find the coordinates of E, the midpoint of AB and of F, the midpoint of AC . b. Prove that the slope of EF 14. The endpoints of segment AB is equal to the slope of are (a, 0) and (a, 0). . BC a. Use congruent triangles to show that P(0, b) and Q(0, c) are both equidistant from the endpoints of g PQ b. Show that . AB is the perpendicular bisector of . AB 8-6 CONCURRENCE OF THE ALTITUDES OF A TRIANGLE The postulates of the coordinate plane and the statements that we have proved about the slopes of perpendicular lines make it possible for us to prove that the three altitudes of a triangle are concurrent, that is, they intersect in one point. Theorem 8.4 The altitudes of a triangle are concurrent. Proof: We can place the triangle anywhere in the coordinate plane. We will place lies on the x-axis and B lies on the y-axis. Let A(a, 0), B(0, b), and it so that C(c, 0) be the vertices of ABC. Let be the altitude from B to be the altitude from A to be the altitude from C to , and . AB , BC AE AC AC BO CF y B(0, b) E F y B(0, b) F E y F E B(0, b) A(a, 0) O C(c, 0) x A(a, 0) O C(c, 0) x x A(a, 0) O C(c, 0) x B is acute B is right B is obtuse In the figures, A and C are acute angles. 14365C08.pgs 7/10/07 8:46 AM Page 318 318 Slopes and Equations of Lines y B(0, b) E F A(a, 0) O C(c, 0) x y B(0, b) F E A(a, 0) O C(c, 0) x y F E We will show that altitudes . and BO CF tudes AE and BO intersect in the same point as alti- Intersection of altitudes and AE BO Intersection of altitudes and BO CF 1. The slope of side 0 2 b c 2 0 5 2b c . BC is 2. The slope of altitude is perpendicular to BC AE , which c , is . b 1. The slope of side 0 2 b a 2 0 5 2b a . AB is 2. The slope of altitude is perpendicular to AB CF , which a , is . b 3. The equation of AE is 3. The equation of CF is ac or ac a or y x . b b AC is a vertical line, a segment is a horizontal line, 4. Since BO of the y-axis since B is on the y-axis. AC is a vertical line, a segment is a horizontal line, 4. Since BO of the y-axis since B is on the y-axis. B(0, b) 5. The equation of g BO is x 0. 5. The equation of g BO is x 0. A(a, 0) O C(c, 0) x AE 6. The coordinates of the and intersection of be found by finding the common solution of their c equations: y x and x 0. b can BO ac b CF 6. The coordinates of the and intersection of be found by finding the common solution of their a equations: y x and x 0. b can BO ac b 7. Since one of the equations is x 0, replace x by 0 in the other equation: 7. Since one of the equations is x 0, replace x by 0 in the other equation: bx 2 ac b b(0) 2 ac b y 5 c 5 c 5 2ac b 8. The coordinates of the AE and BO intersection of 0, 2ac b B are A . bx 2 ac b b(0) 2 ac b y 5 a 5 a 5 2ac b 8. The coordinates of the and CF intersection of 0, 2ac are b B A . BO The altitudes of a triangle are concurrent at 0, 2ac b B is the altitude from C to A . Note: If B is a right angle, from A to CB three altitudes is B(0, b). , and BO CB is the altitude from B to AC AB , AB is the altitude . The intersection of these 14365C08.pgs 7/10/07 8:46 AM Page 319 Concurrence of the Altitudes of a Triangle 319 The point where the altitudes of a triangle intersect is called the orthocenter. EXAMPLE 1 The coordinates of the vertices of PQR are P(0, 0), Q(2, 6), and R(4, 0). Find the coordinates of the orthocenter of the triangle. Solution Let PL be the altitude from P to . QR The slope of QR is 6 2 0 22 2 4 5 6 26 5 21 . The slope of PL is 1. The equation of y x. PL is y 2 0 x 2 0 5 1 or QN be the altitude from Q to Let PR The point of intersection, N, is on the line determined by P and R. . PR The slope of zontal line. Therefore, of a vertical line that has no slope. is 0 since QN is a horiPR is a segment The equation of QN is x 2. y Q(2, 6) L N S(2, 2) P(0, 0) M R(4, 0) is the common solution of the The intersection S of the altitudes equations x 2 and y x. Therefore, the coordinates of the intersection S are (2, 2). By Theorem 8.4, point S is the orthocenter of the triangle or the point where the altitudes are concurrent. and QN PL Answer The orthocenter of PQR is S(2, 2). Alternative Solution Use the result of the proof given in this section. The coordinates of the point of intersection of 0, 2ac the altitudes are b B to apply, Q must lie on the y-axis and P and R must lie on the x-axis. Since P and R already lie on the x-axis, we need only to use a translation to move Q in the horizontal direction to the y-axis. Use the translation (x, y) → (x + 2, y): . In order for this result A y Q(2, 6) Q(0, 6) R(6, 0) x P(2, 0) R(4, 0) P(0, 0) P(0, 0) → P(2, 0) Q(2, 6) → Q(0, 6) R(4, 0) → R(6, 0) Therefore, A(a, 0) P(2, 0) or a 2 B(0, b) Q(0, 6) or b 6 C(c, 0) R(6, 0) or c 6. 14365C08.pgs 7/10/07 8:46 AM Page 320 320 Slopes and Equations of Lines The coordinates of S, the point at which the altitudes of PRQ intersect, are 0, 22(6) 6 B The intersection of the altitudes of PQR is S, the preimage of S(0, 2) under the translation (x, y) → (x + 2, y). Therefore, the coordinates of S are (2, 2). 0, 2ac b B (0, 2) 5 A A Answer The orthocenter of PQR is (2, 2). EXAMPLE 2 The coordinates of the vertices of ABC are A(0, 0), B(3, 4) and C(2, 1). Find the coordinates of the orthocenter of the triangle. Solution The slope of AC is 1 2 0 2 2 0 5 1 2 . Let BD be the altitude from B to . AC The slope of altitude is 2. BD g BD The equation of line 4 2 y 3 2 x 5 22 4 2 y 5 22(3 2 x) is 2y 5 26 1 2x 2 4 y 5 22x 1 10 4 2 1 3 2 2 5 3 is BC 1 5 3 . be the altitude from A to . BC The slope of Let AE The slope of altitude The equation of line AE g AE is 21 3 . is B(3, 4) D C(2, 1) A(0, 0) E S(6, 2) y 2 0 x 2 0 5 21 3 y 5 21 3x The orthocenter S is the common solution of the equations y 2x 10 and y 5 21 . 3x 22x 1 10 5 21 3x 212 3x 5 210 25 3x 5 210 x 5 210 x 5 6 21 23 5 B A The x-coordinate is 6; the y-coordinate is 3(6) 5 22 . Answer The coordinates of the orthocenter are (6, 2) 14365C08.pgs 7/10/07 8:46 AM Page 321 Concurrence of the Altitudes of a Triangle 321 Exercises Writing About Mathematics In 1–2, the vertices of ABC are A(a, 0), B(0, b), and C(c, 0), as shown in the diagrams of the proof of Theorem 8.4. Assume that b 0. 1. Esther said that if A is to the left of the origin and C is to the right of the origin, then the point of intersection of the altitudes is above the origin. Do you agree with Esther? Explain why or why not. 2. Simon said that if A is an obtuse angle, and both A and C are to the right of the origin, then the point of intersection of the altitudes is above the origin. Do you agree with Simon? Explain why or why not. Developing Skills 3. The coordinates of DEF are D(9, 0), E(0, 12), and F(16, 0). a. Show that DEF is a right triangle. b. Show that E is the orthocenter of the triangle. In 4–9, find the coordinates of the orthocenter of each triangle with the given vertices. 4. A(2, 0), B(0, 6), C(3, 0) 6. L(7, 2), M(2, 12), N(11, 2) 8. G(5, 2), H(4, 8), I(5, 1) 5. D(12, 0), E(0, 8), F(6, 0) 7. P(3, 4), Q(1, 8), R(3, 4) 9. J(0, 3), K(3, 4), L(2, 1) Applying Skills 10. Two of the vertices of ABC are A(3, 0) and C(6, 0). The altitudes from these vertices intersect at P(0, 3). g AB g CB a. b. is a line through A, perpendicular to CP . Write the equation of is a line through C, perpendicular to AP . Write the equation of g . AB g CB . c. Find B, the intersection of g AB and g CB . d. Write an equation of the line that contains the altitude from B to . AC e. Show that P is a point on that line. 11. Two of the vertices of ABC are A(2, 2) and C(5, 5). |
The altitudes from these vertices intersect at P(1, 1). g AB g CB a. b. is a line through A, perpendicular to CP . Write the equation of is a line through C, perpendicular to AP . Write the equation of g AB g CB . . c. Find B, the intersection of g AB and g .CB 14365C08.pgs 8/2/07 5:48 PM Page 322 322 Slopes and Equations of Lines d. Write an equation of the line that contains the altitude from B to . AC e. Show that P is a point on that line. Hands-On Activity In this activity we will use a compass and a straightedge to construct the orthocenters of various triangles. For each triangle: a. Graph the triangle on paper or using geometry software. b. Using a compass, straightedge, and pencil, or geometry software, construct the orthocenter. (If using the computer, you are only allowed to use the point, line segment, line, and circle creation tools of your geometry software and no other software tools.) (1) A(3, 0), B(0, 2), C(4, 0) (2) D(4, 7), E(0, 5), F(3, 1) (3) G(4, 2), H(6, 0), I(0, 4) CHAPTER SUMMARY Definitions to Know • The slope, m, of a line that passes through any two points A( ) and , is the ratio of the difference of the y-values of ), where B( these points to the difference of the corresponding x-values. 2 x2 x1, y1 x2, y2 x1 slope of g AB m y2 2 y1 x2 2 x1 • Three or more lines are concurrent if they intersect in one point. • The orthocenter of a triangle is the point at which the altitudes of a trian- gle intersect. Postulate 8.1 A, B, and C lie on the same line if and only if the slope of AB is equal to the slope of BC . Theorems 8.1 If the endpoints of a line segment are , then the coordi- (x1,y1) and x1 1 x2 2 , (x2,y2) y1 1 y2 2 A . B nates of the midpoint of the segment are 8.2 Under a translation, slope is preserved, that is, if line l has slope m, then under a translation, the image of l also has slope m. 8.3 Two non-vertical lines are perpendicular if and only if the slope of one is the negative reciprocal of the other. 8.4 The altitudes of a triangle are concurrent. 14365C08.pgs 7/10/07 8:46 AM Page 323 Chapter Summary 323 VOCABULARY 8-1 Slope • x • y 8-2 y-intercept • x-intercept • Point-slope form of an equation • Slope- intercept form of an equation 8-3 Midpoint formula 8-4 Transformational proof 8-6 Orthocenter REVIEW EXERCISES In 1–3, the coordinates of the endpoints of a line segment are given. a. Find the coordinates of each midpoint. b. Find the slope of each segment. c. Write an equation of the line that is determined by each pair of points. 1. (0, 0) and (6, 4) 4. The coordinates of P are (2, 5) and the coordinates of Q are (6, 1). 2. (3, 2) and (7, 4) 3. (1, 1) and (5, 5) a. What is the slope of g PQ b. What is the equation of ? g PQ ? c. What are the coordinates of the midpoint of ? PQ d. What is the equation of the perpendicular bisector of 5. The vertices of RST are R(2, 2), S(1, 4), and T(7, 1). ? PQ a. Show that RST is a right triangle. b. Find the coordinates of the midpoint of . RT c. Write the equation of the line that contains the median from S. d. Show that the median of the triangle from S is also the altitude from S. e. Prove that RST is an isosceles triangle. 6. Two of the vertices of ABC are A(1, 2) and B(9, 6). The slope of AC is 1 and the slope of 1 3 7. The vertices of DEF are D(1, 1), E(5, 5), F(1, 5). is . What are the coordinates of B? BC a. Find the coordinates of the midpoint of each side of the triangle. b. Find the slope of each side of the triangle. c. Find the slope of each altitude of the triangle. d. Write an equation of the perpendicular bisector of each side of the tri- angle. e. Show that the three perpendicular bisectors intersect in a point and find the coordinates of that point. 14365C08.pgs 7/10/07 8:46 AM Page 324 324 Slopes and Equations of Lines 8. The vertices of ABC are A(7, 1), B(5, 3), and C(3, 5). a. Prove that ABC is a right triangle. b. Let M be the midpoint of AB AMN CMN and use this result to show that M is equidistant from the vertices of ABC. and N be the midpoint of AC . Prove that 9. The vertices of DEF are D(2, 3), E(5, 0), and F(2, 3) a. Find the coordinates of M, the midpoint of DF . b. Show that DE > FE . 10. The coordinates of the vertices of quadrilateral ABCD are A(5, 4), B(1, 6), C(1, 2), and D(4, 1). Show that ABCD has a right angle. Exploration The following exploration may be completed using graph paper or using geometry software. We know that the area of a triangle is equal to one-half the product of the lengths of the base and height. We can find the area of a right triangle in the coordinate plane if the base and height are segments of the x-axis and y-axis. The steps that follow will enable us to find the area of any triangle in the coordinate plane. For example, find the area of a triangle if the vertices have the coordinates (2, 5), (5, 9), and (8, 2). STEP 1. Plot the points and draw the triangle. STEP 2. Through the vertex with the smallest x-coordinate, draw a vertical line. STEP 3. Through the vertex with the largest x-coordinate, draw a vertical line. STEP 4. Through the vertex with the smallest y-coordinate, draw a horizontal line. STEP 5. Through the vertex with the largest y-coordinate, draw a horizontal line. STEP 6. The triangle is now enclosed by a rectangle with horizontal and vertical sides. Find the length and width of the rectangle and the area of the rectangle. STEP 7. The rectangle is separated into four triangles: the given triangle and three triangles that have a base and altitude that are a horizontal and a vertical line. Find the area of these three triangles. STEP 8. Find the sum of the three areas found in step 7. Subtract this sum from the area of the rectangle. The difference is the area of the given triangle. Repeat steps 1 through 8 for each of the triangles with the given vertices. a. (2, 0), (3, 7), (6, 1) b. (3, 6), (0, 4), (5, 1) c. (0, 5), (6, 6), (2, 5) Can you use this procedure to find the area of the quadrilateral with vertices at (0, 2), (5, 2), (5, 3), and (2, 5)? 14541C08.pgs 1/25/08 3:46 PM Page 325 CUMULATIVE REVIEW Part I Cumulative Review 325 Chapters 1–8 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. When the coordinates of A are (2, 3) and of B are (2, 7), AB equals (1) 4 (2) 4 (3) 10 (4) 10 2. The slope of the line whose equation is 2x y 4 is (1) 1 2 (2) 2 (3) 2 (4) 4 3. Which of the following is an example of the associative property for multi- plication? (1) 3(2 5) 3(5 2) (2) 3(2 5) (3 2) 5 4. The endpoints of midpoint of (1) (2, 3) AB AB are (3) 3(2 5) 3(2) 5 (4) 3 (2 5) (3 2) 5 are A(0, 6) and B(4, 0). The coordinates of the (2) (2, 3) (3) (2, 3) (4) (2, 3) 5. In isosceles triangle DEF, DE = DF. Which of the following is true? (1) D E (2) F E (3) D F (4) D E F 6. The slope of line l is . The slope of a line perpendicular to l is (1) 2 3 22 3 (3) 3 2 (4) 23 2 2 3 (2) 7. The coordinates of two points are (0, 6) and (3, 0). The equation of the line through these points is (1) y 2x 6 (2) y 2x 6 1 (3) y 3 2x 1 (4) y x 3 2 8. The converse of the statement “If two angles are right angles then they are congruent” is (1) If two angles are congruent then they are right angles. (2) If two angles are not right angles then they are not congruent. (3) Two angles are congruent if and only if they are right angles. (4) Two angles are congruent if they are right angles. 9. The measure of an angle is twice the measure of its supplement. The mea- sure of the smaller angle is (1) 30 (2) 60 (3) 90 (4) 120 10. Under a reflection in the y-axis, the image of (4, 2) is (2) (4, 2) (3) (4, 2) (1) (4, 2) (4) (2, 4) 14365C08.pgs 7/10/07 8:46 AM Page 326 326 Slopes and Equations of Lines Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. a. Draw ABC in the coordinate plane if the coordinates of A are (3, 1), of B are (1, 5), and of C are (5, 2). b. Under a translation, the image of A is A(1, 1). Draw ABC, the image of ABC under this translation, and give the coordinates of B and C. c. If this translation can be written as (x, y) → (x a, y b), what are the values of a and b? 12. The coordinates of the vertices of DEF are D(1, 6), E(3, 3), and F(1, 2). Is DEF a right triangle? Justify your answer. Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. In the diagram, TQ ' RS h TQ bisects RTS and . Prove that RST is isosceles. 14. The following statements are true: • If Evanston is not the capital of Illinois, then Chicago is not the capital. • Springfield is the capital of Illinois or Chicago is the capital of Illinois. T • Evanston is not the capital of Illinois. Use the laws of logic to prove that Springfield is the capital of Illinois. R Q S 14365C08.pgs 7/10/07 8:46 AM Page 327 Cumulative Review 327 Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Find the coordinates of the point of intersection of the lines whose equa- tions are y 3x 1 and x 2y 5. 16. Given: CGF and DGE bisect each other at G. Prove: CD > FE 14365C09.pgs 7/10/07 8:48 AM Page 328 CHAPTER 9 CHAPTER TABLE OF CONTENTS 9-1 Proving Lines Parallel 9-2 Properties of Parallel Lines 9-3 Parallel Lines in the Coordinate Plane 9-4 The Sum of the Measures of the Angles of a Triangle 9-5 Proving Tria |
ngles Congruent by Angle, Angle, Side 9-6 The Converse of the Isosceles Triangle Theorem 9-7 Proving Right Triangles Congruent by Hypotenuse, Leg 9-8 Interior and Exterior Angles of Polygons Chapter Summary Vocabulary Review Exercises Cumulative Review 328 PARALLEL LINES “If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles.” This statement, Euclid’s fifth postulate, is called Euclid’s parallel postulate. Throughout history this postulate has been questioned by mathematicians because many felt it was too complex to be a postulate. Throughout the history of mathematics, attempts were made to prove this postulate or state a related postulate that would make it possible to prove Euclid’s parallel postulate. Other postulates have been proposed that appear to be simpler and which could provide the basis for a proof of the parallel postulate. The form of the parallel postulate most commonly used in the study of elementary geometry today was proposed by John Playfair (1748–1819). Playfair’s postulate states: Through a point not on a given line there can be drawn one and only one line parallel to the given line. 14365C09.pgs 7/10/07 8:48 AM Page 329 9-1 PROVING LINES PARALLEL Proving Lines Parallel 329 You have already studied many situations involving intersecting lines that lie in the same plane. When all the points or lines in a set lie in a plane, we say that these points or these lines are coplanar. Let us now consider situations involving coplanar lines that do not intersect in one point. DEFINITION Parallel lines are coplanar lines that have no points in common, or have all points in common and, therefore, coincide. The word “lines” in the definition means straight lines of unlimited extent. We say that segments and rays are parallel if the lines that contain them are parallel. We indicate that g CD . The parallel lines is parallel to g CD g AB and g AB g AB g CD by writing extended indefi- nitely never intersect and have no points in common. The parallel lines g AB and g CD may have all points in common, that is, be two different names for the same line. A line is parallel to itself. Thus, g AB g CD . In Chapter 4, we stated the following postulate: g AB g AB , g CD g CD and C A D B A B C D Two distinct lines cannot intersect in more than one point. This postulate, together with the definition of parallel lines, requires that g AB and g : CD one of three possibilities exist for any two coplanar lines, 1. 2. 3. g AB g AB g AB g AB g AB g AB and and and and and and g CD g CD g CD g CD g CD g CD have no points in common. are parallel. have only one point in common. intersect. have all points in common. are the same line. These three possibilities can also be stated in the following postulate: Postulate 9.1 Two distinct coplanar lines are either parallel or intersecting. 14365C09.pgs 7/10/07 8:48 AM Page 330 330 Parallel Lines EXAMPLE 1 If line l is not parallel to line p, what statements can you make about these two lines? Solution Since l is not parallel to p, l and p cannot be the same line, and they have exactly one point in common. Answer Parallel Lines and Transversals g g AB intersects CD C A B D When two lines intersect, four angles are formed that have the same vertex and no common interior points. In this set of four angles, there are two pair of congruent vertical angles and four pair of supplementary adjacent angles. When two lines are intersected by a third line, two such sets of four angles are formed. DEFINITION A transversal is a line that intersects two other coplanar lines in two different points Two lines, l and m, are cut by a transversal, t. Two sets of angles are formed, each containing four angles. Each of these angles has one ray that is a subset of l or of m and one ray that is a subset of t. In earlier courses, we learned names to identify these sets of angles. • The angles that have a part of a ray between l and m are interior angles. Angles 3, 4, 5, 6 are interior angles. • The angles that do not have a part of a ray between l and m are exterior angles. Angles 1, 2, 7, 8 are exterior angles. • Alternate interior angles are on opposite sides of the transversal and do not have a common vertex. Angles 3 and 6 are alternate interior angles, and angles 4 and 5 are alternate interior angles. • Alternate exterior angles are on opposite sides of the transversal and do not have a common vertex. Angles 1 and 8 are alternate exterior angles, and angles 2 and 7 are alternate exterior angles. • Interior angles on the same side of the transversal do not have a common vertex. Angles 3 and 5 are interior angles on the same side of the transversal, and angles 4 and 6 are interior angles on the same side of the transversal. • Corresponding angles are one exterior and one interior angle that are on the same side of the transversal and do not have a common vertex. Angles 1 and 5, angles 2 and 6, angles 3 and 7, and angles 4 and 8 are pairs of corresponding angles. 14365C09.pgs 7/10/07 8:48 AM Page 331 In the diagram shown on page 330, the two lines cut by the transversal are not parallel lines. However, when two lines are parallel, many statements may be postulated and proved about these angles. Proving Lines Parallel 331 Theorem 9.1a If two coplanar lines are cut by a transversal so that the alternate interior angles formed are congruent, then the two lines are parallel. Given and g CD g AB and F, respectively; 1 2. are cut by transversal Prove g AB g CD g EF at points E A C E 1 2 F B D Proof To prove this theorem, we will use an indirect proof. g AB g AB 1. 2. Statements is not parallel to g . CD and g CD transversal are cut by g EF at points . and F, respectively. g g AB CD point P, forming EFP. and intersect at some 4. m1 m2 5. But 1 2. 6. m1 m2 g AB g CD 7. Reasons 1. Assumption. 2. Given. 3. Two distinct coplanar lines are either parallel or intersecting. 4. The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle. 5. Given. 6. Congruent angles are equal in measure. 7. Contradiction in steps 4 and 6. Now that we have proved Theorem 9.1, we can use it in other theorems that also prove that two lines are parallel. Theorem 9.2a If two coplanar lines are cut by a transversal so that the corresponding angles are congruent, then the two lines are parallel. 14365C09.pgs 7/10/07 8:48 AM Page 332 332 Parallel Lines Given g EF intersects g AB and g CD ; 1 5. Prove g AB g CD Proof Statements g AB intersects 1. g EF 1 5 2. 1 3 3. 3 5 g CD g AB 4 Reasons and g ; CD 1. Given. 2. Vertical angles are congruent. 3. Transitive property of congruence. 4. If two coplanar lines are cut by a transversal so that the alternate interior angles formed are congruent, then the two lines are parallel. Theorem 9.3a If two coplanar lines are cut by a transversal so that the interior angles on the same side of the transversal are supplementary, then the lines are parallel. Given intersects g EF of 4. Prove g AB g CD g AB and g CD , and 5 is the supplement E A C 4 3 5 F B D Proof Angle 4 and angle 3 are supplementary since they form a linear pair. If two angles are supplements of the same angle, then they are congruent. Therefore, 3 5. Angles 3 and 5 are a pair of congruent alternate interior angles. If two coplanar lines are cut by a transversal so that the alternate interior angles formed are congruent, then the lines are parallel. Therefore, g AB g CD . Theorem 9.4 If two coplanar lines are each perpendicular to the same line, then they are parallel. Given g AB ⊥ g EF and g CD ⊥ g EF . Prove g AB g CD Strategy Show that a pair of alternate interior angles are congruent. A E B 1 C 2 D F 14365C09.pgs 7/10/07 8:48 AM Page 333 Proving Lines Parallel 333 The proof of Theorem 9.4 is left to the student. (See exercise 10.) Methods of Proving Lines Parallel To prove that two coplanar lines that are cut by a transversal are parallel, prove that any one of the following statements is true: 1. A pair of alternate interior angles are congruent. 2. A pair of corresponding angles are congruent. 3. A pair of interior angles on the same side of the transversal are supplementary. 4. Both lines are perpendicular to the same line. EXAMPLE 2 If mA 100 3x and mB 80 3x, . explain why BC AD D C Solution mA mB 100 3x 80 3x 100 80 3x 3x 180 100 3x 80 3x A B Thus, A and B are supplementary. Since AB the segments are parallel, namely, are cut by transversal to form supplementary interior angles on the same side of the transversal, BC and AD AD BC . EXAMPLE 3 If BD bisects ABC, and BC CD , prove CD h BA . C Proof (1) Since BC CD , CBD D because the base B D angles of an isosceles triangle are congruent. (2) Since BD bisects ABC, CBD DBA because the bisector of an angle divides the angle into two congruent angles. A (3) Therefore, by the transitive property of congruence, DBA D. (4) Then, DBA and D are congruent alternate interior angles when h BA are intersected by transversal and two coplanar lines are cut by a transversal so that the alternate interior angles formed are congruent, then the two lines are parallel. . Therefore, BD CD because if CD h BA 14365C09.pgs 7/10/07 8:48 AM Page 334 334 Parallel Lines Exercises Writing About Mathematics 1. Two lines are cut by a transversal. If 1 and 2 are vertical angles and 1 and 3 are alternate interior angles, what type of angles do 2 and 3 form? 2. Is it true that if two lines that are not parallel are cut by a transversal, then the alternate interior angles are not congruent? Justify your answer. Developing Skills In 3–8, the figure shows eight angles formed when g AB and g CD are cut by transversal . For each of the following, state the theorem or theorems g EF g AB g . CD || that prove E 12 . m3 70 and m5 70 5. m3 60 and m6 120 7. m2 160 and m8 160 4. m2 140 and m6 1 |
40 6. m2 150 and m5 30 8. m4 110 and m7 70 Applying Skills 9. Write an indirect proof of Theorem 9.2a, “If two coplanar lines are cut by a transversal so that the corresponding angles are congruent, then the two lines are parallel.” 10. Prove Theorem 9.4, “If two coplanar lines are each perpendicular to the same line, then they are parallel.” In 11 and 12, ABCD is a quadrilateral. 11. If mA 3x and mB 180 3x. Show that AD . BC 12. If DC ⊥ BC and mADC 90, prove AD . BC 13. If bisect each other at AB and CD point E, prove: a. CEA DEB b. ECA EDB c. CA DB C B E A D D A C B 14. Prove that if two coplanar lines are cut by a transversal, forming a pair of alternate exterior angles that are congruent, then the two lines are parallel. 14365C09.pgs 7/10/07 8:48 AM Page 335 Properties of Parallel Lines 335 9-2 PROPERTIES OF PARALLEL LINES In the study of logic, we learned that a conditional and its converse do not always have the same truth value. Once a conditional statement has been proved to be true, it may be possible to prove that its converse is also true. In this section, we will prove converse statements of some of the theorems proved in the previous section. The proof of these converse statements requires the following postulate and theorem: Postulate 9.2 Through a given point not on a given line, there exists one and only one line parallel to the given line. Theorem 9.5 If, in a plane, a line intersects one of two parallel lines, it intersects the other. Given g AB g CD and g EF intersects g AB at H. Prove g EF intersects g . CD Proof Assume g g EF CD g EF does not intersect g CD . Then . Therefore, through H, a point g AB , two lines, g EF and are g CD not on E A H F C B D g CD each parallel to postulate that states that through a given point not on a given line, one and only one line can be drawn parallel to a given line. Since our assumption leads . This contradicts the to a contradiction, the assumption must be false and its negation, sects g CD must be true. g EF inter- Now we are ready to prove the converse of Theorem 9.1a. Theorem 9.1b If two parallel lines are cut by a transversal, then the alternate interior angles formed are congruent. Given g AB g CD g CD at F. , transversal g EF intersects g AB at E and Prove 14365C09.pgs 7/10/07 8:48 AM Page 336 336 Parallel Lines Proof We can use an indirect proof. Assume 1 is so that not congruent to 2. Construct HEF 2. Since HEF and 2 are congruent alternate interior angles, h EH g HE g CD . But g line through E, and we are given AB contradicts the postulate that states that through a given point not on a given line, there exists one and only one line parallel to the given line. Thus, the assumption is false and 1 2. . This is a D C F g AB g CD 2 H A E B Note that Theorem 9.1b is the converse of Theorem 9.1a. We may state the two theorems in biconditional form: Theorem 9.1 Two coplanar lines cut by a transversal are parallel if and only if the alternate interior angles formed are congruent. Each of the next two theorems is also a converse of a theorem stated in Section 9-1. Theorem 9.2b If two parallel lines are cut by a transversal, then the corresponding angles are congruent. (Converse of Theorem 9.2a) Given g AB g CD and transversal g EF Prove Proof Statements g 1. CD 2. 3 5 g AB and transversal 3. 1 3 4. 1 5 Reasons g EF 1. Given. 2. If two parallel lines are cut by a transversal, then the alternate interior angles formed are congruent. 3. Vertical angles are congruent. 4. Transitive property of congruence. 14365C09.pgs 7/10/07 8:48 AM Page 337 Theorem 9.3b If two parallel lines are cut by a transversal, then two interior angles on the same side of the transversal are supplementary. (Converse of Theorem 9.3a) Properties of Parallel Lines 337 Given g AB g CD and transversal g EF Prove 4 is the supplement of 5. Strategy Show that 3 5 and that 4 is the supplement of 3. If two angles are congruent, then their supplements are congruent. Therefore, 4 is also the supplement of 5. E 34 B D A C 5 F The proof of this theorem is left to the student. (See exercise 18.) Since Theorems 9.2b and 9.3b are converses of Theorems 9.2a and 9.3a, we may state the theorems in biconditional form: Theorem 9.2 Two coplanar lines cut by a transversal are parallel if and only if corresponding angles are congruent. Theorem 9.3 Two coplanar lines cut by a transversal are parallel if and only if interior angles on the same side of the transversal are supplementary. EXAMPLE 1 Transversal g EF intersects g AB g CD and g AB g at G and H, CD , mBGH 3x 20, and respectively. If mGHC 2x 10: a. Find the value of x. c. Find mGHD. b. Find mGHC. A C E G B D H F Solution a. Since g AB g CD and these lines are cut by transversal g EF , the alternate inte- rior angles are congruent: mBGH mGHC 3x 20 2x 10 3x 2x 10 20 x 30 b. mGHC 2x 10 2(30) 10 70 14365C09.pgs 7/10/07 8:48 AM Page 338 338 Parallel Lines c. Since GHC and GHD form a linear pair and are supplementary, mGHD 180 mGHC 180 70 110 Answers a. x 70 b. mGHC 70 c. mGHD 110 Using Theorem 9.1, we may also prove the following theorems: Theorem 9.6 If a transversal is perpendicular to one of two parallel lines, it is perpendicular to the other. Given g AB g CD , g EF ⊥ g AB Prove g EF ⊥ g CD A C E B F D Strategy Show that alternate interior angles are right angles. The proof of this theorem is left to the student. (See exercise 19.) Theorem 9.7 If two of three lines in the same plane are each parallel to the third line, then they are parallel to each other. Given g AB g LM and g CD g LM Prove g AB g CD Proof Draw transversal at H. Since g AB intersecting g LM A , this transversal also intersects Similarly, since sal also intersects . Call this point F. g LM g CD at a point G. , this transver- g EJ g LM g AB g CD AB Since g LM , alternate interior angles formed are congruent. Therefore, AFG GHM. Similarly, since , CGH GHM. By the transitive property of congruence, AFG CGH. Angles AFG and CGH are conare intersected by transversal gruent corresponding angles when g LM g CD g CD g AB and 14365C09.pgs 7/10/07 8:48 AM Page 339 Properties of Parallel Lines 339 g EJ . Therefore, g AB g CD because if two coplanar lines are cut by a transversal so that the corresponding angles formed are congruent, then the two lines are parallel. SUMMARY OF PROPERTIES OF PARALLEL LINES If two lines are parallel: 1. A transversal forms congruent alternate interior angles. 2. A transversal forms congruent corresponding angles. 3. A transversal forms supplementary interior angles on the same side of the transversal. 4. A transversal perpendicular to one line is also perpendicular to the other. 5. A third line in the same plane that is parallel to one of the lines is parallel to the other. EXAMPLE 2 Given: Quadrilateral ABCD, BC DA , and BC DA B Prove: AB CD Proof Use congruent triangles to prove congruent alternate A C D interior angles. Statements 1. BC DA C BC 2. DA 3. BCA DAC AC 4. AC 5. BAC DCA 6. BAC DCA C 7. AB CD A A B B D D Reasons 1. Given. 2. Given. 3. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 4. Reflexive property of congruence. 5. SAS. 6. Corresponding parts of congruent triangles are congruent. 7. If two lines cut by a transversal form congruent alternate interior angles, the lines are parallel. 14365C09.pgs 7/10/07 8:48 AM Page 340 340 Parallel Lines Note: In the diagram for Example 2, you may have noticed that two parallel lines, g BC and g DA , each contained a single arrowhead in the same direction. Such pairs of arrowheads are used on diagrams to indicate that two lines are parallel. Exercises Writing About Mathematics 1. a. Is the inverse of Theorem 9.1a always true? Explain why or why not. b. Is the inverse of Theorem 9.6 always true? Explain why or why not. 2. Two parallel lines are cut by a transversal forming alternate interior angles that are supplementary. What conclusion can you draw about the measures of the angles formed by the parallel lines and the transversal. Justify your answer. Developing Skills g AB In 3–12, g EF are cut by transversal g CD 3. m5 when m3 80. 5. m4 when m5 60. 7. m8 when m3 65. 9. m3 when m3 3x and m5 x 28. 10. m5 when m3 x and m4 x 20. 11. m7 when m1 x 40 and m2 5x 10. 12. m5 when m2 7x 20 and m8 x 100. as shown in the diagram. Find: 4. m2 when m6 150. 6. m7 when m1 75. 8. m5 when m2 130. 12 13. Two parallel lines are cut by a transversal. For each pair of interior angles on the same side of the transversal, the measure of one angle exceeds the measure of twice the other by 48 degrees. Find the measures of one pair of interior angles. 14. Two parallel lines are cut by a transversal. The measure of one of the angles of a pair of corresponding angles can be represented by 42 less than three times the other. Find the measures of the angles of this pair of corresponding angles. 15. In the diagram, and a. If mFGD 110 and mFEC 130, find the measures intersect and at F. g AFB g CD g EF g GF g AB of each of the angles numbered 1 through 9. b. What is the measure of an exterior angle of EFG at F 14365C09.pgs 7/10/07 8:48 AM Page 341 Properties of Parallel Lines 341 c. Is the measure of an exterior angle at F greater than the measure of either of the non- adjacent interior angles? d. What is the sum of the measures of the nonadjacent interior angles of an exterior angle at F? e. What is the sum of the measures of the nonadjacent interior angles of the exterior angle, FGD? f. What is the sum of the measures of the nonadjacent interior angles of the exterior angle, FEC? g. What is the sum of the measures of the angles of EFG? 16. Two pairs of parallel lines are drawn; g ABE g DC and g AD g BC . If mCBE 75, find the measure of each angle of quadrilateral ABCD. Applying Skills D C A B E 17. Prove Theorem 9.3b, “If two parallel lines are cut by a transversal, then two interior angles on the same side of the transversal are supplementary.” 18. Prove Theorem 9.6, “If a transversal is perpendicul |
ar to one of two parallel lines, it is per- pendicular to the other.” 19. Prove that if two parallel lines are cut by a transversal, the alternate exterior angles are con- gruent. 20. Given: ABC, h CE Prove: A B. bisects exterior BCD, and h CE AB . 21. Given: CAB DCA and DCA ECB Prove: a. g AB DCE . b. CAB is the supplement of CBG. 22. The opposite sides of quadrilateral PQRS are parallel, that is, is a right angle, prove that Q, R, and S are right angles PQ RS and QR SP . If P 23. The opposite sides of quadrilateral KLMN are parallel, that is, LM NK . If K is an acute angle, prove that M is an acute angle and that L and N are obtuse angles. KL MN and 14365C09.pgs 7/10/07 8:48 AM Page 342 342 Parallel Lines 9-3 PARALLEL LINES IN THE COORDINATE PLANE In Chapter 6 we stated postulates about horizontal and vertical lines in the coordinate plane. One of these postulates states that each vertical line is perpendicular to each horizontal line. We can use this postulate to prove the following theorem: Theorem 9.8 If two lines are vertical lines, then they are parallel. Proof: Since each vertical line is perpendicular to each horizontal line, each vertical line is perpendicular to the x-axis, a horizontal line. Theorem 9.6 states that if two coplanar lines are each perpendicular to the same line, then they are parallel. Therefore, all vertical lines are parallel. A similar theorem can be proved about horizontal lines: Theorem 9.9 If two lines are horizontal lines, then they are parallel. Proof: Since each horizontal line is perpendicular to each vertical line, each horizontal line is perpendicular to the y-axis, a vertical line. Theorem 9.6 states that if two coplanar lines are each perpendicular to the same line, then they are parallel. Therefore, all horizontal lines are parallel. We know that all horizontal lines have the same slope, 0. We also know that all vertical lines have no slope. Do parallel lines that are neither horizontal nor vertical have the same slope? When we draw parallel lines in the coordinate plane, it appears that this is true. Theorem 9.10a If two non-vertical lines in the same plane are parallel, then they have the same slope. Given l1 l2 Prove The slope of l1 is equal to slope of l2. Proof In the coordinate plane, let the slope of l1 be m 0. Choose any point on l1. Through a given point, one and only one line can be drawn perpendicular to a given line. Through that point, draw k, a line perpendicular to l1. y k O l1 l2 x 14365C09.pgs 7/10/07 8:48 AM Page 343 Parallel Lines in the Coordinate Plane 343 If two lines are perpendicular, the slope of one is the negative reciprocal l2 of the slope of the other. Therefore, the slope of k is . It is given that . l1 Then, k is perpendicular to l2 because if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other. The slope of l2 is the negative reciprocal of the slope of k. The negative reciprocal of is m. Therefore, the slope of l1 is equal to the slope of l2. 21 m 21 m Is the converse of this statement true? We will again use the fact that two lines are perpendicular if and only if the slope of one is the negative reciprocal of the slope of the other to prove that it is. Theorem 9.10b If the slopes of two non-vertical lines in the coordinate plane are equal, then the lines are parallel. k y O l1 l2 x Given Lines l1 and l2 with slope m Prove l1 l2 Proof Choose any point on l1. Through a given point, one and only one line can be drawn perpendicular to a given line. Through that point, draw k, a line perpendicular to l1. The slope of k is 21 since two non-vertical lines are m perpendicular if and only if the slope of one is the negative reciprocal of the slope of the other. But this means that l2 Therefore, l1 parallel. ⊥ k because the slope of l2 is also the negative reciprocal of the slope of k. l2 because two lines perpendicular to the same line are We can write the statements that we have proved as a biconditional: Theorem 9.10 Two non-vertical lines in the coordinate plane are parallel if and only if they have the same slope. EXAMPLE 1 The vertices of quadrilateral ABCD are A(2, 4), B(6, 2), C(2, 6), and D(1, 2). a. Show that two sides of the quadrilateral are parallel. b. Show that the quadrilateral has two right angles. 14365C09.pgs 7/10/07 8:48 AM Page 344 344 Parallel Lines Solution The slope of AB 5 22 2 (2422) 2 2 6 5 8 21 2 2 5 24 2 2 (21) 5 26 BC 5 CD 5 2 2 6 DA 5 24 2 2 24 5 22 23 5 4 . 3 3 5 22 . . The slope of The slope of The slope of a. and BC DA equal slopes. are parallel because they have b. The slope of AB the slope of BC Therefore, B is a right angle. is the negative reciprocal of , so they are perpendicular The slope of perpendicular. Therefore, A is a right angle. is the negative reciprocal of the slope of AB DA , so they are Answers a. DABC b. A and B are right angles. EXAMPLE 2 Write an equation for l1, the line through (2, 5) that is parallel to the line l2 whose equation is 2x y 7. Solution (1) Solve the equation of l2 for y: (2) Find the slope of l2. The slope of a line in slope-intercept form is the coefficient of x: (3) Find the slope of l1, which is equal to the slope of l2: (4) Use the definition of slope to write an equation of l1. Let (x, y) and (2, 5) be two points on l1: Answer y 2x 1 or 2x y 1 2x y 7 y 2x 7 y 2x 7 slope slope of l1 2 slope y1 2 y2 x1 2 x2 y 2 5 x 2 (22) 2 y 5 2(x 2) y 5 –2x 4 y 2x 1 14365C09.pgs 7/10/07 8:48 AM Page 345 Parallel Lines in the Coordinate Plane 345 Exercises Writing About Mathematics 1. If l1 and l2 have the same slope and have a common point, what must be true about l1 and l2? 2. Theorem 9.10 is true for all lines that are not vertical. Do vertical lines have the same slope? Explain your answer. Developing Skills In 3–8, for each pair of lines whose equations are given, tell whether the lines are parallel, perpendicular, or neither parallel nor perpendicular. 3. x y = 7 x y 3 5. x 1 3y 1 2 y 3x 2 7. x = 2 x 5 4. 2x y 5 y 2x 3 6. 2x y 6 2x y = 3 8. x = 2 y 3 In 9–12, write an equation of the line that satisfies the given conditions. 9. Parallel to y 3x 1 with y-intercept 4. 10. Perpendicular to y 3x 1 with y-intercept 4. 11. Parallel to x 2y 4 and through the point (4, 5). 12. Parallel to and 3 units below the x-axis. Applying Skills 13. Quadrilateral ABCD has two pairs of parallel sides, . The coordinates of A are (1, 2), the coordinates of B are (7, 1) and the coordinates of C are (8, 2). and AB CD BC DA a. What is the slope of ? AB b. What is the slope of c. Write an equation for d. What is the slope of ? BC e. What is the slope of f. Write an equation for ? CD g . CD ? AD g . AD g. Use the equation of g CD and the equation of g AD to find the coordinates of D. 14365C09.pgs 7/10/07 8:48 AM Page 346 346 Parallel Lines 14. In quadrilateral ABCD, BC ' AB , DA ' AB , and DA ' DC . The coordinates of A are (1, 1), the coordinates of B are (4, 2), and the coordinates of C are (2, 4). a. What is the slope of b. What is the slope of c. What is the slope of d. What is the slope of ? AB ? BC ? Justify your answer. AD ? Justify your answer. DC e. Write an equation for f. Write an equation for g . DC g . AD g. Use the equation of g DC and the equation of g AD to find the coordinates of D. 15. The coordinates of the vertices of quadrilateral PQRS are P(0, 1), Q(4, 0), R(2, 3), and S(2, 2). a. Show that PQRS has two pairs of parallel sides. b. Show that PQRS does not have a right angle. 16. The coordinates of the vertices of quadrilateral KLMN are K(2, 1), L(4, 3), M(2, 1), and N(1, 2). a. Show that KLMN has only one pair of parallel sides. b. Show that KLMN has two right angles. Hands-On Activity 1 In this activity, we will use a compass and a straightedge, or geometry software to construct a line parallel to a given line through a point not on the line. STEP 1. Given a point, P, not on line, l. Through P, construct a line perpen- dicular to line l. Label this line n. P l STEP 2. Through P, construct a line, p, perpendicular to line n. Result: l p a. Justify the construction given in the procedure. b. In (1)–(3), construct a line parallel to the line through the given point. (1) y 1 4x 1 5 ; 3, 51 2 B A (2) 12x y 19; (12, 4) (3) y 21 9x 2 3 ; (0, 4) Hands-On Activity 2 A midsegment is a segment formed by joining two midpoints of the sides of a triangle. In this activity, we will prove that a midsegment is parallel to the third side of the triangle using coordinate geometry. 1. With a partner or in a small group, complete the following: a. Write the coordinates of a triangle using variables. These coordinates can be any convenient variables. midsegment 14365C09.pgs 7/10/07 8:48 AM Page 347 The Sum of the Measures of the Angles of a Triangle 347 b. Find the midpoints of two sides of the triangle. c. Prove that the midsegment formed is parallel to the third side of the triangle. 2. Compare your proof with the proofs of the other groups. Were different coordinates used? Which coordinates seem easiest to work with? 9-4 THE SUM OF THE MEASURES OF THE ANGLES OF A TRIANGLE In previous courses, you have demonstrated that the sum of the measures of the angles of a triangle is 180 degrees. The congruent angles formed when parallel lines are cut by a transversal make it possible for us to prove this fact. Theorem 9.11 The sum of the measures of the angles of a triangle is 180°. Given ABC Prove mA mB mC 180 D A B E C Proof Statements Reasons 1. Let g DE be the line through B that is parallel to . AC 2. mDBE 180 3. mDBA mABC mCBE 180 4. A DBA and C CBE 5. mA mDBA and mC CBE 1. Through a given point not on a given line, there exists one and only one line parallel to the given line. 2. A straight angle is an angle whose degree measure is 180. 3. The whole is equal to the sum of all its parts. 4. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 5. Congruent angles are equal in measure. 6. mA mABC mC 180 6. Substitution postulate. Ma |
ny corollaries to this important theorem exist. 14365C09.pgs 7/10/07 8:48 AM Page 348 348 Parallel Lines Corollary 9.11a If two angles of one triangle are congruent to two angles of another triangle, then the third angles are congruent. Proof: Let ABC and DEF be two triangles in which A D and B E. Since the sum of the degree measures of the angles of a triangle is 180, then mA mB mC mD mE mF. We use the subtraction postulate to prove that mC mF and therefore, that C F. Corollary 9.11b The acute angles of a right triangle are complementary. Proof: In any triangle ABC, mA mB mC 180. If C is a right angle, mC 90, mA mB 90 180 mA mB 90 Therefore, A and B are complementary. Corollary 9.11c Each acute angle of an isosceles right triangle measures 45°. Proof: In isosceles right triangle ABC, mC 90 and . Therefore, mA = mB. Using Corollary 9.12b, we know that A and B are complementary. Therefore, the measure of each must be 45. AC BC Corollary 9.11d Each angle of an equilateral triangle measures 60°. Proof: In equilateral triangle ABC, mA mB mC. We substitute mA for mB and mC in the equation mA mB mC 180, and then solve the resulting equation: 3mA 180 so mA 60. Corollary 9.11e The sum of the measures of the angles of a quadrilateral is 360°. Proof: In quadrilateral ABCD, we draw , forming two triangles. The sum of AC the measures of the angles of quadrilateral ABCD is the sum of the measures of the angles of the two triangles: D C 180 180 360 A B 14365C09.pgs 7/10/07 8:48 AM Page 349 The Sum of the Measures of the Angles of a Triangle 349 Corollary 9.11f The measure of an exterior angle of a triangle is equal to the sum of the measures of the nonadjacent interior angles. The proof is left to the student. (See exercise 30.) Note: Recall that the Exterior Angle Theorem of Section 7-5 gives an inequality that relates the exterior angle of a triangle to the nonadjacent interior angles: “The measure of an exterior angle of a triangle is greater than the measure of either nonadjacent interior angle.” Corollary 9.11f is a version of the Exterior Angle Theorem involving equality. EXAMPLE 1 The measure of the vertex angle of an isosceles triangle exceeds the measure of each base angle by 30 degrees. Find the degree measure of each angle of the triangle. Solution Let x measure of each base angle. Let x 30 measure of vertex angle. The sum of the measures of the angles of a triangle is 180. x x x 30 180 3x 30 180 3x 150 x 50 x 30 80 Answer The angle measures are 50°, 50°, and 80°. EXAMPLE 2 In ABC, the measures of the three angles are represented by 9x, 3x 6, and 11x 2. Show that ABC is a right triangle. Solution Triangle ABC will be a right triangle if one of the angles is a right angle. Write and equation for the sum of the measures of the angles of ABC. 9x 3x 6 11x 2 180 23x 4 180 23x 184 x 8 14365C09.pgs 7/10/07 8:48 AM Page 350 350 Parallel Lines Substitute x 8 in the representations of the angle measures. 9x 9(8) 72 3x 6 3(8) 6 24 6 18 11x 2 11(8) 2 88 2 90 Answer Triangle ABC is a right triangle because the degree measure of one of its angles is 90. EXAMPLE 3 B is a not a point on DCB and AC > BC g ACD . Ray h CE . Prove that bisects g CE . g AB Solution Given: h CE bisects DCB and AC > BC . Prove: g AB g CE Proof Statements h CE bisects DCB. 1. 2. DCE ECB 3. mDCE mECB AC > BC 4. 5. mCAB mCBA 6. mDCB mCAB mCBA 7. mDCB mDCE mECB 8. mDCE mECB mCAB mCBA 9. mDCE mDCE mCAB mCAB or 2mDCE 2 mCAB 10. mDCE mCAB g CE g AB 11. B C E D A Reasons 1. Given. 2. Definition of an angle bisector. 3. Measures of congruent angles are equal. 4. Given. 5. Isosceles triangle theorem. 6. An exterior angle of a triangle is equal to the sum of the measures of the nonadjacent interior angles. 7. Partition postulate. 8. Substitution postulate (steps 6 and 7). 9. Substitution postulate (steps 3, 5, and 8). 10. Division postulate. 11. If two lines are cut by a transversal forming equal corresponding angles, then the lines are parallel. 14365C09.pgs 7/10/07 8:48 AM Page 351 The Sum of the Measures of the Angles of a Triangle 351 Exercises Writing About Mathematics 1. McKenzie said that if a triangle is obtuse, two of the angles of the triangle are acute. Do you agree with McKenzie? Explain why or why not. 2. Giovanni said that since the sum of the measures of the angles of a triangle is 180, the angles of a triangle are supplementary. Do you agree with Giovanni? Explain why or why not. Developing Skills In 3–6, determine whether the given numbers can be the degree measures of the angles of a triangle. 3. 25, 100, 55 4. 95, 40, 45 5. 75, 75, 40 6. 12, 94, 74 In 7–10, the given numbers are the degree measures of two angles of a triangle. Find the measure of the third angle. 7. 80, 60 8. 45, 85 9. 90, 36 10. 65, 65 In 11–14, the measure of the vertex angle of an isosceles triangle is given. Find the measure of a base angle. 11. 20 12. 90 13. 76 14. 110 In 15–18, the measure of a base angle of an isosceles triangle is given. Find the measure of the vertex angle. 15. 80 16. 20 17. 45 18. 63 19. What is the measure of each exterior angle of an equilateral triangle? In 20–23, the diagram shows ABC and exterior ACD. 20. If mA 40 and mB 20, find mACD and mACB. 21. If mA 40 and mB 50, find mACD and mACB. 22. If mA 40 and mACB 120, find mACD and mB. 23. If mA 40, mB 3x 20, and mACD 5x 10, find mB, mACD, and mACB. Applying Skills D C A B 24. The measure of each base angle of an isosceles triangle is 21° more than the measure of the vertex angle. Find the measure of each angle of the triangle. 14365C09.pgs 7/10/07 8:48 AM Page 352 352 Parallel Lines 25. The measure of an exterior angle at C of isosceles ABC is 110°. If AC = BC, find the mea- sure of each angle of the triangle. 26. The measure of an exterior angle at D of isosceles DEF is 100°. If DE = EF, find the mea- sure of each angle of the triangle. 27. Triangle LMN is a right triangle with M the right angle. If mL 32, find the measure of N and the measure of the exterior angle at N. 28. In ABC, mA 2x 18, mB x 40, and mC 3x – 40. a. Find the measure of each angle of the triangle. b. Which is the longest side of the triangle? 29. The measure of an exterior angle at B, the vertex of isosceles ABC, can be represented by 3x 12. If the measure of a base angle is 2x 2, find the measure of the exterior angle and of the interior angles of ABC. 30. Prove Corollary 9.11f, “The measure of an exterior angle of a triangle is equal to the sum of the measures of the nonadjacent interior angles.” 31. a. In the coordinate plane, graph points A(5, 2), B(2, 2), C(2, 1), D(1, 1). g AB and BDC. b. Draw c. Explain how you know that BDC is an isosceles right triangle. d. What is the measure of BDC? Justify your answer. e. What is the measure of DBA? Justify your answer. 32. Prove that the sum of the measures of the angles of hexagon ABCDEF is 720°. (Hint: draw AD .) 33. ABCD is a quadrilateral with h BD the bisector of ABC and h DB the bisector of ADC. Prove that A C. 9-5 PROVING TRIANGLES CONGRUENT BY ANGLE, ANGLE, SIDE When two angles of one triangle are congruent to two angles of another triangle, the third angles are congruent. This is not enough to prove that the two triangles are congruent. We must know that at least one pair of corresponding sides are congruent. We already know that if two angles and the side between them in one triangle are congruent to the corresponding angles and side in another triangle, then the triangles are congruent by ASA. Now we want to prove angle-angle-side or AAS triangle congruence. This would allow us to conclude that if any two angles and any side in one triangle are congruent to the corresponding angles and side in another triangle, then the triangles are congruent. 14365C09.pgs 7/10/07 8:48 AM Page 353 Proving Triangles Congruent by Angle, Angle, Side 353 Theorem 9.12 If two angles and the side opposite one of them in one triangle are congruent to the corresponding angles and side in another triangle, then the triangles are congruent. (AAS) Given ABC and DEF, A D, C F, and AB > DE Prove ABC DEF C F A B D E Proof Statements Reasons 1. A D 2. C F 3. B E 1. Given. 2. Given. 3. If two angles of one triangle are congruent to two angles of another triangle, then the third angles are congruent. AB > DE 4. 5. ABC DEF 4. Given. 5. ASA. Therefore, when two angles and any side in one triangle are congruent to the corresponding two angles and side of a second triangle, we may say that the triangles are congruent either by ASA or by AAS. The following corollaries can proved using AAS. Note that in every right tri- angle, the hypotenuse is the side opposite the right angle. Corollary 9.12a Two right triangles are congruent if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of the other right triangle. The proof uses AAS and is left to the student. (See exercise 15.) Corollary 9.12b If a point lies on the bisector of an angle, then it is equidistant from the sides of the angle. Recall that the distance from a point to a line is the length of the perpendicular from the point to the line. The proof uses AAS and is left to the student. (See exercise 16.) You now have four ways to prove two triangles congruent: SAS, ASA, SSS, and AAS. A P C B 14365C09.pgs 7/10/07 8:48 AM Page 354 354 Parallel Lines EXAMPLE 1 Prove that the altitudes drawn to the legs of an isosceles triangle from the endpoints of the base are congruent. B Given: Isosceles triangle ABC with BA , BC AE ⊥ BC , D E and CD ⊥ BA . Prove: CD AE A C Proof Statements Reasons 1. In ABC, BA . BC 1. Given. E C C A 2. BAC BCA ⊥ AE , BC 3. CD 4. CDA and AEC are right BA ⊥ angles. A 5. CDA AEC AC 7. DAC ECA S 6. AC 8. CD AE 2. If two sides of a triangle are congruent, the angles opposite these sides are congruent. 3. Given. 4. Perpendicular lines are two lines that intersect to form right angles. 5. All right angles are congruent. 6. Reflexive property of congruence. 7. AAS (steps 2, 5, and 6). 8. Co |
rresponding parts of congruent triangles are congruent. B B A A D EXAMPLE 2 The coordinates of the vertices of ABC are A(6, 0), B(1, 0) and C(5, 2). The coordinates of DEF are D(3, 0), E(8, 0), and F(4, 2). Prove that the triangles are congruent. C(5, 2) y F(4, 2) B(1, 0) O A(6, 0) D(3, 0) E(8, 0) x Solution (1) Prove that the triangles are right triangles. In ABC: 2 2 0 The slope of AC is The slope of CB is 25 2 (26) 5 2 25 2 (21) 5 2 2 2 0 1 5 2 . 24 5 21 2 The slope of . The slope of In DEF: DF is 2 2 0 FE is . 24 5 21 2 . Two lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. Therefore, ACB is a right triangle. Also, DFE is a right triangle. , DFE is a right angle, and , ACB is a right angle, and DF ' FE AC ' CB 14365C09.pgs 7/10/07 8:48 AM Page 355 Proving Triangles Congruent by Angle, Angle, Side 355 (2) Prove that two acute angles are congruent. Two lines are parallel if their slopes are equal. Therefore, . The x-axis is a transversal forming congruent corresponding angles, so CBA and FED are congruent. CB FE (3) Prove that the hypotenuses are congruent. The hypotenuse of ABC is hypotenuse of DEF is DE have the same measure are congruent, and so , and AB |–6 (1)| 5. The AB , and DE |3 8| 5. Line segments that AB > DE . (4) Therefore, ABC DEF because the hypotenuse and an acute angle of one triangle are congruent to the hypotenuse and an acute angle of the other. EXAMPLE 3 Show that if a triangle has two sides and an angle opposite one of the sides congruent to the corresponding sides and angle of another triangle, the triangles may not be congruent. Solution (1) Draw an angle, ABC. A (2) Open a compass to a length that is smaller than AB but larger than the distance from A to . Use the compass to mark two points, D and E, on BC BC . (3) Draw AD (4) In ABD and ABE, and AE . AB AD > AE , AB . In these B D E C A B B, and two triangles, two sides and the angle opposite one of the sides are congruent to the corresponding parts of the other triangle. But ABD and ABE are not congruent. This counterexample proves that SSA is not sufficient to prove triangles congruent. D E C B Note: Triangles in which two sides and an angle opposite one of them are congruent may not be congruent to each other. Therefore, SSA is not a valid proof of triangle congruence. Similarly, triangles in which all three angles are congruent may not be congruent to each other, so AAA is also not a valid proof of triangle congruence. 14365C09.pgs 7/10/07 8:48 AM Page 356 356 Parallel Lines Exercises Writing About Mathematics 1. In Example 3, we showed that SSA cannot be used to prove two triangles congruent. Does this mean that whenever two sides and an angle opposite one of the sides are congruent to the corresponding parts of another triangle the two triangles are not congruent? Explain your answer. 2. In the coordinate plane, points A and C are on the same horizontal line and C and B are on the same vertical line. Are CAB and CBA complementary angles? Justify your answer. Developing Skills In 3–8, each figure shows two triangles. Congruent parts of the triangles have been marked. Tell whether or not the given congruent parts are sufficient to prove that the triangles are congruent. Give a reason for your answer. 3. D C 4. D A B A E C B 5. F E 6. C 7. B C 8. A D A BD Applying Skills A R B Q S P 9. Prove that if two triangles are congruent, then the altitudes drawn from corresponding ver- tices are congruent. 10. Prove that if two triangles are congruent, then the medians drawn from corresponding ver- tices are congruent. 11. Prove that if two triangles are congruent, then the angle bisectors drawn from correspond- ing vertices are congruent. 14365C09.pgs 7/10/07 8:48 AM Page 357 The Converse of the Isosceles Triangle Theorem 357 12. Given: Quadrilateral ABCD with A C and h BD the bisector of ABC. h DB bisects ADC. Prove: 13. Given: AB CD , AB > CD , and AB ' BEC . A Prove: AED and BEC bisect each other. B E C C A B D D 14. a. Use a translation to prove that ABC and DEF in Example 2 are congruent. b. Use two line reflections to prove that ABC and DEF in Example 2 are congruent. 15. Prove Corollary 9.12a, “Two right triangles are congruent if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of the other right triangle.” 16. Prove Corollary 9.12b, “If a point lies on the bisector of an angle, it is equidistant from the sides of the angle.” 17. Prove that if three angles of one triangle are congruent to the corresponding angles of another (AAA), the triangles may not be congruent. (Through any point on side of ABC, draw a line segment parallel to AC BC .) 9-6 THE CONVERSE OF THE ISOSCELES TRIANGLE THEOREM The Isosceles Triangle Theorem, proved in Section 5-3 of this book, is restated here in its conditional form. If two sides of a triangle are congruent, then the angles opposite these sides are congruent. When we proved the Isosceles Triangle Theorem, its converse would have been very difficult to prove with the postulates and theorems that we had available at that time. Now that we can prove two triangles congruent by AAS, its converse is relatively easy to prove. Theorem 9.13 If two angles of a triangle are congruent, then the sides opposite these angles are congruent. 14365C09.pgs 7/10/07 8:48 AM Page 358 358 Parallel Lines Given ABC with A B. Prove CA CB Proof We can use either the angle bisector or the altitude from C to separate the triangle into two congruent triangles. We will use the angle bisector. A Statements Reasons C D B 1. Draw ACB. CD , the bisector of 2. ACD BCD 3. A B CD 4. CD 5. ACD BCD 6. CA CB 1. Every angle has one and only one bisector. 2. An angle bisector of a triangle is a line segment that bisects an angle of the triangle. 3. Given. 4. Reflexive property of congruence. 5. AAS. 6. Corresponding parts of congruent triangles are congruent. The statement of the Isosceles Triangle Theorem (Theorem 5.1) and its con- verse (Theorem 9.14) can now be written in biconditional form: Two angles of a triangle are congruent if and only if the sides opposite these angles are congruent. To prove that a triangle is isosceles, we may now prove that either of the fol- lowing two statements is true: 1. Two sides of the triangle are congruent. 2. Two angles of the triangle are congruent. Corollary 9.13a If a triangle is equiangular, then it is equilateral. Given ABC with A B C. Prove ABC is equilateral. 14365C09.pgs 7/10/07 8:48 AM Page 359 The Converse of the Isosceles Triangle Theorem 359 Proof We are given equiangular ABC. Then since A B, the sides opposite these angles are congruent, that is, AC > AB erty of congruence, AB > BC > CA for the same reason. Therefore, BC > AC . Also, since B C, AC > AB by the transitive prop- , and ABC is equilateral. EXAMPLE 1 In PQR, Q R. If PQ 6x 7 and PR 3x 11, find: a. the value of x b. PQ c. PR Solution a. Since two angles of PQR b. PQ 5 6x 2 7 c. PR 5 3x 1 11 are congruent, the sides opposite these angles are congruent. Thus, PQ PR. 6x 7 3x 11 6x 3x 11 7 3x 18 x 6 Answer 5 6(6) 2 7 5 36 2 7 5 29 Answer 5 3(6) 1 11 5 18 1 11 5 29 Answer EXAMPLE 2 The degree measures of the three angles of ABC are represented by mA x 30, mB 3x, and mC 4x 30. Describe the triangle as acute, right, or obtuse, and as scalene, isosceles, or equilateral. Solution The sum of the degree measures of the angles of a triangle is 180. x 30 3x 4x 30 180 8x 60 180 8x 120 x 15 Substitute x 15 in the representations given for the three angle measures. m/A 5 x 1 30 5 15 1 30 5 45 m/B 5 3x m/C 5 4x 1 30 5 3(15) 5 45 5 4(15) 1 30 5 60 1 30 5 90 Since A and B each measure 45°, the triangle has two congruent angles and therefore two congruent sides. The triangle is isosceles. Also, since one angle measures 90°, the triangle is a right triangle. Answer ABC is an isosceles right triangle. 14365C09.pgs 7/10/07 8:48 AM Page 360 360 Parallel Lines EXAMPLE 3 Given: Quadrilateral ABCD with and h AC bisects DAB. AB CD D C Prove: AD > CD A B Proof Statements Reasons D C AB CD 1. 2. DCA CAB A B h AC bisects DAB. 3. 4. CAB DAC 5. DCA DAC congruence. 6. AD > CD Exercises Writing About Mathematics 1. Given. 2. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 3. Given. 4. A bisector of an angle divides the angle into two congruent parts. 5. Transitive property of 6. If two angles of a triangle are congruent, the sides opposite these angles are congruent. 1. Julian said that the converse of the Isosceles Triangle Theorem could have been proved as a corollary to Theorem 7.3, “If the lengths of two sides of a triangle are unequal, then the measures of the angles opposite these sides are unequal.” Do you agree with Julian? Explain why or why not. 2. Rosa said that if the measure of one angle of a right triangle is 45 degrees, then the triangle is an isosceles right triangle. Do you agree with Rosa? Explain why or why not. 14365C09.pgs 7/10/07 8:48 AM Page 361 The Converse of the Isosceles Triangle Theorem 361 In 3–6, in each case the degree measures of two angles of a triangle are given. a. Find the degree measure of the third angle of the triangle. b. Tell whether the triangle is isosceles or is not isosceles. 3. 70, 40 4. 30, 120 6. 80, 40 5. 50, 65 7. In ABC, mA mC, AB 5x 6, and BC 3x 14. Find the value of x. 8. In PQR, mQ mP, PR 3x, and RQ 2x 7. Find PR and RQ. 9. In MNR, MN NR, mM = 72, and mR 2x. Find the measures of R and of N. 10. In ABC, mA 80 and mB 50. If AB 4x 4, AC 2x 16, and BC 4x 6, find the measure of each side of the triangle. 11. The degree measures of the angles of ABC are represented by x 10, 2x, and 2x 30. Show that ABC is an isosceles triangle. 12. The degree measures of the angles of ABC are represented by x 35, 2x 10, and 3x 15. Show that ABC is an equilateral triangle. 13. The degree measures of the angles of ABC are represented by 3x 18, 4x 9, and 10x. Show that ABC is an isoscele |
s right triangle. 14. What is the measure of each exterior angle of an equilateral triangle? 15. What is the sum of the measures of the exterior angles of any triangle? Applying Skills 16. Given: P is not on › ‹ ABCD and ABP PCD. Prove: BPC is isosceles. P A B C D 17. Given: P is not on AB and PAB PBA. Prove: P is on the perpendicular bisector of AB . P A B 14365C09.pgs 7/10/07 8:48 AM Page 362 362 Parallel Lines 18. Given: h BE angle of ABC, and bisects DBC, an exterior AC . h BE Prove: AB > CB E C 19. Given: P is not on , ABCD PBC PCB, and APB DPC Prove: AP > DP P A B D A B C D 20. Prove Theorem 9.13 by drawing the altitude from C. 9-7 PROVING RIGHT TRIANGLES CONGRUENT BY HYPOTENUSE, LEG We showed in Section 5 of this chapter that, when two sides and an angle opposite one of these sides in one triangle are congruent to the corresponding two sides and angle in another triangle, the two triangles may or may not be congruent. When the triangles are right triangles, however, it is possible to prove that they are congruent. The congruent angles are the right angles, and each right angle is opposite the hypotenuse of the triangle. Theorem 9.14 If the hypotenuse and a leg of one triangle are congruent to the corresponding parts of the other, then the two right triangles are congruent. (HL) Given Right ABC with right angle B and right DEF AC with right angle E, DF EF BC , C F Prove ABC DEF Proof To prove this theorem, we will construct a third triangle, GEF, that shares a common side with DEF and prove that each of the two given triangles is congruent to GEF and, thus, to each other. D A B E We first show that ABC is congruent to GEF: 14365C09.pgs 7/10/07 8:48 AM Page 363 Proving Right Triangles Congruent by Hypotenuse, Leg 363 (1) Since any line segment may be C F EG > AB to G so that extended any required length, extend DE FG (2) GEF and DEF form a linear pair, and DEF is a right angle. Therefore, GEF is a right angle. We are given that B is a right angle. All right angles are congruent, so B GEF. . Draw D A E B . G BC > EF (3) We are also given (4) Therefore, ABC GEF by SAS. We now show that DEF is also congruent to the constructed triangle, GEF: . (5) Since corresponding sides of congru- C F ent triangles are congruent, AC > GF AC > DF tive property of congruence. . Since we are given GF > DF , by the transi- A B D E G (6) If two sides of a triangle are congru- ent, the angles opposite these sides are congruent. In DFG, GF > DF , so D G. Also, DEF GEF since all right angles are congruent. (7) Therefore, DEF GEF by AAS. (8) Therefore, ABC DEF by the transitive property of congruence (steps 4 and 7). This theorem is called the hypotenuse-leg triangle congruence theorem, abbreviated HL. Therefore, from this point on, when the hypotenuse and a leg of one right triangle are congruent to the corresponding parts of a second right triangle, we may say that the triangles are congruent. A corollary of this theorem is the converse of Corollary 9.12b. Corollary 9.14a If a point is equidistant from the sides of an angle, then it lies on the bisector of the angle. Given ABC, h PD ' BA at D, h PF ' BC at F, and PD PF Prove ABP CBP Strategy Use HL to prove PDB PFB. A D P The proof of this theorem is left to the stu- B F C dent. (See exercise 8.) 14365C09.pgs 7/10/07 8:48 AM Page 364 364 Parallel Lines Concurrence of Angle Bisectors of a Triangle In earlier chapters, we saw that the perpendicular bisectors of the sides of a triangle intersect in a point and that the altitudes of a triangle intersect in a point. Now we can prove that the angle bisectors of a triangle intersect in a point. Theorem 9.15 The angle bisectors of a triangle are concurrent. Given ABC with AL the bisector of B, and of C. the bisector of A, CN BM the bisector C Prove , BMAL , and CN intersect in a point, P. M L Proof Let P be the point at which P AL and BM intersect. If a point lies on the bisector of an angle, then it is equidistant from the sides of the angle. Therefore, P is equidisbecause it lies on the bisector of A, and P is equidistant from AB because it lies on the bisector of B. Therefore, P is tant from BC , equidistant from . If a point is equidistant from the sides of AC AB an angle, then it lies on the bisector of the angle. Since P is equidistant from AC bisectors of ABC intersect at a point, P. , then it lies of the bisector of C. Therefore, the three angle and and AC AB , and and BC BC A N B The point where the angle bisectors of a triangle are concurrent is called the incenter. EXAMPLE 1 Given: ABC, AD AB . BC ⊥ , BD AB DC , and C Prove: DAB BCD Proof We can show that ADB and CBD are right triangles and use HL to prove them congruent. A D B 14365C09.pgs 7/10/07 8:48 AM Page 365 Proving Right Triangles Congruent by Hypotenuse, Leg 365 Statements Reasons 1. 2. 3. ⊥ AB BD AB DC ⊥ BD DC 1. Given. 2. Given. 3. If a line is perpendicular to one of two parallel lines it is perpendicular to the other. 4. ABD and CDB are right 4. Perpendicular lines intersect to angles. 5. AD BC form right angles. 5. Given. BD 6. BD 7. ADB CBD 8. DAB BCD 6. Reflexive property of congruence. 7. HL (steps 5 and 6). 8. Corresponding parts of congruent triangles are congruent Exercises Writing About Mathematics 1. In two right triangles, the right angles are congruent. What other pairs of corresponding parts must be known to be congruent in order to prove these two right triangles congruent? 2. The incenter of ABC is P. If PD is the distance from P to AB and Q is any other point on AB , is PD greater than PQ, equal to PQ, or less than PQ? Justify your answer. Developing Skills 3. In ABC, mCAB 40 and mABC 60. The angle bisectors of ABC intersect at P. a. Find mBCA. b. Find the measure of each angle of APB. c. Find the measure of each angle of BPC. d. Find the measure of each angle of CPA. e. Does the bisector of CAB also bisect CPB? Explain your answer. M C P L A N B 14541C09.pgs 1/25/08 3:52 PM Page 366 366 Parallel Lines 4. Triangle ABC is an isosceles right triangle with the right angle at C. Let P be the incenter of ABC. a. Find the measure of each acute angle of ABC. b. Find the measure of each angle of APB. c. Find the measure of each angle of BPC. d. Find the measure of each angle of CPA. e. Does the bisector of ACB also bisect APB? Explain your answer. 5. Triangle ABC is an isosceles triangle with mC 140. Let P be the incenter of ABC. a. Find the measure of each acute angle of ABC. b. Find the measure of each angle of APB. c. Find the measure of each angle of BPC. d. Find the measure of each angle of CPA. e. Does the bisector of ACB also bisect APB? Explain your answer. 6. In RST, the angle bisectors intersect at P. If mRTS = 50, mTPR = 120, and mRPS 115, find the measures of TRS, RST, and SPT. 7. a. Draw a scalene triangle on a piece of paper or using geometry software. Label the trian- gle ABC. b. Using compass and straightedge or geometry software, construct the angle bisectors of the angles of the triangle. Let and be the bisector of A, be the bisector of C, such that L, M, and N are points on the triangle. be the bisector of B, BM CN AL c. Label the incenter P. d. In ABC, does AP = BP = CP? Explain why or why not. e. If the incenter is equidistant from the vertices of DEF, what kind of a triangle is DEF? Applying Skills 8. Prove Corollary 9.14a, “If a point is equidistant from the sides of an angle, then it lies on the bisector of the angle.” 9. Given DB ' ABC and AD ' DC , when is ABD congruent to DBC? Explain. D A B C 10. When we proved that the bisectors of the angles of a triangle intersect in a point, we began AL by stating that two of the angle bisectors, intersect, show that they are not parallel. (Hint: Show that a pair of interior angles on the same side of the transversal cannot be supplementary.) , intersect at P. To prove that they . are cut by transversal AB and and BM BM AL 14365C09.pgs 7/10/07 8:48 AM Page 367 Interior and Exterior Angles of Polygons 367 11. Given: Quadrilateral ABCD, AB ' BD , BD ' DC , and AD > CB . Prove: A C and AD CB 12. In QRS, the bisector of QRS is perpendicular to QS at P. a. Prove that QRS is isosceles. b. Prove that P is the midpoint of QS . 13. Each of two lines from the midpoint of the base of an isosceles triangle is perpendicular to one of the legs of the triangle. Prove that these lines are congruent. 14. In quadrilateral ABCD, A and C are right angles and AB = CD. Prove that: a. AD = BC b. ABD CDB c. ADC is a right angle. 15. In quadrilateral ABCD, ABC and BCD are right angles, and AC = BD. Prove that AB = CD. 16. Given: Prove: h h , APS ABC h PD ' ADE h APS bisects CAE. h with , and ADE , and PB PD. h PB ' ABC , E D S P A B C 9-8 INTERIOR AND EXTERIOR ANGLES OF POLYGONS Polygons Recall that a polygon is a closed figure that is the union of line segments in a plane. Each vertex of a polygon is the endpoint of two line segments. We have proved many theorems about triangles and have used what we know about triangles to prove statements about the sides and angles of quadrilaterals, polygons with four sides. Other common polygons are: • A pentagon is a polygon that is the union of five line segments. • A hexagon is a polygon that is the union of six line segments. • An octagon is a polygon that is the union of eight line segments. • A decagon is a polygon that is the union of ten line segments. • In general, an n-gon is a polygon with n sides. 14365C09.pgs 7/10/07 8:48 AM Page 368 368 Parallel Lines A convex polygon is a polygon in which each of the interior angles measures less than 180 degrees. Polygon PQRST is a convex polygon and a pentagon. A concave polygon is a polygon in which at least one interior angle measures more than 180 degrees. Polygon ABCD is a concave polygon and a quadrilateral. In the rest of this textbook, unless otherwise stated, all polygons are convex Interior Angles of a Polygon A pair of angles whose vertices are the endpoints of a common side are called consecutive angles. And the ver |
tices of consecutive angles are called consecutive vertices or adjacent vertices. For example, in PQRST, P and Q are consecutive angles and P and Q are consecutive or adjacent vertices. Another pair of consecutive angles are T and P. Vertices R and T are nonadjacent vertices. A diagonal of a polygon is a line segment whose endpoints are two nonadjacent vertices. In hexagon ABCDEF, the vertices adjacent to B are A and C and the vertices nonadjacent to B are D, E, and F. Therefore, there are three diago. nals with endpoint B: BF , and , BD BE E F A B D C The polygons shown above have four, five, and six sides. In each polygon, all possible diagonals from a vertex are drawn. In the quadrilateral, two triangles are formed. In the pentagon, three triangles are formed, and in the hexagon, four triangles are formed. Note that in each polygon, the number of triangles formed is two less than the number of sides. • In a quadrilateral: the sum of the measures of the angles is 2(180) 360. • In a pentagon: the sum of the measures of the angles is 3(180) 540. • In a hexagon: the sum of the measures of the angles is 4(180) 720. 14365C09.pgs 7/10/07 8:48 AM Page 369 Interior and Exterior Angles of Polygons 369 In general, the number of triangles into which the diagonals from a vertex separate a polygon of n sides is two less than the number of sides, or n 2. The sum of the interior angles of the polygon is the sum of the interior angles of the triangles formed, or 180(n 2). We have just proved the following theorem: Theorem 9.16 The sum of the measures of the interior angles of a polygon of n sides is 180(n 2)°. Exterior Angles of a Polygon At any vertex of a polygon, an exterior angle forms a linear pair with the interior angle. The interior angle and the exterior angle are supplementary. Therefore, the sum of their measures is 180°. If a polygon has n sides, the sum of the interior and exterior angles of the polygon is 180n. Therefore, in a polygon with n sides: The measures of the exterior angles 180n the measures of the interior angles 180n 180(n 2) 180n 180n 360 360 We have just proved the following theorem: Theorem 9.17 The sum of the measures of the exterior angles of a polygon is 360°. DEFINITION A regular polygon is a polygon that is both equilateral and equiangular. If a triangle is equilateral, then it is equiangular. For polygons that have more than three sides, the polygon can be equiangular and not be equilateral, or can be equilateral and not be equiangular. Equilateral but not equiangular Equiangular but not equilateral Equiangular but not equilateral Equilateral but not equiangular 14365C09.pgs 7/10/07 8:48 AM Page 370 370 Parallel Lines EXAMPLE 1 The measure of an exterior angle of a regular polygon is 45 degrees. a. Find the number of sides of the polygon. b. Find the measure of each interior angle. c. Find the sum of the measures of the interior angles. Solution a. Let n be the number of sides of the polygon. Then the sum of the measures of the exterior angles is n times the measure of one exterior angle. 45n 5 360 n 5 360 45 n 5 8 Answer b. Each interior angle is the supplement of each exterior angle. Measure of each interior angle 180 45 135 Answer c. Use the sum of the measures of the interior angles, 180(n 2). 180(n 2 2) 5 180(8 2 2) 5 180(6) 5 1,080 Answer or Multiply the measure of each interior angle by the number of sides. 8(135) 1,080 Answer Answers a. 8 sides b. 135° c. 1,080° EXAMPLE 2 In quadrilateral ABCD, mA x, mB 2x 12, mC = x 22, and mD 3x. a. Find the measure of each interior angle of the quadrilateral. b. Find the measure of each exterior angle of the quadrilateral. Solution a. mA mB mC mD 18(n 2) x 2x 12 x 22 3x 180(4 2) 7x 10 360 7x 350 x 50 14365C09.pgs 7/10/07 8:48 AM Page 371 Interior and Exterior Angles of Polygons 371 mA x 50 mC x 22 50 22 72 mB 2x 12 2(50) 12 88 mD 3x 3(50) 150 b. Each exterior angle is the supplement of the interior angle with the same vertex. The measure of the exterior angle at A is 180 50 130. The measure of the exterior angle at B is 180 88 92. The measure of the exterior angle at C is 180 72 108. The measure of the exterior angle at D is 180 150 30. Answers a. 50°, 88°, 72°, 150° b. 130°, 92°, 108°, 30° Exercises Writing About Mathematics 1. Taylor said that each vertex of a polygon with n sides is the endpoint of (n 3) diagonals. Do you agree with Taylor? Justify your answer. n 2(n 2 3) 2. Ryan said that every polygon with n sides has Justify your answer. diagonals. Do you agree with Ryan? Developing Skills 3. Find the sum of the degree measures of the interior angles of a polygon that has: a. 3 sides b. 7 sides c. 9 sides d. 12 sides 4. Find the sum of the degree measures of the interior angles of: a. a hexagon b. an octagon c. a pentagon d. a quadrilateral 5. Find the sum of the measures of the exterior angles of a polygon that has: a. 4 sides b. 8 sides c. 10 sides d. 36 sides In 6–14, for each regular polygon with the given number of sides, find the degree measures of: a. one exterior angle b. one interior angle 6. 4 sides 9. 8 sides 12. 20 sides 7. 5 sides 10. 9 sides 13. 36 sides 8. 6 sides 11. 12 sides 14. 42 sides 14365C09.pgs 7/10/07 8:48 AM Page 372 372 Parallel Lines 15. Find the number of sides of a regular polygon each of whose exterior angles contains: a. 30° b. 45° c. 60° d. 120° 16. Find the number of sides of a regular polygon each of whose interior angles contains: a. 90° b. 120° c. 140° d. 160° 17. Find the number of sides a polygon if the sum of the degree measures of its interior angles is: a. 180 e. 1,440 b. 360 f. 2,700 c. 540 g. 1,800 d. 900 h. 3,600 Applying Skills 18. The measure of each interior angle of a regular polygon is three times the measure of each exterior angle. How many sides does the polygon have? 19. The measure of each interior angle of a regular polygon is 20 degrees more than three times the measure of each exterior angle. How many sides does the polygon have? 20. The sum of the measures of the interior angles of a concave polygon is also 180(n 2), where n is the number of sides. Is it possible for a concave quadrilateral to have two interior angles that are both more than 180°? Explain why or why not. 21. From vertex A of regular pentagon ABCDE, two diagonals are drawn, forming three triangles. a. Prove that two of the triangles formed by the diagonals are congruent. b. Prove that the congruent triangles are isosceles. c. Prove that the third triangle is isosceles. 22. From vertex L of regular hexagon LMNRST, three diagonals are drawn, forming four triangles. a. Prove that two of the triangles formed by the diagonals are congruent. b. Prove that the other two triangles formed by the diagonals are congruent. c. Find the measures of each of the angles in each of the four triangles. 23. The coordinates of the vertices of quadrilateral ABCD are A(2, 0), B(0, 2), C(2, 0), and D(0, 2). a. Prove that each angle of the quadrilateral is a right angle. b. Segments of the x-axis and the y-axis are diagonals of the quadrilateral. Prove that the four triangles into which the diagonals separate the quadrilateral are congruent. c. Prove that ABCD is a regular quadrilateral. Hands-On Activity In Section 9-7, we saw that the angle bisectors of a triangle are concurrent in a point called the incenter. In this activity, we will study the intersection of the angle bisectors of polygons. a. Draw various polygons that are not regular of different sizes and numbers of sides. Construct the angle bisector of each interior angle. Do the angle bisectors appear to intersect in a single point? b. Draw various regular polygons of different sizes and numbers of sides. Construct the angle bisector of each interior angle. Do the angle bisectors appear to intersect in a single point? c. Based on the results of part a and b, state a conjecture regarding the intersection of the angle bisector of polygons. 14365C09.pgs 7/10/07 8:48 AM Page 373 Chapter Summary 373 CHAPTER SUMMARY Definitions to Know • Parallel lines are coplanar lines that have no points in common, or have all points in common and, therefore, coincide. • A transversal is a line that intersects two other coplanar lines in two dif- ferent points. • The incenter is the point of intersection of the bisectors of the angles of a triangle. • A convex polygon is a polygon in which each of the interior angles mea- sures less than 180 degrees. • A concave polygon is a polygon in which at least one of the interior angles measures more than 180 degrees. • A regular polygon is a polygon that is both equilateral and equiangular. Postulates Theorems and Corollaries 9.1 9.2 9.1 9.2 9.3 9.4 9.5 9.6 9.7 Two distinct coplanar lines are either parallel or intersecting. Through a given point not on a given line, there exists one and only one line parallel to the given line. Two coplanar lines cut by a transversal are parallel if and only if the alternate interior angles formed are congruent. Two coplanar lines cut by a transversal are parallel if and only if corresponding angles are congruent. Two coplanar lines cut by a transversal are parallel if and only if interior angles on the same side of the transversal are supplementary. If two coplanar lines are each perpendicular to the same line, then they are parallel. If, in a plane, a line intersects one of two parallel lines, it intersects the other. If a transversal is perpendicular to one of two parallel lines, it is perpendicular to the other. If two of three lines in the same plane are each parallel to the third line, then they are parallel to each other. If two lines are vertical lines, then they are parallel. If two lines are horizontal lines, then they are parallel. 9.8 9.9 9.10 Two non-vertical lines in the coordinate plane are parallel if and only if they have the same slope. 9.11 The sum of the measures of the angles of a triangle is 180°. 9.11a If two angles of one triangle are congruent to two angles of another tri- angle, then the third angle |
s are congruent. 9.11b The acute angles of a right triangle are complementary. 9.11c Each acute angle of an isosceles right triangle measures 45°. 9.11d Each angle of an equilateral triangle measures 60°. 9.11e The sum of the measures of the angles of a quadrilateral is 360°. 14365C09.pgs 7/10/07 8:48 AM Page 374 374 Parallel Lines 9.11f The measure of an exterior angle of a triangle is equal to the sum of the 9.12 measures of the nonadjacent interior angles. If two angles and the side opposite one of them in one triangle are congruent to the corresponding angles and side in another triangle, then the triangles are congruent. (AAS) 9.12a Two right triangles are congruent if the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of the other right triangle. 9.12b If a point lies on the bisector of an angle, then it is equidistant from the 9.13 sides of the angle. If two angles of a triangle are congruent, then the sides opposite these angles are congruent. 9.13a If a triangle is equiangular, then it is equilateral. 9.14 If the hypotenuse and a leg of one right triangle are congruent to the corresponding parts of the other, then the two right triangles are congruent. (HL) 9.14a If a point is equidistant from the sides of an angle, then it lies on the bisector of the angle. 9.15 The angle bisectors of a triangle are concurrent. 9.16 The sum of the measures of the interior angles of a polygon of n sides is 180(n 2)°. 9.17 The sum of the measures of the exterior angles of a polygon is 360°. VOCABULARY 9-1 Euclid’s parallel postulate • Playfair’s postulate • Coplanar • Parallel lines • Transversal • Interior angles • Exterior angles • Alternate interior angles • Alternate exterior angles • Interior angles on the same side of the transversal • Corresponding angles 9-3 Midsegment 9-5 AAS triangle congruence 9-7 Hypotenuse-leg triangle congruence theorem (HL) • Incenter 9-8 Pentagon • Hexagon • Octagon • Decagon • n-gon • Convex polygon • Concave polygon • Consecutive angles • Consecutive vertices • Adjacent vertices • Diagonal of a polygon • Regular polygon 14365C09.pgs 7/10/07 8:48 AM Page 375 REVIEW EXERCISES In 1–5, g AB g CD respectively. and these lines are cut by transversal 1. If mAEF 5x and mDFE 75, find x. 2. If mCFE 3y 20 and mAEG 4y 10, find y. 3. If mBEF 5x and mCFE 7x 48, find x. 4. If mDFE y and mBEF 3y 40, find mDFE. A C Review Exercises 375 at points E and F, g GH G E B F D H 5. If mAEF 4x and mEFD 3x 18, find: a. the value of x c. mEFD b. mAEF d. mBEF e. mCFH 6. The degree measure of the vertex angle of an isosceles triangle is 120. Find the measure of a base angle of the triangle. 7. In ABC, A C. If AB 8x 4 and CB 3x 34, find x. 8. In an isosceles triangle, if the measure of the vertex angle is 3 times the measure of a base angle, find the degree measure of a base angle. 9. In a triangle, the degree measures of the three angles are represented by x, x 42, and x 6. Find the angle measures. 10. In PQR, if mP 35 and mQ 85, what is the degree measure of an exterior angle of the triangle at vertex R? 11. An exterior angle at the base of an isosceles triangle measures 130°. Find the measure of the vertex angle. 12. In ABC, if 13. In DEF, if 14. In PQR, AB DE AC DF and mA 70, find mB. and mE 13, find mD. PQ is extended through Q to point T, forming exterior RQT. If mRQT 70 and mR 10, find mP. 15. In ABC, AC BC . The degree measure of an exterior angle at vertex C is represented by 5x 10. If mA 30, find x. 16. The degree measures of the angles of a triangle are represented by x 10, 2x 20, and 3x 10. Find the measure of each angle of the triangle. 17. If the degree measures of the angles of a triangle are represented by x, y, and x y, what is the measure of the largest angle of the triangle? 14365C09.pgs 7/10/07 8:48 AM Page 376 376 Parallel Lines 18. If parallel lines are cut by a transversal so that the degree measures of two corresponding angles are represented by 2x 50 and 3x 20, what is the value of x? 19. The measure of one exterior angle of a regular polygon is 30°. How many sides does the regular polygon have? 20. What is the sum of the degree measures of the interior angles of a polygon with nine sides? 21. Given: Right triangle ABC with C the right angle. Prove: AB AC 22. Given: Prove: and AEB AC BD CED bisect each other at E. 23. P is not on ABCD APB DPC, prove that and PA PB PC PA > PD , and . , , PD are drawn. If PB PC and 24. P is not on ABCD and AB > DC , , and PA PB PA > PD , prove that , and . PC PD are drawn. If PBC PCB 25. Herbie wanted to draw pentagon ABCDE with mA = mB 120 and mC = mD = 150. Is such a pentagon possible? Explain your answer. Exploration The geometry that you have been studying is called plane Euclidean geometry. Investigate a non-Euclidean geometry. How do the postulates of a nonEuclidean geometry differ from the postulates of Euclid? How can the postulates from this chapter be rewritten to fit the non-Euclidean geometry you investigated? What theorems from this chapter are not valid in the nonEuclidean geometry that you investigated? One possible non-Euclidean geometry is the geometry of the sphere suggested in the Chapter 1 Exploration. CUMULATIVE REVIEW Chapters 1–9 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. , which of the following may be false? 1. If M is the midpoint of AB (1) M is between A and B. (2) AM = MB (3) A, B, and M are collinear. g MN (4) , a line that intersects AB at M, is the perpendicular bisector of .AB 14365C09.pgs 7/10/07 8:48 AM Page 377 Cumulative Review 377 2. The statement “If two angles form a linear pair, then they are supplemen- tary” is true. Which of the following statements must also be true? (1) If two angles do not form a linear pair, then they are not supplementary. (2) If two angles are not supplementary, then they do not form a linear pair. (3) If two angles are supplementary, then they form a linear pair. (4) Two angles form a linear pair if and only if they are supplementary. 3. Which of the following is a statement of the reflexive property of equality for all real numbers a, b, and c? (1) a = a (2) If a = b, then b = a. (3) If a = b and b = c, then a = c. (4) If a = b, then ac = bc. 4. Two angles are complementary. If the measure of the larger angle is 10 degrees less than three times the measure of the smaller, what is the measure of the larger angle? (1) 20° (2) 25° (3) 65° (4) 70° 5. Under the transformation + R908 , the image of (2, 5) is (1) (5, 2) (4) (2, 5) 6. An equation of the line through (0, 1) and perpendicular to the line (3) (5, 2) rx-axis (2) (5, 2) x 3y 4 is (1) 3x y 1 (2) x 3y 1 (3) 3x y 1 (4) x 3y 1 7. The coordinates of the midpoint of the line segment whose endpoints are (3, 4) and (5, 6) are (1) (1, 1) (2) (4, 5) (3) (4, 5) (4) (4, 5) 8. If a, b, c, and d are real numbers and a b and c d, which of the follow- ing must be true? (1) a c b d (2) a c b d (3) ac bc c . b a (4) d 9. The measure of each base angle of an isosceles triangle is 5 more than twice the measure of the vertex angle. The measure of the vertex angle is (1) 34° (4) 136.25° (3) 43.75° (2) 73° 10. Which of the following properties is not preserved under a line reflection? (1) distance (2) orientation (3) angle measure (4) midpoint 14365C09.pgs 7/10/07 8:48 AM Page 378 378 Parallel Lines Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. g AD . If mCAB 65, and B is a point that is not on mCBD = 20, and mBCD 135, which is the longest side of ABC? , prove that ABP is 12. If P is a point on the perpendicular bisector of 11. C is a point on AD AB isosceles. Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. ABCD is an equilateral quadrilateral. Prove that the diagonal, AC , bisects DAB and DCB. g AEB g CED g CB and g AD 14. intersect at E and Prove that mDEB mEBC mEDA. . A D E C B Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The measures of the angles of a triangle are in the ratio 3 : 4 : 8. Find the measure of the smallest exterior angle. 16. Write an equation of the perpendicular bisector of are A(1, 2) and B(7, 6). of the endpoints of AB AB if the coordinates 14365C10.pgs 7/10/07 8:50 AM Page 379 QUADRILATERALS Euclid’s fifth postulate was often considered to be a “flaw” in his development of geometry. Girolamo Saccheri (1667–1733) was convinced that by the application of rigorous logical reasoning, this postulate could be proved. He proceeded to develop a geometry based on an isosceles quadrilateral with two base angles that are right angles.This isosceles quadrilateral had been proposed by the Persian mathematician Nasir al-Din al-Tusi (1201–1274). Using this quadrilateral, Saccheri attempted to prove Euclid’s fifth postulate by reasoning to a contradiction. After his death, his work was published under the title Euclid Freed of Every Flaw. Saccheri did not, as he set out to do, prove the parallel postulate, but his work laid the foundations for new geometries. János Bolyai (1802–1860) and Nicolai Lobachevsky (1793–1856) developed a geometry that allowed two lines parallel to the given line, through a point not on a given line. Georg Riemann (1826– |
1866) developed a geometry in which there is no line parallel to a given line through a point not on the given line. CHAPTER 10 CHAPTER TABLE OF CONTENTS 10-1 The General Quadrilateral 10-2 The Parallelogram 10-3 Proving That a Quadrilateral Is a Parallelogram 10-4 The Rectangle 10-5 The Rhombus 10-6 The Square 10-7 The Trapezoid 10-8 Areas of Polygons Chapter Summary Vocabulary Review Exercises Cumulative Review 379 14365C10.pgs 7/10/07 8:50 AM Page 380 380 Quadrilaterals 10-1 THE GENERAL QUADRILATERAL Patchwork is an authentic American craft, developed by our frugal ancestors in a time when nothing was wasted and useful parts of discarded clothing were stitched into warm and decorative quilts. Quilt patterns, many of which acquired names as they were handed down from one generation to the next, were the product of creative and industrious people. In the Lone Star pattern, quadrilaterals are arranged to form larger quadrilaterals that form a star. The creators of this pattern were perhaps more aware of the pleasing effect of the design than of the mathematical relationships that were the basis of the pattern. A quadrilateral is a polygon with four sides. In this chapter we will study the various special quadrilaterals and the properties of each. Let us first name the general parts and state properties of any quadrilateral, using PQRS as an example. • Consecutive vertices or adjacent vertices are vertices that are endpoints of the same side such as P and Q, Q and R, R and S, S and P. Q P R S • Consecutive sides or adjacent sides are sides that have a common endand point, such as and and and , SP QR QR PQ PQ RS RS SP , . , • Opposite sides of a quadrilateral are sides that do not have a common endpoint, such as PQ and RS , SP and QR . • Consecutive angles of a quadrilateral are angles whose vertices are consec- utive, such as P and Q, Q and R, R and S, S and P. • Opposite angles of a quadrilateral are angles whose vertices are not con- secutive, such as P and R, Q and S. • A diagonal of a quadrilateral is a line segment whose endpoints are two nonadjacent vertices of the quadrilateral, such as PR and QS . • The sum of the measures of the angles of a quadrilateral is 360 degrees. Therefore, mP mQ mR mS 360. 10-2 THE PARALLELOGRAM DEFINITION A parallelogram is a quadrilateral in which two pairs of opposite sides are parallel. 14365C10.pgs 7/10/07 8:50 AM Page 381 D C Quadrilateral ABCD is a parallelogram because The symbol for parallelogram ABCD is ~ABCD . The Parallelogram 381 AB CD and BC DA . A B that are parallel in the figure. Note the use of arrowheads, pointing in the same direction, to show sides Theorem 10.1 A diagonal divides a parallelogram into two congruent triangles. Given Parallelogram ABCD with diagonal AC D C Prove ABC CDA A B Proof Since opposite sides of a parallelogram are parallel, alternate interior angles can be proved congruent using the diagonal as the transversal. Statements Reasons 1. ABCD is a parallelogram. 1. Given. 2. AB CD and BC DA 3. BAC DCA and BCA DAC AC > AC 4. 5. ABC CDA 2. A parallelogram is a quadrilateral in which two pairs of opposite sides are parallel. 3. If two parallel lines are cut by a transversal, alternate interior angles are congruent. 4. Reflexive property of congruence. 5. ASA. We have proved that the diagonal AC two congruent triangles. An identical proof could be used to show that divides the parallelogram into two congruent triangles, ABD CDB. divides parallelogram ABCD into BD The following corollaries result from this theorem. Corollary 10.1a Opposite sides of a parallelogram are congruent. Corollary 10.1b Opposite angles of a parallelogram are congruent. The proofs of these corollaries are left to the student. (See exercises 14 and 15.) 14365C10.pgs 7/10/07 8:50 AM Page 382 382 Quadrilaterals We can think of each side of a parallelogram as a segment of a transversal that intersects a pair of parallel lines. This enables us to prove the following theorem. Theorem 10.2 Two consecutive angles of a parallelogram are supplementary. Proof ~ABCD , opposite sides are parallel. If In two parallel lines are cut by a transversal, then two interior angles on the same side of the transversal are supplementary. Therefore, A is supplementary to B, B is supplementary to C, C is supplementary to D, and D is supplementary to A. D C A B Theorem 10.3 The diagonals of a parallelogram bisect each other. Given ~ABCD with diagonals AC and BD intersecting at E. D C Prove AC and BD bisect each other. E A B Proof Statements Reasons 1. AB CD D C 2. BAE DCE and ABE CDE B 3. AB > CD C 4. ABE CDE E E D A A 1. Opposite sides of a parallelogram are parallel. 2. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 3. Opposite sides of a parallelogram are congruent. 4. ASA. 5. AE > CE and BE > DE 5. Corresponding part of congruent triangles are congruent. B 6. E is the midpoint of AC and 6. The midpoint of a line segment of BD . divides the segment into two congruent segments. 7. AC and BD bisect each other. 7. The bisector of a line segment intersects the segment at its midpoint. 14365C10.pgs 7/10/07 8:50 AM Page 383 DEFINITION The distance between two parallel lines is the length of the perpendicular from any point on one line to the other line. The Parallelogram 383 Properties of a Parallelogram 1. Opposite sides are parallel. 2. A diagonal divides a parallelogram into two congruent triangles. 3. Opposite sides are congruent. 4. Opposite angles are congruent. 5. Consecutive angles are supplementary. 6. The diagonals bisect each other. EXAMPLE 1 In ~ABCD , mB exceeds mA by 46 degrees. Find mB. Solution Let x mA. Then x 46 mB. Two consecutive angles of a parallelogram are supplementary. Therefore, mA mB 180 x x 46 180 2x 46 180 2x 134 x 67 mB x 46 67 46 113 Answer mB 113 Exercises Writing About Mathematics 1. Theorem 10.2 states that two consecutive angles of a parallelogram are supplementary. If two opposite angles of a quadrilateral are supplementary, is the quadrilateral a parallelogram? Justify your answer. 2. A diagonal divides a parallelogram into two congruent triangles. Do two diagonals divide a parallelogram into four congruent triangles? Justify your answer. 14365C10.pgs 7/10/07 8:50 AM Page 384 384 Quadrilaterals Developing Skills 3. Find the degree measures of the other three angles of a parallelogram if one angle measures: a. 70 b. 65 c. 90 d. 130 e. 155 f. 168 In 4–11, ABCD is a parallelogram. 4. The degree measure of A is represented by 2x 20 and the degree measure of B by 2x. Find the value of x, of mA, and of mB. 5. The degree measure of A is represented by 2x 10 and the degree measure of B by 3x. Find the value of x, of mA, and of mB. D C A B 6. The measure of A is 30 degrees less than twice the measure of B. Find the measure of each angle of the parallelogram. 7. The measure of A is represented by x 44 and the measure of C by 3x. Find the mea- sure of each angle of the parallelogram. 8. The measure of B is represented by 7x and mD by 5x 30. Find the measure of each angle of the parallelogram. 9. The measure of C is one-half the measure of B. Find the measure of each angle of the parallelogram. 10. If AB 4x 7 and CD 3x 12, find AB and CD. 11. If AB 4x y, BC y 4, CD 3x 6, and DA 2x y, find the lengths of the sides of the parallelogram. 12. The diagonals of 13. The diagonals of ~ABCD ~ABCD intersect at E. If AE 5x 3 and EC 15 x, find AC. intersect at E. If DE 4y 1 and EB 5y 1, find DB. Applying Skills 14. Prove Corollary 10.1a, “Opposite sides of a parallelogram are congruent.” 15. Prove Corollary 10.1b, “Opposite angles of a parallelogram are congruent.” 16. Given: Parallelogram EBFD and parallelogram ABCD with D C F EAB and DCF Prove: EAD FCB AE B 17. Petrina said that the floor of her bedroom is in the shape of a parallelogram and that at least one of the angles is a right angle. Show that the floor of Petrina’s bedroom has four right angles. 18. The deck that Jeremiah is building is in the shape of a quadrilateral, ABCD. The measure of the angle at A is not equal to the measure of the angle at C. Prove that the deck is not in the shape of a parallelogram. 14365C10.pgs 7/10/07 8:50 AM Page 385 Proving That a Quadrilateral Is a Parallelogram 385 19. Quadrilaterals ABCD and PQRS are parallelograms with . Prove that ABCD PQRS or draw a counterexample to show that they may not be congruent. AB > PQ BC > QR and 20. Quadrilaterals ABCD and PQRS are parallelograms with AB > PQ , BC > QR , and B Q. Prove that ABCD PQRS or draw a counterexample to show that they may not be congruent. 10-3 PROVING THAT A QUADRILATERAL IS A PARALLELOGRAM If we wish to prove that a certain quadrilateral is a parallelogram, we can do so by proving its opposite sides are parallel, thus satisfying the definition of a parallelogram. Now we want to determine other ways of proving that a quadrilateral is a parallelogram. Theorem 10.4 If both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram. Given Quadrilateral ABCD with AB > CD , AD > BC D C Prove ABCD is a parallelogram. Proof AC is a diagonal. Triangles ABC and CDA In ABCD, are congruent by SSS. Corresponding parts of congruent triangles are congruent, so BAC DCA and DAC ACB. is a . Alternate interior angles BAC and DCA transversal that cuts DC is also a transversal that cuts are congruent, so AC Alternate interior angles DAC and ACB are congruent so Therefore, ABCD is a parallelogram. and AD AD BC AB AB DC and . BC . AC A B . Theorem 10.5 If one pair of opposite sides of a quadrilateral is both congruent and parallel, the quadrilateral is a parallelogram. Given Quadrilateral ABCD with AB CD and AB > CD D C Prove ABCD is a parallelogram. Proof Since AB is parallel to CD , BAC and DCA are a A B pair of congruent alternate interior angles. Therefore, by SAS, DCA BAC. Corresponding parts of congruent triangles are congruent, so DAC ACB. and congruent alternate interior |
angles DAC and ACB. Therefore, and ABCD is a parallelogram. is a transversal that cuts BC AD BC AD AC , , forming 14365C10.pgs 7/10/07 8:50 AM Page 386 386 Quadrilaterals Theorem 10.6 If both pairs of opposite angles of a quadrilateral are congruent, the quadrilateral is a parallelogram. Given Quadrilateral ABCD with A C and B D D C Prove ABCD is a parallelogram. Proof The sum of the measures of the angles of a quadrilateral B is 360 degrees. Therefore, mA mB mC mD 360. It is given that A C and B D. Congruent angles have equal measures so mA mC and mB mD. A By substitution, mA mD mA mD 360. Then, 2mA 2mD 360 or mA mD 180. Similarly, mA mB 180. If the sum of the measures of two angles is 180, the angles are supplementary. Therefore, A and D are supplementary and A and B are supplementary. Two coplanar lines are parallel if a pair of interior angles on the same side of the transversal are supplementary. Therefore, Quadrilateral ABCD is a parallelogram because it has two pairs of parallel sides. and . AD BC AB DC Theorem 10.7 If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram. Given Quadrilateral ABCD with AE > EC , BE > ED Prove ABCD is a parallelogram. AC and BD intersecting at E, D C E A B Strategy Prove that ABE CDE to show that one pair of opposite sides is congruent and parallel. The proof of Theorem 10.7 is left to the student. (See exercise 15.) SUMMARY To prove that a quadrilateral is a parallelogram, prove that any one of the following statements is true: 1. Both pairs of opposite sides are parallel. 2. Both pairs of opposite sides are congruent. 3. One pair of opposite sides is both congruent and parallel. 4. Both pairs of opposite angles are congruent. 5. The diagonals bisect each other. 14365C10.pgs 7/10/07 8:50 AM Page 387 EXAMPLE 1 Proving That a Quadrilateral Is a Parallelogram 387 Given: ABCD is a parallelogram. D F C E is on AB , F is on DC , and EB > DF . Prove: DE FB Proof We will prove that EBFD is a parallelogram. A E B Statements Reasons 1. ABCD is a parallelogram. 1. Given. 2. AB DC 3. EB DF 4. EB > DF 5. EBFD is a parallelogram. 6. DE FB 2. Opposite sides of a parallelogram are parallel. 3. Segments of parallel lines are parallel. 4. Given. 5. If one pair of opposite sides of a quadrilateral is both congruent and parallel, the quadrilateral is a parallelogram. 6. Opposite sides of a parallelogram are parallel. Exercises Writing About Mathematics 1. What statement and reason can be added to the proof in Example 1 to prove that DE > FB ? 2. What statement and reason can be added to the proof in Example 1 to prove that DEB BFD? Developing Skills In 3–7, in each case, the given is marked on the figure. Tell why each quadrilateral ABCD is a parallelogram. 3. D C 4. D C 5. D C 6. D C 7 14365C10.pgs 7/10/07 8:50 AM Page 388 388 Quadrilaterals 8. ABCD is a quadrilateral with AB CD and A C. Prove that ABCD is a parallelo- gram. 9. PQRS is a quadrilateral with P R and P the supplement of Q. Prove that PQRS is a parallelogram. 10. DEFG is a quadrilateral with DF drawn so that FDE DFG and GDF EFD. Prove that DEFG is a parallelogram. 11. ABCD is a parallelogram. E is the midpoint of AB and F is the midpoint of CD . Prove that AEFD is a parallelogram. 12. EFGH is a parallelogram and J is a point on h EF such that F is the midpoint of EJ . Prove that FJGH is a parallelogram. CD 13. ABCD is a parallelogram. The midpoint of is R, and the midpoint of of a. Prove that APS CRQ and that BQP DSR. b. Prove that PQRS is a parallelogram. is S. DA AB is P, the midpoint of BC is Q, the midpoint 14. A quadrilateral has three right angles. Is the quadrilateral a parallelogram? Justify your answer. Applying Skills 15. Prove Theorem 10.7, “If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.” 16. Prove that a parallelogram can be drawn by joining the endpoints of two line segments that bisect each other. 17. The vertices of quadrilateral ABCD are A(2, 1), B(4, 2), C(8, 2), and D(2, 5). Is ABCD a parallelogram? Justify your answer. 18. Farmer Brown’s pasture is in the shape of a quadrilateral, PQRS. The pasture is crossed by two diagonal paths, QS paths do not bisect each other. and PR . The quadrilateral is not a parallelogram. Show that the 19. Toni cut two congruent scalene triangles out of cardboard. She labeled one triangle ABC and the other ABC so that ABC ABC. She placed the two triangles next to each other so that A coincided with B and B coincided with A. Was the resulting quadrilateral ACBC a parallelogram? Prove your answer. 20. Quadrilateral ABCD is a parallelogram with M the midpoint of AB and N the midpoint of CD . a. Prove that AMND and MBCN are parallelograms. b. Prove that AMND and MBCN are congruent. 14365C10.pgs 7/10/07 8:50 AM Page 389 10-4 THE RECTANGLE The Rectangle 389 DEFINITION A rectangle is a parallelogram containing one right angle. If one angle, A, of a parallelogram ABCD is a right angle, then ~ABCD is a rectangle. Any side of a rectangle may be called the base of the rectangle. Thus, if side , is called the is taken as the base, then either consecutive side, AD BC or AB altitude of the rectangle. Since a rectangle is a special kind of parallelogram, a rectangle has all the properties of a parallelogram. In addition, we can prove two special properties for the rectangle. Theorem 10.8 All angles of a rectangle are right angles. Given Rectangle ABCD with A a right angle. Prove B, C, and D are right angles. D A C B Proof By definition, rectangle ABCD is a parallelogram. Opposite angles of a parallelogram are congruent, so A C. Angle A is a right angle so C is a right angle. Consecutive angles of a parallelogram are supplementary. Therefore, since A and B are supplementary and A is right angle, B is also a right angle. Similarly, since C and D are supplementary and C is a right angle, D is also a right angle. Theorem 10.9 The diagonals of a rectangle are congruent. Given ABCD is a rectangle. Prove AC > BD Strategy Prove DAB CBA. D A C B The proof of Theorem 10.9 is left to the student. (See exercise 12.) 14365C10.pgs 7/10/07 8:50 AM Page 390 390 Quadrilaterals Properties of a Rectangle 1. A rectangle has all the properties of a parallelogram. 2. A rectangle has four right angles and is therefore equiangular. 3. The diagonals of a rectangle are congruent. Proving That a Quadrilateral Is a Rectangle We prove that a quadrilateral is a rectangle by showing that it has the special properties of a rectangle. For example: Theorem 10.10 If a quadrilateral is equiangular, then it is a rectangle. Given Quadrilateral ABCD with A B C D. D C Prove ABCD is a rectangle. Proof Quadrilateral ABCD is a parallelogram because the opposite angles are congruent. The sum of the measures of the angles of a quadrilateral is 360 degrees. Thus, the measure of each of the four congruent angles is 90 degrees. Therefore, ABCD is a rectangle because it is a parallelogram with a right angle. A B Theorem 10.11 If the diagonals of a parallelogram are congruent, the parallelogram is a rectangle. Given Parallelogram ABCD with AC > BD Prove ABCD is a rectangle. Strategy Prove that DAB CBA. Therefore, DAB and CBA are both congruent and supplementary. D A C B The proof of Theorem 10.11 is left to the student. (See exercise 13.) SUMMARY To prove that a quadrilateral is a rectangle, prove that any one of the following statements is true: 1. The quadrilateral is a parallelogram with one right angle. 2. The quadrilateral is equiangular. 3. The quadrilateral is a parallelogram whose diagonals are congruent. 14365C10.pgs 7/10/07 8:50 AM Page 391 EXAMPLE 1 Given: ABCD is a parallelogram with mA mD. Prove: ABCD is a rectangle. Proof We will prove that ABCD has a right angle. The Rectangle 391 D A C B Statements Reasons 1. ABCD is a parallelogram. 1. Given. 2. AB CD 3. A and D are supplementary. 4. mA mD 180 5. mA mD 6. mA mA 180 or 2mA 180 7. mA 90 2. A parallelogram is a quadrilateral in which two pairs of opposite sides are parallel. 3. Two consecutive angles of a parallelogram are supplementary. 4. Supplementary angles are two angles the sum of whose measures is 180. 5. Given. 6. A quantity may be substituted for its equal. 7. Division postulate. 8. ABCD is a rectangle. 8. A rectangle is a parallelogram with a right angle. EXAMPLE 2 The lengths of diagonals of a rectangle are represented by 7x centimeters and 5x 12 centimeters. Find the length of each diagonal. Solution The diagonals of a rectangle are congruent and therefore equal in length. 7x 5x 12 2x 12 x 6 7x 5 7(6) 5x 1 12 5 5(6) 1 12 5 42 5 30 1 12 5 42 Answer The length of each diagonal is 42 centimeters. 14365C10.pgs 7/10/07 8:50 AM Page 392 392 Quadrilaterals Exercises Writing About Mathematics 1. Pauli said that if one angle of a parallelogram is not a right angle, then the parallelogram is not a rectangle. Do you agree with Pauli? Explain why or why not. 2. Cindy said that if two congruent line segments intersect at their midpoints, then the quadrilateral formed by joining the endpoints of the line segments in order is a rectangle. Do you agree with Cindy? Explain why or why not. Developing Skills In 3–10, the diagonals of rectangle ABCD intersect at E. 3. Prove that ABE is isosceles. 4. AC 4x 6 and BD 5x 2. Find AC, BD, AE, and BE. 5. AE y 12 and DE 3y 8. Find AE, DE, AC, and BD. 6. BE 3a 1 and ED 6a 11. Find BE, ED, BD, and AC. 7. AE x 5 and BD 3x 2. Find AE, BD, AC, and BE. 8. mCAB 35. Find mCAD, mACB, mAEB, and AED. 9. mAEB 3x and mDEC x 80. Find mAEB, mDEC, mCAB, and mCAD. 10. mAED y 10 and mAEB 4y 30. Find mAED, mAEB, mBAC, and D A E mCAD. C B Applying Skills 11. Write a coordinate proof of Theorem 10.8, “All angles of a rectangle are right angles.” Let the vertices of the rectangle be A(0, 0), B(b, 0), C(b, c), and D(0, c). 12. Prove Theorem 10.9, “The diagonals of a rectangle are congruent.” 13. Prove Theorem 10.11, “If the diagonals of a parallelogram are congruent, the parallelogram is |
a rectangle.” 14. If PQRS is a rectangle and M is the midpoint of 15. The coordinates of the vertices of ABCD are A(2, 0), B(2, 2), C(5, 4), and D(1, 6). , prove that PM > QM RS . a. Prove that ABCD is a rectangle. b. What are the coordinates of the point of intersection of the diagonals? c. The vertices of ABCD are A(0, 2), B(2, 2), C(4, 5), and D(6, 1). Under what specific transformation is ABCD the image of ABCD? d. Prove that ABCD is a rectangle. 14365C10.pgs 7/10/07 8:50 AM Page 393 The Rhombus 393 16. The coordinates of the vertices of PQRS are P(2, 1), Q(1, 3), R(5, 1), and S(2, 5). a. Prove that PQRS is a parallelogram. b. Prove that PQRS is not a rectangle. c. PQRS is the image of PQRS under T–3,–3 PQRS? d. Prove that PQRS is congruent to PQRS. + ry = x. What are the coordinates of 17. Angle A in quadrilateral ABCD is a right angle and quadrilateral ABCD is not a rectangle. Prove that ABCD is not a parallelogram. 18. Tracy wants to build a rectangular pen for her dog. She has a tape measure, which enables her to make accurate measurements of distance, but has no way of measuring angles. She places two stakes in the ground to represent opposite corners of the pen. How can she find two points at which to place stakes for the other two corners? 19. Archie has a piece of cardboard from which he wants to cut a rectangle with a diagonal that measures 12 inches. On one edge of the cardboard Archie marks two points that are less than 12 inches apart to be the endpoints of one side of the rectangle. Explain how Archie can use two pieces of string that are each 12 inches long to find the other two vertices of the rectangle. 10-5 THE RHOMBUS DEFINITION A rhombus is a parallelogram that has two congruent consecutive sides. If the consecutive sides AB > AD ), then (that is, if and AB ~ABCD is a rhombus. AD of parallelogram ABCD are congruent Since a rhombus is a special kind of parallelogram, a rhombus has all the properties of a parallelogram. In addition, we can prove three special properties for the rhombus. Theorem 10.12 All sides of a rhombus are congruent. Given ABCD is a rhombus with AB > DA . D C Prove AB > BC > CD > DA Proof By definition, rhombus ABCD is a parallelogram. It is AB > DA given that and are congruent, so transitive property of congruence, AB > CD . Opposite sides of a parallelogram BC > DA . Using the AB > BC > CD > DA . A B 14365C10.pgs 7/10/07 8:50 AM Page 394 394 Quadrilaterals Theorem 10.13 The diagonals of a rhombus are perpendicular to each other. Given Rhombus ABCD Prove AC ' BD D C Proof By Theorem 10.12, all sides of a rhombus are congruent. AC . Point B is equidistant from the endpoints Segments that are congruent are equal. Consider the diagonal A and C since BA BC. Point D is also equidistant from the endpoints A and C since DA DC. If two points are each equidistant from the endpoints of a line segment, the points determine the perpendicular bisector of the line segment. Therefore, AC ' BD A B . Theorem 10.14 The diagonals of a rhombus bisect its angles. Given Rhombus ABCD Prove bisects DAB and DCB and AC and CBA. DB bisects CDA Strategy Show that the diagonals separate the rhombus into four A congruent triangles. D C E B The proof of this theorem is left to the student. (See exercise 16.) Properties of a Rhombus 1. A rhombus has all the properties of a parallelogram. 2. A rhombus is equilateral. 3. The diagonals of a rhombus are perpendicular to each other. 4. The diagonals of a rhombus bisect its angles. Methods of Proving That a Quadrilateral Is a Rhombus We prove that a quadrilateral is a rhombus by showing that it has the special properties of a rhombus. 14365C10.pgs 7/10/07 8:50 AM Page 395 Theorem 10.15 If a quadrilateral is equilateral, then it is a rhombus. The Rhombus 395 Given Quadrilateral ABCD with AB > BC > CD > DA D C Prove ABCD is a rhombus. Proof AB > BC > CD > DA It is given that in ABCD, . Since both pairs of opposite sides are congruent, ABCD is a parallelogram. Two consecutive sides of parallelogram ABCD are congruent, so by definition, ABCD is a rhombus. A B Theorem 10.16 If the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus. Given Parallelogram ABCD with AC ' BD D C Prove ABCD is a rhombus. Strategy The diagonals divide parallelogram ABCD into four triangles. Prove that two of the triangles that share a common side are congruent. Then use the fact that corresponding parts of congruent triangles are congruent to show that parallelogram ABCD has two congruent consecutive sides. E A B The proof of this theorem is left to the student. (See exercise 17.) SUMMARY To prove that a quadrilateral is a rhombus, prove that any one of the following statements is true: 1. The quadrilateral is a parallelogram with two congruent consecutive sides. 2. The quadrilateral is equilateral. 3. The quadrilateral is a parallelogram whose diagonals are perpendicular to each other. EXAMPLE 1 PQRS is a rhombus and mPQR 60. Prove that divides the rhombus into two equilateral triangles. PR P Q 60° S R 14365C10.pgs 7/10/07 8:50 AM Page 396 396 Quadrilaterals Proof Since all sides of a rhombus are congruent, we know that PQ > RQ . Thus, PQR is isosceles and its base angles are equal in measure. Let mPRQ mRPQ x. x x 60 180 2x 60 180 2x 120 x 60 Therefore, mPRQ 60, mRPQ 60, and mPQR 60. Triangle PQR is equilateral since an equiangular triangle is equilateral. Since opposite angles of a rhombus have equal measures, mRSP 60. By similar reasoning, RSP is equilateral. P Q 60° S R EXAMPLE 2 Given: ABCD is a parallelogram. D C AB 2x 1, DC 3x 11, AD x 13. Prove: ABCD is a rhombus. Proof (1) Since ABCD is a parallelogram, opposite sides are equal in length: (2) Substitute x 12 to find the : AD lengths of sides and AB A B DC AB 3x 11 2x 1 x 12 AB 2x 1 2(12) 1 25 AD x 13 12 13 25 (3) Since ABCD is a parallelogram with two congruent consecutive sides, ABCD is a rhombus. Exercises Writing About Mathematics 1. Rochelle said that the diagonals of a rhombus separate the rhombus into four congruent right triangles. Do you agree with Rochelle? Explain why or why not. 14365C10.pgs 7/10/07 8:50 AM Page 397 The Rhombus 397 2. Concepta said that if the lengths of the diagonals of a rhombus are represented by d1 and . Do you agree with Concepta? d2, then a formula for the area of a rhombus is A Explain why or why not. 1 2d1d2 Developing Skills In 3–10, the diagonals of rhombus ABCD intersect at E. 3. Name four congruent line segments. 4. Name two pairs of congruent line segments. D C 5. Name a pair of perpendicular line segments. E 6. Name four right angles. 7. Under a rotation of 90o about E, does A map to B? does B map to C? Justify your answer. A B 8. Does rhombus ABCD have rotational symmetry under a rotation of 90o about E? 9. Under a reflection in E, name the image of A, of B, of C, of D, and of E. 10. Does rhombus ABCD have point symmetry under a reflection in E? 11. In rhombus PQRS, mP is 120. a. Prove that the diagonal b. If PR 24 cm, what is the length of each side of the rhombus? separates the rhombus into two equilateral triangles. PR In 12–15, tell whether each conclusion follows from the given premises. If not, draw a counterexample. 12. In a parallelogram, opposite sides are congruent. A rhombus is a parallelogram. Conclusion: In a rhombus, opposite sides are congruent. 13. In a rhombus, diagonals are perpendicular to each other. A rhombus is a parallelogram. Conclusion: In a parallelogram, diagonals are perpendicular to each other. 14. The diagonals of a rhombus bisect the angles of the rhombus. A rhombus is a parallelogram. Conclusion: The diagonals of a parallelogram bisect its angles. 15. Consecutive angles of a parallelogram are supplementary. A rhombus is a parallelogram. Conclusion: Consecutive angles of a rhombus are supplementary. Applying Skills 16. Prove Theorem 10.14, “The diagonals of a rhombus bisect its angles.” 17. Prove Theorem 10.16, “If the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus.” 14365C10.pgs 7/10/07 8:50 AM Page 398 398 Quadrilaterals 18. The vertices of quadrilateral ABCD are A(1, 1), B(4, 0), C(5, 5), and D(0, 4). a. Prove that ABCD is a parallelogram. b. Prove that the diagonals of ABCD are perpendicular. c. Is ABCD a rhombus? Justify your answer. 19. If a diagonal separates a quadrilateral KLMN into two equilateral triangles, prove that KLMN is a rhombus with the measure of one angle equal to 60 degrees. 20. Prove that the diagonals of a rhombus separate the rhombus into four congruent right triangles. 21. Prove that if the diagonals of a quadrilateral are the perpendicular bisectors of each other, the quadrilateral is a rhombus. 22. Prove that if the diagonals of a parallelogram bisect the angles of the parallelogram, then the parallelogram is a rhombus. 23. Let P be any point on diagonal BD of rhombus ABCD. Prove that AP > CP . 24. The vertices of quadrilateral ABCD are A(2, 1), B(2, 4), C(2, 1), D(2, 4). a. Prove that ABCD is a parallelogram. b. Find the coordinates of E, the point of intersection of the diagonals. C PD A B c. Prove that the diagonals are perpendicular. d. Is ABCD a rhombus? Justify your answer. 25. ABCD is a parallelogram. The midpoint of AB is M, the midpoint of CD is N, and AM AD. a. Prove that AMND is a rhombus. b. Prove that MBCN is a rhombus. c. Prove that AMND is congruent to MBCN. Hands-On Activity In this activity, you will construct a rhombus given the diagonal. You may use geometry software or compass and straightedge. 1. First draw a segment measuring 12 centimeters. This will be the diagonal of the rhombus. The endpoints of this segment are opposite vertices of the rhombus. 2. Now construct the perpendicular bisector of this segment. 3. Show that you can choose any point on that perpendicular bisector as a third vertex of the rhombus. 4. How can you determine the fourth vertex of the rhombus? 5. Compare the rhombus you constructed with rhom |
buses constructed by your classmates. How are they alike? How are they different? 14365C10.pgs 7/10/07 8:50 AM Page 399 10-6 THE SQUARE DEFINITION A square is a rectangle that has two congruent consecutive sides. The Square 399 If consecutive sides and ), then rectangle ABCD is a square. AD AB if AB > AD of rectangle ABCD are congruent (that is, Theorem 10.17 A square is an equilateral quadrilateral. Given ABCD is a square with AB > BC . Prove AB > BC > CD > DA Proof A square is a rectangle and a rectangle is a parallelo- gram, so ABCD is a parallelogram. It is given that AB > BC gruent, so tive property of congruence, . Opposite sides of a parallelogram are conAB > CD BC > DA . Using the transiAB > BC > CD > DA . and D A C B Theorem 10.18 A square is a rhombus. Given Square ABCD Prove ABCD is a rhombus. Proof A square is an equilateral quadrilateral. If a quadrilateral is equilateral, then it is a rhombus. Therefore, ABCD is a rhombus. Properties of a Square 1. A square has all the properties of a rectangle. 2. A square has all the properties of a rhombus. Methods of Proving That a Quadrilateral Is a Square We prove that a quadrilateral is a square by showing that it has the special properties of a square. 14365C10.pgs 7/10/07 8:50 AM Page 400 400 Quadrilaterals Theorem 10.19 If one of the angles of a rhombus is a right angle, then the rhombus is a square. Given ABCD is a rhombus with A a right angle. Prove ABCD is a square. Strategy Show that ABCD is a rectangle and that AB > BC . The proof of this theorem is left to the student. (See exercise 12.) SUMMARY To prove that a quadrilateral is a square, prove either of the following statements: 1. The quadrilateral is a rectangle in which two consecutive sides are congruent. 2. The quadrilateral is a rhombus one of whose angles is a right angle. EXAMPLE 1 Given: Quadrilateral ABCD is equilateral and ABC is a D right angle. Prove: ABCD is a square. Proof Statements A Reasons C B 1. ABCD is equilateral. 1. Given 2. ABCD is a rhombus. 2. If a quadrilateral is equilateral, then it is a rhombus. 3. ABC is a right angle. 3. Given. 4. ABCD is a square. 4. If one of the angles of a rhombus is a right angle, then the rhombus is a square. 14365C10.pgs 7/10/07 8:50 AM Page 401 EXAMPLE 2 In square PQRS, Find mPSQ. SQ is a diagonal. Solution A square is a rhombus, and the diagonal of a rhom- bus bisects its angles. Therefore, the diagonal bisects PSR. Since a square is a rectangle, PSR is a right angle and mPSR 90. Therefore, mPSQ 45. SQ 1 2(90) The Square 401 S R P Q Exercises Writing About Mathematics 1. Ava said that the diagonals of a square separate the square into four congruent isosceles right triangles. Do you agree with Ava? Justify your answer. 2. Raphael said that a square could be defined as a quadrilateral that is both equiangular and equilateral. Do you agree with Raphael? Justify your answer. Developing Skills In 3–6, the diagonals of square ABCD intersect at M. AC is the perpendicular bisector of 3. Prove that 4. If AC 3x 2 and BD 7x 10, find AC, BD, AM, BM. 5. If AB a b, BC 2a, and CD 3b 5, find AB, BC, CD, . BD C D M and DA. 6. If mAMD x 2y and mABC 2x y, find the values of x and y. B A In 7–11, tell whether each conclusion follows from the given premises. If not, draw a counterexample. 7. If a quadrilateral is a square, then all sides are congruent. If all sides of a quadrilateral are congruent then it is a rhombus. Conclusion: If a quadrilateral is a square, then it is a rhombus. 8. A diagonal of a parallelogram separates the parallelogram into two congruent triangles. A square is a parallelogram. Conclusion: A diagonal of a square separates the square into two congruent triangles. 9. If a quadrilateral is a square, then all angles are right angles. If a quadrilateral is a square, then it is a rhombus. Conclusion: In a rhombus, all angles are right angles. 14365C10.pgs 7/10/07 8:50 AM Page 402 402 Quadrilaterals 10. If a quadrilateral is a square, then it is a rectangle. If a quadrilateral is a rectangle, then it is a parallelogram. Conclusion: If a quadrilateral is a square, then it is a parallelogram. 11. If a quadrilateral is a square, then it is equilateral. If a quadrilateral is a square, then it is a rectangle. Conclusion: If a quadrilateral is equilateral, then it is a rectangle. Applying Skills 12. Prove Theorem 10.19, “If one of the angles of a rhombus is a right angle, then the rhombus is a square.” 13. Prove that the diagonals of a square are perpendicular to each other. 14. Prove that the diagonals of a square divide the square into four congruent isosceles right triangles. 15. Two line segments, AEC and BED , are congruent. Each is the perpendicular bisector of the other. Prove that ABCD is a square. 16. Prove that if the midpoints of the sides of a square are joined in order, another square is formed. 17. The vertices of quadrilateral PQRS are P(1, 1), Q(4, 2), R(7, 1), and S(4, 4). a. Prove that the diagonals of the quadrilateral bisect each other. b. Prove that the diagonals of the quadrilateral are perpendicular to each other. c. Is the quadrilateral a square? Justify your answer. d. The vertices of PQRS are P(1, 1), Q(2, 4), R(1, 7), and S(4, 4). Under what specific transformation is PQRS the image of PQRS? 18. The vertices of quadrilateral ABCD are A(3, 2), B(1, 2), C(5, 2), and D(1, 6). a. Prove that the diagonals of the quadrilateral bisect each other. b. Prove that the diagonals of the quadrilateral are perpendicular to each other. c. Is the quadrilateral a square? Justify your answer. d. The vertices of ABCD are A(6, 3), B(2, 1), C(2, 3), and D(2, 7). Under what specific transformation is ABCD the image of ABCD? 10-7 THE TRAPEZOID DEFINITION A trapezoid is a quadrilateral in which two and only two sides are parallel. D C A B If AB DC and AD trapezoid. The parallel sides, the nonparallel sides, AD is not parallel to and , are called the legs of the trapezoid. , then quadrilateral ABCD is a , are called the bases of the trapezoid; and DC AB BC BC 14365C10.pgs 7/10/07 8:50 AM Page 403 The Trapezoid 403 The Isosceles Trapezoid and Its Properties DEFINITION An isosceles trapezoid is a trapezoid in which the nonparallel sides are congruent. T S Q R TS QR If and QT > RS , then QRST is an isosceles trapezoid. The angles whose vertices are the endpoints of a base are called base angles. Here, Q and R are one pair of base angles because Q and R are endpoints of base . Also, T and S are a second pair of base angles because T and S are endpoints of base QR TS . Proving That a Quadrilateral Is an Isosceles Trapezoid We prove that a quadrilateral is an isosceles trapezoid by showing that it satisfies the conditions of the definition of an isosceles trapezoid: only two sides are parallel and the nonparallel sides are congruent. We may also prove special theorems for an isosceles trapezoid. Theorem 10.20a If a trapezoid is isosceles, then the base angles are congruent. Given Isosceles trapezoid QRST with QR ST and TQ > SR Prove Q R T S Q P R Proof Statements Reasons T S 1. Through S, draw a line parallel that intersects at P: QR to QT SP QT QR ST Q P 2. R 3. QPST is a parallelogram. T S 4. QT > SP Q P R 5. QT > SR 6. SP > SR 1. Through a given point, only one line can be drawn parallel to a given line. 2. Given. 3. A parallelogram is a quadrilateral with two pairs of parallel sides. 4. Opposite sides of a parallelogram are congruent. 5. Given. 6. Transitive property of congruence. 14365C10.pgs 7/10/07 8:50 AM Page 404 404 Quadrilaterals (Continued) Statements 7. SPR R T S 8. Q SPR Q P R Reasons 7. If two sides of a triangle are congruent, the angles opposite these sides are congruent. 8. If two parallel lines are cut by a transversal, the corresponding angles are congruent. 9. Q R 9. Transitive property of congruence. We have proved Theorem 10.20a for Q R but S and T are also congruent base angles. We often refer to Q and R as the lower base angles and S and T as the upper base angles. The proof of this theorem for S and T is left to the student. (See exercise 15.) Theorem 10.20b If the base angles of a trapezoid are congruent, then the trapezoid is isosceles. Given Trapezoid QRST with QR ST and Q R T S Prove QT > RS Strategy Draw SP TQ . Prove SPR R. Then use the Q P R converse of the isosceles triangle theorem. The proof of this theorem is left to the student. (See exercise 16.) Theorems 10.20a and 10.20b can be written as a biconditional. Theorem 10.20 A trapezoid is isosceles if and only if the base angles are congruent. We can also prove theorems about the diagonals of an isosceles trapezoid. Theorem 10.21a If a trapezoid is isosceles, then the diagonals are congruent. Given Isosceles trapezoid ABCD with AB CD and D C AD > BC Prove AC > BD A B 14365C10.pgs 7/10/07 8:50 AM Page 405 The Trapezoid 405 Proof We will show DAB CBA. It is given that in trapezoid ABCD, . It is given that ABCD is an isosceles trapezoid. In an isosceles trapezoid, base angles are congruent, so DAB CBA. By the reflexive property, AB > AB . Therefore, DAB CBA by SAS. Corresponding parts of congruent triangles are congruent, so AD > BC AC > BD . Theorem 10.21b If the diagonals of a trapezoid are congruent, then the trapezoid is isosceles. Given Trapezoid ABCD with AB CD and AC > BD D C Prove AD > BC Strategy Draw and DE ' AB CF ' AB DEB and CFA are congruent by HL. Therefore, CAB DBA. Now, prove that ACB BDA by SAS. Then are congruent corresponding parts of congruent triangles. BC . First prove that A E F B AD and The proof of this theorem is left to the student. (See exercise 17.) Theorems 10.21a and 10.21b can also be written as a biconditional. Theorem 10.21 A trapezoid is isosceles if and only if the diagonals are congruent. Recall that the median of a triangle is a line segment from a vertex to the midpoint of the opposite sides. A triangle has three medians. A trapezoid has only one median, and it joins two midpoints. D C median A B DEFINITION The median of a trapezo |
id is a line segment whose endpoints are the midpoints of the nonparallel sides of the trapezoid. We can prove two theorems about the median of a trapezoid. Theorem 10.22 The median of a trapezoid is parallel to the bases. Given Trapezoid ABCD with midpoint of AD AB CD , M the , and N the midpoint of Prove MN AB and MN CD BC D C M A N B 14365C10.pgs 7/10/07 8:50 AM Page 406 406 Quadrilaterals Proof We will use a coordinate proof. Place the trapezoid so that the parallel sides are on horizontal lines. Place on the x-axis with (0, 0) the coordinates of A and (b, 0) the coordinates of y D(d, e) M C(c, e) N A(0, 0) B(b, 0) x AB B, and b 0. Place CD on a line parallel to the x-axis. Every point on a line parallel to the x-axis has the same y-coordinate. Let (c, e) be the coordinates of C and (d, e) be the coordinates of D, and c, d, e 0. Since M is the midpoint of AD 0 1 d 2 , the coordinates of M are , 0 1 e 2 Since N is the midpoint of BC , 0 1 e 2 , the coordinates of N are Points that have the same y-coordinate are on the same horizontal line. is a horizontal line. All horizontal lines are parallel. Therefore, MN CD MN and Therefore, MN AB . Theorem 10.23 The length of the median of a trapezoid is equal to one-half the sum of the lengths of the bases. Given Trapezoid ABCD with AB CD , M the , and N the midpoint of midpoint of CD AD Prove MN 1 2(AB 1 CD) y D(d, e) C(c, e) M N A(0, 0) B(b, 0) x Strategy Use a coordinate proof. Let the coordinates of A, B, C, D, M, and N be those used in the proof of Theorem 10.22. The length of a horizontal line segment is the absolute value of the difference of the x-coordinates of the endpoints. The proof of this theorem is left to the student. (See exercise 19.) EXAMPLE 1 Proof The coordinates of the vertices of ABCD are A(0, 0), B(4, 1), C(5, 2), and D(2, 6). Prove that ABCD is a trapezoid. Slope of AB Slope of BC Slope of CD Slope of DA 4 5 21 4 21 2 0 4 2 0 5 21 2 2 (21 26 1 5 3 23 5 24 3 22 5 3 0 2 6 14365C10.pgs 7/10/07 8:50 AM Page 407 The Trapezoid 407 The slopes of AB BC and and CD are equal. Therefore, DA are not equal. Therefore, The slopes of lel. Because quadrilateral ABCD has only one pair of parallel sides, it is a trapezoid. and and DA CD BC AB are parallel. are not paral- EXAMPLE 2 In quadrilateral ABCD, mA 105, mB 75, and mC 75. a. Is ABCD a parallelogram? Justify your answer. b. Is ABCD a trapezoid? Justify your answer. c. Is ABCD an isosceles trapezoid? Justify your answer. d. If AB x, BC 2x 1, CD 3x 8, and DA x 1, find the measure of each side of the quadrilateral. Solution a. One pair of opposite angles of ABCD are A and C. Since these angles are not congruent, the quadrilateral is not a parallelogram. The quadrilateral does not have two pairs of parallel sides. C 75° D 105° A 75° B b. A and B are interior angles on the same side of transversal AB and they are supplementary. Therefore, eral has only one pair of parallel sides and is therefore a trapezoid. c. Because B and C are congruent base angles of the trapezoid, the and AD BC are parallel. The quadrilat- trapezoid is isosceles. d. The congruent legs of the trapezoid are AB CD AB and . CD x 3x 8 2x 8 x 4 AB x 4 BC 2x 1 CD 3x 8 2(4) 1 8 1 7 3(4) 8 12 8 4 DA x 1 4 1 5 14541C10.pgs 1/25/08 3:47 PM Page 408 408 Quadrilaterals Exercises Writing About Mathematics 1. Can a trapezoid have exactly one right angle? Justify your answer. 2. Can a trapezoid have three obtuse angles? Justify your answer. Developing Skills In 3–8, ABCD is an isosceles trapezoid, AB DC , and AD > BC . 3. If mADC 110, find: a. mBCD b. mABC c. mDAB. 4. If AD 3x 7 and BC 25, find the value of x. 5. If AD 2y 5 and BC y 3, find AD. 6. If mDAB 4x 5 and mABC 3x 15, find the measure of each angle of the trapezoid. 7. If mADC 4x 20 and mDAB 8x 20, find the measure of each angle of the trapezoid. 8. The perimeter of ABCD is 55 centimeters. If AD DC BC and AB 2AD, find the measure of each side of the trapezoid. A B D C In 9–14, determine whether each statement is true or false. Justify your answer with an appropriate definition or theorem, or draw a counterexample. 9. In an isosceles trapezoid, nonparallel sides are congruent. 10. In a trapezoid, at most two sides can be congruent. 11. In a trapezoid, the base angles are always congruent. 12. The diagonals of a trapezoid are congruent if and only if the nonparallel sides of the trape- zoid are congruent. 13. The sum of the measures of the angles of a trapezoid is 360°. 14. In a trapezoid, there are always two pairs of supplementary angles. Applying Skills 15. In Theorem 10.20a, we proved that the lower base angles of QRST, Q and R, are con- gruent. Use this fact to prove that the upper base angles of QRST, S and T, are congruent. 16. Prove Theorem 10.20b, “If the base angles of a trapezoid are congruent, then the trapezoid is isosceles.” 17. a. Prove Theorem 10.21b, “If the diagonals of a trapezoid are congruent, then the trapezoid is isosceles.” b. Why can’t Theorem 10.21b be proved using the same method as in 10.21a? 14365C10.pgs 7/10/07 8:50 AM Page 409 18. Prove Theorem 10.23, “The length of the median of a trapezoid is equal to one-half the sum of the lengths of the bases.” Areas of Polygons 409 19. Let the coordinates of the vertices of ABCD be A(2, 6), B(6, 2), C(0, 8), and D(2, 4). . DA a. Find the slopes of , and CD AB BC , , b. Prove that ABCD is a trapezoid. c. Find the coordinates of E and F, the midpoints of the nonparallel sides. d. Find the slope of EF . e. Show that the median is parallel to the bases. 20. Prove that the diagonals of a trapezoid do not bisect each other. 21. Prove that if the diagonals of a trapezoid are unequal, then the trapezoid is not isosceles. 22. Prove that if a quadrilateral does not have a pair of consecutive angles that are supplemen- tary, the quadrilateral is not a trapezoid. 10-8 AREAS OF POLYGONS DEFINITION The area of a polygon is the unique real number assigned to any polygon that indicates the number of non-overlapping square units contained in the polygon’s interior. We know that the area of a rectangle is the product of the lengths of two adjacent sides. For example, the rectangle to the right contains mn unit squares or has an area of m n square units. In rectangle ABCD, AB b, the length of the base, and BC h, the length of the altitude, a line segment perpendicular to the base. Area of ABCD (AB)(BC) bh The formula for the area of every other polygon can be derived from this formula. In order to derive the formulas for the areas of other polygons from the formula for the area of a rectangle, we will use the following postulate. Postulate 10.1 The areas of congruent figures are equal 14365C10.pgs 7/10/07 8:50 AM Page 410 410 Quadrilaterals EXAMPLE 1 ABCD is a parallelogram and E is a point on that if DC b and DE h, the area of parallelogram ABCD bh. such that AB DE ' AB . Prove g AB such that Proof Let F be a point on g CF ' AB to the same line are parallel, Therefore, EFCD is a parallelogram with a right angle, that is, a rectangle. . Since two lines perpendicular DE CF . Perpendicular lines intersect to form D b h C h A E B F right angles. Therefore, DEA and CFB are right angles and DEA and CFB are right triangles. In these right triangles, DE > CF because the opposite sides of a parallelogram are congruent. Therefore, DEA CFB by HL. AD > BC and The base of rectangle EFCD is DC. Since ABCD is a parallelogram, DC AB b. The height is DE h. Therefore: Area of rectangle EFCD (DC)(DE) bh Area of parallelogram ABCD area of AED area of trapezoid EBCD Area of rectangle EFCD area of BFC area of trapezoid EBCD Area of AED area of BFC Therefore, the area of parallelogram ABCD is equal to the area of rectangle EFCD or bh. Exercises Writing About Mathematics 1. If ABCD and PQRS are rectangles with AB PQ and BC QR, doABCD and PQRS have equal areas? Justify your answer. 2. If ABCD and PQRS are parallelograms with AB PQ and BC QR, do ABCD and PQRS have equal areas? Justify your answer. Developing Skills 3. Find the area of a rectangle whose vertices are (0, 0), (8, 0), (0, 5), and (8, 5). 4. a. Draw ABC. Through C, draw a line parallel to g AB , and through B, draw a line parallel g AC to . Let the point of intersection of these lines be D. b. Prove that ABC DBC. c. Let E be a point on g AB such that g CE ' AB of Example 1 to prove that the area of ABC bh. 1 2 . Let AB b and CE h. Use the results 14365C10.pgs 7/10/07 8:50 AM Page 411 5. Find the area of a triangle whose vertices are (1, 1), (7, 1), and (3, 5). 6. a. Draw trapezoid ABCD. Let E and F be points on g AB such that g CE ' AB g DF ' AB . and Areas of Polygons 411 b. Prove that CE DF. c. Let AB b1, CD b2, and CE DF h. Prove that the area of trapezoid ABCD is h 2(b1 1 b2) . 7. Find the area of a trapezoid whose vertices are (2, 1), (2, 2), (2, 7), and (2, 4). 8. a. Draw rhombus ABCD. Let the diagonals of ABCD intersect at E. b. Prove that ABE CBE CDE ADE. c. Let AC d1 and BD d2. Prove that the area of ABE d. Prove that the area of rhombus ABCD . 1 2d1d2 1 . 8d1d2 9. Find the area of a rhombus whose vertices are (0, 2), (2, 1), (4, 2), and (2, 5). Applying Skills 10. a. The vertices of ABCD are A(2, 1), B(2, 2), C(6, 1), and D(2, 4). Prove that ABCD is a rhombus. b. Find the area of ABCD. 11. a. The vertices of ABCD are A(2, 2), B(1, 1), C(4, 2), and D(1, 5). Prove that ABCD is a square. b. Find the area of ABCD. c. Find the coordinates of the vertices of ABCD, the image of ABCD under a reflec- tion in the y-axis. d. What is the area of ABCD? e. Let E and F be the coordinates of the fixed points under the reflection in the y-axis. Prove that AEAF is a square. f. What is the area of AEAF? 12. KM is a diagonal of parallelogram KLMN. The area of KLM is 94.5 square inches. a. What is the area of parallelogram KLMN? b. If MN 21.0 inches, what is the length of NP to KL ? , the perpendicular line segment from N 13. ABCD is a parallelogram and S and T are two points on g CD . Prove that the area of ABS is eq |
ual to the area of ABT. 14. The vertices of ABCD are A(1, 2), B(4, 2), C(4, 6), and D(4, 2). Draw the polygon on graph paper and draw the diagonal, a. Find the area of DBC. b. Find the area of DBA. c. Find the area of polygon ABCD. . DB 14365C10.pgs 8/2/07 5:51 PM Page 412 412 Quadrilaterals 15. The altitude to a base DH area of a trapezoid is equal to the product, (DH)(EF). of trapezoid ABCD is AB and the median is EF . Prove that the 16. The vertices of polygon PQRS are P(1, 2), Q(9, 1), R(8, 4), and S(3, 4). Draw a vertical line through P that intersects a horizontal line through S at M and a horizontal line through Q at N. Draw a vertical line through Q that intersects a horizontal line through R at L. a. Find the area of rectangle NQLM. b. Find the areas of PNQ, QLR, and SMP. c. Find the area of polygon PQRS. 17. Find the area of polygon ABCD if the coordinates of the vertices are A(5, 0), B(8, 2), C(8, 8), and D(0, 4). Hands-On Activity 1. Draw square ABCD with diagonals that intersect at E. Let AC d. Represent BD, AE, EC, BE, and ED in terms of d. 2. Express the area of ACD and of ACB in terms of d. 3. Find the area of ABCD in terms of d. 4. Let AB s. Express the area of ABCD in terms of s. 5. Write an equation that expresses the relationship between d and s. 6. Solve the equation that you wrote in step 5 for d in terms of s. 7. Use the result of step 6 to express the length of the hypotenuse of an isosceles right triangle in terms of the length of a leg. CHAPTER SUMMARY Definitions to Know • A parallelogram is a quadrilateral in which two pairs of opposite sides are parallel. • The distance between two parallel lines is the length of the perpendicular from any point on one line to the other line. • A rectangle is a parallelogram containing one right angle. • A rhombus is a parallelogram that has two congruent consecutive sides. • A square is a rectangle that has two congruent consecutive sides. • A trapezoid is a quadrilateral in which two and only two sides are parallel. • An isosceles trapezoid is a trapezoid in which the nonparallel sides are congruent. • The area of a polygon is the unique real number assigned to any polygon that indicates the number of non-overlapping square units contained in the polygon’s interior. 14365C10.pgs 7/10/07 8:50 AM Page 413 Postulate 10.1 The areas of congruent figures are equal. Chapter Summary 413 Theorems and Corollaries 10.1 A diagonal divides a parallelogram into two congruent triangles. 10.1a Opposite sides of a parallelogram are congruent. 10.1b Opposite angles of a parallelogram are congruent. 10.2 Two consecutive angles of a parallelogram are supplementary. 10.3 The diagonals of a parallelogram bisect each other. 10.4 If both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram. If one pair of opposite sides of a quadrilateral is both congruent and parallel, the quadrilateral is a parallelogram. If both pairs of opposite angles of a quadrilateral are congruent, the quadrilateral is a parallelogram. If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram. 10.5 10.6 10.7 10.8 All angles of a rectangle are right angles. 10.9 The diagonals of a rectangle are congruent. 10.10 If a quadrilateral is equiangular, then it is a rectangle. 10.11 If the diagonals of a parallelogram are congruent, the parallelogram is a rectangle. 10.12 All sides of a rhombus are congruent. 10.13 The diagonals of a rhombus are perpendicular to each other. 10.14 The diagonals of a rhombus bisect its angles. 10.15 If a quadrilateral is equilateral, then it is a rhombus. 10.16 If the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus. 10.17 A square is an equilateral quadrilateral. 10.18 A square is a rhombus. 10.19 If one of the angles of a rhombus is a right angle, then the rhombus is a square. 10.20 A trapezoid is isosceles if and only if the base angles are congruent. 10.21 A trapezoid is isosceles if and only if the diagonals are congruent. 10.22 The median of a trapezoid is parallel to the bases. 10.23 The length of the median of a trapezoid is equal to one-half the sum of the lengths of the bases. VOCABULARY 10-1 Quadrilateral • Consecutive vertices of a quadrilateral • Adjacent vertices of a quadrilateral • Consecutive sides of a quadrilateral • Adjacent sides of a quadrilateral • Opposite sides of a quadrilateral • Consecutive angles of a quadrilateral • Opposite angles of a quadrilateral • Diagonal of a quadrilateral 10-2 Parallelogram • Distance between two parallel lines 14365C10.pgs 7/10/07 8:50 AM Page 414 414 Quadrilaterals 10-4 Rectangle 10-5 Rhombus 10-6 Square 10-7 Trapezoid • Bases of a trapezoid • Legs of a trapezoid • Isosceles trapezoid • Base angles of a trapezoid • Lower base angles • Upper base angles • Median of a trapezoid 10-8 Area of a polygon REVIEW EXERCISES 1. Is it possible to draw a parallelogram that has only one right angle? Explain why or why not. 2. The measure of two consecutive angles of a parallelogram are represented by 3x and 5x 12. Find the measure of each angle of the parallelogram. 3. Point P is on side BC of rectangle ABCD. If AB BP, find mAPC. 4. Quadrilateral ABCD is a parallelogram, M is a point on point on CD . If DM ' AB and BN ' CD AB , prove that AMD CNB. , and N is a 5. Quadrilateral ABCD is a parallelogram, M is a point on , prove that AND CMB. CM AN point on CD . If AB , and N is a 6. The diagonals of rhombus ABCD intersect at E. a. Name three angles that are congruent to EAB. b. Name four angles that are the complements of EAB. 7. The diagonals of parallelogram PQRS intersect at T. If PTQ is isosceles with PTQ the vertex angle, prove that PQRS is a rectangle. 8. Point P is the midpoint of side BC of rectangle ABCD. Prove that AP > DP . 9. A regular polygon is a polygon that is both equilateral and equiangular. a. Is an equilateral triangle a regular polygon? Justify your answer. b. Is an equilateral quadrilateral a regular polygon? Justify your answer. 10. The diagonals of rhombus ABEF intersect at G and the diagonals of rhombus BCDE intersect at H. Prove that BHEG is a rectangle. F E D G B H C A 14365C10.pgs 7/10/07 8:50 AM Page 415 11. The diagonals of square ABEF intersect at G and the diagonals of square BCDE intersect at H. Prove that BHEG is a square. Review Exercises 415 F A E D G H B C 12. Points A(3, 2), B(3, 2), C(5, 3), and D(1, 3) are the vertices of quadrilateral ABCD. a. Plot these points on graph paper and draw the quadrilateral. b. What kind of quadrilateral is ABCD? Justify your answer. c. Find the area of quadrilateral ABCD. 13. The vertices of quadrilateral DEFG are D(1, 1), E(4, 1), F(1, 3), and G(2, 1). a. Is the quadrilateral a parallelogram? Justify your answer. b. Is the quadrilateral a rhombus? Justify your answer. c. Is the quadrilateral a square? Justify your answer. d. Explain how the diagonals can be used to find the area of the quadrilat- eral. 14. The coordinates of the vertices of quadrilateral ABCD are A(0, 0), B(2b, 0), C(2b 2d, 2a), and D(2d, 2a). a. Prove that ABCD is a parallelogram. b. The midpoints of the sides of ABCD are P, Q, R, and S. Find the coor- dinates of these midpoints. c. Prove that PQRS is a parallelogram. 15. The area of a rectangle is 12 square centimeters and the perimeter is 16 centimeters. a. Write an equation for the area of the rectangle in terms of the length, x, and the width, y. b. Write an equation for the perimeter of the rectangle in terms of the length, x, and the width, y. c. Solve the equations that you wrote in a and b to find the length and the width of the rectangle. 16. Each of the four sides of quadrilateral ABCD is congruent to the corresponding side of quadrilateral PQRS and A P. Prove that ABCD and PQRS are congruent quadrilaterals or draw a counterexample to prove that they may not be congruent. 14365C10.pgs 8/2/07 5:52 PM Page 416 416 Quadrilaterals Exploration In Chapter 5 we found that the perpendicular bisectors of the sides of a triangle intersect in a point called the circumcenter. Do quadrilaterals also have circumcenters? In this activity, you will explore the perpendicular bisectors of the sides of quadrilaterals. You may use compass and straightedge or geometry software. a. Construct the perpendicular bisectors of two adjacent sides of each of the following quadrilaterals. b. Construct the third and fourth perpendicular bisectors of the sides of each of the quadrilaterals. For which of the above quadrilaterals is the intersection of the first two perpendicular bisectors the same point as the intersection of the last two perpendicular bisectors? c. When all four perpendicular bisectors of a quadrilateral intersect in the same point, that point is called the circumcenter. Of the specific quadrilaterals studied in this chapter, which types do you expect to have a circumcenter and which types do you expect not to have a circumcenter? d. For each of the quadrilaterals above that has a circumcenter, place the point of your compass on the circumcenter and the pencil on any of the vertices. Draw a circle. e. Each circle drawn in d is called the circumcircle of the polygon. Based on your observations, write a definition of circumcircle. 14365C10.pgs 7/10/07 8:50 AM Page 417 CUMULATIVE REVIEW Part I Cumulative Review 417 CHAPTERS 1–10 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The measure of an exterior angle at vertex B of isosceles ABC is 70°. The measure of a base angle of the triangle is (3) 55° (2) 70° (1) 110° (4) 35° 2. The measure of A is 12 degrees less than twice the measure of its com- plement. The measure of the A is (1) 34° (2) 56° (3) 64° (4) 116° 3. The slope of the line determined by A(2, 3) and B(1, 3) is (1) –2 (2) 21 2 (3) 1 2 (4) 2 4. What is the slope of a line that is parallel to the line whose equation is 3x y 5? (1) –3 (2) 25 3 (3) 5 3 (4) 3 5 |
. The coordinates of the image of A(3, 2) under a reflection in the x-axis are (1) (3, 2) (2) (3, 2) (3) (3, 2) (4) (2, 3) 6. The measures of two sides of a triangle are 8 and 12. The measure of the third side cannot be (1) 16 (2) 12 (3) 8 (4) 4 7. The line segment is the median and the altitude of ABC. Which of BD the following statements must be false? (1) (2) BDA is a right triangle. (3) mA 90 (4) B is equidistant from A and C. 8. What is the equation of the line through (0, 1) and perpendicular to the bisects . AC BD line whose equation is y 2x 5? (1) y 2x 1 (2) x 2y 2 0 (3) 2y 1 x (4) 2x y 2 0 9. Which of the following transformations is not a direct isometry? (1) line reflection (2) point reflection (3) translation (4) rotation 10. If AC > DF and A D, which of the following is not sufficient to prove that ABC DEF? (1) AB > DE (2) BC > EF (3) C F (4) B E 14365C10.pgs 7/10/07 8:50 AM Page 418 418 Quadrilaterals Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The measures of the angles of a triangle can be represented by 3x, 4x 5, and 5x 17. Find the measure of each angle of the triangle. 12. In the diagram, g AB g CD . Prove that mx my mz Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. The bases, AB and DE , of two isosceles triangles, ABC and DEF, are congruent. If CAB FDE, prove that ABC DEF. 14. Points A, B, C, and D are on a circle with center at O and diameter AB CD OC and separate the trapezoid into three equilateral triangles. (Hint: All radii AOB and . Prove that AO > CD . ABCD is a trapezoid with OD of a circle are congruent.) Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The vertices of quadrilateral KLMN are K(3, 1), L(2, 0), M(6, 4), and N(2, 6). Show that KLMN is a trapezoid. 16. The coordinates of the vertices of ABC are A(2, 2), B(4, 0), and C(4, 2). Find the coordinates of the vertices of ABC, the image of ABC ry=x. under the composition ry-axis + 14365C11.pgs 8/2/07 5:53 PM Page 419 THE GEOMETRY OF THREE DIMENSIONS Bonaventura Cavalieri (1598–1647) was a follower of Galileo and a mathematician best known for his work on areas and volumes. In this aspect, he proved to be a forerunner of the development of integral calculus. His name is associated with Cavalieri’s Principle which is a fundamental principle for the determination of the volume of a solid. Cavalieri’s Principle can be stated as follows: Given two geometric solids and a plane, if every plane parallel to the given plane that intersects both solids intersects them in surfaces of equal areas, then the volumes of the two solids are equal. This means that two solids have equal volume when their corresponding cross-sections are in all cases equal. CHAPTER 11 CHAPTER TABLE OF CONTENTS 11-1 Points, Lines, and Planes 11-2 Perpendicular Lines and Planes 11-3 Parallel Lines and Planes 11-4 Surface Area of a Prism 11-5 Volume of a Prism 11-6 Pyramids 11-7 Cylinders 11-8 Cones 11-9 Spheres Chapter Summary Vocabulary Review Exercises Cumulative Review 419 14365C11.pgs 7/12/07 1:04 PM Page 420 420 The Geometry of Three Dimensions 11-1 POINTS, LINES, AND PLANES In this text, we have been studying points and lines in a plane, that is, the geometry of two dimensions. But the world around us is three-dimensional. The geometry of three dimensions is called solid geometry. We begin this study with some postulates that we can accept as true based on our observations. We know that two points determine a line. How many points are needed to determine a plane? A table or chair that has four legs will sometimes be unsteady on a flat surface. But a tripod or a stool with three legs always sits firmly. This observation suggests the following postulate. Postulate 11.1 There is one and only one plane containing three non-collinear points. For a set of the three non-collinear points that determine a plane, each pair of points determines a line and all of the points on that line are points of the plane. C B A Postulate 11.2 A plane containing any two points contains all of the points on the line determined by those two points. These two postulates make it possible for us to prove the following theorems. Theorem 11.1 There is exactly one plane containing a line and a point not on the line. Given Line l and point P not on l. Prove There is exactly one plane containing l and P. B l A P Proof Choose two points A and B on line l. The three points, A, B, and P, determine one and only one plane. If the two points A and B on line l are on the plane, then all of the points of l are on the plane, that is, the plane contains line l. Therefore, there is exactly one plane that contains the given line and point. 14365C11.pgs 7/12/07 1:04 PM Page 421 Theorem 11.2 If two lines intersect, then there is exactly one plane containing them. Points, Lines, and Planes 421 This theorem can be stated in another way: Theorem 11.2 Two intersecting lines determine a plane. Given Lines l and m intersecting at point P. Prove There is exactly one plane containing l and m. l B m P A Proof Choose two points, A on line l and B on line m. The three points, A, B, and P, determine one and only one plane. A plane containing any two points contains all of the points on the line determined by those two points. Since the two points A and P on line l are on the plane, then all of the points of l are on the plane. Since the two points B and P on line m are on the plane, then all of the points of m are on the plane. Therefore, there is exactly one plane that contains the given intersecting lines. The definition of parallel lines gives us another set of points that must lie in a plane. DEFINITION Parallel lines in space are lines in the same plane that have no points in common. This definition can be written as a biconditional: Two lines are parallel if and only if they are coplanar and have no points in common. If g AB and g CD are par- allel lines in space, then they determine a plane if and only if they are two dis- tinct lines. We have seen that intersecting lines lie in a plane and parallel lines lie in a plane. For lie in a plane. Also, example, in the diagram, g are interCD secting lines so they lie in a plane. But there are some pairs of lines that do not intersect and so g AB g AB g CD g and BF g AB and are not parallel. In the diagram, g CG g AB are neither intersecting nor parallel lines. g EH are called skew lines. g CG g BF and and are another pair of skew lines. H D C g AB and E A B G F 14365C11.pgs 7/12/07 1:05 PM Page 422 422 The Geometry of Three Dimensions DEFINITION Skew lines are lines in space that are neither parallel nor intersecting. EXAMPLE 1 Does a triangle determine a plane? Solution A triangle consists of three line segments and the three non-collinear points that are the endpoints of each pair of segments. Three non-collinear points determine a plane. That plane contains all of the points of the lines determined by the points. Therefore, a triangle determines a plane. Exercises Writing About Mathematics 1. Joel said that another definition for skew lines could be two lines that do not lie in the same plane. Do you agree with Joel? Explain why or why not. 2. Angelina said that if AC and BD intersect at a point, then A, B, C, and D lie in a plane and form a quadrilateral. Do you agree with Angelina? Explain why or why not. Developing Skills g AB 3. is parallel to g CD and AB CD. Prove that A, B, C, and D must lie in a plane and form a trapezoid. g AB g CD and gram. g AD g BC . Prove that A, B, C, and D must lie in a plane and form a parallelo- BED AEC C, and D must lie in a plane and form a square. and each segment is the perpendicular bisector of the other. Prove that A, B, 4. 5. In 6–9, use the diagram at the right. 6. Name two pairs of intersecting lines. 7. Name two pairs of skew lines. 8. Name two pairs of parallel lines. 9. Which pairs of lines that you named in exercises 6, 7, and 8 are not lines in the same plane? H G D C E F A B 14365C11.pgs 7/12/07 1:05 PM Page 423 Perpendicular Lines and Planes 423 10. Let p represent “Two lines are parallel.” Let q represent “Two lines are coplanar.” Let r represent “Two lines have no point in common.” a. Write the biconditional “Two lines are parallel if and only if they are coplanar and have no points in common” in terms of p, q, r, and logical symbols. b. The biconditional is true. Show that q is true when p is true. Applying Skills 11. A photographer wants to have a steady base for his camera. Should he choose a base with four legs or with three? Explain your answer. 12. Ken is building a tool shed in his backyard. He begins by driving four stakes into the ground to be the corners of a rectangular floor. He stretches strings from two of the stakes to the opposite stakes and adjusts the height of the stakes until the strings intersect. Explain how the strings assure him that the four stakes will all be in the same plane. 13. Each of four equilateral triangles has a common side with each of the three other triangles and form a solid called a tetrahedron. Prove that the triangles are congruent. D C A B 11-2 PERPENDICULAR LINES AND PLANES Look at the floor, walls, and ceiling of the classroom. Each of these surfaces can be represented by a plane. Many of t |
hese planes intersect. For example, each wall intersects the ceiling and each wall intersects the floor. Each intersection can be represented by a line segment. This observation allows us to state the following postulate. Postulate 11.3 If two planes intersect, then they intersect in exactly one line. The Angle Formed by Two Intersecting Planes Fold a piece of paper. The part of the paper on one side of the crease represents a half-plane and the crease represents the edge of the half-plane. The folded paper forms a dihedral angle. 14365C11.pgs 7/31/07 1:19 PM Page 424 424 The Geometry of Three Dimensions DEFINITION A dihedral angle is the union of two half-planes with a common edge. Each half-plane of a dihedral angle can be compared to a ray of an angle in a plane (or a plane angle) and the edge to the vertex of a plane angle. If we choose some point on the edge of a dihedral angle and draw, from this point, a ray in each half-plane perpendicular to the edge, we have drawn a plane angle. DEFINITION The measure of a dihedral angle is the measure of the plane angle formed by two rays each in a different half-plane of the angle and each perpendicular to the common edge at the same point of the edge. EE plane angle whose measure is the same as that of the dihedral angle can be drawn at any points on the edge of the dihedral angle. Each plane angle of a dihedral angle has the same measure. In the figure, planes p and g AB h AC g ' AB . In plane p, q intersect at . The measure of the dihedral angle is equal to the measure of CAD. Also, in plane p, h and in plane q, . The measure of the dihedral angle is equal BE to the measure of EBF. and in plane q, g ' AB g ' AB h BF h AD g ' AB DEFINITION Perpendicular planes are two planes that intersect to form a right dihedral angle. In the diagram, QST is a right angle. In plane m, h ST n, . The dihedral angle formed by half-planes of planes m and n with has the same measure as QST. Therefore, the dihedral angle is a right h ' SR g RS h SQ h ' SR and in plane edge angle, and m n. ' The floor and a wall of a room usually form a right dihedral angle. Look at the line that is the intersection of two adjacent walls of the classroom. This line intersects the ceiling in one point and intersects the floor in one point. This observation suggests the following theorem. Theorem 11.3 If a line not in a plane intersects the plane, then it intersects in exactly one point. 14365C11.pgs 7/12/07 1:05 PM Page 425 Perpendicular Lines and Planes 425 Given Line l is not in plane p and l intersects p. Prove Line l intersects p in exactly one point. Proof Use an indirect proof. Assume that line l intersects the plane in two points. Then all of the points on line l lie in plane p, that is, the line lies in the plane. Because this contradicts the hypothesis that line l is not in plane p, the assumption must be false. A line not in a plane that intersects the plane, intersects it in exactly one point. p l A Again, look at the corner of the classroom in the figure on the previous page. The line that is the intersection of two adjacent walls intersects the ceiling so that the line is perpendicular to any line in the ceiling through the point of intersection. We say that this line is perpendicular to the plane of the ceiling. DEFINITION A line is perpendicular to a plane if and only if it is perpendicular to each line in the plane through the intersection of the line and the plane. A plane is perpendicular to a line if the line is perpendicular to the plane. It is easy to demonstrate that a line that is perpendicular to one line in a plane may not be perpendicular to the plane. For example, fold a rectangular sheet of paper. Draw a ray perpendicular to the crease with its endpoint on the crease. Keep the half of the folded sheet that does not contain the ray in contact with your desk. This is the plane. Move the other half to different positions. The ray that you drew is always perpendicular to the crease but is not always perpendicular to the plane. Based on this observation, we can state the following postulate. Postulate 11.4 At a given point on a line, there are infinitely many lines perpendicular to the given line. In order to prove that a line is perpendicular to a plane, the definition requires that we show that every line through the point of intersection is perpendicular to the given line. However, it is possible to prove that if line l is known to be perpendicular to each of two lines in plane p that intersect at point A, then l is perpendicular to plane p at A. 14365C11.pgs 7/12/07 1:05 PM Page 426 426 The Geometry of Three Dimensions Theorem 11.4 If a line is perpendicular to each of two intersecting lines at their point of intersection, then the line is perpendicular to the plane determined by these lines. Given A plane p determined by g AP and g BP , two lines that intersect at P. Line l such that g g l ' AP l ' BP and . Prove l ⊥ p Proof To begin, let R and S be points on l such that P is the mid RS and . Since it is given that g l ' BP point of g , RPA, SPA, RPB, l ' AP and SPB are right angles and therefore congruent. Then let other line through P in plane p. Draw theorem, we need to show three different pairs of congruent triangles: RPA SPA, RPB SPB, and RPQ SPQ. However, to establish the last congruence we must prove that RAB SAB and RAQ SAQ. intersecting g AB g PT g PT be any at Q. To prove this 1) RPA SPA by SAS and AR > AS . (2) RPB SPB by SAS and BR > BS . (3) RAB SAB by SSS and RAB SAB. 14365C11.pgs 7/12/07 1:05 PM Page 427 Perpendicular Lines and Planes 427 4) RAQ SAQ by by SAS since RAB RAQ and SAB SAQ, RQ > SQ and . (5) RPQ SPQ by SSS and RPQ SPQ. Now since RPQ and SPQ are a congruent linear pair of angles, they are can be any line in p through P, l , that is, right angles, and g l ' PQ . Since g PQ g PT is perpendicular to every line in plane p through point P. Theorem 11.5a If two planes are perpendicular to each other, one plane contains a line perpendicular to the other plane. Given Plane p ⊥ plane q Prove A line in p is perpendicular to q and a line in q is perpendicular to p. Proof If planes p and q are perpendicular to each other then they form a right dihedral angle. Let g AB be C p q A B D g AC g ' AB the edge of the dihedral angle. In plane p, cong AD struct right dihedral angle, CAD is a right angle and p, , are each perpendicular to , and in plane q, construct g AD g AB g AC and g ' AB g AC . Therefore, . Since p and q from a g ' AD . Two lines in plane g AD . Similarly, g AB and g AD , are each perpendicular to . Therefore, ' p g AC two lines in plane q, g AC ' q . 14365C11.pgs 7/12/07 1:05 PM Page 428 428 The Geometry of Three Dimensions The converse of Theorem 11.5a is also true. Theorem 11.5b If a plane contains a line perpendicular to another plane, then the planes are perpendicular. Given g AC in plane p and g AC ' q Prove p ⊥ q Proof Let g AB be the line of intersection of planes p and g ' AB . Since is per- g AD g AC q. In plane q, draw pendicular to q, it is perpendicular to any line through A in q. Therefore, g AC g AC . Thus, CAD is the plane angle of the g ' AD g ' AB and also C p q A B D dihedral angle formed by planes p and q. Since CAD is a right angle. Therefore, the dihedral angle is a right angle, and p and q are perpendicular planes. is perpendicular to , g AC g AD This theorem and its converse can be stated as a biconditional. Theorem 11.5 Two planes are perpendicular if and only if one plane contains a line perpendicular to the other. We know that in a plane, there is only one line perpendicular to a given line at a given point on the line. In space, there are infinitely many lines perpendicular to a given line at a given point on the line. These perpendicular lines are all in the same plane. However, only one line is perpendicular to a plane at a given point. Theorem 11.6 Through a given point on a plane, there is only one line perpendicular to the given plane. 14365C11.pgs 7/12/07 1:05 PM Page 429 Given Plane p and g AB ' p at A. Prove g AB is the only line perpendicular to p at A. Proof Use an indirect proof. Assume that there exists another line, ' p at A. Points A, B, and C determine a plane, q, g AC that intersects plane p at g . Therefore, in AD g ⊥ AC g AD g AB g ' AD and plane q, given plane, there is only one line perpendicular to a given line at a given point. Our assumption is false, and there is only one line perpendicular to a given plane at a given point. . But in a Perpendicular Lines and Planes 429 q p B A C D As we noted above, in space, there are infinitely many lines perpendicular to a given line at a given point. Any two of those intersecting lines determine a plane perpendicular to the given line. Each of these pairs of lines determine the same plane perpendicular to the given line. Theorem 11.7 Through a given point on a line, there can be only one plane perpendicular to the given line. Given Any point P on g . AB Prove There is only one plane perpendicular to g AB . Proof Use an indirect proof. Assume that there are two planes, m and n, g AB g APB , . Choose that are each perpendicular to any point Q in m. Since m ⊥ g AP g PQ ⊥ . Points A, P, and Q determine a g PR plane p that intersects plane n in a line Since n ⊥ plane p, g APB ⊥ g PR g AP . Therefore, in , g PQ . But in a g AP g AP g PR and ⊥ ⊥ . A P B A plane, at a given point there is one and only P one line perpendicular to a given line. Our assumption must be false, and there is only one plane perpendicular to at P. B g AB m Q n R Q R m n p 14365C11.pgs 7/12/07 1:05 PM Page 430 430 The Geometry of Three Dimensions Theorem 11.8 If a line is perpendicular to a plane, then any line perpendicular to the given line at its point of intersection with the given plane is in the plane. Given g AB ' plane p at A and g AB g ' AC . Prove g AC is in plane p. Proof Points A, B, and C determine a plane q. g . Plane q intersects plane p in a line, AD g AB g ' AD because is perp |
endicular to g AB every line in p through A. It is given that g AB . Therefore, g ' AC g AC and in plane q are perpendicular to at A. But at a g AD g AB q B C A D p given point in a plane, only one line can be drawn perpendicular to a given line. Therefore, g AD g AC that is, C is on . Since section of planes p and q, and g AD are the same line, g AD g AC is the inter- is in plane p. Theorem 11.9 If a line is perpendicular to a plane, then every plane containing the line is perpendicular to the given plane. Given Plane p with on p. g AB ' p at A, and C any point not Prove The plane q determined by A, B, and C is per- pendicular to p. Proof Let the intersection of p and q be g AD , so g AD is the edge of the dihedral angle formed by p and g AE q. Let g AD be a line in p that is perpendicular to g AB is perpendicular to g AB . Since ' p every line in p through A. Therefore, g ' AD g AB , q B A C E D p and g AB g ' AE . BAE is a plane angle whose measure is the measure of the dihedral angle. Since g AB q ⊥ p. , mBAE 90. Therefore, the dihedral angle is a right angle, and g ' AE 14365C11.pgs 7/12/07 1:05 PM Page 431 Perpendicular Lines and Planes 431 EXAMPLE 1 Show that the following statement is false. Two planes perpendicular to the same plane have no points in common. Solution Recall that a statement that is sometimes true and sometimes false is regarded to be false. Consider the adjacent walls of a room. Each wall is perpendicular to the floor but the walls intersect in a line. This counterexample shows that the given statement is false. EXAMPLE 2 Planes p and q intersect in g ' AB . In p, . If mCAD 90, is g AD in q, g AB g g ' AB AC p ' q ? and Solution Since in p, g AC g ' AB , and in q, g AD g ' AB , CAD is a plane angle whose measure is equal to the measure of the dihedral angle formed by the planes. Since CAD is not a right angle, then the dihedral angle is not a right angle, and the planes are not perpendicular. D q p A C B EXAMPLE 3 Given: Line l intersects plane p at A, and l is not perpendicular to p. l Prove: There is at least one line through A in plane p that is not perpendicular to l. A p Proof Use an indirect proof. g AB Let and g AC be two lines through A in g l ' AC g l ' AB p. Assume that Therefore, l ⊥ p because if a line is perpendicular to each of two lines at their point of intersection, then the line is perpendicular to the plane determined by and . these lines. But it is given that l is not perpendicular to p. Therefore, our assump- tion is false, and l is not perpendicular to at least one of the lines g AB and g .AC 14365C11.pgs 7/12/07 1:05 PM Page 432 432 The Geometry of Three Dimensions Exercises Writing About Mathematics 1. Carmen said if two planes intersect to form four dihedral angles that have equal measures, then the planes are perpendicular to each other. Do you agree with Carmen? Explain why or why not. 2. Each of three lines is perpendicular to the plane determined by the other two. a. Is each line perpendicular to each of the other two lines? Justify your answer. b. Name a physical object that justifies your answer. Developing Skills In 3–11, state whether each of the statements is true or false. If it is true, state a postulate or theorem that supports your answer. If it is false, describe or draw a counterexample. 3. At a given point on a given line, only one line is perpendicular to the line. 4. If A is a point in plane p and B is a point not in p, then no other point on g AB is in plane p. 5. A line perpendicular to a plane is perpendicular to every line in the plane. 6. A line and a plane perpendicular to the same line at two different points have no points in common. 7. Two intersecting planes that are each perpendicular to a third plane are perpendicular to each other. 8. If g AB is perpendicular to plane p at A and g AB is in plane q, then p ⊥ q. 9. At a given point on a given plane, only one plane is perpendicular to the given plane. 10. If a plane is perpendicular to one of two intersecting lines, it is perpendicular to the other. 11. If a line is perpendicular to one of two intersecting planes, it is perpendicular to the other. Applying Skills 12. Prove step 1 of Theorem 11.4. 13. Prove step 3 of Theorem 11.4. 14. Prove step 5 of Theorem 11.4. 15. Prove that if a line segment is perpendicular to a plane at the midpoint of the line segment, then every point in the plane is equidistant from the endpoints of the line segment. Given: ⊥ plane p at M, the midpoint of AB point in plane p. AB , and R is any p Prove: AR BR R R p M M A B 14365C11.pgs 7/12/07 1:05 PM Page 433 16. Prove that if two points are each equidistant from the endpoints of a line segment, then the line segment is perpendicular to the plane determined by the two points and the midpoint of the line segment. Given: M is the midpoint of RA > RB , and , AB SA > SB . Prove: is perpendicular to the plane AB determined by M, R, and S. Parallel Lines and Planes 433 p R A M B S 17. Equilateral triangle ABC is in plane p and g AD is perpendicular to plane p. Prove that BD > CD . g g AB AC AB AC, prove that ABD ACD. and intersect at A and determine plane p. 18. g AD is perpendicular to plane p at A. If 19. Triangle QRS is in plane p, ST is perpendicular to plane p, and QTS RTS. Prove that TQ > TR . 20. Workers who are installing a new telephone pole position the pole so that it is perpendicu- lar to the ground along two different lines. Prove that this is sufficient to prove that the telephone pole is perpendicular to the ground. 21. A telephone pole is perpendicular to the level ground. Prove that two wires of equal length attached to the pole at the same point and fastened to the ground are at equal distances from the pole. 11-3 PARALLEL LINES AND PLANES Look at the floor, walls, and ceiling of the classroom. Each of these surfaces can be represented by a plane. Some of these surfaces, such as the floor and the ceiling, do not intersect. These can be represented as portions of parallel planes. DEFINITION Parallel planes are planes that have no points in common. A line is parallel to a plane if it has no points in common with the plane. 14365C11.pgs 7/12/07 1:05 PM Page 434 434 The Geometry of Three Dimensions EXAMPLE 1 Plane p intersects plane q in g CD . Prove that if g AB g AB and and intersect, then planes q and r are plane r in g CD not parallel. Proof Let E be the point at which g AB and g intersect. Then E is a point on q CD and E is a point on r. Therefore, planes q and r intersect in at least one point and are not parallel. A q r C B D p E Theorem 11.10 If a plane intersects two parallel planes, then the intersection is two parallel lines. Given Plane p intersects plane m at g CD , m n. g AB and plane n at B Prove g AB g CD Proof Use an indirect proof. g AB Lines g CD and are two lines of plane p. Two A m p n C D lines in the same plane either intersect or are par- allel. If g AB is not parallel to g AB g CD , then it is a point of plane m. Since E is a point of is a point of is a point of plane n. But m n and have no points in common. Therefore, are two lines in the same plane that do not intersect, and and g AB g CD , then it g AB g CD . g CD then they intersect in some point E. Since E In a plane, two lines perpendicular to a given line are parallel. Can we prove that two lines perpendicular to a given plane are also parallel? 14365C11.pgs 7/12/07 1:05 PM Page 435 Theorem 11.11 Two lines perpendicular to the same plane are parallel. Parallel Lines and Planes 435 Given Plane p, line g LA ⊥ p at A, and line g MB ⊥ p at B. q Prove g LA g MB Proof We will construct line , and show that g LA to g NB g MB at B that is parallel and g NB are the same line. (1) Since it is given that g LA ⊥ p at A, g LA is per- L N M p A C B D ⊥ pendicular to any line in p through A, so g LA . Let q be the plane determined by g ' AB dihedral angle. g AB g AC draw . In plane p, . Then LAC is a right angle, and p and q form a right and g AB g LA (2) At point B, there exists a line g NB in q that is parallel to g LA . If one of two parallel lines is perpendicular to a third line, then the other is perpendicu⊥ lar to the third line, that is, since g ' AB , then g NB g AB g LA . (3) Draw g BD g ' AB in p. Because p and q form a right dihedral angle, NBD is a right angle, and so g NB g ' BD . (4) Therefore, g NB is perpendicular to two lines in p at B (steps 2 and 3), so g NB is perpendicular to p at B. (5) But it is given that g MB ⊥ p at B and there is only one line perpendicular to a given plane at a given point. Therefore, line, and g LA g MB . g MB and g NB are the same We have shown that two lines perpendicular to the same plane are parallel. Since parallel lines lie in the same plane, we have just proved the following corollary to this theorem: Corollary 11.11a Two lines perpendicular to the same plane are coplanar. 14365C11.pgs 7/12/07 1:05 PM Page 436 436 The Geometry of Three Dimensions Theorem 11.12a If two planes are perpendicular to the same line, then they are parallel. Given Plane p ⊥ g AB at A and q ⊥ g AB at B. Prove p q Proof Use an indirect proof. Assume that p is not parallel to q. Then p and q intersect in a line. Let R be any point on the line of intersection. Then A, B, and R determine a plane, s. In plane s, g AR g ' AB and g BR g ' AB . But two lines p s R A B q in a plane that are perpendicular to the same line are parallel. Therefore, our assumption must be false, and p q. Theorem 11.12b If two planes are parallel, then a line perpendicular to one of the planes is perpendicular to the other. Given Plane p parallel to plane q, and intersecting plane q at B g AB ⊥ plane p and Prove g AB ⊥ plane q Proof To prove this theorem, we will construct two lines and g FB g BE g AB in q that are both perpendicular to . From this, we will conclude that g AB p q A B pendicular to q. (1) Let C be a point in p. Let r be the plane determined by A, B, and C intersecting q at g AC ⊥ plane p. Therefore, are parallel, g AB g |
BE g BE Then, in plane r, g AB g ' BE . . Since p and q . It is given that g ' AC g AB . is per- C p E q A B r 14365C11.pgs 7/12/07 1:05 PM Page 437 (2) Let D be a point in p. Let s be the plane determined by A, B, and D . Since p and q . It is given g BF g BF ⊥ plane p. Therefore, . Then, in plane s, intersecting q at g AD are parallel, g AB g ' AD g ' BF . that g AB g AB Parallel Lines and Planes 437 D A p s F B q (3) If a line is perpendicular to each of two intersecting lines at their point of intersection, then the line is perpendicular to the plane determined by these lines. Therefore, g AB ⊥ plane q. Theorem 11.12a and 11.2b are converse statements. Therefore, we may write these two theorems as a biconditional. Theorem 11.12 Two planes are perpendicular to the same line if and only if the planes are parallel. Let p and q be two parallel planes. From A in p, draw AB ⊥ q at B. Therefore, AB ⊥ p at A. The distance from p to q is AB. p q A B DEFINITION The distance between two planes is the length of the line segment perpendicular to both planes with an endpoint on each plane. 14365C11.pgs 7/12/07 1:05 PM Page 438 438 The Geometry of Three Dimensions Theorem 11.13 Parallel planes are everywhere equidistant. Given Parallel planes p and q, with each perpendicular to p and q with an endpoint on each plane. and BD AC Prove AC BD Proof Two lines perpendicular to the same plane are both parallel and coplanar. Therefore, AC BD and lie on the same plane. That plane intersects parallel planes p and q in parallel lines g CD . In the plane of g BD g AC and , ABDC is a and g AB parallelogram with a right angle, that is, a rectan- gle. Therefore, AC BD. AC and BD are congruent and EXAMPLE 2 Line l is perpendicular to plane p and line l is not perpendicular to plane q. Is p q? Solution Assume that p q. If two planes are parallel, then a line perpendicular to one is perpendicular to the other. Therefore, since l is perpendicular to plane p, l must be perpendicular to plane q. This contradicts the given statement that l is not perpendicular to q, and the assumption is false. Therefore, p is not parallel to q EXAMPLE 3 Given: ' g AB and AB CD. plane p at A, g CD ' plane p at C, B D Prove: A, B, C, and D are the vertices of a paral- lelogram. A C p 14365C11.pgs 7/12/07 1:06 PM Page 439 Parallel Lines and Planes 439 Proof Two lines perpendicular to the same plane are parallel and coplanar. Therefore, since it is given that are each perpendicular to p, they are parallel and coplanar. Since AB CD and segments of equal length are congruent, ABCD is a quadrilateral with one pair of sides congruent and parallel. Therefore, ABCD is a parallelogram. g CD g AB and Exercises Writing About Mathematics 1. Two planes are perpendicular to the same plane. Are the planes parallel? Justify your answer. 2. Two planes are parallel to the same plane. Are the planes parallel? Justify your answer. Developing Skills In 3–9, each of the given statements is sometimes true and sometimes false. a. Give an example from the diagram to show that the statement can be true. b. Give a counterexample from the diagram to show that the statement can be false. In the diagram, each quadrilateral is a rectangle. 3. If two planes are perpendicular, a line parallel to one plane is perpendicular to the other. 4. Two planes parallel to the same line are parallel to each other. 5. Two lines perpendicular to the same line are parallel. 6. Two lines that do not intersect are parallel. Two planes perpendicular to the same plane are parallel to each other. 8. Two lines parallel to the same plane are parallel. 9. If two lines are parallel, then a line that is skew to one line is skew to the other. Applying Skills 10. ABC is an isosceles triangle with base AC and point E on . Prove that ADE is isosceles. BC in plane p. Plane q p through point D on AB 11. Plane p is perpendicular to g PQ Prove that AQ > BQ . at Q and two points in p, A and B, are equidistant from P. 14365C11.pgs 7/12/07 1:06 PM Page 440 440 The Geometry of Three Dimensions 12. Noah is building a tool shed. He has a rectangular floor in place and wants to be sure that the posts that he erects at each corner of the floor as the ends of the walls are parallel. He erects each post perpendicular to the floor. Are the posts parallel to each other? Justify your answer. 13. Noah wants the flat ceiling on his tool shed to be parallel to the floor. Two of the posts are 80 inches long and two are 78 inches long. Will the ceiling be parallel to the floor? Justify your answer. What must Noah do to make the ceiling parallel to the floor? 11-4 SURFACE AREA OF A PRISM Polyhedron In the plane, a polygon is a closed figure that is the union of line segments. In space, a polyhedron is a figure that is the union of polygons. DEFINITION A polyhedron is a three-dimensional figure formed by the union of the surfaces enclosed by plane figures. The portions of the planes enclosed by a plane figure are called the faces of the polyhedron. The intersections of the faces are the edges of the polyhedron and the intersections of the edges are the vertices of the polyhedron. DEFINITION A prism is a polyhedron in which two of the faces, called the bases of the prism, are congruent polygons in parallel planes. Examples of prisms The surfaces between corresponding sides of the bases are called the lateral sides of the prism and the common edges of the lateral sides are called the lateral edges. An altitude of a prism is a line segment perpendicular to each of the bases with an endpoint on each base. The height of a prism is the length of an altitude. altitude lateral side base base lateral edge 14365C11.pgs 7/12/07 1:06 PM Page 441 Surface Area of a Prism 441 Since the bases are parallel, the corresponding sides of the bases are congruent, parallel line segments. Therefore, each lateral side has a pair of congruent, parallel sides, the corresponding edges of the bases, and are thus parallelograms. The other pair of sides of these parallelograms, the lateral edges, are also congruent and parallel. Therefore, we can make the following statement: The lateral edges of a prism are congruent and parallel. DEFINITION A right prism is a prism in which the lateral sides are all perpendicular to the bases. All of the lateral sides of a right prism are rectangles. Using graph paper, cut two 7-by-5 rectangles and two 7-by-4 rectangles. Use tape to join the 7-by-4 rectangles to opposite sides of one of the 7-by-5 rectangles along the congruent edges. Then join the other 7-by-5 rectangle to the congruent edges forming four of the six sides of a prism. Place the prism on your desk on its side so that one pair of congruent rectangles are the bases and the lateral edges are perpendicular to the bases. Are the opposite faces parallel? What would be the shape and size of the two missing sides? Then move the top base so that the lateral edges are not perpendicular to the bases. The figure is no longer a right prism. Are the opposite faces parallel? What would be the shape and size of the two missing sides? Cut two more 7-by-5 rectangles and two parallelograms that are not rectangles. Let the lengths of two of the sides of the parallelograms be 7 and the length of the altitude to these sides be 4. Join the parallelograms to opposite sides of one of the rectangles along congruent sides. Then join the other rectangle to congruent edges forming four of the six sides of a prism. Place the prism on your desk so that the rectangles are the bases and the lateral edges are perpendicular to the bases. Are the opposite faces parallel? Is the prism a right prism? What would be the shape and size of the two missing lateral sides? Move the top base of the prism so that the sides are not perpendicular to the bases. Are the opposite faces parallel? What would be the shape and size of the two missing sides? Now turn this prism so that the parallelograms are the bases and the edges of the rectangles are perpendicular to the bases. What is the shape of the two missing sides? Is this a right prism? The solids that you have made are called parallelepipeds. 14365C11.pgs 7/12/07 1:06 PM Page 442 442 The Geometry of Three Dimensions DEFINITION A parallelepiped is a prism that has parallelograms as bases. Examples of parallelepipeds Rectangular Solids DEFINITION A rectangular parallelepiped is a parallelepiped that has rectangular bases and lateral edges perpendicular to the bases. A rectangular parallelepiped is usually called a rectangular solid. It is the most common prism and is the union of six rectangles. Any two parallel rectangles of a rectangular solid can be the bases. In the figure, ABCDEFGH is a rectangular solid. The bases ABCD and EFGH are congruent rectangles with AB EF CD GH 7 and BC FG DA HE 5. Two of the lateral sides are rectangles ABFE and DCGH whose dimensions are 7 by 4. The other two lateral sides are BCGF and ADHE whose dimensions are 5 by 4 • The area of each base: 7 5 35 • The area of each of two lateral sides: 7 4 28 • The area of each of the other two lateral sides: 5 4 20 The lateral area of the prism is the sum of the areas of the lateral faces. The total surface area is the sum of the lateral area and the areas of the bases. • The lateral area of the prism is 2(28) 2(20) 96. • The area of the bases are each 2(35) 70. • The surface area of the prism is 96 70 166. 14365C11.pgs 7/12/07 1:06 PM Page 443 Surface Area of a Prism 443 EXAMPLE 1 The bases of a right prism are regular hexagons. The length of each side of a base is 5 centimeters and the height of the prism is 8 centimeters. Describe the number, shape, and size of the lateral sides. Solution A hexagon has six sides. Because the base is a regular hexagon, it has six congruent sides, and therefore, the prism has six lateral sides. Since it is a right prism, the lateral sides are rectangles. The length of each of two edges of a lateral side is the length of an edge of a base, 5 centimeters. The length |
of each of the other two edges of a lateral side is the height of the prism, 8 centimeters. 8 cm 5 cm Answer There are six lateral sides, each is a rectangle that is 8 centimeters by 5 centimeters. EXAMPLE 2 The bases of a right prism are equilateral triangles. The length of one edge of a base is 4 inches and the height of the prism is 5 inches. a. How many lateral sides does this prism have and what is their shape? b. What is the lateral area of the prism? Solution a. Because this is a prism with a triangular base, the prism has three lateral sides. Because it is a right prism, the lateral sides are rectangles. b. For each rectangular side, the length of one pair of edges is the length of an edge of the base, 4 inches. The height of the prism, 5 inches, is the length of a lateral edge. Therefore, the area of each lateral side is 4(5) or 20 square inches. The lateral area of the prism is 3(20) 60 square inches. Answers a. 3 b. 60 square inches EXAMPLE 3 The lateral sides of a prism are five congruent rectangles. Prove that the bases are equilateral pentagons. Proof The five lateral sides are congruent rectangles. Two parallel sides of each rectangle are lateral edges. The other two parallel sides of each rectangle are edges of the bases. Each edge of a base is a side of a rectangle. Therefore, the base has five sides. The corresponding sides of the congruent rectangles are congruent. Therefore, the edges of each base are congruent. The bases are equilateral pentagons. 14365C11.pgs 7/12/07 1:06 PM Page 444 444 The Geometry of Three Dimensions Exercises Writing About Mathematics 1. Cut a 12-by-5 rectangle from graph paper, fold it into three 4-by-5 rectangles and fasten the two sides of length 5 with tape. Then cut a 16-by-5 rectangle from graph paper, fold it into four 4-by-5 rectangles and fasten the two sides of length 5 with tape. Let the open ends of each figure be the bases of a prism. a. What is the shape of a base of the prism formed from the 12-by-5 paper? Can the shape of this base be changed? Explain your answer. b. What is the shape of a base of the prism formed from the 16-by-5 paper? Can the shape of this base be changed? Explain your answer. c. Are both figures always right prisms? Explain your answer. 2. A prism has bases that are rectangles, two lateral faces that are rectangles and two lateral faces that are parallelograms that are not rectangles. a. Is an altitude of the solid congruent to an altitude of one of the rectangular faces? Explain your answer. b. Is an altitude of the solid congruent to an altitude of one of the faces that are parallel- ograms? Explain your answer. Developing Skills In 3–6, find the surface area of each of the rectangular solid with the given dimensions. 3. 5.0 cm by 8.0 cm by 3.0 cm 4. 15 in. by 10.0 in. by 2.0 ft 5. 2.5 ft by 8.0 ft by 12 ft 6. 56.3 cm by 18.7 cm by 0.500 m 7. The bases of a prism are right triangles whose edges measure 9.00 centimeters, 40.0 centimeters, and 41.0 centimeters. The lateral sides of the prism are perpendicular to the bases. The height of the prism is 14.5 centimeters. a. What is the shape of the lateral sides of the prism? b. What are the dimensions of each lateral side of the prism? c. What is the total surface area of the prism? 8. The bases of a right prism are isosceles triangles. The lengths of the sides of the bases are 5 centimeters, 5 centimeters, and 6 centimeters. The length of the altitude to the longest side of a base is 4 centimeters. The height of the prism is 12 centimeters. a. What is the shape of the lateral sides of the prism? b. What are the dimensions of each lateral side of the prism? c. What is the total surface area of the prism? 9. How many faces does a parallelepiped have? Justify your answer. 14365C11.pgs 7/12/07 1:06 PM Page 445 10. A prism has bases that are trapezoids. Is the prism a parallelepiped? Justify your answer. 11. The length of an edge of a cube is 5.20 inches. What is the total surface area of the cube to Surface Area of a Prism 445 the nearest square inch? Applying Skills 12. The bases of a parallelepiped are ABCD and EFGH, and ABCD EFGH. Prove that AE BF CG DH and that AE BF CG DH. 13. The bases of a prism are ABC and DEF, and ABC DEF. The line through A perpendicular to the plane of ABC intersects the plane of DEF at D, and the line through B perpendicular to the plane of ABC intersects the plane of DEF at E. a. Prove that the lateral faces of the prism are rectangles. b. When are the lateral faces of the prism congruent polygons? Justify your answer. 14. A right prism has bases that are squares. The area of one base is 81 square feet. The lateral area of the prism is 144 square feet. What is the length of the altitude of the prism? 15. Show that the edges of a parallelepiped form three sets of parallel line segments. 16. The lateral faces of a parallelepiped are squares. What must be the shape of the bases? Justify your answer. 17. The lateral faces of a parallelepiped are squares. One angle of one of the bases is a right angle. Prove that the parallelepiped is a cube, that is, a rectangular parallelepiped with congruent faces. 18. The walls, floor, and ceiling of a room form a rectangular solid. The total surface area of the room is 992 square feet. The dimensions of the floor are 12 feet by 20 feet. a. What is the lateral area of the room? b. What is the height of the room? Hands-On Activity Let the bases of a prism be ABCD and ABCD. Use the prisms that you made out of graph paper for page 441 to demonstrate each of the following. 1. When AAr is perpendicular to the planes of the bases, the lateral faces are rectangles and the height of the each lateral face is the height of the prism. 2. When a line through A perpendicular to the bases intersects the plane of ABCD at a point on , two of the lateral faces are rectangles and two are parallelograms. The height g ArBr of the prism is the height of the lateral faces that are parallelograms but the height of the rectangles is not equal to the height of the prism. 3. When a line through A perpendicular to the bases intersects the plane of ABCD at a point that is not on a side of ABCD, then the lateral faces are parallelograms and the height of the prism is not equal to the heights of the parallelogram. 14365C11.pgs 7/12/07 1:06 PM Page 446 446 The Geometry of Three Dimensions 11-5 VOLUME OF A PRISM A cube whose edges each measure 1 centimeter is a unit of volume called a cubic centimeter. If the bases of a rectangular solid measure 8 centimeters by 5 centimeters, we know that the area of a base is 8 5 or 40 square centimeters and that 40 cubes each with a volume of 1 cubic centimeter can fill one base. If the height of the solid is 3 centimeters, we know that we can place 3 layers with 40 cubic centimeters in each layer to fill the rectangular solid. The volume of the solid is 40 3 or 120 cubic centimeters. The volume of the rectangular solid is the area of the base times the height. This can be applied to any prism and suggests the following postulate. 8 5 3 Postulate 11.5 The volume of a prism is equal to the area of the base times the height. If V represents the volume of a prism, B represents the area of the base, and h the height of the prism, then: V Bh H G E A F C 10 cm D 8 cm B 15 cm P L 15 cm R N 16 cm 10 cm Q M The figure shows two prisms. One is a parallelepiped with parallelograms ABCD and EFGH as bases and rectangular faces ABFE, BCGF, CDHG, and DAEH. If AB is 15 centimeters and the length of the altitude from D to is 8 centimeters, then the area of the base ABCD is 15 8 or 120 square centimeters. If BF, the height of the parallelepiped, is 10 centimeters, then: AB Volume of the parallelepiped 5 Bh 5 120 3 10 5 1,200 cubic centimeters 14365C11.pgs 7/12/07 1:06 PM Page 447 Volume of a Prism 447 The other prism has bases that are triangles, LMN and PQR. If LM is 16 centimeters and the length of the altitude to is 15 centimeters, then the area of a base is (16)(15) or 120 square centimeters. If the height of this prism is 10 centimeters, then: LM 1 2 Volume of the triangular prism 5 Bh 5 120 3 10 5 1,200 cubic centimeters Note that for these two prisms, the areas of the bases are equal and the heights of the prisms are equal. Therefore, the volumes of the prisms are equal. This is true in general since volume is defined as the area of the base times the height of the prism. The terms “base” and “height” are used in more than one way when describing a prism. For example, each of the congruent polygons in parallel planes is a base of the prism. The distance between the parallel planes is the height of the prism. In order to find the area of a base that is a triangle or a parallelogram, we use the length of a base and the height of the triangle or parallelogram. When finding the area of a lateral face that is a parallelogram, we use the length of the base and the height of that parallelogram. Care must be taken in distinguishing to what line segments the words “base” and “height” refer. EXAMPLE 1 The bases of a right prism are ABC and ABC with D a point on AD ' BC Find the volume of the prism. , CB , AB 10 cm, AC 10 cm, BC 12 cm, AD 8 cm, and BB 15 cm. Solution Since this is a right prism, all of the lateral faces are rectangles and the height of the prism, AA, is the height of each face. Each base is an isosceles triangle. The length of the base of the isosceles triangle is BC 12 cm, and the length of the altitude to the base of the triangle is AD 8 cm. C C 10 cm 12 cm D 8 cm A 10 cm B B 15 cm A Area of ABC 48 1 2(BC)(AD) 1 2(12)(8) Since the prism is a right prism, the height of the prism is BB 15. Volume of the prism 5 (area of a base)(height of the prism) 5 (48)(15) 5 720 cubic centimeters Answer 14365C11.pgs 7/12/07 1:06 PM Page 448 448 The Geometry of Three Dimensions Exercises Writing About Mathematics 1. Zoe said that if two solids have equal volumes and equal heights, then they must have con- gruent bases. Do you agree with Zo |
e? Justify your answer. 2. Piper said that the height of a prism is equal to the height of each of its lateral sides only if all of the lateral sides of the prism are rectangles. Do you agree with Piper? Explain why or why not. Developing Skills In 3–7, find the volume of each prism. 3. The area of the base is 48 square feet and the height is 18 inches. 4. The prism is a rectangular solid whose dimensions are 2.0 feet by 8.5 feet by 1.6 feet. 5. One base is a right triangle whose legs measure 5 inches and 7 inches. The height of the prism is 9 inches. 6. One base is a square whose sides measure 12 centimeters and the height of an altitude is 75 millimeters. 7. One base is parallelogram ABCD and the other is parallelogram ABCD, AB 47 cm, AB AAr ' AB , AAr ' AD 56 cm, , and the length of the perpendicular from D to AA 19 cm. Applying Skills 8. A fish tank in the form of a rectangular solid is to accommodate 6 fish, and each fish needs at least 7,500 cubic centimeters of space. The dimensions of the base are to be 30 centimeters by 60 centimeters. What is the minimum height that the tank needs to be? 9. A parallelepiped and a rectangular solid have equal volume and equal height. The bases of the rectangular solid measure 15 centimeters by 24 centimeters. If the length of one side of a base of the parallelepiped measures 20 centimeters, what must be the length of the altitude to that base? 10. A prism whose bases are triangles and one whose bases are squares have equal volume and equal height. Triangle ABC is one base of the triangular prism and PQRS is one base of the and AB PQ , what is the ratio of AB to square prism. If CD? Justify your answer. is the altitude from C to CD AB 11. Prove that a plane that lies between the bases of a triangular prism and is parallel to the bases intersects the lateral sides of the prism in a triangle congruent to the bases. 12. Prove that the lateral area of a right prism is equal to the perimeter of a base times the height of the prism. 14365C11.pgs 7/12/07 1:06 PM Page 449 11-6 PYRAMIDS A pyramid is a solid figure with a base that is a polygon and lateral faces that are triangles. Each lateral face shares a common edge with the base and a common edge with two other lateral faces. All lateral edges meet in a point called the vertex. The altitude of a pyramid is the perpendicular line segment from the vertex to the base ( in the diagram on the right.) The height of a pyramid is the length of the altitude. PC Pyramids 449 P vertex altitude C Examples of pyramids Regular Pyramids P regular pyramid A regular pyramid is a pyramid whose base is a regular polygon and whose altitude is perpendicular to the base at its center. The lateral edges of a regular polygon are congruent. Therefore, the lateral faces of a regular pyramid are isosceles triangles. The length of the altitude of a triangular lateral face of a regular pyramid, PB, is the slant height of the pyramid. slant height B Surface Area and Volume of a Pyramid D C D C A B A B The figure shows a prism and a pyramid that have congruent bases and equal heights. If we were to fill the pyramid with water and empty the water into the prism, we would need to do this three times to fill the prism. Thus, since the volume of a prism is given by Bh: Volume of a pyramid 1 3 Bh The lateral area of a pyramid is the sum of the areas of the faces. The total surface area is the lateral area plus the area of the bases. 14541C11.pgs 1/25/08 3:53 PM Page 450 450 The Geometry of Three Dimensions EXAMPLE 1 A regular pyramid has a square base and four lateral sides that are isosceles triangles. The length of an edge of the base is 10 centimeters and the height of the pyramid is 12 centimeters. The length of the altitude to the base of each lateral side is 13 centimeters. a. What is the total surface area of the pyramid? b. What is the volume of the pyramid? Solution Let e be the length of a side of the square base: e 10 cm Let hp be the height of the pyramid: hp 12 cm Let hs be the slant height of the pyramid: hs a. The base is a square with e 10 cm. 13 cm 12 cm 13 cm The area of the base is e2 (10)2 100 cm2. 10 cm Each lateral side is an isosceles triangle. The length of each base, e, is 10 centimeters and the height, hs, is 13 centimeters. 1 The area of each lateral side is 2ehs The total surface area of the pyramid is 100 4(65) 360 cm2. 1 2(10)(13) 65 cm2. b. The volume of the prism is one-third of the area of the base times the height of the pyramid. 3Bhp 3(100)(12) V 5 1 5 1 5 400 cm3 Answers a. 360 cm2 b. 400 cm3 Properties of Regular Pyramids The base of a regular pyramid is a regular polygon and the altitude is perpendicular to the base at its center. The center of a regular polygon is defined as the point that is equidistant to its vertices. In a regular polygon with three sides, an equilateral triangle, we proved that the perpendicular bisector of the sides of the triangle meet in a point that is equidistant from the vertices of the triangle. In a regular polygon with four sides, a square, we know that the diagonals are congruent. Therefore, the point at which the diagonals bisect each other is equidistant from the vertices. In Chapter 13, we will show that for any regular polygon, a point equidistant from the vertices exists. For now, we can use this fact to show that the lateral sides of a regular pyramid are isosceles triangles. 14541C11.pgs 1/25/08 3:53 PM Page 451 Pyramids 451 E D C M B A EM For example, consider a regular pyramid with square ABCD for a base and vertex E. The diagonals of ABCD intersect at M and AM BM CM DM. is perpendicular to the base, it is perpendicSince ular to any line in the base through M. Therefore, and EMA is a right angle. Also, EM ' MA and EMB is a right angle. Since the diagEM ' MB onals of a square are congruent and bisect each other, , EMA EMB MA > MB by SAS, and angles. Similar reasoning will lead us to conclude that ED > EA Therefore, we can make the following statement: . Then since EA > EB EM > EM because they are corresponding parts of congruent tri, and . A similar proof can be given for any base that is a regular polygon. EB > EC , EC > ED E The lateral faces of a regular pyramid are isosceles triangles. In the regular pyramid with base ABCD and vertex E, AB BC CD DA and AE BE CE DE D C Therefore, ABE BCE CDE DAE, that is, the lateral faces of the pyramid are congruent. The lateral faces of a regular pyramid are congruent. A B EXAMPLE 2 A regular pyramid has a base that is the hexagon is 2.5 cenABCDEF and vertex at V. If the length timeters, and the slant height of the pyramid is 6 centimeters, find the lateral area of the pyramid. AB V Solution The slant height of the pyramid is the height of a lat- E eral face. Therefore: Area of nABV 5 1 5 1 5 15 2bh 2(2.5)(6) 2 cm2 F D 6 c m C A 2.5 cm B The lateral faces of the regular pyramid are congruent. Therefore, they have equal areas. There are six lateral faces. Lateral area of the pyramid 5 6 15 2 B A 5 45 cm2 Answer 14365C11.pgs 7/12/07 1:06 PM Page 452 452 The Geometry of Three Dimensions Exercises Writing About Mathematics 1. Martin said that if the base of a regular pyramid is an equilateral triangle, then the foot of the altitude of the pyramid is the point at which the altitudes of the base intersect. Sarah said that it is the point at which the medians intersect. Who is correct? Justify your answer. 2. Are the lateral faces of a pyramid always congruent triangles? Explain your answer. Developing Skills In 3–5, the information refers to a regular pyramid. Let e be the length of an edge of the base and hs be the slant height. Find lateral area of each pyramid. 3. The pyramid has a square base; e 12 cm, hs 10 cm 4. The pyramid has a triangular base; e 8.0 ft, hs 10 ft 5. The pyramid has a base that is a hexagon; e 48 cm, hs 32 cm In 6–9, the information refers to a regular pyramid. Let e be the length of an edge of the base and hp be the height of the pyramid. Find the volume of each pyramid. 6. The area of the base is 144 square centimeters; hp 7. The area of the base is 27.6 square inches; hp 8. The pyramid has a square base; e 2 ft, hp 9. The pyramid has a square base; e 22 cm, hp 10. The volume of a pyramid is 576 cubic inches and the height of the pyramid is 18 inches. Find 5.0 in. 14 cm 1.5 ft 12 cm the area of the base. Applying Skills 11. A tetrahedron is a solid figure made up of four congruent equilateral triangles. Any one of the triangles can be considered to be the base and the other three to be the lateral sides of a regular pyramid. The length of a side of a triangle is 10.7 centimeters, the slant height is 9.27 centimeters, and the height of the prism is 8.74 centimeters. a. Find the area of the base of the tetrahedron. b. Find the lateral area of the tetrahedron. c. Find the total surface area of the tetrahedron. d. Find the volume of the tetrahedron. 12. When Connie camps, she uses a tent that is in the form of a regular pyramid with a square base. The length of an edge of the base is 9 feet and the height of the tent at its center is 8 feet. Find the volume of the space enclosed by the tent. 13. Prove that the lateral edges of a regular pyramid with a base that is an equilateral triangle are congruent. 14365C11.pgs 7/12/07 1:06 PM Page 453 14. Let F be the vertex of a pyramid with square base ABCD. If AF > CF , prove that the pyra- mid is regular. 15. Prove that the altitudes of the lateral faces of a regular pyramid with a base that is an equi- lateral triangle are congruent. Cylinders 453 16. Let p be the perimeter of the base of a regular pyramid and hs be the slant height. Prove that the lateral area of a regular pyramid is equal to 1 . 2phs 17. a. How does the lateral area of a regular pyramid change when both the slant height and the perimeter are doubled? tripled? Use the formula derived in exercise 16. b. How does the volume of a regular pyramid with a triangle for a base change when both the sides of the base and the |
height of the pyramid are doubled? tripled? 11-7 CYLINDERS P PPr A prism has bases that are congruent polygons in parallel planes. What if the bases were congruent closed curves instead of polygons? Let be a line segment joining corresponding points of two congruent curves. Imagine the moves along surface generated as the curves, always joining corresponding points of the bases. The solid figure formed by the congruent parallel curves and the surface that joins them is called a cylinder. PPr P P P base The closed curves form the bases of the cylinder and the surface that joins the bases is the lateral surface of the cylinder. The altitude of a cylinder is a line segment perpendicular to the bases with endpoints on the bases. The height of a cylinder is the length of an altitude. lateral surface altitude base The most common cylinder is one that has bases that are congruent circles. This cylinder is a circular cylinder. If the line segment joining the centers of the circular bases is perpendicular to the bases, the cylinder is a right circular cylinder. Circular cylinder Right circular cylinder 14365C11.pgs 7/12/07 1:06 PM Page 454 454 The Geometry of Three Dimensions Surface Area and Volume of a Circular Cylinder r The label on a cylindrical can of soup is a rectangle whose length is the circumference of the base of the can and whose width is the height of the can. This label is equal in area to the lateral surface of the cylindrical can. In Exercise 12 of Section 11-5, you proved that the lateral area, A, of a right prism is the product of the perimeter, p, of the prism and the height, hp, of the prism (A php). The circumference of the base of a cylinder corresponds to the perimeter of the base of a prism. Therefore, we can say that the area of the lateral surface of a right circular cylinder is equal to the circumference of the circular base times the height of the cylinder. SOUP SOUP 2r h If a right circular cylinder has bases that are circles of radius r and height h, then: The lateral area of the cylinder 2prh The total surface area of the cylinder 2prh 2pr 2 The volume of the cylinder Bh pr 2h Note: The volume of any circular cylindar is pr2h. Jenny wants to build a right circular cylinder out of cardboard with bases that have a radius of 6.0 centimeters and a height of 14 centimeters. a. How many square centimeters of cardboard are needed for the cylinder to the nearest square centimeter? b. What will be the volume of the cylinder to the nearest cubic centimeter? EXAMPLE 1 Solution a. A 5 2prh 1 2pr2 5 2p(6.0)(14) 1 2p(6.0)2 5 168p 172p 5 240p cm2 Express this result as a rational approximation rounded to the nearest square centimeter. 240p 753.9822369 754 cm2 V 5 pr2h b. 5 p(6.0)2(14) 5 504p cm3 14365C11.pgs 7/12/07 1:06 PM Page 455 Express this result as a rational approximation rounded to the nearest cubic centimeter. 504p 1,583.362697 1,584 cm3 Cylinders 455 Answers a. 754 cm2 b. 1,584 cm3 Exercises Writing About Mathematics 1. Amy said that if the radius of a circular cylinder were doubled and the height decreased by one-half, the volume of the cylinder would remain unchanged. Do you agree with Amy? Explain why or why not. 2. Cindy said that if the radius of a right circular cylinder were doubled and the height decreased by one-half, the lateral area of the cylinder would remain unchanged. Do you agree with Cindy? Explain why or why not. Developing Skills In 3–6, the radius of a base, r, and the height, h, of a right circular cylinder are given. Find for each cylinder: a. the lateral area, b. the total surface area, c. the volume. Express each measure as an exact value in terms of p. 3. r 34.0 cm, h 60.0 cm 4. r 4.0 in., h 12 in. 5. r 18.0 in., h 2.00 ft 6. r 1.00 m, h 75.0 cm 7. The volume of a right circular cylinder is 252 cubic centimeters and the radius of the base is 3.6 centimeters. What is the height of the cylinder to the nearest tenth? 8. The volume of a right circular cylinder is 586 cubic centimeters and the height of the cylin- der is 4.6 centimeters. What is the radius of the base to the nearest tenth? 9. The areas of the bases of a cylinder are each 124 square inches and the volume of the cylinder is 1,116 cubic inches. What is the height, h, of the cylinder? h 10. A circular cylinder has a base with a radius of 7.5 centimeters and a height of 12 centimeters. A rectangular prism has a square base and a height of 8.0 centimeters. If the cylinder and the prism have equal volumes, what is the length of the base of the prism to the nearest tenth? 14365C11.pgs 7/12/07 1:06 PM Page 456 456 The Geometry of Three Dimensions Applying Skills 11. A can of beets has a top with a diameter of 2.9 inches and a height of 4.2 inches. What is the volume of the can to the nearest tenth? 12. A truck that delivers gasoline has a circular cylindrical storage space. The diameter of the bases of the cylinder is 11 feet, and the length (the height of the cylinder) is 17 feet. How many whole gallons of gasoline does the truck hold? (Use 1 cubic foot 7.5 gallons.) 13. Karen makes pottery on a potter’s wheel. Today she is making vases that are in the shape of a circular cylinder that is open at the top, that is, it has only one base. The base has a radius of 4.5 centimeters and is 0.75 centimeters thick. The lateral surface of the cylinder will be 0.4 centimeters thick. She uses 206 cubic centimeters of clay for each vase. a. How much clay is used for the base of the vase to the nearest tenth? b. How much clay will be used for the lateral surface of the vase to the nearest tenth? c. How tall will a vase be to the nearest tenth? d. What will be the area of the lateral surface to the nearest tenth? (Use the value of the height of the vase found in part c.) 14. Mrs. Taggart sells basic cookie dough mix. She has been using circular cylindrical containers to package the mix but wants to change to rectangular prisms that will pack in cartons for shipping more efficiently. Her present packaging has a circular base with a diameter of 4.0 inches and a height of 5.8 inches. She wants the height of the new package to be 6.0 inches and the dimensions of the base to be in the ratio 2 : 5. Find, to the nearest tenth, the dimensions of the new package if the volume is to be the same as the volume of the cylindrical containers. 11-8 CONES vertex Q altitude slant height O O base P Think of g OQ perpendicular to plane p at O. Think of a point P on plane p. Keeping point Q fixed, move P through a circle on p with center at O. The surface generated by g PQ is a right circular conical surface. Note that a conical surface extends infinitely. In our discussion, we will consider the part of the conical surface generated by PQ from plane p to Q, called a right circular cone. The point Q is the vertex of the cone. The circle in plane p with radius OP is the base of the cone, is the altitude of the cone, OQ is the height of the cone, and PQ is the slant height of the cone. OQ Q O P p 14365C11.pgs 7/12/07 1:06 PM Page 457 Cones 457 We can make a model of a right circular cone. Draw a large circle on a piece of paper and draw two radii. Cut out the circle and remove the part of the circle between the two radii. Join the two cut edges of the remaining part of the circle with tape. Surface Area and Volume of a Cone For a pyramid, we proved that the lateral area is equal to one-half the product of the perimeter of the base and the slant height. A similar relationship is true for a cone. The lateral area of a cone is equal to one-half the product of the circumference of the base and the slant height. Let L be the lateral area of the cone, C be the circumference of the base, S be the total surface area of the cone, hs be the slant height, and r be the radius of the base. Then: L 1 2Chs 1 2(2pr) hs prhs S L pr2 prhs pr2 hs hc r We can also use the relationship between the volume of a prism and the volume of a pyramid to write a formula for the volume of a cone. The volume of a cone is equal to one-third the product of the area of the base and the height of the cone. Let V be the volume of the cone, B be the area of the base with radius r, and hc be the height of the cone. Then: V 1 3Bhc 1 3pr 2hc hs hc r EXAMPLE 1 A right circular cone has a base with a radius of 10 inches, a height of 24 inches, and a slant height of 26 inches. Find the exact values of: a. the lateral area b. the area of the base c. the total surface area of the cone 26 in. 24 in. 10 in. Solution The radius of the base, r, is 10 inches. Therefore, the diameter of the base is 20 inches, and the circumference of the base, C, is 20p, the slant height, hs, is 26 inches, and the height of the cone, hc, is 24 inches. a. Lateral area 2Chs 2(20p)(26) 5 1 5 1 5 260p in.2 Answer 14365C11.pgs 7/12/07 1:06 PM Page 458 458 The Geometry of Three Dimensions b. Area of the base 5 p(10)2 5 100p cm2 Answer c. Total surface area 5 260p 1100p 5 360p in.2 Answer 26 in. 24 in. 10 in. EXAMPLE 2 A cone and a cylinder have equal volumes and equal heights. If the radius of the base of the cone is 3 centimeters, what is the radius of the base of the cylinder? Solution Let r be the radius of the base of the cylinder and h be the height of both the cylinder and the cone. Volume of the cylinder pr2h 1 Volume of the cone 3p(3)2h Volume of the cylinder Volume of the cone pr2h 1 3p(3)2h r2 3 r 3 " Answer Exercises Writing About Mathematics 1. Elaine said that if a pyramid and a cone have equal heights and bases that have equal areas, then they have equal lateral areas. Do you agree with Elaine? Justify your answer. 2. Josephus said that if two cones have equal heights and the radius of one cone is equal to the diameter of the other, then the volume of the larger cone is twice the volume of the smaller. Do you agree with Josephus? Explain why or why not. Developing Skills In 3–6, the radius of the base, r, the slant height of the cone, hs, and the height of the cone, hc, are given. Find: a. the lateral area of |
the cone, b. the total surface area of the cone, c. the volume of the cone. Express each measure as an exact value in terms of p and rounded to the nearest tenth. 3. r 3.00 cm, hs 4. r 5.00 cm, hs 5. r 24 cm, hs 6. r 8.00 cm, hs 5.00 cm, hc 13.0 cm, hc 4.00 cm 12.0 cm 10.0 cm, hc 25 cm, hc 6.00 cm 7.0 cm 14365C11.pgs 7/12/07 1:06 PM Page 459 Spheres 459 7. The volume of a cone is 127 cubic inches and the height of the cone is 6.0 inches. What is the radius of the base to the nearest tenth? 8. The volume of a cone is 56 cubic centimeters and the area of the base is 48 square centime- ters. What is the height of the cone to the nearest tenth? 9. The area of the base of a cone is equal to the area of the base of a cylinder, and their volumes are equal. If the height of the cylinder is 2 feet, what is the height of the cone? Applying Skills 10. The highway department has a supply of road salt for use in the coming winter. The salt forms a cone that has a height of 10 feet and a circular base with a diameter of 12 feet. How many cubic feet of salt does the department have stored? Round to the nearest foot. 11. The spire of the city hall is in the shape of a cone that has a circular base that is 20 feet in diameter. The slant height of the cone is 40 feet. How many whole gallons of paint will be needed to paint the spire if a gallon of paint will cover 350 square feet? 12. A cone with a height of 10 inches and a base with a radius of 6 inches is cut into two parts by a plane parallel to the base. The upper part is a cone with a height of 4 inches and a base with a radius of 2.4 inches. Find the volume of the lower part, the frustum of the cone, in terms of p. 13. When a cone is cut by a plane perpendicular to the base through the center of the base, the cut surface is a triangle whose base is the diameter of the base of the cone and whose altitude is the altitude of the cone. If the radius of the base is equal to the height of the cone, prove that the cut surface is an isosceles right triangle. 11-9 SPHERES hs hc hc r hc r In a plane, the set of all points at a given distance from a fixed point is a circle. In space, this set of points is a sphere. DEFINITION A sphere is the set of all points equidistant from a fixed point called the center. The radius of a sphere is the length of the line segment from the center of the sphere to any point on the sphere. 14365C11.pgs 7/31/07 1:19 PM Page 460 460 The Geometry of Three Dimensions O O O P p p p No points in common 1 point in common Infinitely many points in common If the distance of a plane from the center of a sphere is greater than the radius of the sphere, the plane will have no points in common with the sphere. If the distance of a plane from the center of a sphere is equal to the radius of the sphere, the plane will have one point in common with the sphere. If the distance of a plane from the center of a sphere is less than the radius of the sphere, the plane will have infinitely many points in common with the sphere. We can prove that these points form a circle. Recall the definition of a circle. A circle is the set of all points in a plane equidistant from a fixed point in the plane called the center. Theorem 11.14a The intersection of a sphere and a plane through the center of the sphere is a circle whose radius is equal to the radius of the sphere. Given A sphere with center at O and radius r. Plane p through O intersects the sphere. Prove The intersection is a circle with radius r. O A p Proof Let A be any point on the intersection. Since A is on the sphere, OA r. Therefore, every point on the intersection is at the same distance from O and the intersection is a circle with radius r. DEFINITION A great circle of a sphere is the intersection of a sphere and a plane through the center of the sphere. Theorem 11.14b If the intersection of a sphere and a plane does not contain the center of the sphere, then the intersection is a circle. Given A sphere with center at O plane p intersecting the sphere at A and B. Prove The intersection is a circle. O B AC p 14365C11.pgs 7/12/07 1:06 PM Page 461 Proof Statements Reasons 1. Draw a line through O, perpendicular to plane p at C. 1. Through a given point there is one line perpendicular to a given plane. Spheres 461 2. OCA and OCB are right angles. 3. OA > OB OC > OC 4. 5. OAC OBC 6. CA > CB 7. The intersection is a circle. 2. A line perpendicular to a plane is perpendicular to every line in the plane through the intersection of the line and the plane. 3. A sphere is the set of points in space equidistant from a fixed point. 4. Reflexive property. 5. HL. 6. Corresponding sides of congruent triangles are congruent. 7. A circle is the set of all points in a plane equidistant from a fixed point. We can write Theorems 11.14a and 11.14b as a single theorem. Theorem 11.14 The intersection of a plane and a sphere is a circle. In the proof of Theorem 11.14b, we drew right triangle OAC with OA the radius of the sphere and AC the radius of the circle at which the plane and the sphere intersect. Since OCA is the right angle, it is the largest angle of OAC and OA AC. Therefore, a great circle, whose radius is equal to the radius of the sphere, is larger than any other circle that can be drawn on the sphere. We have just proved the following corollary: Corollary 11.14a A great circle is the largest circle that can be drawn on a sphere. Let p and q be any two planes that intersect the sphere with center at O. In the proof of Theorem 11.14b, the radius of the circle is the length of a leg of a right triangle whose hypotenuse is the radius of the sphere and whose other leg is the distance from the center of the circle to the plane. This suggests that if two planes are equidistant from the center of a sphere, they intersect the sphere in congruent circles. 14365C11.pgs 7/12/07 1:06 PM Page 462 462 The Geometry of Three Dimensions Theorem 11.15 If two planes are equidistant from the center of a sphere and intersect the sphere, then the intersections are congruent circles. Given A sphere with center at O inter- p sected by planes p and q, OA OB, OA ' p OB ' q and . Prove The intersections are congruent circles. A O C Proof Let C be any point on the intersection with p and D be any point on the intersection with q. Then OA OB and OC OD (they are both radii of the sphere). Therefore, OAC and OBD are congruent right triangles by HL. Since the corresponding sides of congruent triangles are congruent, the radii of the circles, AC and BD, are equal and the circles are congruent. D B q Surface Area and Volume of a Sphere The formulas for the surface area and volume of a sphere are derived in advanced courses in mathematics. We can state and make use of these formulas. The surface area of a sphere is equal to the area of four great circles. Let S be the surface area of a sphere of radius r. Then the surface area of the sphere is: S 4pr 2 The volume of a sphere is equal to four-thirds the product of p and the cube of the radius. Let V be the volume of a sphere of radius r. Then the volume of the sphere is: V 4 3pr 3 Find the surface area and the volume of a sphere whose radius is 5.25 centimeters to the nearest centimeter. S 5 4pr2 5 4p(5.25)2 5 110.25p < 346.3605901 cm2 EXAMPLE 1 Solution 14365C11.pgs 7/12/07 1:06 PM Page 463 When we round to the nearest centimeter to express the answer, S 346 cm2. Answer Spheres 463 V 5 4 3pr3 5 4 3p(5.25)3 5 192.9375p < 606.1310326 cm3 When we round to the nearest centimeter to express the answer, V 606 cm3. Answer Exercises Writing About Mathematics 1. Meg said that if d is the diameter of a sphere, then the surface area of a sphere is equal to pd2. Do you agree with Meg? Justify your answer. 2. Tim said that if the base a cone is congruent to a great circle of a sphere and the height of the cone is the radius of the sphere, then the volume of the cone is one-half the volume of the sphere. Do you agree with Tim? Justify your answer. Developing Skills In 3–6, find the surface area and the volume of each sphere whose radius, r, is given. Express each answer in terms of p and as a rational approximation to the nearest unit. 3. r 7.50 in. 4. r 13.2 cm 5. r 2.00 ft 6. r 22.3 cm 7. Find the radius of a sphere whose surface area is 100p square feet. 8. Find, to the nearest tenth, the radius of a sphere whose surface area is 84 square centimeters. 9. Find, to the nearest tenth, the radius of a sphere whose volume is 897 cubic inches. 10. Express, in terms of p, the volume of a sphere whose surface area is 196p square inches. Applying Skills 11. A vase is in the shape of a sphere with a radius of 3 inches. How many whole cups of water will come closest to filling the vase? (1 cup 14.4 cubic inches) 12. The radius of a ball is 5.0 inches. The ball is made of a soft foam that weighs 1 ounce per 40 cubic inches. How much does the ball weigh to the nearest tenth? 14365C11.pgs 8/2/07 5:54 PM Page 464 464 The Geometry of Three Dimensions 13. The diameter of the earth is about 7,960 miles. What is the surface area of the earth in terms of p? 14. The diameter of the moon is about 2,160 miles. What is the surface area of the moon in terms of p? 15. A cylinder has a base congruent to a great circle of a sphere and a height equal to the diameter of the sphere. If the radius of the sphere is 16 centimeters, compare the lateral area of the cylinder and the surface area of the sphere. 16. A cylinder has a base congruent to a great circle of a sphere and a height equal to the diameter of the sphere. If the diameter of the sphere is r, compare the lateral area of the cylinder and the surface area of the sphere. Hands-On Activity A symmetry plane is a plane that divides a solid into two congruent parts. For each solid shown below, determine the number of symmetry planes and describe their position relative to the solid. a. d. b. c. Rectangular prism Prism with isosceles triangular base Regular pyramid with square base e. f. Cube Cube S |
phere Right circular cylinder CHAPTER SUMMARY Definitions to Know • Parallel lines in space are lines in the same plane that have no points in common. • Skew lines are lines in space that are neither parallel nor intersecting. • A dihedral angle is the union of two half-planes with a common edge. • The measure of a dihedral angle is the measure of the plane angle formed by two rays each in a different half-plane of the angle and each perpendicular to the common edge at the same point of the edge. 14365C11.pgs 8/2/07 5:54 PM Page 465 Chapter Summary 465 • Perpendicular planes are two planes that intersect to form a right dihedral angle. • A line is perpendicular to a plane if and only if it is perpendicular to each line in the plane through the intersection of the line and the plane. • A plane is perpendicular to a line if the line is perpendicular to the plane. • Parallel planes are planes that have no points in common. • A line is parallel to a plane if it has no points in common with the plane. • The distance between two planes is the length of the line segment perpen- dicular to both planes with an endpoint on each plane. • A polyhedron is a three-dimensional figure formed by the union of the surfaces enclosed by plane figures. • The portions of the planes enclosed by a plane figure are called the faces of the polyhedron. • The intersections of the faces are the edges of the polyhedron, and the intersections of the edges are the vertices of the polyhedron. • A prism is a polyhedron in which two of the faces, called the bases of the prism, are congruent polygons in parallel planes. • The sides of a prism that are not bases are called the lateral sides. • The union of two lateral sides is a lateral edge. • If the line segments joining the corresponding vertices of the bases of a prism are perpendicular to the planes of the bases, then the prism is a right prism. • The altitude of a prism is a line segment perpendicular to each of the bases with an endpoint on each base. • The height of a prism is the length of an altitude. • A parallelepiped is a prism that has parallelograms as bases. • A rectangular parallelepiped is a parallelepiped that has rectangular bases and lateral edges perpendicular to the bases. • The lateral area of the prism is the sum of the areas of the lateral faces. • The total surface area of a solid figure is the sum of the lateral area and the areas of the bases. • A pyramid is a solid figure with a base that is a polygon and lateral faces that are triangles. • A regular pyramid is a pyramid whose base is a regular polygon and whose altitude is perpendicular to the base at its center. • The length of the altitude of a triangular lateral face of a regular pyramid is the slant height of the pyramid. • A cylinder is a solid figure formed by congruent parallel curves and the surface that joins them. 14365C11.pgs 8/2/07 5:54 PM Page 466 466 The Geometry of Three Dimensions • If the line segment joining the centers of circular bases of a cylinder is per- pendicular to the bases, the cylinder is a right circular cylinder. • Let OP be a line segment perpendicular to a plane at O and A be a point on a circle in the plane with center at O. A right circular cone is the solid figure that is the union of a circular base and the surface generated by line segment as P moves around the circle. AP • A sphere is the set of all points equidistant from a fixed point called the center. • The radius of a sphere is the length of the line segment from the center of the sphere to any point on the sphere. • A great circle of a sphere is the intersection of a sphere and a plane through the center of the sphere. Postulates 11.1 There is one and only one plane containing three non-collinear points. 11.2 A plane containing any two points contains all of the points on the line determined by those two points. If two planes intersect, then they intersect in exactly one line. 11.3 11.4 At a given point on a line, there are infinitely many lines perpendicular Theorems and Corollaries 11.5 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 to the given line. The volume of a prism is equal to the area of the base times the height. There is exactly one plane containing a line and a point not on the line. Two intersecting lines determine a plane. If a line not in a plane intersects the plane, then it intersects in exactly one point. If a line is perpendicular to each of two intersecting lines at their point of intersection, then the line is perpendicular to the plane determined by these lines. Two planes are perpendicular if and only if one plane contains a line perpendicular to the other. Through a given point on a plane, there is only one line perpendicular to the given plane. Through a given point on a line, there can be only one plane perpendicular to the given line. If a line is perpendicular to a plane, then any line perpendicular to the given line at its point of intersection with the given plane is in the plane. If a line is perpendicular to a plane, then every plane containing the line is perpendicular to the given plane. If a plane intersects two parallel planes, then the intersection is two parallel lines. 11.11 Two lines perpendicular to the same plane are parallel. 11.11a Two lines perpendicular to the same plane are coplanar. 11.12 Two planes are perpendicular to the same line if and only if the planes are parallel. 11.13 Parallel planes are everywhere equidistant. 14541C11.pgs 1/25/08 3:53 PM Page 467 Vocabulary 467 11.14 The intersection of a plane and a sphere is a circle. 11.14a A great circle is the largest circle that can be drawn on a sphere. 11.15 If two planes are equidistant from the center of a sphere and intersect the sphere, then the intersections are congruent circles. Formulas L lateral area S surface area C circumference of the base of a cone V volume p perimeter of a base r radius of a base of a cylinder or a cone; radius of a sphere B area of a base hs hp hc height of a lateral surface height of a prism or a pyramid height of a cylinder or a cone Right Prism L phs S L 2B V Bhp Regular Pyramid 1 L phs 2 S L B 1 V 3Bhp Right Circular Cylinder L 2prhc S 2prhc V Bhc 2pr 2 pr 2hc Cone L V 1 2Chs 1 3Bhc prhs 1 3pr 2hc Sphere S 4pr 2 V 4 3pr 3 VOCABULARY 11-1 Cavalieri’s Principle • Solid geometry • Parallel lines in space • Skew lines 11-2 Dihedral angle • Plane angle • Measure of a dihedral angle • Perpendicular planes • Line perpendicular to a plane • Plane perpendicular to a line 11-3 Parallel planes • Line parallel to a plane • Distance between two planes 11-4 Polyhedron • Faces of a polyhedron • Edges of a polyhedron • Vertices of a polyhedron • Prism • Bases of a prism • Lateral sides of a prism • Lateral edge of a prism • Altitude of a prism • Height of a prism • Right prism • Parallelepiped • Rectangular parallelepiped • Rectangular solid • Lateral area • Total surface area 11-5 Cubic centimeter 11-6 Pyramid • Vertex of a pyramid • Altitude of a pyramid • Height of a pyramid • Regular pyramid • Slant height of a pyramid 11-7 Cylinder • Base of a cylinder • Lateral surface of a cylinder • Altitude of a cylinder • Height of a cylinder • Right circular cylinder 14365C11.pgs 7/12/07 1:06 PM Page 468 468 The Geometry of Three Dimensions 11-8 Right circular conical surface • Right circular cone • Vertex of a cone • Base of a cone • Altitude of a cone • Height of a cone • Slant height of a cone • Frustum of a cone 11-9 Sphere • Center of a sphere • Radius of a sphere • Great circle of a sphere • Symmetry plane REVIEW EXERCISES In 1–16, answer each question and state the postulate, theorem, or definition that justifies your answer or draw a counterexample. 1. Lines g CD . Is g AB g EF lines? and g CD intersect at E. A line, g EF , is perpendicular to g AB and to perpendicular to the plane determined by the intersecting 2. Plane p is perpendicular to g AB at B. Plane q intersects g AB at B. Can q be g ? AB perpendicular to g RS g RT g AB p, can and 4. Lines 3. A line, , is perpendicular to plane p at R. If T is a second point not on be perpendicular to plane p? g LM are each perpendicular to plane p. Are g AB and g LM coplanar? g AB 5. A line is in plane q and g AB is perpendicular to plane p. Are planes p and q perpendicular? 6. Two planes, p and q, are perpendicular to each other. Does p contain a line perpendicular to q? g AB 7. A line is perpendicular to plane p at B and g BC g ' AB . Is g BC in plane p? 8. A line g RS is perpendicular to plane p and g RS is in plane q. Is plane p per- pendicular to plane q? 9. Plane r intersects plane p in g AB intersect g ? CD g AB and plane r intersects plane q in g CD . Can 10. Planes p and q are each perpendicular to g AB . Are p and q parallel? 11. Two lateral edges of a prism are AB and 12. Two lateral edges of a prism are AB and CD CD g and . Can AB . Is AB CD? g CD intersect? 13. Two prisms have equal heights and bases with equal areas. Do the prisms have equal volumes? 14365C11.pgs 7/12/07 1:06 PM Page 469 Review Exercises 469 14. A prism and a pyramid have equal heights and bases with equal areas. Do the prism and the pyramid have equal volumes? 15. Two planes intersect a sphere at equal distances from the center of the sphere. Are the circles at which the planes intersect the sphere congruent? 16. Plane p intersects a sphere 2 centimeters from the center of the sphere and plane q contains the center of the sphere and intersects the sphere. Are the circles at which the planes intersect the sphere congruent circles? In 17–22, find the lateral area, total surface area, and volume of each solid figure to the nearest unit. 17. The length of each side of the square base of a rectangular prism is 8 cen- timeters and the height is 12 centimeters. 18. The height of a prism with bases that are right triangles is 5 inches. The lengths of the sides of the bases are 9, 12, and 15 inches. 19. The base of a rectangular solid measures 9 feet by 7 feet and the height o |
f the solid is 4 feet. 20. A pyramid has a square base with an edge that measures 6 inches. The slant height of a lateral side is 5 inches and the height of the pyramid is 4 inches. 21. The diameter of the base of a cone is 10 feet, its height is 12 feet, and its slant height is 13 feet. 22. The radius of the base of a right circular cylinder is 7 centimeters and the height of the cylinder is 9 centimeters. 23. A cone and a pyramid have equal volumes and equal heights. Each side of the square base of the pyramid measures 5 meters. What is the radius of the base of the cone? Round to the nearest tenth. 24. Two prisms with square bases have equal volumes. The height of one prism is twice the height of the other. If the measure of a side of the base of the prism with the shorter height is 14 centimeters, find the measure of a side of the base of the other prism in simplest radical form. 25. Ice cream is sold by street vendors in containers that are right circular cylinders. The base of the cylinder has a diameter of 5 inches and the cylinder has a height of 6 inches. a. Find, to the nearest tenth, the amount of ice cream that a container can hold. b. If a scoop of ice cream is a sphere with a diameter of 2.4 inches, find, to the nearest tenth, the amount of ice cream in a single scoop. c. If the ice cream is packed down into the container, how many whole scoops of ice cream will come closest to filling the container? 14365C11.pgs 8/2/07 5:55 PM Page 470 470 The Geometry of Three Dimensions Exploration A regular polyhedron is a solid, all of whose faces are congruent regular polygons with the sides of the same number of polygons meeting at each vertex. There are five regular polyhedra: a tetrahedron, a cube, an octahedron, a dodecahedron, and an icosahedron. These regular polyhedra are called the Platonic solids. Tetrahedron Cube Octahedron Dodecahedron Icosahedron a. Make a paper model of the Platonic solids as follows: (1) Draw the diagrams below on paper. (2) Cut out the diagrams along the solid lines and fold along the dotted lines. (3) Tape the folded sides together. Note: You may wish to enlarge to the diagrams for easier folding. Cube Tetrahedron Dodecahedron Octahedron Icosahedron 14365C11.pgs 7/12/07 1:06 PM Page 471 Cumulative Review 471 b. Using the solids you constructed in part a, fill in the table below: Number of vertices Number of faces Number of edges Tetrahedron Cube Octahedron Dodecahedron Icosahedron Do you observe a relationship among the number of vertices, faces, and edges for each Platonic solid? If so, state this relationship. c. Research the five Platonic solids and investigate why there are only five regular polyhedra. Share your findings with your classmates. CUMULATIVE REVIEW Chapters 1–11 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. If the measures of the angles of a triangle are represented by x, 2x – 20, and 2x, what is the measure of the smallest angle? (1) 40 (3) 80 (2) 60 (4) 90 2. In ABC, mA 40 and the measure of an exterior angle at B is 130. The triangle is (1) scalene and acute (2) scalene and right (3) isosceles and right (4) isosceles and acute 3. The coordinates of the midpoint of a segment with endpoints at (2, –3) and (6, 1) are (1) (4, 2) (2) (4, 2) (3) (4, 2) (4) (2, 1) 4. The lengths of the diagonals of a rhombus are 8 centimeters and 12 cen- timeters. The area of the rhombus is (1) 24 cm2 (2) 32 cm2 (3) 48 cm2 (4) 96 cm2 5. Two parallel lines are cut by a transversal. The measure of one interior angle is x 7 and the measure of another interior angle on the same side of the transversal is 3x 3. What is the value of x? (1) 5 (2) 12 (4) 51 (3) 44 14365C11.pgs 7/12/07 1:06 PM Page 472 472 The Geometry of Three Dimensions 6. If “Today is Monday” is true and “It is May 5” is false, which of the follow- ing is true? (1) Today is Monday and it is May 5. (2) If today is Monday, then it is May 5. (3) Today is Monday only if it is May 5. (4) Today is Monday or it is May 5. 7. Which of the following do not always lie in the same plane? (1) three points (2) two parallel lines (3) two intersecting lines (4) three parallel lines 8. What is the slope of a line perpendicular to the line whose equation is x 2y 3? (1) 1 2 (2) 2 (3) 21 2 (4) 2 9. A quadrilateral has diagonals that are not congruent and are perpendicu- lar bisectors of each other. The quadrilateral is a (1) square (2) rectangle (3) trapezoid (4) rhombus 10. The base of a right prism is a square whose area is 36 square centimeters. The height of the prism is 5 centimeters. The lateral area of a prism is (4) 180 cm2 (1) 30 cm2 (3) 120 cm2 (2) 60 cm2 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. A leg, AB , of isosceles ABC is congruent to a leg, DE , of isosceles DEF. The vertex angle, B, of isosceles ABC is congruent to the vertex angle, E, of isosceles DEF. Prove that ABC DEF. 12. In triangle ABC, altitude CD bisects C. Prove that the triangle is isosceles. Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Quadrilateral BCDE is a parallelogram and B is the midpoint of ABC . Prove that ABDE is a parallelogram. 14365C11.pgs 7/12/07 1:06 PM Page 473 14. Line segment ABC is perpendicular to plane p at B, the midpoint of ABC . Prove that any point on p is equidistant from A and C. Cumulative Review 473 Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Write the equation of the median from B to AC of ABC if the coordi- nates of the vertices are A(3, 2), B(1, 4), and C(5, 2). 16. The coordinates of the endpoints of ArBr coordinates of the endpoints of . tion + R908 rx-axis AB , the image of are A(1, 3) and B(5, 1). Find the under the composi- AB 14365C12.pgs 7/10/07 8:56 AM Page 474 CHAPTER 12 CHAPTER TABLE OF CONTENTS 12-1 Ratio and Proportion 12-2 Proportions Involving Line Segments 12-3 Similar Polygons 12-4 Proving Triangles Similar 12-5 Dilations 12-6 Proportional Relations Among Segments Related to Triangles 12-7 Concurrence of the Medians of a Triangle 12-8 Proportions in a Right Triangle 12-9 Pythagorean Theorem 12-10 The Distance Formula Chapter Summary Vocabulary Review Exercises Cumulative Review 474 RATIO, PROPORTION, AND SIMILARITY The relationship that we know as the Pythagorean Theorem was known by philosophers and mathematicians before the time of Pythagoras (c. 582–507 B.C.). The Indian mathematician Baudha¯yana discovered the theorem more than 300 years before Pythagoras. The Egyptians made use of the 3-4-5 right triangle to determine a right angle. It may have been in Egypt, where Pythagoras studied, that he become aware of this relationship. Ancient sources agree that Pythagoras gave a proof of this theorem but no original documents exist from that time period. Early Greek statements of this theorem did not use the algebraic form c2 a2 b2 with which we are familiar. Proposition 47 in Book 1 of Euclid’s Elements states the theorem as follows: “In right angled triangles, the square on the side subtending the right angle is equal to the sum of the squares on the sides containing the right angle.” Euclid’s proof drew squares on the sides of the right triangle and proved that the area of the square drawn on the hypotenuse was equal to the sum of the areas of the squares drawn on the legs. There exist today hundreds of proofs of this theorem. 14365C12.pgs 7/10/07 8:56 AM Page 475 12-1 RATIO AND PROPORTION Ratio and Proportion 475 People often use a computer to share pictures with one another. At first the pictures may be shown on the computer screen as many small frames. These small pictures can be enlarged on the screen so that it is easier to see detail. Or a picture may be printed and then enlarged. Each picture, whether on the screen or printed, is similar to the original. When a picture is enlarged, the dimensions of each shape are in proportion to each other and with angle measure remaining the same. Shapes that are related in this way are said to be similar. In this chapter we will review what we already know about ratio and pro- portion and apply those ideas to geometric figures. The Meaning of Ratio DEFINITION The ratio of two numbers, a and b, where b is not zero, is the number .a b The ratio can also be written as a : b. The two triangles ABC and DEF have a b the same shape but not the same size: AB 20 millimeters and DE 10 millimelengths by ters. We can compare these means of a ratio, 20 10 or 20 : 10. Since a ratio, like a fraction, is a comparison of two numbers by division, a ratio can be simplified by dividing each term of the ratio by a common factor. Therefore, the ratio of AB to DE can be written as 10 : 5 or as 4 : 2 or as 2 : 1. A ratio is in simplest form when the terms of the ratio have no common factor greater than 1. E F D B C A When the numbers represent lengths such as AB and DE, the lengths must be expressed in terms of the same unit of measure for the ratio to be meaningful. For example, if AB had been given as 2 centimeters, it would have been necessary to change 2 centimeters to 20 millimeters before writing the ratio of AB to DE. Or we could have changed the length of , 10 millimeters, to 1 centiDE meter before writing the ratio of |
AB to DE as 2 : 1. • When using millimeters, the ratio 20 mm : 10 mm 2 : 1. • When using centimeters, the ratio 2 cm : 1 cm 2 : 1. A ratio can also be used to express the relationship among three or more numbers. For example, if the measures of the angles of a triangle are 45, 60, and 75, the ratio of these measures can be written as 45 : 60 : 75 or, in lowest terms, 3 : 4 : 5. 14365C12.pgs 7/10/07 8:56 AM Page 476 476 Ratio, Proportion, and Similarity When we do not know the actual values of two or more measures that are in a given ratio, we use a variable factor to express these measures. For example, if the lengths of the sides of a triangle are in the ratio 3 : 3 : 4, we can let x be the greatest common factor of the measures of the sides. Then the measures of the sides may be expressed as 3x, 3x, and 4x. If the perimeter of the triangle is 120 centimeters, this use of the variable x allows us to write and solve an equation. 3x 3x 4x 120 10x 120 x 12 The measures of the sides of the triangle are 3(12), 3(12), and 4(12) or 36 centimeters, 36 centimeters, and 48 centimeters. The Meaning of Proportion Since the ratio 12 : 16 is equal to the ratio 3 : 4, we may write 16 5 3 12 4 . The equa- is called a proportion. The proportion can also be written as 16 5 3 12 tion 4 12 : 16 3 : 4. DEFINITION A proportion is an equation that states that two ratios are equal. b 5 c a d The proportion can be written also as a : b c : d. The four numbers a, b, c, and d are the terms of the proportion. The first and fourth terms, a and d, are the extremes of the proportion, and the second and third terms, b and c, are the means. C extremes a : b c : d means B Theorem 12.1 In a proportion, the product of the means is equal to the product of the extremes. Given b 5 c a d with b 0 and d 0 Prove ad bc 14365C12.pgs 7/10/07 8:56 AM Page 477 Proof We can give an algebraic proof of this theorem. Ratio and Proportion 477 Statements b 5 c a d a 5 bd bd b B b(ad) 5 d b c d B A d(bc) A 1. 2. 3. 4. 1(ad) 1(bc) 5. ad bc Reasons 1. Given. 2. Multiplication postulate. 3. Associative property of multiplication. 4. A quantity may be substituted for its equal. 5. Multiplicative identity. Corollary 12.1a In a proportion, the means may be interchanged. Given Prove Proof with b 0, c 0, and It is given that postulate of equality. Therefore, and that c 0. Then Corollary 12.1b In a proportion, the extremes may be interchanged. Given Prove Proof with a 0, b 0, and It is given that postulate of equality. Therefore, and that a 0. Then by the multiplication by the multiplication These two corollaries tell us that: If 3 : 5 12 : 20, then 3: 12 5 : 20 and 20 : 5 12 : 3. Any two pairs of factors of the same number can be the means and the extremes of a proportion. For example, since 2(12) 3(8), 2 and 12 can be the means of a proportion and 3 and 8 can be the extremes. We can write several proportions: 3 2 5 12 8 8 2 5 12 3 3 12 5 2 8 8 12 5 2 3 14365C12.pgs 7/10/07 8:56 AM Page 478 478 Ratio, Proportion, and Similarity The four proportions at the bottom of page 477 demonstrate the following corollary: Corollary 12.1c If the products of two pairs of factors are equal, the factors of one pair can be the means and the factors of the other the extremes of a proportion. The Mean Proportional DEFINITION If the two means of a proportion are equal, either mean is called the mean proportional between the extremes of the proportion. In the proportion 6 5 6 2 18 mean proportional is also called the geometric mean. , 6 is the mean proportional between 2 and 18. The EXAMPLE 1 Solve for x: x 1 1 5 9 27 2 Solution Use that the product of the means is equal to the product of the extremes. x 1 1 5 9 27 2 9(x 1) 27(2) 9x 9 54 9x 45 x 5 Check x 1 1 5 9 27 2 5 1 1 5? 9 27 2 6 5? 9 27 2 9 2 5 9 2 Substitute 5 for x. Simplify. ✔ Answer x 5 EXAMPLE 2 Find the mean proportional between 9 and 8. Solution Let x represent the mean proportional. 9 x 5 x 8 x2 72 x 6 72 " 6 36 2 " " 66 2 " Note that there are two solutions, one positive and one negative. Answer 66 2 " 14365C12.pgs 7/10/07 8:56 AM Page 479 EXAMPLE 3 The measures of an exterior angle of a triangle and the adjacent interior angle are in the ratio 7 : 3. Find the measure of the exterior angle. Ratio and Proportion 479 Solution An exterior angle and the adjacent interior angle are supplementary. Let 7x the measure of the exterior angle, and 3x the measure of the interior angle. 7x 3x 180 10x 180 x 18 7x 126 Answer The measure of the exterior angle is 126°. Exercises Writing About Mathematics 1. Carter said that a proportion can be rewritten by using the means as extremes and the extremes as means. Do you agree with Carter? Explain why or why not. 2. Ethan said that the mean proportional will be a rational number only if the extremes are both perfect squares. Do you agree with Ethan? Explain why or why not. Developing Skills In 3–8, determine whether each pair of ratios can form a proportion. 3. 6 : 15, 4 : 10 6. 10 : 15, 8 : 12 4. 8 : 7, 56 : 49 7. 9 : 3, 16 : 4 5. 49 : 7, 1 : 7 8. 3a : 5a, 12 : 20 (a 0) In 9–11, use each set of numbers to form two proportions. 9. 30, 6, 5, 1 10. 18, 12, 6, 4 11. 3, 10, 15, 2 12. Find the exact value of the geometric mean between 10 and 40. 13. Find the exact value of the geometric mean between 6 and 18. 14365C12.pgs 7/10/07 8:56 AM Page 480 480 Ratio, Proportion, and Similarity In 14–19, find the value of x in each proportion. 14. 4 : x 10 : 15 x 1 1 5 8 12 16. x 18. 3x : 15 20 : x Applying Skills 15. 9 8 5 x 36 17. 12 : x x : 75 19. x 3 : 6 4 : x 2 20. B is a point on ABC such that AB : BC 4 : 7. If AC 33, find AB and BC. 21. A line segment 48 centimeters long is divided into two segments in the ratio 1 : 5. Find the measures of the segments. 22. A line segment is divided into two segments that are in the ratio 3 : 5. The measure of one segment is 12 centimeters longer than the measure of the other. Find the measure of each segment. 23. The measures of the sides of a triangle are in the ratio 5 : 6 : 7. Find the measure of each side if the perimeter of the triangle is 72 inches. 24. Can the measures of the sides of a triangle be in the ratio 2 : 3 : 7? Explain why or why not. 25. The length and width of a rectangle are in the ratio 5 : 8. If the perimeter of the rectangle is 156 feet, what are the length and width of the rectangle? 26. The measures of two consecutive angles of a parallelogram are in the ratio 2 : 7. Find the measure of each angle. 12-2 PROPORTIONS INVOLVING LINE SEGMENTS C D A E B The midpoint of any line segment divides the segment into two congruent parts. In ABC, let D be the midpoint of . Draw the . midsegment, and E be the midpoint of AC BC DE The line segment joining the midpoints of ABC forms a new triangle, • D is the midpoint of DEC. What are the ratios of the sides of these triangles? DC AC 5 1 2 EC BC 5 1 . 2 1 and 2AB 1 . Therefore, DC 2AC 1 . Therefore, EC 2BC , it appears that DE • E is the midpoint of If we measure and and AC BC . AB AB DE DE and . It also . We can prove these last two observations as a theorem AB 5 1 DE 2 appears that called the midsegment theorem. Theorem 12.2 A line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side. 14541C12.pgs 1/25/08 3:49 PM Page 481 Proportions Involving Line Segments 481 Given ABC, D is the midpoint of AC , and E is the y C(2a, 2c) midpoint of BC . Prove DE AB and DE 1 2AB E(a b, c) D(a, c) Proof We will use a coordinate proof for this theorem. The triangle can be placed at any convenient position. We will place A at the origin and B on the x-axis. Let the coordinates of the vertices of ABC be A(0, 0), B(2b, 0), and C(2a, 2c). Then: B(2b, 0) A(0, 0) x • The coordinates of D are • The coordinates of E are 0 2 0 2b 2 0 • The slope of AB is • The slope of DE is c 2 c a 1 b 2 a , 2c 1 0 2 B , 2c 1 0 2 B 2a 1 0 2 A 2a 1 2b 2 A 0. AB 0. (a, c). (a + b, c). is a horizontal line segment. DE is a horizontal line segment. Therefore, AB and DE are parallel line segments because horizontal line segments are parallel. The length of a horizontal line segment is the absolute value of the differ- ence of the x-coordinates of the endpoints. AB 5 2b 2 0 5 2b and DE 5 (a 1 b) 2 a 5 b Therefore, DE = 1 2AB . 1 Now that we know that our observations are correct, that DE 2AB , we know that A EDC and B DEC because AB DE that they are corresponding angles of parallel lines. We also know that C C. Therefore, for ABC and DEC, the corresponding angles are congruent and the ratios of the lengths of corresponding sides are equal. and C Again, in ABC, let D be the midpoint of and E be the midpoint of . AC DE Now let F be the midpoint of and G be . We can derive . Draw the midpoint of FG the following information from the segments formed: BC DC . Draw EC F D G E A B • FC • GC 2DC 5 1 1 2 A 2EC 5 1 1 2 A 1 1 2DE 2 A 1 2AC 1 2BC 1 2AB B B 5 1 4AC or 5 1 4BC or FC AC 5 1 4 GC BC 5 1 4 FG AB 5 1 4 • FG (by Theorem 12.2) 4AB • Let AC 4x. Then AD 2x, DC 2x, DF x, and FC x. or 5 1 B 14365C12.pgs 7/10/07 8:56 AM Page 482 482 Ratio, Proportion, and Similarity C F D G E A B • Let BC 4y. Then BE 2y, EC 2y, EG y, and GC x. • Therefore, Also, BG 2y y 3y. y FC AF 5 x 3x 5 1 3y 5 1 . 3 3 FC AF 5 GC GC 1 are each equal to , and 3 BG BG BC FC AF AC GC BG 5 G divide into parts whose ratios form a proportion. Since and and proportionally because these points separate the segments . We say that the points F and DEFINITION Two line segments are divided proportionally when the ratio of the lengths of the parts of one segment is equal to the ratio of the lengths of the parts of the other. The points D and E also divide AC points also separate the segments into parts whose ratios form a proportion. proportionally because these and BC 2x 5 1 DC 5 2x AD . DC 5 BE AD EC Therefore, and BE EC 5 2y 2y 5 1 Theorem 12.3a If two line segments are divided proportionally, then the ratio of th |
e length of a part of one segment to the length of the whole is equal to the ratio of the corresponding lengths of the other segment. Given ABC and DEF with AB BC 5 DE EF . Prove AC 5 DE AB DF Proof 1. 2. Statements BC 5 DE AB EF (AB)(EF) (BC)(DE) 3. (AB)(EF) (BC)(DE) (AB)(DE) (AB)(DE) 4. (AB)(EF DE) (DE)(BC AB) (AB)(DF) (DE)(AC) AC 5 DE AB DF 6. 5. A D B C E F Reasons 1. Given. 2. The product of the means equals the product of the extremes. 3. Addition postulate. 4. Distributive property. 5. Substitution postulate. 6. If the products of two pairs of factors are equal, one pair of factors can be the means and the other the extremes of a proportion. 14365C12.pgs 7/10/07 8:56 AM Page 483 Proportions Involving Line Segments 483 Theorem 12.3b If the ratio of the length of a part of one line segment to the length of the whole is equal to the ratio of the corresponding lengths of another line segment, then the two segments are divided proportionally. The proof of this theorem is left to the student. (See exercise 21.) Theorems 12.3a and 12.3b can be written as a biconditional. Theorem 12.3 Two line segments are divided proportionally if and only if the ratio of the length of a part of one segment to the length of the whole is equal to the ratio of the corresponding lengths of the other segment. EXAMPLE 1 In PQR, S is the midpoint of of RP 7x 5 ST 4x 2 SR 2x 1 PQ 9x 1 and T is the midpoint RQ PQ . Find ST, RP, SR, RQ, PQ, and TQ. Q T S P Solution The length of the line joining the midpoints of two sides of a triangle is equal to one-half the length of the third side. R 4x 2 2(4x 2) 1 2(7x 1 5) 2 (7x 1 5) 1 2 B A 8x 4 7x 5 x 9 ST 4(9) 2 36 2 34 RP 7(9) 5 63 5 68 SR 2(9) 1 18 1 19 RQ 2SR 2(19) 38 PQ 9(9) 1 81 1 82 1 TQ 2(82) 1 2PQ 41 EXAMPLE 2 and ABC do B and E divide DEF ABC and DEF proportionally? are line segments. If AB 10, AC 15, DE 8, and DF 12, Solution If AB 10 and AC 15, then: If DE 8 and DF 12, then: BC 5 15 2 10 5 5 AB : BC 5 10 : 5 5 2 : 1 EF 5 12 2 8 5 4 DE : EF 5 8 : 4 5 2 : 1 Since the ratios of AB : BC and DE : EF are equal, B and E divide DEF proportionally. ABC and 14365C12.pgs 7/10/07 8:56 AM Page 484 484 Ratio, Proportion, and Similarity EXAMPLE 3 In the diagram, ADEC sides of ABC. If AD DE EC and 1 BF FG GC, prove that EG . 2DF BFGC and are two Solution Since D and E are on and DE EC, A E is the midpoint of ADEC . DEC C G E D F B Since F and G are on BFGC and FG GC, G is the midpoint of FGC . is the line segment joining the midpoints of two sides of the tri, a line segment joining the midpoints of two sides In DFC, EG angle. By Theorem 12.2, of a triangle, is parallel to the third side, length of the third side. Therefore, EG = EG DF 1 2DF . , and its length is one-half the Exercises Writing About Mathematics 1. Explain why the midpoints of two line segments always divide those segments proportionally. 2. Points B, C, D, and E divide ABCDEF AB : BF 1 : 5. Do you agree with Emily? Explain why or why not. into five equal parts. Emily said that DF and N is the midpoint of EF . Developing Skills In 3–10, M is the midpoint of 3. Find DE if MN 9. 4. Find MN if DE 17. 5. Find DM if DF 24. 6. Find NF if EF 10. 7. Find DM : DF. 8. Find DP : PF if P is the midpoint of 9. Find mFMN if mD 76. 10. Find mENM if mE 42. MF . F M D N E 11. The length of the diagonal of a rectangle is 12 centimeters. What is the measure of a line segment that joins the midpoints of two consecutive sides of the rectangle? 14365C12.pgs 7/10/07 8:56 AM Page 485 In 12–15, the line segments PQ QR. ABC and PQR are divided proportionally by B and Q. AB BC and Proportions Involving Line Segments 485 12. Find PQ when AB 15, BC 25, and QR 35. 13. Find BC when AB 8, PQ 20, and PR 50. 14. Find AC when AB 12, QR 27, and BC PQ. 15. Find AB and BC when AC 21, PQ 14, and QR 35. 16. Line segment KLMN is divided by L and M such that KL : LM : MN 2 : 4 : 3. Find: a. KL : KN b. LN : MN c. LM : LN d. KM : LN 17. Line segment ABC is divided by B such that AB : BC 2 : 3 and line segment DEF is divided by E such that DE : EF 2 : 3. Show that AB : AC DE : DF. Applying Skills 18. The midpoint the sides of ABC are L, M, and N. C a. Prove that quadrilateral LMCN is a paral- lelogram. b. If AB 12, BC 9, and CA 15, what is M the perimeter of LMCN? N B L 19. In right triangle ABC, the midpoint of the hypotenuse AB is M and the midpoints of the legs are P and Q. Prove that quadrilateral PMQC is a rectangle. A 20. In right triangle ABC, the midpoint of the hypotenuse AB is M, the midpoint of BC is P, and the midpoint of CA is Q. D is a point on a. Prove that QADM is a rectangle. g PM such that PM MD. b. Prove that c. Prove that M is equidistant from the vertices of ABC. CM > AM . 21. Prove Theorem 12.3b, “If the ratio of the length of a part of one line segment to the length of the whole is equal to the ratio of the corresponding lengths of another line segment, then the two segments are divided proportionally.” 22. The midpoints of the sides of quadrilateral ABCD are M, N, P, and Q. Prove that quadrilat- eral MNPQ is a parallelogram. (Hint: Draw .)AC 14365C12.pgs 7/10/07 8:56 AM Page 486 486 Ratio, Proportion, and Similarity 12-3 SIMILAR POLYGONS Two polygons that have the same shape but not the same size are called similar polygons. In the figure to the right, ABCDE PQRST. The symbol is read “is similar to.” These polygons have the same shape because their corresponding angles are congruent and the ratios of the lengths of their corresponding sides are equal DEFINITION Two polygons are similar if there is a one-to-one correspondence between their vertices such that: 1. All pairs of corresponding angles are congruent. 2. The ratios of the lengths of all pairs of corresponding sides are equal. When the ratios of the lengths of the corresponding sides of two polygons are equal, as shown in the example above, we say that the corresponding sides of the two polygons are in proportion. The ratio of the lengths of corresponding sides of similar polygons is called the ratio of similitude of the polygons. The number represented by the ratio of similitude is called the constant of proportionality. Both conditions mentioned in the definition must be true for polygons to be similar. D 4 A C N 6 60 S M 7 60 R 6 B K 9 L P 10 Q Rectangle ABCD is not similar to parallelogram KLMN. The corre, but the corresponding angles are not 6 5 6 4 9 sponding sides are in proportion, congruent. Parallelogram KLMN is not similar to parallelogram PQRS. The corresponding angles are congruent but the corresponding sides are not in proportion, 6 9 Recall that a mathematical definition is reversible: 2 7 10 . 14365C12.pgs 7/10/07 8:56 AM Page 487 Similar Polygons 487 If two polygons are similar, then their corresponding angles are congruent and their corresponding sides are in proportion. and If two polygons have corresponding angles that are congruent and corre- sponding sides that are in proportion, then the polygons are similar. Since triangles are polygons, the definition given for two similar polygons will apply also to two similar triangles. In the figures to the right, ABC ABC. We can draw the following conclusions about the two triangles: C 22 14 A 12 B C 11 7 A 6 B A A AB : ArBr 5 12 : 6 B B BC : BrCr 5 14 : 7 C C CA : CrAr 5 22 : 11 The ratio of similitude for the triangles is 2 : 1. Equivalence Relation of Similarity The relation “is similar to” is true for polygons when their corresponding angles are congruent and their corresponding sides are in proportion. Thus, for a given set of triangles, we can test the following properties: 1. Reflexive property: ABC ABC. (Here, the ratio of the lengths of corresponding sides is 1 : 1.) 2. Symmetric property: If ABC DEF, then DEF ABC. 3. Transitive property: If ABC DEF, and DEF RST, then ABC RST. These properties for any similar geometric figures can be stated as postulates. Postulate 12.1 Any geometric figure is similar to itself. (Reflexive property) Postulate 12.2 A similarity between two geometric figures may be expressed in either order. (Symmetric property) 14365C12.pgs 7/10/07 8:56 AM Page 488 488 Ratio, Proportion, and Similarity Postulate 12.3 Two geometric figures similar to the same geometric figure are similar to each other. (Transitive property) EXAMPLE 1 In right triangle ABC, mA 67.4, AB 13.0, BC 12.0, and CA 5.00. In right triangle DEF, mE 22.6, DE 19.5, EF 18.0, and FD 7.50. Prove that ABC DEF. Proof Triangles ABC and DEF are right triangles. The angles opposite the longest sides are right angles. Therefore, mC 90, mF 90, and C F. The acute angles of a right triangle are complementary. Therefore, mB 90 mA 90 67.4 22.6, and B E. Similarly, mD 90 mE 90 22.6 67.4, and A D. 19.5 5 2 3 DE 5 13.0 AB BC EF 5 12.0 Since the corresponding angles are congruent and the ratios of the lengths of corresponding sides are equal, the triangles are similar. CA FD 5 5.00 7.50 5 2 3 18.0 5 2 3 Exercises Writing About Mathematics 1. Are all squares similar? Justify your answer. 2. Are any two regular polygons similar? Justify your answer. Developing Skills 3. What is the ratio of the lengths of corresponding sides of two congruent polygons? 4. Are all congruent polygons similar? Explain your answer. 5. Are all similar polygons congruent? Explain your answer. 6. What must be the constant of proportionality of two similar polygons in order for the poly- gons to be congruent? 7. The sides of a triangle measure 4, 9, and 11. If the shortest side of a similar triangle mea- sures 12, find the measures of the remaining sides of this triangle. 8. The sides of a quadrilateral measure 12, 18, 20, and 16. The longest side of a similar quadri- lateral measures 5. Find the measures of the remaining sides of this quadrilateral. 14365C12.pgs 7/10/07 8:56 AM Page 489 9. Triangle ABC ABC, and their ratio of similitude is 1 : 3. If the measures of the sides of ABC are represented by a, b, and c, represent the measures of the sides of the larger triangle, ABC. Provin |
g Triangles Similar 489 Applying Skills 10. Prove that any two equilateral triangles are similar. 11. Prove that any two regular polygons that have the same number of sides are similar. 12. In ABC, the midpoint of is M and the midpoint of is N. BC AC a. Show that ABC MNC. b. What is their ratio of similitude? 13. In ABC, the midpoint of AC is M, the midpoint of MC is P, the midpoint of BC is N, and NC is Q. the midpoint of a. Show that ABC PQC. b. What is their ratio of similitude? 14. Show that rectangle ABCD is similar to rectangle EFGH if EF 5 BC AB FG . 15. Show that parallelogram KLMN is similar to parallelogram PQRS if mK mP and PQ 5 LM KL QR . 12-4 PROVING TRIANGLES SIMILAR We have proved triangles similar by proving that the corresponding angles are congruent and that the ratios of the lengths of corresponding sides are equal. It is possible to prove that when some of these conditions exist, all of these conditions necessary for triangles to be similar exist. DE with Hands-On Activity For this activity, you may use a compass and ruler, or geometry software. STEP 1. Draw any triangle, ABC. AB 5 3 DE STEP 2. Draw any line 1 STEP 3. Construct GDE A and HED B. Let F be the intersection , that is, DE 3AB. h h . DG EH a. Find the measures of AC 5 3 CB 5 3 EF DF b. Is 1 1 c. Is DEF ABC? d. Repeat this construction using a different ratio of similitude. Are the trian- , and and ? Is DF AC EF BC of ? , , . gles similar? 14365C12.pgs 7/10/07 8:56 AM Page 490 490 Ratio, Proportion, and Similarity Our observations from the activity on page 489 seem to suggest the follow- ing postulate of similarity. Postulate 12.4 For any given triangle there exists a similar triangle with any given ratio of similitude. We can also prove the angle-angle or AA triangle similarity theorem. Theorem 12.4 Two triangles are similar if two angles of one triangle are congruent to two corresponding angles of the other. (AA) Given ABC and ABC with A A and B B C C Prove ABC ABC C Proof Statement 1. Draw LMN ABC with Reason 1. Postulate of similarity. LM AB 5 ArBr . 2. L A and M B AB 2. If two triangles are similar, then their corresponding angles are congruent. L M 3. A A and B B 4. L A and M B 3. Given. 4. Substitution postulate. LM AB 5 ArBr 5. 6. (AB)(AB) (AB)(LM) AB 7. AB LM 8. ABC LMN 9. ABC LMN 10. ABC ABC 5. Step 1. 6. In any proportion, the product of the means is equal to the product of the extremes. 7. Division postulate. 8. ASA (steps 4, 7). 9. If two triangles are congruent, then they are similar. 10. Transitive property of similarity (steps 1, 9). 14365C12.pgs 7/10/07 8:56 AM Page 491 We can also prove other theorems about similar triangles by construction, such as the side-side-side or SSS similarity theorem. Proving Triangles Similar 491 Theorem 12.5 Two triangles are similar if the three ratios of corresponding sides are equal. (SSS) C Given ABC and ABC with ArBr 5 BC AB BrCr 5 CA CrAr . C Prove ABC ABC Proof We will construct a third triangle DEC that is similar to both ABC and ABC. By the transitive property of similarity, we can conclude that ABC ABC Let AC AC. Choose point D on so that DC AC. Choose point BrCr ArCr E on gruent, so CDE A and C C. Therefore, ABC DEC by AA. If two polygons are similar, then their corresponding sides are in proportion, so . Corresponding angles of parallel lines are con- DE ArBr so that A B . CrD CrAr 5 CrE CrBr Substituting CD CA into the CA proportion gives . CrAr 5 CrE CrBr CrAr 5 BC CA BrCr CrE CrBr 5 BC BrCr We are given that so by the transitive property, . Therefore, (CE)(BC) (CB)(BC) or CE BC. C C A B D A E B By similar reasoning, we find that DE AB. Therefore, DEC ABC by SSS and DEC ABC. Then by the transitive property of similarity, ABC ABC. Theorem 12.6 Two triangles are similar if the ratios of two pairs of corresponding sides are equal and the corresponding angles included between these sides are congruent. (SAS) 14365C12.pgs 7/10/07 8:56 AM Page 492 492 Ratio, Proportion, and Similarity Given ABC and ABC with B B ArBr 5 BC AB BrCr and C D C Prove ABC ABC Strategy The proof follows the same pattern as the previous theorem. Let BC BC. Choose so that BD BC. Choose point D on DE ArCr point E on so that use the given ratios to prove EB AB and EBD ABC by SAS. BrCr ArBr . First prove that ABC EBD. Then A B A B E We refer to Theorem 12.6 as the side-angle-side or SAS similarity theorem. The proof of this theorem will be left to the student. (See exercise 17.) As a consequence of these proofs, we have shown the following theorem to be true. Theorem 12.7a If a line is parallel to one side of a triangle and intersects the other two sides, then the points of intersection divide the sides proportionally. The converse of this theorem is also true. Theorem 12.7b If the points at which a line intersects two sides of a triangle divide those sides proportionally, then the line is parallel to the third side. Given ABC with CA 5 CE CD CB Prove DE AB Proof Since CA 5 CE CD CB and C C, ABC DEC by SAS. Corresponding angles of similar triangles are congruent, so CDE A. As these are congruent DE AB corresponding angles, . C D A E B Theorems 12.7a and 12.7b can be written as a biconditional. Theorem 12.7 A line is parallel to one side of a triangle and intersects the other two sides if and only if the points of intersection divide the sides proportionally. 14365C12.pgs 7/10/07 8:56 AM Page 493 Proving Triangles Similar 493 EXAMPLE 1 The lengths of the sides of PQR are PQ 15 cm, QR 8 cm, and RP 12 cm. If PQR DEF and the length of the smallest side of DEF is 6 centimeters, find the measures of the other two sides of DEF. Solution Since the smallest side of PQR is , EF 6 cm. corresponds to QR EF QR and R PQ QR EF 5 DE 6 5 15 8 DE 8DE 5 90 DE 5 90 8 5 45 4 5 111 4 QR EF 5 RP FD 8 6 5 12 FD 8FD 5 72 FD 5 72 8 5 9 Answer DE 111 4 cm and FD 9 cm EXAMPLE 2 12 cm P D 8 cm Q 15 cm F 6 cm E FD at H and FE at G. F 8 G 12 E H 30 D In DEF, a line is drawn parallel to If FG 8, GE 12, and FD 30, find FH and HD. that intersects DE Solution Since GH is parallel to DE , H and G divide FD and FE proportionally, that is, FH : HD FG : GE. Let x FH. Then HD FD – FH 30 x. Check HD 5 FG FH GE 18 5? 8 12 12 2 3 5 2 3 ✔ HD 5 FG FH GE x 30 2 x 5 8 12 12x 8(30 x) 12x 240 8x 20x 240 x 12 30 x 30 12 18 Answer FH 12 and HD 18 14365C12.pgs 7/10/07 8:56 AM Page 494 494 Ratio, Proportion, and Similarity EXAMPLE 3 Given: Prove: and BFGC are two sides of ADEC ABC with AD DE EC and BF FG GC. DC 5 BC AC FC and ABC DFC C E G D A F B Proof We are given ADEC and AD DE EC. Then AC AD DE EC. By the substitution postulate, AC AD AD AD 3AD and DC DE EC AD AD 2AD. BFGC and BF FG GC. Then BC BF FG GC. We are also given By the substitution postulate, BC BF BF BF 3BF and FC FG FC BF BF 2BF. BC AC 2AD 5 3 DC 5 3AD FC 5 3BF 2 In ABC and DFC, by SAS. 2BF 5 3 2 and C C. Therefore, ABC DFC and DC 5 BC AC FC DC 5 BC AC FC . Therefore, Then, . Exercises Writing About Mathematics 1. Javier said that if an acute angle of one right triangle is congruent to an acute angle of another right trangle, the triangles are similar. Do you agree with Javier? Explain why or why not. 2. Fatima said that since two triangles can be proven similar by AA, it follows that two trian- gles can be proven similar by SS. Explain why Fatima is incorrect. Developing Skills In 3–15, D is a point on drawn to scale.) AC and E is a point on BC of ABC such that DE AB . (The figure is not 3. Prove that ABC DEC. 4. If CA 8, AB 10, and CD 4, find DE. 5. If CA 24, AB 16, and CD 9, find DE. 6. If CA 16, AB 12, and CD 12, find DE. 7. If CE 3, DE 4, and CB 9, find AB. 8. If CD 8, DA 2, and CB 7.5, find CE. 9. If CD 6, DA 4, and DE 9, find AB. 10. If CA 35, DA 10, and CE 15, find EB. C D E A B 14365C12.pgs 7/10/07 8:56 AM Page 495 Dilations 495 11. If CA 48, DA 12, and CE 30, find EB. 12. If CD 15, DA 9, and DE 10, find AB. 13. If CE 20, EB 10, and AB 45, find DE. 14. If CD x, DE x, DA 5, and AB 14, find DE. 15. If CD 6, DE x, DA x 1, and AB 6, DE and DA. Applying Skills 16. Complete the proof of Theorem 12.5 (SSS) by showing that DE AB. 17. Prove Theorem 12.6, “Two triangles are similar if the ratios of two pairs of corresponding sides are equal and the corresponding angles included between these sides are congruent. (SAS)” 18. Triangle ABC is an isosceles right triangle with mC 90 and CD bisects C and inter- sects AB at D. Prove that ABC ACD. 19. Quadrilateral ABCD is a trapezoid with AB CD . The diagonals AC and BD intersect at E. Prove that ABE CDE. 20. Lines g AEB and g CED intersect at E and DAE BCE. Prove that ADE CBE. 21. In parallelogram ABCD on the right, g AE ' BC and D F C g AF ' CD . Prove that ABE ADF. 22. In the coordinate plane, the points A(1, 2), B(3, 2), C(3, 6), D(2, 6), and E(2, 8) are the vertices of ABC and CDE. Prove that ABC CDE. 23. In the coordinate plane, the points P(1, 1), Q(3, 3), R(3, 5), and S(1, 5) are the vertices of PQS and QRS and PQ QS . Prove that PQS QRS. 2 2 " E A B 24. In the coordinate plane, the points O(0, 0), A(4, 0), and B(0, 6) are the coordinates of OAB. The coordinates of C are (4, 3), and D is the midpoint of OAB CDA. AB . Prove that 25. A pyramid with a triangular base is cut by a plane p parallel to the base. Prove that the tri- angle formed by the intersection of plane p with the lateral faces of the pyramid is similar to the base of the pyramid. 12-5 DILATIONS In Chapter 6, we learned about dilations in the coordinate plane. In this section, we will continue to study dilations as they relate to similarity. Recall that a dilation is a transformation in the plane that preserves angle measure but not distance. 14365C12.pgs 7/10/07 8:56 AM Page 496 496 Ratio, Proportion, and Similarity A dilation of k is a transformation of the plane such that: 1. The image of point O, the center of dilation, is O. 2. When k is positive and the image of P is P, then h OP and h OPr are the same ray and OP kOP. 3. When k is negative and the image of P is P, then h OP |
and h OPr are oppo- site rays and OP kOP. When k 1, the dilation is called an enlargement. When 0 k 1, the dilation is called a contraction. Recall also that in the coordinate plane, under a dilation of k with the cen- ter at the origin: P(x, y) → P(kx, ky) or Dk(x, y) (kx, ky) For example, the image of ABC is ABC under a dilation of . The vertices of ABC are A(2, 6), B(6, 4), and C(4, 0). Under a dilation of , the rule is 1 2 1 2 A (x,y) 5 2x, 1 1 D1 2y B 2 A(2, 6) → A(1, 3) B(6, 4) → B(3, 2) C(4, 0) → C(2, 0) Notice that ABC and ABC appear to be similar. We can use a general triangle to prove that for any dilation, the image of a triangle is a similar triangle Let ABC be any triangle in the coordinate plane with A(a, 0), B(b, d), and C(c, e). Under a dilation of k through the origin, the image of ABC is ABC, and the coordinates of ABC, are A(ka, 0), B(kb, kd), and C(kc, ke). y C(kc, ke) C(c, e) B(kb, kd) B(b, d) A(a, 0) A(ka, 0) x O 14365C12.pgs 7/10/07 8:56 AM Page 497 Slope of Slope of b 2 a AB 5 d 2 0 b 2 a 5 d ArBr 5 kd 2 0 kb 2 ka AB ArBr . Slope of Slope of c 2 a AC 5 e 2 0 c 2 a 5 e ArCr 5 ke 2 0 kc 2 ka AC ArCr . Therefore, Therefore, Therefore, Dilations 497 Slope of BC 5 d 2 e b 2 c Slope of BrCr 5 kd 2 ke kb 2 kc BC BrCr . We have shown that AB ArBr and AC ArCr . Therefore, because they are corresponding angles of parallel lines: mOAB mOAB mOAC mOAC mOAB mOAC mOAB mOAC mBAC mBAC In a similar way we can prove that ACB ACB, and so ABC ABC by AA. Therefore, under a dilation, angle measure is preserved but distance is not preserved. Under a dilation of k, distance is changed by the factor k. We have proved the following theorem: Theorem 12.8 Under a dilation, angle measure is preserved. We will now prove that under a dilation, midpoint and collinearity are preserved. Theorem 12.9 Under a dilation, midpoint is preserved. Proof: Under a dilation Dk: A(a, c) → A(ka, kc) B(b, d) → (kb, kd) , c 1 d → M 2 ka 1 b 2 a 1 b 2 M A , kc 1 d 2 B ArBr B B A A The coordinates of the midpoint of are: ka 1 kb 2 , kc 1 kd 2 or ka 1 b 2 , kc 1 d 2 A B Therefore, the image of M is the midpoint , and midpoint is of the image of AB preserved 14365C12.pgs 7/10/07 8:56 AM Page 498 498 Ratio, Proportion, and Similarity Theorem 12.10 Under a dilation, collinearity is preserved. Proof: Under a dilation Dk: A(a, c) → A(ka, kc) B(b, d) → B(kb, kd) P(p, q) → P(kp, kq) Since P is on to the slope of AB PB , the slope of . Therefore: AP is equal slope of AP 5 slope of PB ArBr if and only if the slope P will be on of ArPr is equal to the slope of PrBr . A P A P B y O B x slope of ArPr 5? kc 2 kq ka 2 kp 5? slope of PrBr kq 2 kd kp 2 kb is true, Since Thus, since we have shown that the slope of P is on and collinearity is preserved ArBr k k A or kc 2 kq ka 2 kp 5 ArPr kd 2 kq kb 2 kp is equal to the slope of PrBr , EXAMPLE 1 The coordinates of parallelogram EFGH are E(0, 0), F(3, 0), G(4, 2), and H(1, 2). Under D3, the image of EFGH is EFGH. Show that EFGH is a parallelogram. Is parallelism preserved? Solution D3(x, y) (3x, 3y). Therefore, E(0, 0), F(9, 0), G(12, 6), and H(3, 6). slope of ErFr 5 0 2 0 9 2 0 and slope of HrGr 5 6 2 6 12 2 3 slope of and slope of 5 0 FrGr 5 6 2 0 12 2 9 5 6 3 5 2 5 0 ErHr Since the slopes of the opposite sides of EFGH are equal, the opposite sides are parallel and EFGH is a parallelogram. Parallelism is preserved because the images of parallel lines are parallel. 14365C12.pgs 7/10/07 8:56 AM Page 499 EXAMPLE 2 Dilations 499 Find the coordinates of Q, the image of Q(–3, 7) under the composition of transformations, + D1 . ry-axis 2 Solution Perform the transformations from right to left. The transformation at the right is to be performed first: (23,7) 5 D1 2 23 2 , 7 2 B A Then perform the transformation on the left, using the result of the first transformation: 5 ry-axis A 23 2 , 7 2 B 2, 7 3 2 B A Answer Q 2, 7 3 2 B A Exercises Writing About Mathematics 1. Under Dk, k 0, the image of ABC is ABC. Is ArBr 5 BC AB BrCr 5 AC ArCr ? Justify your answer. 2. Under a dilation, the image of A(3, 3) is A(4, 5) and the image of B(4, 1) is B(6, 1). What are the coordinates of the center of dilation? Developing Skills In 3–6, use the rule (x, y) → 3. (9, 6) 3x, 1 1 3y A 4. (5, 0) B to find the coordinates of the image of each given point. 5. (18, 3) 6. (1, 7) In 7–10, find the coordinates of the image of each given point under D3. 10. 9. (4, 7) 8. (2, 13) 7. (8, 8) 3, 5 1 8 B A In 11–14, each given point is the image under D2. Find the coordinates of each preimage. 11. (4, 2) 12. (6, 8) 13. (3, 2) 14. (20, 11) In 15–20, find the coordinates of the image of each given point under the given composition of transformations. 15. 18. D3 ry-axis + rx-axis(2, 3) + D3(1, 2) 16. 19. (4, 3) R1808 T2,3 + D221 + D101 3 2 (0, 0) 17. D5 3 20. D22 + T5,3(1, 1) + ry 5 x(23, 25) 14365C12.pgs 7/10/07 8:56 AM Page 500 500 Ratio, Proportion, and Similarity In 21–24, each transformation is the composition of a dilation and a reflection in either the x-axis or the y-axis. In each case, write a rule for composition of transformations for which the image of A is A. A 2, 29 9 21. A(3, 3) → A 2 B 22. A(5, 1) → A(20, 4) 23. A(20, 12) → A(5, 3) 24. A(50, 35) → A(10, 7) 25. In the diagram, ABC is the image of ABC. Identify three specific transformations, or compositions of transformations, that can map ABC to ABC. Justify your answer. B C A y A C x B Applying Skills 26. If the coordinates of points A and B are (0, 5) and (5, 0), respectively, and A and B are the images of these points under D3, what type of quadrilateral is answer. A ABArBr ? Justify your 27. Prove that if the sides of one angle are parallel to the sides of another angle, the angles are congruent. h BC h ED , g BEG h EF h BA , and Given: Prove: ABC DEF H B C D E F G 28. The vertices of rectangle ABCD are A(2, 3), B(4, 3), C(4, 1), and D(2, 1). a. Find the coordinates of the vertices of ABCD, the image of ABCD under D5. b. Show that ABCD is a parallelogram. c. Show that ABCD ABCD. d. Show that ABC ABC. 14365C12.pgs 7/10/07 8:56 AM Page 501 Dilations 501 29. The vertices of octagon ABCDEFGH are A(2, 1), B(1, 2), C(1, 2), D(2, 1), E(2, 1), F(1, 2), G(1, 2), H(2, 1). a. Draw ABCDEFGH on graph paper. b. Draw ABCDEFGH, the image of ABCDEFGH under D3, on graph paper and write the coordinates of its vertices. c. Find HA, BC, DE, FG. d. Find HA, BC, DE, FG. e. If AB CD EF GH f. Are ABCDEFGH and ABCDEFGH similar polygons? Justify your answer. , find AB, CD, EF, GH. 2 " 30. Let the vertices of ABC be A(2, 3), B(2, 1), and C(3, 1). a. Find the area of ABC. b. Find the area of the image of ABC under D3. c. Find the area of the image of ABC under D4. d. Find the area of the image of ABC under D5. e. Make a conjecture regarding how the area of a figure under a dilation Dk is related to the constant of dilation k. 31. Complete the following to prove that dilations preserve parallelism, that is, if g AB g CD , then the images of each line under a dilation Dk are also parallel. a. Let AB b. Let and CD AB be two vertical segments with endpoints A(a, b) D(c, b 1 d) B(a, b 1 d) , and , Under the dilation Dk, show that the CrDr images C(c, b) are also parallel. ArBr and , . and CD be two nonvertical , B(c, d) parallel segments with endpoints , and , A(a, b) D(c 1 e, d) show that the images CrDr are also parallel. C(a 1 e, b) . Under the dilation Dk, ArBr and y O B(a, bd) D(c, bd) A(a, b) C(c, b) y O B(c, d) D(ce, d) A(a, b) C(ae, b) x x 14365C12.pgs 7/10/07 8:56 AM Page 502 502 Ratio, Proportion, and Similarity 12-6 PROPORTIONAL RELATIONS AMONG SEGMENTS RELATED TO TRIANGLES We have seen that, if two triangles are similar, their corresponding sides are in proportion. Other corresponding segments such as the altitudes, medians, and angle bisectors in similar triangles are also in proportion. Theorem 12.11 If two triangles are similar, the lengths of corresponding altitudes have the same ratio as the lengths of any two corresponding sides. Given ABC ABC with the ratio of BD ' AC similitude k : 1, BC a, BC a, BD h, and BD h. , BrDr ' ArCr Prove hr 5 a h ar Proof Statements Reasons 1. ABC ABC 2. C C BD ' AC 3. 4. BDC BDC and BrDr ' ArCr 5. DBC DBC 6. 7. ar 5 k a 1 h hr 5 a ar 8. hr 5 a h ar 5 k 1 1. Given. 2. If two triangles are similar, then their corresponding angles are congruent. 3. Given. 4. Perpendicular lines form right angles and all right angles are congruent. 5. AA. 6. Given. 7. If two triangles are similar, then their corresponding sides are in proportion. 8. Transitive property. We can prove related theorems for medians and angle bisectors of similar triangles. 14365C12.pgs 7/10/07 8:56 AM Page 503 Proportional Relations Among Segments Related to Triangles 503 Theorem 12.12 If two triangles are similar, the lengths of corresponding medians have the same ratio as the lengths of any two corresponding sides. Given ABC ABC with the ratio of AC similitude k : 1,M is the midpoint , M is the midpoint of ArCr , of BC a, BC a, BM m, and BM m Prove mr 5 a m ar 5 k 1 Strategy Here we can use SAS to prove BCM BCM. Theorem 12.13 If two triangles are similar, the lengths of corresponding angle bisectors have the same ratio as the lengths of any two corresponding sides. Given ABC ABC with the ratio of similitude k : 1,E is the point at which the bisector of B intersects , E is the point at which the AC bisector of B intersects ArCr , BC a, BC a, BE e, and BE = e Prove er 5 a e ar 5 k 1 Strategy Here we can use that halves of congruent angles are congruent and AA to prove BCE BCE. The proofs of Theorems 12.12 and 12.13 are left to the student. (See exer- cises 10 and 11.) 14365C12.pgs 7/10/07 8:56 AM Page 504 504 Ratio, Proportion, and Similarity EXAMPLE 1 Two triangles are similar. The sides of the smaller triangle have lengths of 4 meters, 6 meters, and 8 meters. The perimeter of the larger triangle is 63 meters. Find the length of the shortest side of the larger triangle. Solution (1) In the smaller tri |
angle, find the perimeter, p: p 4 6 8 18 (2) Let k be the constant of proportionality of the larger triangle to the smaller triangle. Let the measures of the sides of the larger triangle be a, b, and c. Set up proportions and solve for a, b, and c: 4 5 k a 1 a 5 4k 6 5 k b 1 b 5 6k 8 5 k c 1 c 5 8k (3) Solve for k: 4k 6k 8k 63 18k 63 k 3.5 (4) Solve for a, b, and c: a 4k 4(3.5) 14 b 6k 6(3.5) 21 c 8k 8(3.5) 28 Answer The length of the shortest side is 14 meters. EXAMPLE 2 Given: g AFB g CGD , AED and BEC intersect at E, and g EF ' AFB . Prove: ABE DCE and DC 5 EF AB EG . A F B E C G D 14365C12.pgs 7/10/07 8:56 AM Page 505 Proportional Relations Among Segments Related to Triangles 505 Proof g AFB Statements g CGD 1. 2. EAB EDC and EBA ECD 3. ABE DCE g EF ' AFB g EG ' CGD 4. 5. is an altitude from E in is an altitude from 6. 7. EF ABE. EG E in DCE. DC 5 EF AB EG Reasons 1. Given. 2. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. 3. AA. 4. Given. 5. If a line is perpendicular to one of two parallel lines, it is perpendicular to the other. 6. Definition of an altitude of a triangle. 7. If two triangles are similar, the lengths of corresponding altitudes have the same ratio as the lengths of any two corresponding sides. Exercises Writing About Mathematics 1. The lengths of the corresponding sides of two similar triangles are 10 and 25. Irena said that the ratio of similitude is 2 : 5. Jeff said that it is 2 5 : 1 . Who is correct? Justify your answer. 2. Maya said that if the constant of proportionality of two similar triangles is k, then the ratio 1 1 k k . Do you agree with Maya? Explain 1 5 3k 1 1 1 k of the perimeters will be 3k : 1 because why or why not. Developing Skills 3. The ratio of similitude in two similar triangles is 5 : 1. If a side in the larger triangle measures 30 centimeters, find the measure of the corresponding side in the smaller triangle. 4. If the lengths of the sides of two similar triangles are in the ratio 4 : 3, what is the ratio of the lengths of a pair of corresponding altitudes, in the order given? 5. The lengths of two corresponding sides of two similar triangles are 18 inches and 12 inches. If an altitude of the smaller triangle has a length of 6 inches, find the length of the corresponding altitude of the larger triangle. 14365C12.pgs 7/10/07 8:56 AM Page 506 506 Ratio, Proportion, and Similarity 6. The constant of proportionality of two similar triangles is . If the length of a median in the 4 5 larger triangle is 15 inches, find the length of the corresponding median in the smaller triangle. 7. The ratio of the lengths of the corresponding sides of two similar triangles is 6 : 7. What is the ratio of the altitudes of the triangles? 8. Corresponding altitudes of two similar triangles have lengths of 9 millimeters and 6 millimeters. If the length of a median of the larger triangle is 24 millimeters, what is the length of a median of the smaller triangle? 9. In meters, the sides of a triangle measure 14, 18, and 12. The length of the longest side of a similar triangle is 21 meters. a. Find the ratio of similitude of the two triangles. b. Find the lengths of the other two sides of the larger triangle. c. Find the perimeter of each triangle. d. Is the ratio of the perimeters equal to the ratio of the lengths of the sides of the triangle? Applying Skills 10. Prove Theorem 12.12, “If two triangles are similar, the lengths of corresponding medians have the same ratio as the lengths of any two corresponding sides.” 11. Prove Theorem 12.13, “If two triangles are similar, the lengths of corresponding angle bisec- tors have the same ratio as the lengths of any two corresponding sides.” 12. Prove that if two parallelograms are similar, then the ratio of the lengths of the correspond- ing diagonals is equal to the ratio of the lengths of the corresponding sides. 13. Prove that if two triangles are similar, then the ratio of their areas is equal to the square of their ratio of similitude. 14. The diagonals of a trapezoid intersect to form four triangles that have no interior points in common. a. Prove that two of these four triangles are similar. b. Prove that the ratio of similitude is the ratio of the length of the parallel sides. 12-7 CONCURRENCE OF THE MEDIANS OF A TRIANGLE We proved in earlier chapters that the altitudes of a triangle are concurrent and that the angle bisectors of a triangle are concurrent. If we draw the three medians of a triangle, we see that they also seem to intersect in a point. This point is called the centroid of the triangle. Centroid 14365C12.pgs 7/10/07 8:56 AM Page 507 Theorem 12.14 Any two medians of a triangle intersect in a point that divides each median in the ratio 2 : 1. Concurrence of the Medians of a Triangle 507 Given and AM BN intersect at P. are medians of ABC that C Prove AP : MP BP : NP 2 : 1 N P M A B Proof Statements Reasons 1. and AM of ABC. BN are the medians 1. Given. 2. M is the midpoint of N is the midpoint of BC . AC and 2. The median of a triangle is a line segment from a vertex to the midpoint of the opposite side. 3. Draw MN MN AB 4. . 3. Two points determine a line. 5. MNB ABN and NMA BAM 6. MNP ABP 7. MN 1 2AB 8. 2MN AB 9. AB : MN 2 : 1 10. AP : MP BP : NP 2 : 1 4. The line joining the midpoints of two sides of a triangle is parallel to the third side. 5. Alternate interior angles of parallel lines are congruent. 6. AA. 7. The length of the line joining the midpoints of two sides of a triangle is equal to one-half of the length of the third side. 8. Multiplication postulate. 9. If the products of two pairs of factors are equal, the factors of one pair can be the means and the factors of the other the extremes of a proportion. 10. If two triangles are similar, the ratios of the lengths of the corresponding sides are equal. 14365C12.pgs 7/10/07 8:56 AM Page 508 508 Ratio, Proportion, and Similarity Theorem 12.15 The medians of a triangle are concurrent. Given , AM BN ABC. , and CL are medians of Prove , AM BN , and CL are concurrent. C N P P M BN and be AM in the ratio 2 : 1, Proof Let the intersection of P. Then P divides AM that is, AP : PM 2 : 1. Let at P. Then P divides AM ratio 2 : 1, that is, AP : PM 2 : 1. Both P and P are on the same line seg, and divide that line segment in the ratio 2 : 1. Therefore, P and P ment, are the same point and the three medians of ABC are concurrent. intersect in the CL AM AM A L B EXAMPLE 1 Find the coordinates of the centroid of the triangle whose vertices are A(3, 6), B(9, 0), and C(9, 0). Solution A(3, 6) y N(3, 3) P(1, 2) 1 B(9, 0) M(0, 0) O 1 C(9, 0) x (1) Find the coordinates of the midpoint, M, of BC and of the midpoint, N, of AC : coordinates of M coordinates of N 5 29 1 9 2 A 5 (0, 0) , 0 1 0 2 5 23 1 9 2 A 5 (3, 3) , 6 1 0 2 B B 14365C12.pgs 7/10/07 8:56 AM Page 509 (2) Find the equation of g AM g : Equation of AM 23 2 0 y x 5 22 y 5 22x Concurrence of the Medians of a Triangle 509 and the equation of g BN . g Equation of : BN y 2 0 x 2 (29) 5 0 2 3 29 2 3 x 1 9 5 1 4 4y 5 x 1 9 g AM y (3) Find the coordinates of P, the point of intersection of and g BN : Substitute y 2x into the equation 4y x 9, and solve for x. Then find the corresponding value of y. 4(2x) x 9 8x x 9 9x 9 x 1 y 2x y 2(1) y 2 Answer The coordinates of the centroid are P(1, 2). We can verify the results of this example by showing that P is a point on the median from C: (1) The coordinates of L, the midpoint of AB , are: (2) The equation of g A is: CL 0 2 3 9 2 (26 21 5 5y 5 2x 1 9 23 1 (29) 2 , 6 1 0 2 (6, 3) B (3) P(1, 2) is a point on g CL : 5y 5 2x 1 9 5(2) 5? 2(21) 1 9 10 5 10 ✔ Exercises Writing About Mathematics 1. If AM and BN are two medians of ABC that intersect at P, is P one of the points on AM that separate the segment into three congruent parts? Explain your answer. 2. Can the perpendicular bisector of a side of a triangle ever be the median to a side of a trian- gle? Explain your answer. 14365C12.pgs 7/10/07 8:56 AM Page 510 510 Ratio, Proportion, and Similarity Developing Skills In 3–10, find the coordinates of the centroid of each triangle with the given vertices. 3. A(3, 0), B(1, 0), C(1, 6) 5. A(3, 3), B(3, 3), C(3, 9) 7. A(1, 1), B(3, 1), C(1, 7) 9. A(2, 5), B(0, 1), C(10, 1) 4. A(5, 1), B(1, 1), C(1, 5) 6. A(1, 2), B(7, 0), C(1, 2) 8. A(6, 2), B(0, 0), C(0, 10) 10. A(1, 1), B(17, 1), C(5, 5) Applying Skills 11. The coordinates of a vertex of ABC are A(0, 6), and AB AC. a. If B and C are on the x-axis and BC 4, find the coordinates of B and C. b. Find the coordinates of the midpoint M of AB and of the midpoint N of AC . c. Find the equation of g CM g d. Find the equation of BN e. Find the coordinates of the centroid of ABC. . . 12. The coordinates of the midpoint of AB of ABC are M(3, 0) and the coordinates of the centroid are P(0, 0). If ABC is isosceles and AB 6, find the coordinates of A, B, and C. 12-8 PROPORTIONS IN A RIGHT TRIANGLE Projection of a Point or of a Line Segment on a Line Whenever the sun is shining, any object casts a shadow. If the sun were directly overhead, the projection of an object would be suggested by the shadow of that object. DEFINITION The projection of a point on a line is the foot of the perpendicular drawn from that point to the line. The projection of a segment on a line, when the segment is not perpendicular to the line, is the segment whose endpoints are the projections of the endpoints of the given line segment on the line. 14365C12.pgs 7/10/07 8:56 AM Page 511 In the figure, MN Proportions in a Right Triangle 511 is the projection of g PQ is P. If PR ' AB g PQ g . PQ on , the pro- The projection of R on jection of PR on g PQ is P. R A P M B N Q Proportions in the Right Triangle C A D B AB CD ' AB AD In the figure, ABC is a right triangle, with the right angle at C. Altitude is so that two smaller triangles are formed, ACD and drawn to hypotenuse , CDA and CDB are right angles. The projection of CBD. Since . We want to pr |
ove is on AC AB that the three right triangles, ABC, ACD, and CBD, are similar triangles and, because they are similar triangles, the lengths of corresponding sides are in proportion. and the projection of BD CD AB BC on is Theorem 12.16 The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to each other and to the original triangle. Given ABC with ACB a right angle and altiat D. CD ' AB tude C Prove ABC ACD CBD Proof Statements 1. ACB is a right angle. CD ' AB 2. 3. ADC and BDC are right angles. 4. ACB ADC BDC 5. A A and B B 6. ABC ACD and ABC CBD 7. ABC ACD CBD A D B Reasons 1. Given. 2. Given. 3. Perpendicular lines intersect to form right angles. 4. All right angles are congruent. 5. Reflexive property of congruence. 6. AA. 7. Transitive property of similarity. 14365C12.pgs 7/10/07 8:56 AM Page 512 512 Ratio, Proportion, and Similarity Now that we have proved that these triangles are similar, we can prove that the lengths of corresponding sides are in proportion. Recall that if the means of a proportion are equal, either mean is called the mean proportional between the extremes. Corollary 12.16a The length of each leg of a right triangle is the mean proportional between the length of the projection of that leg on the hypotenuse and the length of the hypotenuse. Given ABC with ACB a right angle and altiat D CD ' AB tude Prove AC 5 AC AB AD and BC 5 BC AB BD Proof The lengths of the corresponding sides of similar triangles are in proportion. Therefore, since ABC ACD, AC 5 AC AB AD BC 5 BC AB . BD and since ABC CBD, C A D B Corollary 12.16b The length of the altitude to the hypotenuse of a right triangle is the mean proportional between the lengths of the projections of the legs on the hypotenuse. Proof: The lengths of the corresponding sides of similar triangles are in proportion. Therefore, since ACD CBD, . CD 5 CD AD BD EXAMPLE 1 Solution In right triangle ABC, altitude and DB 18 cm, find: a. AC b. BC c. CD CD is drawn to hypotenuse AB . If AD 8 cm AB 5 AD 1 DB 5 8 1 18 5 26 C A 8 cm D 18 cm B 14365C12.pgs 7/10/07 8:56 AM Page 513 Proportions in a Right Triangle 513 Since CD is the altitude to the hypotenuse of right ABC, then: AC 5 AC AB AD AC 5 AC 26 8 (AC)2 5 208 BC 5 BC AB BD BC 5 BC 26 18 (BC)2 5 468 CD 5 CD AD DB CD 5 CD 8 18 (CD)2 5 144 AC 5 208 " BC 5 468 " 5 16 13 5 144 CD 5 " 5 12 13 " 5 4 " " 13 36 " " 13 5 6 " Answers a. 4 13 " cm b. 6 13 " cm c. 12 cm EXAMPLE 2 The altitude to the hypotenuse of right triangle ABC separates the hypotenuse into two segments. The length of one segment is 5 inches more than the measure of the other. If the length of the altitude is 6 inches, find the length of the hypotenuse. Solution Let x the measure of the shorter segment. Then x 5 the measure of the longer segment. (1) The length of the altitude is the mean proportional between the lengths of the segments of the hypotenuse: (2) Set the product of the means equal to the product of the extremes: (3) Write the equation in standard form: (4) Factor the left side: (5) Set each factor equal to 0 and solve for x. Reject the negative root: (6) The length of the hypotenuse is the sum of the lengths of the segments: Answer The length of the hypotenuse is 13 inches. C 6 in(x 5) 36 x2 5x 36 x2 5x 36 0 (x 4)(x 9 reject x x 5 4 4 5 13 in. 14365C12.pgs 7/10/07 8:56 AM Page 514 514 Ratio, Proportion, and Similarity Exercises Writing About Mathematics 1. When altitude is drawn to the hypotenuse of right triangle ABC, it is possible that ACD and BCD are congruent as well as similar. Explain when ACD BCD. CD 2. The altitude to the hypotenuse of right RST separates the hypotenuse, RS , into two con- gruent segments. What must be true about RST? Developing Skills In 3–12, ABC is a right triangle with ACB the right angle. Altitude each case find the required length. CD intersects AB at D. In 3. If AD 3 and CD 6, find DB. 5. If AC 10 and AD 5, find AB. 7. If AD 4 and DB 9, find CD. 9. If AD 3 and DB 27, find CD. 11. If DB 8 and AB 18, find BC. 4. If AB 8 and AC 4, find AD. 6. If AC 6 and AB 9, find AD. 8. If DB 4 and BC 10, find AB. 10. If AD 2 and AB 18, find AC. 12. If AD 3 and DB 9, find AC. Applying Skills In 13–21, the altitude to the hypotenuse of a right triangle divides the hypotenuse into two segments. 13. If the lengths of the segments are 5 inches and 20 inches, find the length of the altitude. 14. If the length of the altitude is 8 feet and the length of the shorter segment is 2 feet, find the length of the longer segment. 15. If the ratio of the lengths of the segments is 1: 9 and the length of the altitude is 6 meters, find the lengths of the two segments. 16. The altitude drawn to the hypotenuse of a right triangle divides the hypotenuse into two segments of lengths 4 and 5. What is the length of the altitude? 17. If the length of the altitude to the hypotenuse of a right triangle is 8, and the length of the hypotenuse is 20, what are the lengths of the segments of the hypotenuse? (Let x and 20 x be the lengths of the segments of the hypotenuse.) 18. The altitude drawn to the hypotenuse of a right triangle divides the hypotenuse into seg- ments of lengths 2 and 16. What are the lengths of the legs of the triangle? 19. In a right triangle whose hypotenuse measures 50 centimeters, the shorter leg measures 30 centimeters. Find the measure of the projection of the shorter leg on the hypotenuse. 20. The segments formed by the altitude to the hypotenuse of right triangle ABC measure 8 inches and 10 inches. Find the length of the shorter leg of ABC. 21. The measures of the segments formed by the altitude to the hypotenuse of a right triangle are in the ratio 1 : 4. The length of the altitude is 14. a. Find the measure of each segment. b. Express, in simplest radical form, the length of each leg. 14365C12.pgs 7/10/07 8:56 AM Page 515 Pythagorean Theorem 515 12-9 PYTHAGOREAN THEOREM The theorems that we proved in the last section give us a relationship between the length of a legs of a right triangle and the length of the hypotenuse. These proportions are the basis for a proof of the Pythagorean Theorem, which was studied in earlier courses. Theorem 12.17a If a triangle is a right triangle, then the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides (the legs). Given ABC is a right triangle with ACB the right angle, c is the length of the hypotenuse, a and b are the lengths of the legs. Prove c2 a2 b2 A b c C a B Proof Statements Reasons A b C cx D x a B 1. ABC is a right triangle with ACB the right angle. CD ' AB . 2. Draw Let BD x and AD c x. 3. c a 5 a x and b 5 b c c 2 x 4. cx a2 and c(c x) b2 c2 cx b2 5. cx c2 cx a2 b2 c2 a2 b2 1. Given. 2. From a point not on a given line, one and only one perpendicular can be drawn to the given line. 3. The length of each leg of a right triangle is the mean proportional between the length of the projection of that leg on the hypotenuse and the length of the hypotenuse. 4. In a proportion, the product of the means is equal to the product of the extremes. 5. Addition postulate. 14365C12.pgs 7/10/07 8:56 AM Page 516 516 Ratio, Proportion, and Similarity The Converse of the Pythagorean Theorem If we know the lengths of the three sides of a triangle, we can determine whether the triangle is a right triangle by using the converse of the Pythagorean Theorem. Theorem 12.17b If the square of the length of one side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle. Given ABC with AB c, BC a, CA b, and c2 a2 b2 C b a F b a Prove ABC is a right triangle with C the right A c B D E angle. Proof Draw DEF with EF a, FD b, and F a right angle. Then DE 2 a2 b2, DE 2 c2 and DE c. Therefore, ABC DEF by SSS. Corresponding angles of congruent triangles are congruent, so C F and C is a right angle. Triangle ABC is a right triangle. We can state Theorems 12.17a and 12.17b as a single theorem. Theorem 12.17 A triangle is a right triangle if and only if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides. EXAMPLE 1 What is the length of the altitude to the base of an isosceles triangle if the length of the base is 18 centimeters and the length of a leg is 21 centimeters? Round your answer to the nearest centimeter. Solution The altitude to the base of an isosceles triangle is perpendicular to the base and is the is the altitude to the base AC , CD bisects the base. In ABC, AB hypotenuse of right triangle ACD, AD 9.0 cm, and AC 21 cm. AD2 CD2 AC2 92 CD2 212 81 CD2 441 CD2 360 CD 10 5 6 360 5 19 9.0 cm 21 cm 10 36 D A C " " " " B Answer The length of the altitude is approximately 19 centimeters. 14365C12.pgs 7/10/07 8:56 AM Page 517 Pythagorean Theorem 517 EXAMPLE 2 When a right circular cone is cut by a plane through the vertex and perpendicular to the base of the cone, the cut surface is an isosceles triangle. The length of the hypotenuse of the triangle is the slant height of the cone, the length one of the legs is the height of the cone, and the length of the other leg is the radius of the base of the cone. If a cone has a height of 24 centimeters and the radius of the base is 10 centimeters, what is the slant height of the cone? Solution Use the Pythagorean Theorem: (hs)2 (hc)2 r2 (hs)2 242 102 (hs)2 676 26 cm Answer hs hs hc r Pythagorean Triples When three integers can be the lengths of the sides of a right triangle, this set of numbers is called a Pythagorean triple. The most common Pythagorean triple is 3, 4, 5: 32 42 52 If we multiply each number of a Pythagorean triple by some positive integer x, then the new triple created is also a Pythagorean triple because it will satisfy the relation a2 b2 c2. For example: If {3, 4, 5} is a Pythagorean triple, then {3x, 4x, 5x} is also a Pythagorean triple for a similar triangle where the ratio of similitude o |
f the second triangle to the first triangle is x : 1. Let x 2. Then {3x, 4x, 5x} {6, 8, 10} and 62 82 102. Let x 3. Then {3x, 4x, 5x} {9, 12, 15}, and 92 122 152. Let x 10. Then {3x, 4x, 5x} {30, 40, 50}, and 302 402 502. Here are other examples of Pythagorean triples that occur frequently: {5, 12, 13} or, in general, {5x, 12x, 13x} where x is a positive integer. {8, 15, 17} or, in general, {8x, 15x, 17x} where x is a positive integer. The 45-45-Degree Right Triangle The legs of an isosceles right triangle are congruent. The measure of each acute angles of an isosceles right triangle is 45°. If two triangles are isosceles right triangles then they are similar by AA. An isosceles right triangle is called a 4545-degree right triangle. 14365C12.pgs 7/10/07 8:56 AM Page 518 518 Ratio, Proportion, and Similarity x s x When a diagonal of a square is drawn, the square is separated into two isosceles right triangles. We can express the length of a leg of the isosceles right triangle in terms of the length of the hypotenuse or the length of the hypotenuse in terms of the length of a leg. Let s be the length of the hypotenuse of an isosceles right triangle and x be the length of each leg. Use the Pythagorean Theorem to set up two equalities. Solve one for x and the other for s: Solve for x: a2 1 b2 5 c2 x2 1 x2 5 s2 2x2 5 s2 x2 5 s2 2 Solve for s: c2 5 a2 1 b2 s2 5 x2 1 x2 s2 5 2x2 " s2 2 s2 2 ? 2 2 $ $ 2 2 s The 30-60-Degree Right Triangle An altitude drawn to any side of an equilateral triangle bisects the base and separates the triangle into two congruent right triangles. Since the measure of each angle of an equilateral triangle is 60°, the measure of one of the acute angles of a right triangle formed by the altitude is 60° and the measure of the other acute angle is 30°. Each of the congruent right triangles formed by drawing an altitude to a side of an equilateral triangle is called a 30-60-degree right triangle. If two triangles are 30-60-degree right triangles, then they are similar by AA. In the diagram, ABC is an equilateral triangle with s the length of each side and h the length of an altitude. Then s is the length of the hypotenuse of the 30-60-degree triangle and h CD ' AB , is the length of a leg. In the diagram, s AC s, AD , and CD h. 2 a2 b2 c2 C 30° s h A 2 s 2 B s2 4 h2 s2 h2 h2 4 4s2 3 4s2 3 h " 2 s 60° s 2 A D B 14365C12.pgs 7/10/07 8:56 AM Page 519 EXAMPLE 3 Solution Pythagorean Theorem 519 In right triangle ABC, the length of the hypotenuse, , is 6 centimeters and the length of one leg is 3 cenAB timeters. Find the length of the other leg. B a2 1 b2 5 c2 a2 1 (3)2 5 (6)2 a2 1 9 5 36 a2 5 27 a 5 27 5 " 6 cm 3 Answer C 3 cm A 9 ? 3 5 3 " " Note: The measure of one leg is one-half the measure of the hypotenuse and the length of the other leg is times the length of the hypotenuse. Therefore, this triangle is a 30-60-degree right triangle. We can use a calculator to verify this. " 2 3 Recall that tan A = find the measure of A. opp adj 5 BC AC 5 3 3 3 5 " 3 " . Use a graphing calculator to ENTER: 2nd TAN1 2nd ¯ 3 ENTER The calculator will return 60 as mA. Therefore, mB 30. Exercises Writing About Mathematics 1. Ira said that if the lengths of the sides of an obtuse triangle ABC are a, b, and c with c opposite the obtuse angle, then c2 a2 b2. Do you agree with Ira? Explain why or why not. (Hint: Make use of the altitude from one of the acute angles.) 2. Sean said that if the measures of the diagonals of a parallelogram are 6 and 8 and the mea- sure of one side of the parallelogram is 5 then the parallelogram is a rhombus. Do you agree with Sean? Explain why or why not. Developing Skills In 3–8, in each case the lengths of three sides of a triangle are given. Tell whether each triangle is a right triangle. 3. 6, 8, 10 4. 7, 8, 12 5. 5, 7, 8 6. 15, 36, 39 7. 14, 48, 50 8. 2, 2 , 4 3 " 14365C12.pgs 7/10/07 8:56 AM Page 520 520 Ratio, Proportion, and Similarity 9. Find, to the nearest tenth of a centimeter, the length of a diagonal of a square if the mea- sure of one side is 8.0 centimeters. 10. Find the length of the side of a rhombus whose diagonals measure 40 centimeters and 96 centimeters. 11. The length of a side of a rhombus is 10 centimeters and the length of one diagonal is 120 millimeters. Find the length of the other diagonal. 12. The length of each side of a rhombus is 13 feet. If the length of the shorter diagonal is 10 feet, find the length of the longer diagonal. 13. Find the length of the diagonal of a rectangle whose sides measure 24 feet by 20 feet. 14. The diagonal of a square measures 12 feet. a. What is the exact measure of a side of the square? b. What is the area of the square? 15. What is the slant height of a cone whose height is 36 centimeters and whose radius is 15 centimeters? 16. One side of a rectangle is 9 feet longer than an adjacent side. The length of the diagonal is 45 feet. Find the dimensions of the rectangle. 17. One leg of a right triangle is 1 foot longer than the other leg. The hypotenuse is 9 feet longer than the shorter leg. Find the length of the sides of the triangle. Applying Skills 18. Marvin wants to determine the edges of a rectangular garden that is to be 10 feet by 24 feet. He has no way of determining the measure of an angle but he can determine lengths very accurately. He takes a piece of cord that is 60 feet long and makes a mark at 10 feet and at 34 feet from one end. Explain how the cord can help him to make sure that his garden is a rectangle. 19. A plot of land is in the shape of an isosceles trapezoid. The lengths of the parallel sides are 109 feet and 95 feet. The length of each of the other two sides is 25 feet. What is the area of the plot of land? 20. From a piece of cardboard, Shanti cut a semicircle with a radius of 10 inches. Then she used tape to join one half of the diameter along which the cardboard had been cut to the other half, forming a cone. What is the height of the cone that Shanti made? 21. The lengths of two adjacent sides of a parallelogram are 21 feet and 28 feet. If the length of a diagonal of the parallelogram is 35 feet, show that the parallelogram is a rectangle. 22. The lengths of the diagonals of a parallelogram are 140 centimeters and 48 centimeters. The length of one side of the parallelogram is 74 centimeters. Show that the parallelogram is a rhombus. 14365C12.pgs 7/10/07 8:56 AM Page 521 The Distance Formula 521 23. A young tree is braced by wires that are 9 feet long and fastened at a point on the trunk of the tree 5 feet from the ground. Find to the nearest tenth of a foot how far from the foot of the tree the wires should be fastened to the ground in order to be sure that the tree will be perpendicular to the ground. 24. The length of one side of an equilateral triangle is 12 feet. What is the distance from the centroid of the triangle to a side? (Express the answer in simplest radical form.) 12-10 THE DISTANCE FORMULA When two points in the coordinate plane are on the same vertical line, they have the same x-coordinate and the distance between them is the absolute value of the difference of their y-coordinates. In the diagram, the coordinates of A are (4, 8) and the coordinates of C are (4, 2). y A(4, 8) CA 5 8 2 2 5 6 When two points in the coordinate plane are on the same horizontal line, they have the same y-coordinate and the distance between them is the absolute value of the difference of their x-coordinates. In the diagram, the coordinates of B are (1, 2) and the coordinates of C are (4, 2). CB 5 1 2 4 5 3 1 O B(1, 2) 1 C(4, 2) x In ABC, C is a right angle and AB is the hypotenuse of a right triangle. Using the Pythagorean Theorem, we can find AB: AB2 5 CA2 1 CB2 AB2 5 62 1 32 AB2 5 36 1 9 AB 5 45 " AB 5 3 " 5 This example suggests a method that can be used to find a formula for the length of any segment in the coordinate plane. 14365C12.pgs 7/10/07 8:56 AM Page 522 522 Ratio, Proportion, and Similarity Let B(x1, y1) and A(x2, y2) be any two points in the coordinate plane. From A draw a vertical line and from B draw a horizontal line. Let the intersection of these two lines be C. The coordinates of C are (x2, y1). Let AB c, CB a x2 , and CA b y2 . Then, y1 x1 y O A(x2, y2) c b a B(x1, y1) C(x2, y1) x c2 a2 b2 x1 c2 x2 c (x2 2 x1)2 1 (y2 2 y1)2 2 y2 y1 2 " This result is called the distance formula. If the endpoints of a line segment in the coordinate plane are B(x1, y1) and A(x2, y2), then: AB " (x2 2 x1)2 1 (y2 2 y1)2 EXAMPLE 1 The coordinates of the vertices of quadrilateral ABCD are A(1, 3), B(6, 4), C(5, 3), and D(2, 4). y D a. Prove that ABCD is a rhombus. b. Prove that ABCD is not a square. 1 1 O A C x B Solution a. AB 5 " (6 2 (21))2 1 (24 2 (23))2 BC 5 5 5 5 " " " (7)2 1 (21)2 49 1 1 50 5 5 5 (5 2 6)2 1 (3 2 (24))2 (21)2 1 (7)2 " " 1 1 49 " 50 " (22 2 5)2 1 (4 2 3)2 DA 5 (27)2 1 (1)2 " " CD 5 5 5 5 49 1 1 " 50 " (21 2 (22))2 1 (23 2 4)2 (1)2 1 (27)2 " " 1 1 49 " 50 " 5 5 5 14365C12.pgs 7/10/07 8:56 AM Page 523 The Distance Formula 523 The lengths of the sides of the quadrilateral are equal. Therefore, the quadrilateral is a rhombus. b. If a rhombus is a square, then it has a right angle. METHOD 1 AC 5 5 5 5 " " " " (5 2 (21))2 1 (3 2 (23))2 If B is a right angle, then (6)2 1 (6)2 36 1 36 72 AC2 5 AB2 1 BC2 2 1 2 5? 50 72 A " B B 72 2 50 1 50 ✘ A " 50 2 B A " Therefore, ABC is not a right triangle, B is not a right angle and ABCD is not a square. METHOD 2 slope of AB 5 24 2 (23) 6 2 (21) slope of BC 5 3 2 (24) 5 2 6 5 21 7 5 21 7 5 7 21 5 27 AB The slope of Therefore, AB rhombus is not a square. is not equal to the negative reciprocal of the slope of . BC , B is not a right angle and the is not perpendicular to BC EXAMPLE 2 Prove that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices using the distance formula. y B(0, 2b) Proof We will use a coordinate proof. The triangle can be placed at any convenient position. Let right triangle ABC have vertices A(2a, 0), B(0, 2b), |
and C(0, 0). Let M be the midpoint of the hypotenuse . The coordinates of M, the midpoint of AB AB , are , 0 1 2b 2 2a 1 0 2 A 5 (a, b) B M(a, b) O C(0, 0) A(2a, 0) x 14365C12.pgs 7/10/07 8:56 AM Page 524 524 Ratio, Proportion, and Similarity y B(0, 2b) Then, since M is the midpoint of formula: AB , AM BM, and using the distance M(a, b) AM 5 BM 5 (a 2 0)2 1 (b 2 2b)2 " CM 5 " (a 2 0)2 1 (b 2 0)2 x C(0, 0) A(2a, 0) O 5 5 a2 1 (2b)2 " a2 1 b2 " 5 a2 1 b2 " Therefore, the midpoint of the hypotenuse is equidistant from the vertices of the triangle. EXAMPLE 3 Prove that the medians to the base angles of an isosceles triangle are congruent. Given: Isosceles ABC with vertices A(2a, 0), B(0, 2b), C(2a, 0). Let M be the midpoint of AB . and N be the midpoint of BC y B(0, 2b) M N Prove: CM > AN Proof The coordinates of M are The coordinates of N are A(2a, 0) O C(2a, 0) x 22a 1 0 2 , 0 1 2b 2 A The length of B CM 5 (2a, b) . is 2a 1 0 2 , 0 1 2b 2 A The length of B AN is 5 (a, b) . (2a 2 2a)2 1 (b 2 0)2 (22a 2 a)2 1 (0 2 b)2 " 5 (23a)2 1 b2 " " 5 (23a)2 1 (2b)2 " 5 9a2 1 b2 " CM AN; therefore, CM > AN . 5 9a2 1 b2 " Exercises Writing About Mathematics 1. Can the distance formula be used to find the length of a line segment when the endpoints of the segment are on the same horizontal line or on the same vertical line? Justify your answer. 2. Explain why x2 x1 2 (x2 x1)2. 14365C12.pgs 7/10/07 8:56 AM Page 525 The Distance Formula 525 Developing Skills In 3–10, the coordinates of the endpoints of in simplest form. AB are given. In each case, find the exact value of AB 3. A(1, 2), B(4, 6) 5. A(3, 2), B(5, 4) 7. A(1, 2), B(3, 4) 9. A(6, 2), B(1, 3) 4. A(–1, 6), B(4, 6) 6. A(0, 2), B(3, 1) 8. A(–5, 2), B(1, 6) 10. A(–3, 3), B(3, 3) 11. The coordinates of A are (0, 4) and the x-coordinate of B is 5. What is the y-coordinate of B if AB 13? (Two answers are possible.) 12. The coordinates of M are (2, 1) and the y-coordinate of N is 5. What is the x-coordinate of N if MN = 3 ? (Two answers are possible.) 5 " 13. The vertices of a quadrilateral are A(0, 2), B(5, 2), C(8, 2), D(3, 2). Prove that the quadrilateral is a rhombus using the distance formula. 14. The vertices of a triangle are P(1, 1), Q(7, 1), and R(3, 3). a. Show that PQR is an isosceles triangle. b. Show that PQR is a right triangle using the Pythagorean Theorem. c. Show that the midpoint of the hypotenuse is equidistant from the vertices. 15. The vertices of a triangle are L(1, 1), M(7, 3), and N(2, 2). a. Show that LMN is a scalene triangle. b. Show that LMN is a right triangle using the Pythagorean Theorem. c. Show that the midpoint of MN is equidistant from the vertices. 16. The vertices of DEF are D(2, 3), E(5, 0), and F(2, 3). Show that 17. The coordinates of the vertices of BAT are B(2, 7), A(2, 1), and T(11, 4). DE > FE . a. Find the equation of the line that is the altitude from B to AT . b. Find the coordinates of D, the foot of the perpendicular or the foot of the altitude from part a. c. Find the length of the altitude BD . 18. The coordinates of the vertices of EDF are E(2, 0), D(4, 0), and F a. Find ED, DF, and FE. b. Is EDF equilateral? Justify your answer. 1, 3 3 " . B A 19. The vertices of ABC are A(1, 1), B(4, 1), and C(2, 4). a. Find the coordinates of the vertices of ABC, the image of ABC under the trans- formation D2. b. Show that distance is not preserved under the dilation. 14365C12.pgs 7/10/07 8:56 AM Page 526 526 Ratio, Proportion, and Similarity c. Show that ABC ABC using SSS. d. Use part c to show that the angle measures of ABC are preserved under the dilation. 20. The vertices of quadrilateral ABCD are A(2, 0), B(3, 1), C(4, 1), and D(3, 2). a. Show that ABCD is a parallelogram using the distance formula. b. Find the coordinates of the vertices of quadrilateral ABCD, the image of ABCD under the transformation D3. c. Show that ABCD is a parallelogram using the distance formula. d. Use part c to show that the images of the parallel segments of ABCD are also parallel under the dilation. 21. The vertices of quadrilateral ABCD are A(2, 2), B(2, 0), C(3, 3), and D(1, 1). Use the distance formula to prove that ABCD is a parallelogram but not a rhombus. 22. The vertices of ABC are A(0, 2), B(4, 6), and C(2, 4). Prove that ABC is an isosceles right triangle using the Pythagorean Theorem. Applying Skills 23. Use the distance formula to prove that (a, 0), (0, b) and (a, 0) are the vertices of an isosce- les triangle. 24. The vertices of square EFGH are E(0, 0), F(a, 0), G(a, a), and H(0, a). Prove that the diagonals of a square, EG and FH , are congruent and perpendicular. 25. The vertices of quadrilateral ABCD are A(0, 0), B(b, c), C(b a, c), and D(a, 0). Prove that if a2 b2 c2 then ABCD is a rhombus. 26. Prove the midpoint formula using the distance formula. Let P and Q have coordinates (x1, y1) and (x2, y2), respectively. Let M have coordinates point of if and only if PM MQ. PQ A x1 1 x2 2 y1 1 y2 2 , . M is the mid- B a. Find PM. b. Find MQ. c. Show that PM MQ. 27. The vertices of WX are W(w, y) and X(x, z). a. What are the coordinates of b. Show, using the distance formula, that WX is k times the length of WX. under the dilation Dk? , the image of WX WrXr 14365C12.pgs 8/2/07 5:57 PM Page 527 CHAPTER SUMMARY Definitions to Know a • The ratio of two numbers, a and b, where b is not zero, is the number . b • A proportion is an equation that states that two ratios are equal. Chapter Summary 527 • In the proportion , the first and fourth terms, a and d, are the extremes of the proportion, and the second and third terms, b and c, are the means. b 5 c a d • If the means of a proportion are equal, the mean proportional is one of the means. • Two line segments are divided proportionally when the ratio of the lengths of the parts of one segment is equal to the ratio of the lengths of the parts of the other. • Two polygons are similar if there is a one-to-one correspondence between their vertices such that: 1. All pairs of corresponding angles are congruent. 2. The ratios of the lengths of all pairs of corresponding sides are equal. • The ratio of similitude of two similar polygons is the ratio of the lengths of corresponding sides. • A dilation of k is a transformation of the plane such that: 1. The image of point O, the center of dilation, is O. 2. When k is positive and the image of P is P, then h OP and h OPr are the same ray and OP kOP . 3. When k is negative and the image of P is P, then h OP and h OPr are opposite rays and OP kOP. • The projection of a point on a line is the foot of the perpendicular drawn from that point to the line. • The projection of a segment on a line, when the segment is not perpendicular to the line, is the segment whose endpoints are the projections of the endpoints of the given line segment on the line. • A Pythagorean triple is a set of three integers that can be the lengths of the sides of a right triangle. Postulates 12.1 Any geometric figure is similar to itself. (Reflexive property) 12.2 A similarity between two geometric figures may be expressed in either order. (Symmetric property) 12.3 Two geometric figures similar to the same geometric figure are similar to each other. (Transitive property) 12.4 For any given triangle there exist a similar triangle with any given ratio of similitude. 14365C12.pgs 7/10/07 8:56 AM Page 528 528 Ratio, Proportion, and Similarity Theorems 12.1 In a proportion, the product of the means is equal to the product of the extremes. In a proportion, the means may be interchanged. 12.1a 12.1b In a proportion, the extremes may be interchanged. 12.1c If the products of two pairs of factors are equal, the factors of one pair can be the means and the factors of the other the extremes of a proportion. 12.3 12.4 12.2 A line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side. Two line segments are divided proportionally if and only if the ratio of the length of a part of one segment to the length of the whole is equal to the ratio of the corresponding lengths of the other segment. Two triangles are similar if two angles of one triangle are congruent to two corresponding angles of the other. (AA) Two triangles are similar if the three ratios of corresponding sides are equal. (SSS) Two triangles are similar if the ratios of two pairs of corresponding sides are equal and the corresponding angles included between these sides are congruent. (SAS) 12.5 12.6 12.7 A line is parallel to one side of a triangle and intersects the other two sides if and only if the points of intersection divide the sides proportionally. 12.8 Under a dilation, angle measure is preserved. 12.9 Under a dilation, midpoint is preserved. 12.10 Under a dilation, collinearity is preserved. 12.11 12.12 If two triangles are similar, the lengths of corresponding altitudes have the same ratio as the lengths of any two corresponding sides. If two triangles are similar, the lengths of corresponding medians have the same ratio as the lengths of any two corresponding sides. If two triangles are similar, the lengths of corresponding angle bisectors have the same ratio as the lengths of any two corresponding sides. 12.14 Any two medians of a triangle intersect in a point that divides each 12.13 median in the ratio 2 : 1. 12.15 The medians of a triangle are concurrent. 12.16 The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to each other and to the original triangle. 12.16a The length of each leg of a right triangle is the mean proportional between the length of the projection of that leg on the hypotenuse and the length of the hypotenuse. 12.16b The length of the altitude to the hypotenuse of a right triangle is the mean proportional between the lengths of the projections of the legs on the hypotenuse. 12.17 A triangle is a right triangle if and only if the square of the length of the longest side is |
equal to the sum of the squares of the lengths of the other two sides. 14365C12.pgs 7/10/07 8:56 AM Page 529 Formulas In the coordinate plane, under a dilation of k with the center at the origin: Review Exercises 529 P(x, y) → P(kx, ky) or Dk(x, y) (kx, ky) If x is the length of a leg of an isosceles right triangle and s is the length of the hypotenuse, then: x s and s 2 " 2 x 2 " If s is the length of a side of an equilateral triangle and h is the length of an altitude then: If the endpoints of a line segment in the coordinate plane are B(x1, y1) and A(x2, y2), then: h 3 2 s " AB " (x2 2 x1)2 1 (y2 2 y1)2 VOCABULARY 12-1 Similar • Ratio of two numbers • Proportion • Extremes • Means • Mean proportional • Geometric mean 12-2 Midsegment theorem • Line segments divided proportionally 12-3 Similar polygons • Ratio of similitude • Constant of proportionality 12-4 Postulate of similarity • AA triangle similarity • SSS similarity theorem • SAS similarity theorem 12-5 Dilation • Enlargement • Contraction • Constant of dilation, k 12-7 Centroid 12-8 Projection of a point on a line • Projection of a segment on a line 12-9 Pythagorean Theorem • Pythagorean triple • 45-45-degree right triangle • 30-60-degree right triangle 12-10 Distance formula • Foot of an altitude REVIEW EXERCISES 1. Two triangles are similar. The lengths of the sides of the smaller triangle are 5, 6, and 9. The perimeter of the larger triangle is 50. What are the lengths of the sides of the larger triangle? 2. The measure of one angle of right ABC is 67° and the measure of an angle of right LMN is 23°. Are the triangles similar? Justify your answer. 14365C12.pgs 7/10/07 8:56 AM Page 530 530 Ratio, Proportion, and Similarity 3. A line parallel to side AB of ABC intersects AC at E and BC at F. If EC 12, AC 20, and AB 15, find EF. 4. A line intersects side AC FC 5, BC 15, prove that EFC ABC. of ABC at E and BC at F. If EC 4, AC 12, 5. The altitude to the hypotenuse of a right triangle divides the hypotenuse into two segments. If the length of the altitude is 12 and the length of the longer segment is 18, what is the length of the shorter segment? 6. In LMN, L is a right angle, b. Find MN. a. Find LP. LP is an altitude, MP 8, and PN 32. c. Find ML. d. Find NL. 7. The length of a side of an equilateral triangle is 18 centimeters. Find, to the nearest tenth of a centimeter, the length of the altitude of the triangle. 8. The length of the altitude to the base of an isosceles triangle is 10.0 cen- timeters and the length of the base is 14.0 centimeters. Find, to the nearest tenth of a centimeter, the length of each of the legs. 9. The coordinates of the endpoints of PQ are P(2, 7) and Q(8, 1). a. Find the coordinates of the endpoints of PrQr under the composition D2 + rx-axis b. What is the ratio . PQ PrQr ? c. What are the coordinates of M, the midpoint of d. What are the coordinates of M, the image of M under PQ ? D2 + rx-axis ? e. What are the coordinates of N, the midpoint of f. Is M the midpoint of ? Justify your answer. PrQr PrQr ? 10. If g AB g CD and AD and BC intersect at E, prove that ABE DCE. 11. A line intersects AC at E and at F. If ABC EFC, g AB g EF . prove that BC 12. Find the length of the altitude to the bases of isosceles trapezoid KLMN if KL 20 cm, MN 38 cm, and KN 15 cm. 13. Find the length of a side of a rhombus if the measures of the diagonals of the rhombus are 30 inches and 40 inches. 14365C12.pgs 7/10/07 8:56 AM Page 531 Review Exercises 531 14. The length of a side of a rhombus is 26.0 centimeters and the length of one diagonal is 28.0 centimeters. Find to the nearest tenth the length of the other diagonal. 15. The coordinates of the vertices of RST are R(4, 1), S(3, 2), and T(2, 1). a. Find the length of each side of the triangle in simplest radical form. b. Prove that the triangle is a right triangle. 16. The vertices of ABC are A(0, 0), B(4, 3), and C(0, 5). a. Prove that ABC is isosceles. b. The median to BC is AD . Find the coordinates of D. c. Find AD and DB. d. Prove that is the altitude to 17. The vertices of ABC are A(2, 1), B(2, 1), and C(0, 3). AD BC using the Pythagorean Theorem. a. Find the coordinates of ABC, the image of ABC under D3. b. Find, in radical form, the lengths of the sides of ABC and of ABC. c. Prove that ABC ABC. d. Find the coordinates of P, the centroid of ABC. CP PM 5 2 1 e. Let M be the midpoint of f. Find the coordinates of P, the centroid of ABC. g. Is P the image of P under D3? h. Let M be the midpoint of ArBr . Prove that . Prove that AB . CrP PMr 5 2 1 . 18. A right circular cone is cut by a plane through a diameter of the base and the vertex of the cone. If the diameter of the base is 20 centimeters and the height of the cone is 24 centimeters, what is the slant height of the cone? Exploration A line parallel to the shorter sides of a rectangle can divide the rectangle into a square and a smaller rectangle. If the smaller rectangle is similar to the given rectangle, then both rectangles are called golden rectangles and the ratio of the lengths of their sides is called the golden ratio. The golden ratio is 1 1 A : 2. 5 " B 14365C12.pgs 7/10/07 8:56 AM Page 532 532 Ratio, Proportion, and Similarity Follow the steps to construct a golden rectangle. You may use compass and straightedge or geometry software. STEP 1. Draw square ABCD. STEP 2. Construct E, the midpoint of AB . STEP 3. Draw the ray h AB . STEP 4. With E as the center and radius EC, draw an arc that h AB . Call this intersects point F. STEP 5. Draw the ray h DC . D C G A E 1 B F STEP 6. Construct the line perpendicular to of this line with be point G. h DC h AB through F. Let the intersection a. Let AB BC 2 and EB 1. Find EC EF, AF AE EF, and BF EF EB. Express each length as an irrational number in simplest radical form. b. Show that AFGD and FGCB are golden rectangles by showing that they are similar, that is, that AD 5 FG AF BF . c. Repeat steps 1 through 6 using a different square. Let AB x. Complete parts a and b. Do you obtain the same ratio? d. Research the golden rectangle and share your findings with your classmates. CUMULATIVE REVIEW Chapters 1–12 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The length and width of a rectangle are 16 centimeters and 12 centimeters. What is the length of a diagonal of the rectangle? (3) 25 cm (2) 20 cm (1) 4 cm (4) 4 cm 7 " 2. The measure of an angle is 12 degrees more than twice the measure of its supplement. What is the measure of the angle? (2) 56 (1) 26 (3) 64 (4) 124 3. What is the slope of the line through A(2, 8) and B(4, 1)? (1) 23 2 (2) 3 2 (3) 22 3 (4) 2 3 14365C12.pgs 7/10/07 8:56 AM Page 533 Cumulative Review 533 4. The measures of the opposite angles of a parallelogram are represented by 2x 34 and 3x 12 . Find the value of x. (1) 22 (2) 46 (3) 78 (4) 126 5. Which of the following do not determine a plane? (1) three lines each perpendicular to the other two (2) two parallel lines (4) a line and a point not on it (3) two intersecting lines 6. Which of the following cannot be the measures of the sides of a triangle? (1) 5, 7, 8 (2) 3, 8, 9 (3) 5, 7, 7 (4) 2, 6, 8 7. Under a rotation of 90° about the origin, the image of the point whose coordinates are (3, 2) is the point whose coordinates are (1) (2, 3) (3) (3, 2) (2) (2, 3) (4) (2, 3) 8. If a conditional statement is true, which of the following must be true? (1) converse (2) inverse 9. In the figure, side AB of ABC is extended through B to D. If mCBD 105 and mA 53, what is the measure of C? (1) 22 (2) 52 (3) 75 (4) 158 (3) contrapositive (4) biconditional C A B D 10. A parallelogram with one right angle must be (1) a square. (2) a rectangle. (3) a rhombus. (4) a trapezoid. Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Given: ABCD with AB > CD , E not on › ‹ , and ABCD BE > CE . Prove: AE > DE . 12. Given: perpendicular to plane p at R, points A and B in plane p, and RS RA > RB . l s _ _ l e _ _ Prove: SA > SB _ _ l s _ _ l e 14365C12.pgs 7/10/07 8:56 AM Page 534 534 Ratio, Proportion, and Similarity Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. The radius of the base of a right circular cone is one-half the slant height of the cone. The radius of the base is 2.50 feet. a. Find the lateral area of the cone to the nearest tenth of a square foot. b. Find the volume of the cone to the nearest tenth of a cubic foot. 14. The coordinates of the vertices of ABC are A(1, 0), B(4, 0), and C(2, 6). a. Write an equation of the line that contains the altitude from C to AB . b. Write an equation of the line that contains the altitude from B to AC . c. What are the coordinates of D, the intersection of the altitudes from C and from B? d. Write an equation for g AD . e. Is g AD perpendicular to BC , that is, does g AD contain the altitude from A to BC ? Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. In the diagram, ABCD is a rectangle and ADE is an isosceles triangle , a median to . If with AD BC of AE > DE EF of ADE, is extended to intersect at G, prove that G is the midpoint BC . E D F A C G B 16. The coordinates of the vertices of ABC are A(2, 5), B(3, 1), and C(6, 4). a. Find the coordin |
ates of ABC, the image of ABC under the com- position ry 5 x b. Show that ry 5 x + rx-axis . + rx-axis(x,y) R90°(x, y). 14365C13.pgs 7/12/07 3:56 PM Page 535 GEOMETRY OF THE CIRCLE Early geometers in many parts of the world knew that, for all circles, the ratio of the circumference of a circle to its diameter was a constant. Today, we write C p d 5 p to represent this constant. Euclid established that the ratio of the area of a circle to the square of its diame- , but early geometers did not use the symbol A d2 5 k ter was also a constant, that is, . How do these constants, p and k, relate to one another? Archimedes (287–212 B.C.) proposed that the area of a circle was equal to the area of a right triangle whose legs have lengths equal to the radius, r, and the circumference, C, of a circle. Thus A rC. He used indirect proof and the areas of inscribed and circumscribed polygons to prove his conjecture and to prove 1 2 310 71 , p ,3 1 7 that . Since this inequality can be written as 3.140845 . . . 3.142857 . . . , Archimedes’ approximation was correct to two decimal places. p Use Archimedes’ formula for the area of a circle and d 2r to show that and the facts that A and that 4k = pr2 C d 5 p .p CHAPTER 13 CHAPTER TABLE OF CONTENTS 13-1 Arcs and Angles 13-2 Arcs and Chords 13-3 Inscribed Angles and Their Measures 13-4 Tangents and Secants 13-5 Angles Formed by Tangents, Chords, and Secants 13-6 Measures of Tangent Segments, Chords, and Secant Segments 13-7 Circles in the Coordinate Plane 13-8 Tangents and Secants in the Coordinate Plane Chapter Summary Vocabulary Review Exercises Cumulative Review 535 14365C13.pgs 7/12/07 3:56 PM Page 536 536 Geometry of the Circle 13-1 ARCS AND ANGLES In Chapter 11, we defined a sphere and found that the intersection of a plane and a sphere was a circle. In this chapter, we will prove some important relationships involving the measures of angles and line segments associated with circles. Recall the definition of a circle. DEFINITION A circle is the set of all points in a plane that are equidistant from a fixed point of the plane called the center of the circle. If the center of a circle is point O, the circle is called circle O, written in symbols as (O . B A radius of a circle (plural, radii) is a line segment from the center of the circle to any point of the circle. The term radius is used to mean both the line segment and the length of the line segment. If A, B, and C are points of circle O, then are radii of the circle. Since the definition of a circle states that all points of the circle are equidistant from its center, OA OB OC. Thus, because equal line segments are congruent. We can state what we have just proved as a theorem. OA > OB > OC , and , OA OB OC O A C Theorem 13.1 All radii of the same circle are congruent. A circle separates a plane into three sets of points. If we let the length of the radius of circle O be r, then: • Point C is on the circle if OC r. • Point D is outside the circle if OD r. • Point E is inside the circle if OE r. C r O E D The interior of a circle is the set of all points whose distance from the cen- ter of the circle is less than the length of the radius of the circle. The exterior of a circle is the set of all points whose distance from the cen- ter of the circle is greater than the length of the radius of the circle. Central Angles Recall that an angle is the union of two rays having a common endpoint and that the common endpoint is called the vertex of the angle. DEFINITION A central angle of a circle is an angle whose vertex is the center of the circle. 14365C13.pgs 7/12/07 3:56 PM Page 537 Arcs and Angles 537 In the diagram, LOM and MOR are central angles because the vertex of each angle is point O, the center of the circle. L O M R Types of Arcs An arc of a circle is the part of the circle between two points on the circle. In the diagram, A, B, C, and D are points on circle O and AOB intersects the circle at two distinct points, A and B, separating the circle into two arcs Minor arc ( D ABX ) O A Major arc ( D )ACBX 1. If mAOB 180, points A and B and the points of the circle in the inte- rior of AOB make up minor arc AB, written as ABX . 2. Points A and B and the points of the circle not in the interior of AOB make up major arc AB. A major arc is usually named by three points: the two endpoints and any other point on the major arc. Thus, the major arc with endpoints A and B is written as ACBX or 3. If mAOC 180, points A and C separate circle O into two equal parts, ABCX each of which is called a semicircle. In the diagram above, name two different semicircles. and . ADBX ADCX Degree Measure of an Arc An arc of a circle is called an intercepted arc, or an arc intercepted by an angle, if each endpoint of the arc is on a different ray of the angle and the other points of the arc are in the interior of the angle. DEFINITION The degree measure of an arc is equal to the measure of the central angle that intercepts the arc. 14365C13.pgs 7/12/07 3:56 PM Page 538 538 Geometry of the Circle 80 G F 80 O E mGFEX 5 180 In circle O, GOE is a straight angle, mGOE 180, and mFOG 80. . Also, since Therefore, the degree measure of is 80°, written as FGX , mFEX 5 180 2 80 5 100 and mFGX 5 80 mFEGX 5 100 1 180 5 280 Therefore, mGFX 1 mFEGX 5 80 1 280 5 360 1. The degree measure of a semicircle is 180. Thus: 2. The degree measure of a major arc is equal to 360 minus the degree mea- sure of the minor arc having the same endpoints. Do not confuse the degree measure of an arc with the length of an arc. The degree measure of every circle is 360 but the circumference of a circle is 2p times the radius of the circle. Example: in circle O, OA 1 cm, and in circle O, OA 1.5 cm. In both circles, the degree measure of the circle is 360° but the circumference of circle O is 2p centimeters, and the circumference of circle O is 3p centimeters. O O A A Congruent Circles, Congruent Arcs, and Arc Addition DEFINITION Congruent circles are circles with congruent radii. If OC > OrCr , then circles O and O are congruent. C O C O DEFINITION Congruent arcs are arcs of the same circle or of congruent circles that are equal in measure. 14365C13.pgs 7/31/07 1:15 PM Page 539 Arcs and Angles 539 If mCDX 5 mCrDrX and O is not congruent to circle O, then and 60, then CDX have the same degree measure. (O > (Or CsDsX CDX > CrDrX is not congruent to . However, if circle CDX CsDsX even if D 60 C O C O 60 D 60 C D O Postulate 13.1 Arc Addition Postulate BCX and mABCX BCX ABX If point and no other points in common, then ABX m . m are two arcs of the same circle having a common endand ABCXBCX ABX The arc that is the sum of two arcs may be a minor arc, a major arc, or semimBCX 5 40 , circle. For example, A, B, C, and D are points of circle O, mABX 5 90 , and h OB and 1. Minor arc: Also, BCX h are opposite rays. OD ABX ACX mACX 5 mABX 1 mBCX 5 90 1 40 5 130 2. Major arc: BCDXABX ABDX mABDX 5 mABX 1 mBCDX = Also, 40 B C 90 O A 5 90 1 180 5 270 D h 3. Semicircle: Since OB BCX BCDXCDX Thus, and h OD are opposite rays, BOD is a straight angle. mBCX 1 mCDX 5 mBCDX 5 180 , a semicircle. Also, . 14365C13.pgs 7/31/07 1:15 PM Page 540 540 Geometry of the Circle Theorem 13.2a In a circle or in congruent circles, if central angles are congruent, then their intercepted arcs are congruent Given Circle O circle O, AOB COD, and AOB AOB. ABX > ArBrX ABX > CDX and . B B A C A O O Prove Proof D It is given that AOB COD and AOB AOB. Therefore, mAOB mCOD mAOB because congruent angles have equal measures. Then since the degree measure of an arc is equal to the degree meamABX 5 mCDX 5 mArBrX . It is sure of the central angle that intercepts that arc, also given that circle O and circle O are congruent circles. Congruent arcs are defined as arcs of the same circle or of congruent circles that are equal in measure. Therefore, since their measures are equal, ABX > ArBrX ABX > CDX and . The converse of this theorem can be proved by using the same definitions and postulates. Theorem 13.2b In a circle or in congruent circles, central angles are congruent if their intercepted arcs are congruent. Theorems 13.2a and 13.2b can be written as a biconditional. Theorem 13.2 In a circle or in congruent circles, central angles are congruent if and only if their intercepted arcs are congruent. EXAMPLE 1 h OA Let and h OB a. mBOC b. be opposite rays and mAOC 75. Find: mBACX mACX mABX mBCX d. e. c. Solution a. m/BOC 5 m/AOB 2 m/AOC 5 180 2 75 5 105 A 75° C O B D 14541C13.pgs 1/25/08 3:48 PM Page 541 b. m c. m d. m mAOC 75 mBOC 105 ACX BCX ABX mAOB 180 mBACX 5 mBDAX 1 mACX e. 5 180 1 75 5 255 Arcs and Angles 541 mBACX 5 360 2 mBCX or 5 360 2 105 5 255 Answers a. 105 b. 75 c. 105 d. 180 e. 255 Exercises Writing About Mathematics 1. Kay said that if two lines intersect at the center of a circle, then they intercept two pairs of congruent arcs. Do you agree with Kay? Justify your answer. 2. Four points on a circle separate the circle into four congruent arcs: it true that g AC g ' BD ? Justify your answer. CDXBCXABX , , , and DAX . Is Developing Skills In 3–7, find the measure of the central angle that intercepts an arc with the given degree measure. 3. 35 4. 48 5. 90 6. 140 7. 180 In 8–12, find the measure of the arc intercepted by a central angle with the given measure. 8. 60 9. 75 10. 100 11. 120 12. 170 In 13–22, the endpoints of measure. 15. 13. mBOC mBCX mDAX mBCDX 19. 21. mAOC 17. AOC are on circle O, mAOB 89, and mCOD 42. Find each mABX 14. 16. mDOA 18. mBOD mDABX 22. mADCX 20. C 89° O 42° D B A 14541C13.pgs 1/25/08 3:48 PM Page 542 542 Geometry of the Circle 23. In 23–32, P, Q, S, and R are points on circle O, mPOQ 100, mQOS 110, and mSOR 35. Find each measure. mPQX mSRX mRPX 27. 29. mQOR mSRPX mQSX 24. 26. mROP mPQSX mQSRX mRPQX 110° 100° 35° 25. 28. 32. 31. 30. Q O R P S Applying Skills 33. Given: Circle O with ABX > CDX . Prove: AOB COD A O B D C 34. Given: and AB are on circle O. CD intersect at O, and the endpoints of Prov |
e: AC > BD 35. In circle O, AOB ' COD . Find mACX 36. Points A, B, C, and D lie on circle O, and a square. Hands-On Activity and mADCX AC ' BD . AB and CD A C O D B at O. Prove that quadrilateral ABCD is For this activity, you may use compass, protractor, and straightedge, or geometry software. 1. Draw circle O with a radius of 2 inches and circle O with a radius of 3 inches. 2. Draw points A and B on the circle O so that mArBrX 60. so that 3. Show that AOB AOB. mABX 60 and points A and B on circle O 14365C13.pgs 7/12/07 3:56 PM Page 543 Arcs and Chords 543 13-2 ARCS AND CHORDS DEFINITION A chord of a circle is a line segment whose endpoints are points of the circle. A diameter of a circle is a chord that has the center of the circle as one of its points. AB and , are chords of circle O. In the diagram, Since O is a point of is a diameter. Since OA and OC are the lengths of the radius of circle O, OA OC . and O is the midpoint of AOC If the length of the radius of circle O is r, and the AOC AOC AOC length of the diameter is d, then d 5 AOC 5 OA 1 OC 5 r 1 r 5 2r d 2r That is: ABX , and major The endpoints of a chord are points on a circle and, therefore, determine two arcs of a circle, a minor arc and , central AOB, a major arc. In the diagram, chord are all determined by points A minor and B. We proved in the previous section that in a circle, congruent central angles intercept congruent arcs. Now we can prove that in a circle, congruent central angles have congruent chords and that congruent arcs have congruent chords. ABX AB C O A B A B O Theorem 13.3a In a circle or in congruent circles, congruent central angles have congruent chords. Given (O > (Or and COD AOB AOB A B A B Prove CD > AB > ArBr C O D O 14541C13.pgs 1/25/08 3:48 PM Page 544 544 Geometry of the Circle Proof We will show that COD AOB AOB by SAS. A B A B It is given that AOB COD and AOB AOB. Therefore, AOB COD AOB by the ArOr transitive property of congruence. Since the radii of congruent circles, these segments are all congruent: D , AO DO BO CO O , , , C O , and BrOr are DO > CO > AO > BO > ArOr > BrOr Therefore, by SAS, COD AOB AOB. Since corresponding parts of congruent triangles are congruent, CD > AB > ArBr . The converse of this theorem is also true. Theorem 13.3b In a circle or in congruent circles, congruent chords have congruent central angles. Given (O > (Or and CD > AB > ArBr Prove COD AOB AOB C D A O A B O Strategy This theorem can be proved in a manner similar to Theorem 13.3a: prove that COD AOB AOB by SSS. B The proof of Theorem 13.3b is left to the student. (See exercise 23.) Theorems 13.3a and 13.3b can be stated as a biconditional. Theorem 13.3 In a circle or in congruent circles, two chords are congruent if and only if their central angles are congruent. Since central angles and their intercepted arcs have equal degree measures, we can also prove the following theorems. Theorem 13.4a In a circle or in congruent circles, congruent arcs have congruent chords. Given (O > (Or and CDX > ABX > ArBrX Prove CD > AB > ArBr B B A O O A C D 14365C13.pgs 7/31/07 1:43 PM Page 545 Arcs and Chords 545 Proof First draw line segments from O to A, B, C, D, A, and B. Congruent arcs have congruent central angles. Therefore, COD AOB AOB. In a circle or in congruent circles, two chords are congruent if and only if their central angles are congruent. Therefore, CD > AB > ArBr . The converse of this theorem is also true. Theorem 13.4b In a circle or in congruent circles, congruent chords have congruent arcs. Given Prove CD > AB > ArBr and (O > (Or CDX > ABX > ArBrX Strategy First draw line segments from O to A, B, C, D, A and B. B B A O O A C D Prove COD AOB AOB by SSS. Then use congruent central angles to prove congruent arcs. The proof of Theorem 13.4b is left to the student. (See exercise 24.) Theorems 13.4a and 13.4b can be stated as a biconditional. Theorem 13.4 In a circle or in congruent circles, two chords are congruent if and only if their arcs are congruent. EXAMPLE 1 mABX In circle O, mBCX a. Find 35, mBOC 110, and mCDX . and AOD is a diameter. Solution a. b. Explain why AB = CD. mBCX 5 m/BOC 5 110 mCDX 5 180 2 mABX 2 mBCX 5 180 2 35 2 110 5 35 A 35 B O 110 D C b. In a circle, arcs with equal measure are congruent. Therefore, since mABX mCDX their arcs are congruent. Therefore, 35 and ABX 35, CDX AB > CD and AB = CD. . In a circle, chords are congruent if 14365C13.pgs 7/12/07 3:56 PM Page 546 546 Geometry of the Circle Chords Equidistant from the Center of a Circle We defined the distance from a point to a line as the length of the perpendicular from the point to the line. The perpendicular is the shortest line segment that can be drawn from a point to a line. These facts can be used to prove the following theorem. Theorem 13.5 A diameter perpendicular to a chord bisects the chord and its arcs. Given Diameter of circle O, chord , and COD ACX > BCX AB ADX > BDX . Prove AE > BE , , and AB ' CD at E. A B C E O D Proof Statements Reasons and . OB 1. Two points determine a line. 1. Draw OA AB ' CD 2. 3. AEO and BEO are right angles. 4. OA > OB OE > OE 5. 6. AOE BOE 7. AE > BE 8. AOE BOE ACX > BCX 9. 10. AOD is the supplement of AOE. BOD is the supplement of BOE. 11. AOD BOD ADX > BDX 12. 2. Given. 3. Perpendicular lines intersect to form right angles. 4. Radii of a circle are congruent. 5. Reflexive property of congruence. 6. HL. 7. Corresponding parts of congruent triangles are congruent. 8. Corresponding parts of congruent triangles are congruent. 9. In a circle, congruent central angles have congruent arcs. 10. If two angles form a linear pair, then they are supplementary. 11. Supplements of congruent angles are congruent. 12. In a circle, congruent central angles have congruent arcs. 14365C13.pgs 7/31/07 1:16 PM Page 547 Since a diameter is a segment of a line, the following corollary is also true: Arcs and Chords 547 Corollary 13.5a A line through the center of a circle that is perpendicular to a chord bisects the chord and its arcs. An apothem of a circle is a perpendicular line segment from the center of a circle to the midpoint of a chord. The term apothem also refers to the length of the segment. In the diagram, E is the midpoint of chord in circle O, AB ' CD , or OE, is the apothem. , and AB OE A C E O D Theorem 13.6 The perpendicular bisector of the chord of a circle contains the center of the circle. Given Circle O, and chord AB with midpoint M and perpendicular bisector k. Prove Point O is a point on k. Proof In the diagram, M is the midpoint of chord AB in circle O. A M O k B B Then, AM = MB and AO = OB (since these are radii). Points O and M are each equidistant from the endpoints of AB . Two points that are each equidis- tant from the endpoints of a line segment determine the perpendicular bisec- tor of the line segment. Therefore, g OM Through a point on a line there is only one perpendicular line. Thus, is the perpendicular bisector of g OM AB . and k are the same line, and O is on k, the perpendicular bisector of AB . Theorem 13.7a If two chords of a circle are congruent, then they are equidistant from the center of the circle. Given Circle O with AB > CD , OE ' AB , and OF ' CD . Prove OE > OF Proof A line through the center of a circle that is perpendicular to a chord bisects the chord and its arcs. Therefore, g OE and g OF bisect the congru- ent chords EB > FD OB > OD AB and CD . Since halves of congruent segments are congruent, OB . Draw . Therefore, OBE ODF by HL, and . Since and and OD OD OB OE > OF . are radii of the same circle, A C E O F B D 14365C13.pgs 7/12/07 3:56 PM Page 548 548 Geometry of the Circle The converse of this theorem is also true. Theorem 13.7b If two chords of a circle are equidistant from the center of the circle, then the chords are congruent. Given Circle O with OE ' AB , OF ' CD , and OE > OF . Prove AB > CD are radii of the same . Therefore, OBE ODF by HL, and and OD . A line through the center of a circle that is perpendicular to a OB OD . Since and OB OB > OD Proof Draw circle, EB > FD chord bisects the chord. Thus, gruent segments are congruent, AE > EB and AB > CD . CF > FD . Since doubles of con- A C E O F B D We can state theorems 13.7a and 13.7b as a biconditional. Theorem 13.7 Two chords are equidistant from the center of a circle if and only if the chords are congruent. What if two chords are not equidistant from the center of a circle? Which is the longer chord? We know that a diameter contains the center of the circle and is the longest chord of the circle. This suggests that the shorter chord is farther from the center of the circle. (1) Let and AB AB CD. CD be two chords of circle O and (2) Draw OE ' AB and OF ' CD . A E C O F B D (3) A line through the center of a circle that is perpendicular to the chord bisects the chord. Therefore, 1 2CD 5 FD . (4) The distance from a point to a line is the length of the perpendicular 1 2AB 5 EB and from the point to the line. Therefore, OE is the distance from the center , and OF is the distance from the center of the circle of the circle to . to CD AB Recall that the squares of equal quantities are equal, that the positive square roots of equal quantities are equal, and that when an inequality is multiplied by a negative number, the inequality is reversed. 14365C13.pgs 7/12/07 3:56 PM Page 549 (5) Since AB CD: Arcs and Chords 549 AB , CD 2AB , 1 1 2CD EB , FD EB2 , FD2 2EB2 . 2FD2 (6) Since OBE and ODF are right triangles and OB = OD: OB2 5 OD2 OE2 1 EB2 5 OF2 1 FD2 (7) When equal quantities are added to both sides of an inequality, the order of the inequality remains the same. Therefore, adding the equal quantities from step 6 to the inequality in step 5 gives: OE2 1 EB2 2 EB2 . OF2 1 FD2 2 FD2 OE2 . OF2 OE . OF Therefore, the shorter chord is farther from the center of the circle. We have just proved the following theorem: Theorem 13.8 In a circle, if the lengths of two chords are unequal, then the shorter chord is farther from the cente |
r. EXAMPLE 2 In circle O, mABX 90 and OA 6. a. Prove that AOB is a right triangle. b. Find AB. c. Find OC, the apothem to . AB Solution a. If mABX 90, then mAOB 90 because the measure of an arc is equal to the measure of the central angle that intercepts the arc. Since AOB is a right angle, AOB is a right triangle. A 6 O 90° C B 14365C13.pgs 7/12/07 3:56 PM Page 550 550 Geometry of the Circle A 6 O 90° C B b. Use the Pythagorean Theorem for right AOB. Since OA and OB are radii, OB OA 6. AB2 OA2 OB2 AB2 62 62 AB2 36 36 AB2 72 72 5 36 " " , AB OC ' AB 2 5 6 2 " and bisects AB . Therefore, AB " c. Since OC is the apothem to . In right OCA, 2 OC2 1 AC2 AC 5 3 " 5 OA2 2 5 62 OC2 1 2 3 " B A 18 5 36 OC2 1 5 18 OC2 " Note: In the example, since AOB is an isosceles right triangle, mAOB is 45. Therefore AOC is also an isosceles right triangle and OC AC. " " OC 5 18 5 2 5 3 2 9 " Polygons Inscribed in a Circle If all of the vertices of a polygon are points of a circle, then the polygon is said to be inscribed in the circle. We can also say that the circle is circumscribed about the polygon. In the diagram: A 1. Polygon ABCD is inscribed in circle O. B 2. Circle O is circumscribed about polygon ABCD. D O C In an earlier chapter we proved that the perpendicular bisectors of the sides of a triangle meet at a point and that that point is equidistant from the verg tices of the triangle. In the diagram, PN are the perpendicular bisectors of the sides of ABC. Every point on the perpendicular bisectors of a line segment is equidistant from the endpoints of the line segment. Therefore, PA PB PC and A, B, and C are points on a circle with center at P, that is, any triangle can be inscribed in a circle. g , PL g PM , and C P L M B N A 14365C13.pgs 7/12/07 3:56 PM Page 551 EXAMPLE 3 Prove that any rectangle can be inscribed in a circle. Proof Let ABCD be any rectangle. The diagonals of a rectangle Arcs and Chords 551 D AC > BD and AC BD. Since a rec- BD AC AE EC and BD BE ED. Halves of equal are congruent, so tangle is a parallelogram, the diagonals of a rectangle intersect at E, then and bisect each other. If 1 2 quantities are equal. Therefore, AE EC BE ED and the vertices of the rectangle are equidistant from E. Let E be the center of a circle with radius AE. The vertices of ABCD are on the circle and ABCD is inscribed in the circle. AC 1 2 A E C B Exercises Writing About Mathematics 1. Daniela said that if a chord is 3 inches from the center of a circle that has a radius of 5 inches, then a 3-4-5 right triangle is formed by the chord, its apothem, and a radius. Additionally, the length of the chord is 4 inches. Do you agree with Daniela? Explain why or why not. 2. Two angles that have the same measure are always congruent. Are two arcs that have the same measure always congruent? Explain why or why not. Developing Skills In 3–7, find the length of the radius of a circle whose diameter has the given measure. 3. 6 in. 4. 9 cm 5. 3 ft 6. 24 mm 7. " 24 cm In 8–12, find the length of the diameter of a circle whose radius has the given measure. 8. 5 in. 9. 12 ft 10. 7 cm 11. 6.2 mm 12. yd 5 " 13. In circle O, AOB is a diameter, AB 3x 13, and AO 2x 5. Find the length of the radius and of the diameter of the circle. In 14–21, DCOE is a diameter of circle O, AB is a chord of the circle, and OD ' AB at C. 14. If AB 8 and OC 3, find OB. 16. If OC 20 and OB 25, find AB. 15. If AB 48 and OC 7, find OB. 17. If OC 12 and OB 18, find AB. 14365C13.pgs 7/12/07 3:56 PM Page 552 552 Geometry of the Circle 18. If AB 18 and OB 15, find OC. 19. If AB 20 and OB 15, find OC. 20. If mAOB 90, and AB 30 OB and DE. , find 2 " 21. If mAOB 60, and AB 30, find OB and OC. 22. In circle O, chord LM is 3 centimeters from the center and chord RS is 5 centimeters from the center. Which is the longer chord? Applying Skills 23. Prove Theorem 13.3b, “In a circle or in congruent circles, congruent chords have congruent central angles.” 24. Prove Theorem 13.4b, “In a circle or in congruent circles, congruent chords have congruent arcs.” 25. Diameter AOB bisects chord CD of circle O intersects chord CD and is perpendicular to chord at E and bisects CD . CDX at B. Prove that AOB 26. The radius of a spherical ball is 13 centimeters. A piece that has a plane surface is cut off of the ball at a distance of 12 centimeters from the center of the ball. What is the radius of the circular faces of the cut pieces? 27. Triangle ABC is inscribed in circle O. The distance from the center of the circle to AB is greater than the distance from the center of the circle to ter of the circle to Which is the largest angle of ABC? Justify your answer. BC is greater than the distance from the center of the circle to AC . BC , and the distance from the cen- 13-3 INSCRIBED ANGLES AND THEIR MEASURES B In the diagram, ABC is an angle formed by two chords that have a common endpoint on the circle. A DEFINITION C An inscribed angle of a circle is an angle whose vertex is on the circle and whose sides contain chords of the circle. We can use the fact that the measure of a central angle is equal to the measure of its arc to find the relationship between ABC and the measure of its arc, .ACX 14365C13.pgs 7/12/07 3:56 PM Page 553 CASE 1 One of the sides of the inscribed angle contains a diameter of the circle. Inscribed Angles and Their Measures 553 B x x O 2x A 2x C OA Consider first an inscribed angle, ABC, with a diameter of circle O. . Then AOB is an isosceles triangle and mOAB mOBA x. Draw Angle AOC is an exterior angle and mAOC x x 2x. Since AOC is a central angle, mAOC . Therefore, mABC x mACX 5 2x BC 1 . 2mACX We have shown that when one of the sides of an inscribed angle contains a diameter of the circle, the measure of the inscribed angle is equal to one-half the measure of its intercepted arc. Is this true for angles whose sides do not contain the center of the circle? B CASE 2 The center of the circle is in the interior of the angle. Let ABC be an inscribed angle in which the center of the circle is in the A O C interior of the angle. Draw BOD , a diameter of the circle. Then: mABD 1 2mADX and mDBC 1 2mDCX D Therefore: m/ABC 5 m/ABD 1 m/DBC 2mDCX 2mADX 1 1 5 1 2(mADX 1 mDCX) 5 1 2mACX 5 1 B O D A C CASE 3 The center of the circle is not in the interior of the angle. Let ABC be an inscribed angle in which the center of the circle is not in the interior of the angle. Draw BOD 1 2mADX , a diameter of the circle. Then: and mDBC 1 2mDCX mABD Therefore: m/ABC 5 m/ABD 2 m/DBC 2mDCX 2mADX 2 1 2(mADX 2 mDCX) 2mACX 5 1 5 1 5 1 These three possible positions of the sides of the circle with respect to the center the circle prove the following theorem: Theorem 13.9 The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc. There are two statements that can be derived from this theorem. 14365C13.pgs 7/12/07 3:56 PM Page 554 554 Geometry of the Circle Corollary 13.9a An angle inscribed in a semicircle is a right angle. Proof: In the diagram, ABC is inscribed in semicircle semicircle whose degree measure is 180°. Therefore, is a diameter of circle O, and is a . Also ADCX ABCX AOC mABC 2 mADCX 5 1 5 1 2(180) 5 90 B A C O D Since any triangle can be inscribed in a circle, the hypotenuse of a triangle can be the diameter of a circle with the midpoint of the hypotenuse the center of the circle. Corollary 13.9b If two inscribed angles of a circle intercept the same arc, then they are congruent. Proof: In the diagram, ABC and ADC are inscribed angles and each angle intercepts ACX and mADC . Since ABC and ADC have equal measures, they are congruent. . Therefore, mABC 2mACX 2mACX 1 1 EXAMPLE Triangle ABC is inscribed in circle O, mB 70, and mACX a. b. mA c. mC d. mABX B A D C mBCX 100. Find: Solution a. If the measure of an inscribed angle is one-half the measure of its intercepted arc, then the measure of the intercepted arc is twice the measure of the inscribed angle. A 2mACX m/B 5 1 2m/B 5 mACX 2(70) 5 mACX 140 5 mACX B 70 100 C 14365C13.pgs 7/12/07 3:56 PM Page 555 Inscribed Angles and Their Measures 555 c. m/C 5 180 2 (m/A 1 m/B) 5 180 2 (50 1 70) 5 180 2 120 5 60 b. m/A 5 1 2mBCX 2(100) 5 1 5 50 d. mABX 5 2m/C 5 2(60) 5 120 Answers a. 140° b. 50° c. 60° d. 120° SUMMARY Type of Angle Degree Measure Example Central Angle The measure of a central angle is equal to the measure of its intercepted arc. Inscribed Angle The measure of an inscribed angle is equal to one-half the measure of its intercepted arc. B A A 1 m1 mABX B 1 m/1 5 1 2m ABX Exercises Writing About Mathematics 1. Explain how you could use Corollary 13.9a to construct a right triangle with two given line segments as the hypotenuse and one leg. 14365C13.pgs 7/12/07 3:56 PM Page 556 556 Geometry of the Circle 2. In circle O, ABC is an inscribed angle and mACX 50. In circle O, PQR is an inscribed 50. Is ABC PQR if the circles are not congruent circles? Justify mPRX angle and your answer. Developing Skills In 3–7, B is a point on circle O not on 3. 88 4. 72 ACX 5. 170 ACX , an arc of circle O. Find mABC for each given m . 6. 200 7. 280 In 8–12, B is a point on circle O not on , an arc of circle O. Find m ACX for each given mABC. 8. 12 9. 45 11. 95 12. 125 ACX 10. 60 13. Triangle ABC is inscribed in a circle, mA 80 and 88. Find: mACX mBCX a. b. mB c. mC mABX d. mBACX e. 14. Triangle DEF is inscribed in a circle, DE > EF , and B mEFX 100. Find: a. mD mDEX b. c. mF d. mE mDFX e. A 80 88 C D A E 100 F B AC In 15–17, chords and 15. If mB 42 and mAEB 104, find: mBCX mADX a. mA BD b. c. intersect at E in circle O. d. mD e. mC E D C 16. If AB DC and mB 40, find: 17. If a. mD mADX mDCX a. mADX b. mABX b. mA mBCX c. c. mB 100, 110, and 18. Triangle ABC is inscribed in a circle and mBCX mABX mCAX b. a. c. d. mA e. mDEC 96, find: mBCX d. mAEB e. mC mABX mBCX : d. mA mCAX e. mB : 2 : 3 : 7. Find: f. mC 14365C13.pgs 7/12/07 3:57 PM Page 557 19. Triangle RST is inscribed in a circle and mSTX mTRX mRSX b. a. c. Inscribed An |
gles and Their Measures 557 mRSX 5 mSTX 5 mTRX . Find: d. mR e. mS f. mT Applying Skills 20. In circle O, LM and RS intersect at P. a. Prove that LPR SPM. b. If LP 15 cm, RP 12 cm, and SP 10 cm, find MP. 21. Triangle ABC is inscribed in a circle. If mABX 100 and mBCX isosceles. 22. Parallelogram ABCD is inscribed in a circle. mABCX mADCX . = a. Explain why mABCX mADCX . and b. Find c. Explain why parallelogram ABCD must be a rectangle. R P M L O S 130, prove that ABC is A D O B C D G F E Ex. 23 Ex. 24 Ex. 25 23. Triangle DEF is inscribed in a circle and G is any point not on mDEX 1 mEFX 5 mFGDX , show that DEF is a right triangle. DEFX . If 24. In circle O, ASA. AOC and BOD are diameters. If AB > CD , prove that ABC DCB by 25. Chords ABX 26. Prove that a trapezoid inscribed in a circle is isosceles. of circle O intersect at E. If and BD AC CDX , prove that ABC DCB. and major are arcs of circle O and AB CD . Prove that ADCX and major ADCX are arcs of circle O and ADX > BCX . Prove C B O 27. Minor ABCX ADX > BCX ABCX AB CD . 28. Minor that . 29. In circle O, AOC and BOD are diameters. Prove that AB CD . D A 14365C13.pgs 7/12/07 3:57 PM Page 558 558 Geometry of the Circle 30. Points A, B, C, D, E, and F are on circle O, AB CD EF , . ABCD and CDEF are trapezoids. Prove that CB ED and CAX > BDX > DFX > ECX . 31. Quadrilateral ABCD is inscribed in circle O, and . Prove that ABCD is not a parallelogram. CDX to ABX is not congruent 13-4 TANGENTS AND SECANTS In the diagram, line p has no points in common with the circle. Line m has one point in common with the circle. Line m is said to be tangent to the circle. Line k has two points in common with the circle. Line k is said to be a secant of the circle. P O A DEFINITION A tangent to a circle is a line in the plane of the circle that intersects the circle in one and only one point. DEFINITION A secant of a circle is a line that intersects the circle in two points. Let us begin by assuming that at every point on a circle, there exists exactly one tangent line. We can state this as a postulate. Postulate 13.2 At a given point on a given circle, one and only one line can be drawn that is tangent to the circle. OP Let P be any point on circle O and be a radius to that point. If line m containing points P and Q is perpendicular to , then OQ OP because the perpendicular is the shortest distance from a point to a line. Therefore, every point on the line except P is outside of circle O and line m must be tangent to the circle. This establishes the truth of the following theorem. OP Q m P O 14365C13.pgs 7/12/07 3:57 PM Page 559 Tangents and Secants 559 Theorem 13.10a If a line is perpendicular to a radius at a point on the circle, then the line is tangent to the circle. The converse of this theorem is also true. Theorem 13.10b If a line is tangent to a circle, then it is perpendicular to a radius at a point on the circle. Given Line m is tangent to circle O at P. Prove Line m is perpendicular to OP . Proof We can use an indirect proof. m b P O OP at P since, at a given point on a given line, one and only one line can Assume that m is not perpendicular to . OP Then there is some line b that is perpendicular to be drawn perpendicular to the given line. Then by Theorem 13.10a, b is a tangent to circle O at P. But this contradicts the postulate that states that at a given point on a circle, one and only one tangent can be drawn. Therefore, our assumption is false and its negation must be true. Line m is perpendicular to OP . We can state Theorems 13.10a and 13.10b as a biconditional. Theorem 13.10 A line is tangent to a circle if and only if it is perpendicular to a radius at its point of intersection with the circle. Common Tangents DEFINITION A common tangent is a line that is tangent to each of two circles. In the diagram, g AB is tangent to circle O at A and to circle O at B. Tangent is said to be a common internal tangent because the tangent intersects the line segment joining the centers of the circles. g AB A O O B 14365C13.pgs 7/12/07 3:57 PM Page 560 560 Geometry of the Circle In the diagram, g is tangent to cirCD cle P at C and to circle P at D. Tangent g is said to be a common external CD tangent because the tangent does not intersect the line segment joining the centers of the circles. P C P D The diagrams below show that two circles can have four, three, two, one, or no common tangents. 4 common tangents 3 common tangents 2 common tangents 1 common tangent No common tangents Two circles are said to be tangent to each other if they are tangent to the is tangent to circle O and to same line at the same point. In the diagram, circle O at T. Circles O and O are tangent externally because every point of one of the circles, except the point of tangency, is an external point of the other circle. g ST g MN In the diagram, is tangent to circle P and to circle P at M. Circles P and P are tangent internally because every point of one of the circles, except the point of tangency, is an internal point of the other circle. T O O S EXAMPLE Given: Circles O and O with a common internal , tangent to circle O at A and OOr tangent, circle O at B, and C the intersection of and g AB g AB . Prove: AC BC 5 OC OrC 14365C13.pgs 7/12/07 3:57 PM Page 561 Proof We will use similar triangles to prove the segments proportional. Tangents and Secants 561 g AB is tangent to circle O at A and circle O at B. A line tangent to a Line circle is perpendicular to a radius drawn to the point of tangency. Since perpendicular lines intersect to form right angles and all right angle are congruent, OAC OBC. Also, OCA OCB because vertical angles are congruent. Therefore, OCA OCB by AA. The lengths of corresponding sides of similar triangles are proportional. Therefore, . BC 5 OC AC OrC EXAMPLE 2 Circle O is tangent to g AB at A, O is tangent to g AB at B, and OOr intersects g AB at C. a. Prove that . b. If AC 8, AB 12, and OA 9, find OB. AC BC 5 OA OrB Solution a. We know that OAB OBA because they are right angles and that OCA OCB because they are vertical angles. Therefore, BC 5 OA AC OCA OCB by AA and OrB . b. AB 5 AC 1 BC 12 5 8 1 BC 4 5 BC BC 5 OA AC OrB 4 5 9 8 OrB 8OrB 5 36 OrB 5 36 8 5 9 2 Answer Tangent Segments DEFINITION A tangent segment is a segment of a tangent line, one of whose endpoints is the point of tangency. In the diagram, PQ segments of the tangents and g PQ PR and O from P. are tangent g PR to circle P Q O R Theorem 13.11 Tangent segments drawn to a circle from an external point are congruent. 14365C13.pgs 7/12/07 3:57 PM Page 562 562 Geometry of the Circle Given g PQ tangent to circle O at Q and g PR tangent to circle O at R. P Q O R Prove PQ > PR Proof Draw OQ , RP OP OR and , and . Since are OR . Since are tangent to the circle at Q and and OQ OQ > OR both radii of the same circle, QP R, OQP and ORP are both right angles, and OPQ and OPR are right is the hypotenuse of both OPQ and OPR. Therefore, triangles. Then OPQ OPR by HL. Corresponding parts of congruent triangles are congruent, so PQ > PR OP . The following corollaries are also true. Corollary 13.11a If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents. Given g PQ tangent to circle O at Q and g PR tangent to circle O at R. Prove h PO bisects RPQ. Strategy Use the proof of Theorem 13.11 to show that angles OPQ and RPO are congruent. P Q O R Corollary 13.11b If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle whose vertex is the center of the circle and whose rays are the two radii drawn to the points of tangency. Given g PQ tangent to circle O at Q and g PR tan- gent to circle O at R. Prove h OP bisects QOR. Strategy Use the proof of Theorem 13.11 to show that angles QOP and ROP are congruent. P Q O R 14365C13.pgs 7/12/07 3:57 PM Page 563 Tangents and Secants 563 The proofs of Corollaries 13.11a and 13.11b are left to the student. (See exercises 15 and 16.) A Polygon Circumscribed About a Circle A polygon is circumscribed about a circle if each side of the polygon is tangent to the circle. When a polygon is circumscribed about a circle, we also say that the circle is inscribed in the polygon. For example, in the diais tangent to gram, AB circle O at F, is tangent to circle O at H. Therefore, ABCD is circumscribed about circle O and circle O is inscribed in quadrilateral ABCD. is tangent to circle O at G, is tangent to circle O at E, DA CD BC If ABC is circumscribed about circle O, then we know that are the , OA bisectors of the angles of ABC, and O is the point at which the angle bisectors of the angles of a triangle intersect. , and OB OC EXAMPLE 3 C BC CA , and are tangent to circle O at D, E, and , AB F, respectively. If AF 6, BE 7, and CE 5, find the perimeter of ABC. Solution Tangent segments drawn to a circle from an exter- nal point are congruent. AD AF 6 BD BE 7 CF CE 5 Therefore AB 5 AD 1 BD 5 6 1 7 5 13 BC 5 BE 1 CE 5 7 1 5 5 12 CA 5 CF 1 AF 5 5 1 6 5 11 Perimeter 5 AB 1 BC 1 CA 5 13 1 12 1 11 5 36 Answer 14365C13.pgs 7/12/07 3:57 PM Page 564 564 Geometry of the Circle EXAMPLE 4 Point P is a point on a line that is tangent to circle O at R, P is 12.0 centimeters from the center of the circle, and the length of the tangent segment from P is 8.0 centimeters. a. Find the exact length of the radius of the circle. b. Find the length of the radius to the nearest tenth. Solution a. P is 12 centimeters from the center of circle O; OP 12. R 8 cm P The length of the tangent segment is 8 centimeters; RP 8. A line tangent to a circle is perpendicular to the radius drawn to the point of tangency; OPR is a right triangle. O 12 cm RP2 OR2 OP2 82 OP2 122 64 OP2 144 OP2 80 b. Use a calculator to evaluate OP 80 5 16 " " 5 5 4 " 5 " 4 . 5 " ENTER: 4 2nd ¯ 5 ENTER DISPLAY: 8.94427191 To the nearest tenth, OP 8.9. Answers |
a. 4 5 " cm b. 8.9 cm Exercises Writing About Mathematics 1. Line l is tangent to circle O at A and line m is tangent to circle O at B. If AOB is a diame- ter, does l intersect m? Justify your answer. 2. Explain the difference between a polygon inscribed in a circle and a circle inscribed in a polygon. 14365C13.pgs 7/12/07 3:57 PM Page 565 Tangents and Secants 565 Developing Skills In 3 and 4, ABC is circumscribed about circle O and D, E, and F are points of tangency. 3. If AD 5, EB 5, and CF 10, find the lengths of the sides of the triangle and show that the triangle is isosceles. 4. If AF 10, CE 20, and BD 30, find the lengths of the sides of the triangle and show that the triangle is a right triangle. A F D PQ In 5–11, T and R. is tangent to circle O at P, SQ is tangent to circle O at S, and OQ intersects circle O at C E B P 5. If OP 15 and PQ 20, find: a. OQ b. SQ c. TQ 6. If OQ 25 and PQ 24, find: a. OP b. RT c. RQ 7. If OP 10 and OQ 26, find: a. PQ b. RQ c. TQ 8. If OP 6 and TQ 13, find: a. OQ b. PQ c. SQ 9. If OS 9 and RQ 32, find: a. OQ b. SQ c. PQ 10. If PQ 3x, SQ 5x 8, and OS x 1, find: a. PQ b. SQ c. OS d. OQ 11. If SQ 2x, OS 2x 2, and OQ 3x 1, find: a. x b. SQ c. OS d. OQ 12. The sides of ABC are tangent to a circle at D, E, and F. If DB 4, BC 7, and the perimeter of the triangle is 30, find. BE b. EC c. CF d. AF e. AC f. AB 13. Line g RP is tangent to circle O at P and the circle at M, the midpoint of OR OR intersects . If RP 3.00 cm, A F C find the length of the radius of the circle: a. in radical form b. to the nearest hundredth O P 14. Points E, F, G, and H are the points of tangency to circle O of EFX FGX , AB BC is 70°, and DA is 50°. Find: , respectively. The measure of , and CD GHX of a. mEOF b. mFOG f. mEAO g. mEAH k. the sum of the measures of the angles of quadrilateral ABCD c. mGOH d. mHOE i. mGCF h. mFBE is 80°, of e. mAOE j. mHDG M , R E O B F GC A H D 14365C13.pgs 7/12/07 3:57 PM Page 566 566 Geometry of the Circle Applying Skills 15. Prove Corollary 13.11a, “If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents.” 16. Prove Corollary 13.11b, “If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle whose vertex is the center of the circle and whose rays are the two radii drawn to the points of tangency.” g PQ are tangent to circle O at Q and R. 17. Lines and Q P g PR a. Prove that PQR PRQ. OP intersecting RQ at S and prove that QS RS b. Draw and OP ' QR . c. If OP 10, SQ 4 and OS SP, find OS and SP. O R 18. Tangents AC and BC to circle O are perpendicular to each other at C. Prove: AC > AO a. b. OC " c. AOBC is a square. 2OA 19. Isosceles ABC is circumscribed about circle O. The points of tangency of the legs, AB and , are D and F, and the point of tangency of the base, 20. Line is a common external tangent to circle . BC AC point of g AB g O and circle O. AB and to circle O at B, and OA OB. g AB a. Prove that g OOr . is tangent to circle O at A is not parallel to g OOr and g AB . b. Let C be the intersection of Prove that OAC OBC. BC , is E. Prove that E is the mid- C A O B O c. If , and BC 12, find AC, AB, OC, OC, and OO. OA OrB 5 2 3 g AB is tangent to circle O at A and to circle O at B, and is a common internal tangent to circles O and O. 21. Line g AB OA OB. The intersection of a. Prove that OC OC. b. Prove that AC BC. OOr and g AB is C. O A C B O 14365C13.pgs 7/12/07 3:57 PM Page 567 Angles Formed by Tangents, Chords, and Secants 567 Hands-On Activity Consider any regular polygon. Construct the angle bisectors of each interior angle. Since the interior angles are all congruent, the angles formed are all congruent. Since the sides of the regular polygon are all congruent, congruent isosceles triangles are formed by ASA. Any two adjacent triangles share a common leg. Therefore, they all share the same vertex. Since the legs of the triangles formed are all congruent, the vertex is equidistant from the vertices of the regular polygon. This common vertex is the center of the regular polygon. In this Hands-On Activity, we will use the center of a regular polygon to inscribe a circle in the polygon. a. Using geometry software or compass, protractor, and straightedge, construct a square, a regular pentagon, and a regular hexagon. For each figure: (1) Construct the center P of the regular polygon. (The center is the intersection of the angle bisectors of a regular polygon.) (2) Construct an apothem or perpendicular from P to one of the sides of regular polygon. (3) Construct a circle with center P and radius equal to the length of the apothem. b. Prove that the circles constructed in part a are inscribed inside of the polygon. Prove: (1) The apothems of each polygon are all congruent. (2) The foot of each apothem is on the circle. (3) The sides of the regular polygon are tangent to the circle. c. Let r be the distance from the center to a vertex of the regular polygon. Since the center is equidistant from each vertex, it is possible to circumscribe a circle about the polygon with radius r. Let a be the length of an apothem and s be the length of a side of the regular polygon. How is the radius, r, of the circumscribed circle related to the radius, a, of the inscribed circle? 13-5 ANGLES FORMED BY TANGENTS, CHORDS, AND SECANTS Angles Formed by a Tangent and a Chord g AB AC CD is a diameter. When is tangent to circle O at A, is In the diagram, AD is drawn, a chord, and ADC is a right angle because it is an angle inscribed in a semicircle, and ACD is the complement of CAD. , BAC is a right angle, and DAB is the Also, complement of CAD. Therefore, since complements of the same angle are congruent, ACD DAB. We can conclude that since mACD = , then mDAB = 2mADX CA ' AB 1 1 . 2mADX A O B C D 14365C13.pgs 7/12/07 3:57 PM Page 568 568 Geometry of the Circle We can state what we have just proved on page 567 as a theorem. Theorem 13.12 The measure of an angle formed by a tangent and a chord that intersect at the point of tangency is equal to one-half the measure of the intercepted arc. Angles Formed by Two Intersecting Chords D O B E A C We can find how the measures of other angles and their intercepted arcs are intersect in the related. For example, in the diagram, two chords AB is drawn. Angle AED is an exterior angle of DEB. interior of circle O and DB Therefore, and CD mAED mBDE mDBE 2mDAX 2mBCX 1 1 2(mBCX 1 mDAX ) 1 1 BCX Notice that is the arc intercepted by BEC and is the arc intercepted by AED, the angle vertical to BEC. We can state this relationship as a theorem. DAX Theorem 13.13 The measure of an angle formed by two chords intersecting within a circle is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Angles Formed by Tangents and Secants We have shown how the measures of angles whose vertices are on the circle or within the circle are related to the measures of their intercepted arcs. Now we want to show how angles formed by two tangents, a tangent and a secant, or two secants, all of which have vertices outside the circle, are related to the measures of the intercepted arcs. A Tangent Intersecting a Secant In the diagram, g PTQ and g PRS is a tangent to circle O at R is a secant that intersects the circle at T is drawn. Then SRQ is an and at Q. Chord exterior angle of PRQ. RQ S Q R O T P 14365C13.pgs 7/12/07 3:57 PM Page 569 Angles Formed by Tangents, Chords, and Secants 569 mRQP mP mSRQ mP mSRQ mRQP mP mP 2mRQX 2mRTX 2(mRQX 2 mRTX) 1 1 1 Two Intersecting Secants In the diagram, g PTR that intersects the circle at R and T, and is a secant to circle O g PQS is a secant to circle O that intersects the circle is drawn. Then RQS at Q and S. Chord is an exterior angle of RQP. RQ mPRQ mP mRQS T R O Q S mP mRQS mPRQ mP mP 2mRSX 2mQTX 2(mRSX 2 mQTX ) 1 1 1 Two Intersecting Tangents In the diagram, tangent to the circle at Q, and T is a point on major g PRS is tangent to circle O at R, g is PQ RQX . is drawn. Then SRQ is an exterior angle of RQ Chord RQP. R S O Q P P mPQR mP mSRQ T mP mSRQ mPQR mP mP 2mRTQX 2mRQX 2(mRTQX 2 mRQX ) 1 1 1 For each pair of lines, a tangent and a secant, two secants, and two tangents, the steps necessary to prove the following theorem have been given: Theorem 13.14 The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs. 14365C13.pgs 7/12/07 3:57 PM Page 570 570 Geometry of the Circle EXAMPLE 1 A tangent and a secant are drawn to circle O from point P. The tangent intersects the circle at Q and mQRX : mRSX : mSQX 5 3 : 5 : 7 the secant at R and S. If , find: Q P O R mQRX a. d. mQRS mRSX b. e. mRQP mSQX c. f. mP S Solution Let mQRX 3x, mRSX 5x, and mSQX 7x. 3x 1 5x 1 7x 5 360 15x 5 360 x 5 24 mQRX 5 3x a. mRSX 5 5x b. 5 3(24) 5 72 5 5(24) 5 120 c. mSQX 5 7x 7(24) 5 168 d. m/QRS 5 1 5 1 2mSQX 2(168) e. m/RQP 5 1 5 1 2mQRX 2(72) f. 2(mSQX 2 mQRX ) m/P 5 1 5 1 2(168 2 72) 5 84 5 36 5 48 Answers a. 72° b. 120° c. 168° d. 84° e. 36° f. 48° Note: m/QRS 5 m/RQP 1 m/P 5 36 1 48 5 84 EXAMPLE 2 Two tangent segments, RP R. If mR is 70, find the measure of the minor arc into which the circle is divided. and RQ PQX , are drawn to circle O from an external point and of the major arc PSQX 14365C13.pgs 7/12/07 3:57 PM Page 571 Angles Formed by Tangents, Chords, and Secants 571 P R Q S Solution The sum of the minor arc and the major arc with the same endpoints is 360. Let x = mPQX . Then 360 x = mPSQX . mR 70 1 2(mPSQX 2 mPQX) 1 2(360 2 x 2 x) 1 2(360 2 2x) 70 70 180 x x 110 360 x 360 110 250 Answer mPQX 110 and mPSQX 250 SUMMARY Type of Angle Degree Measure Example Formed by a Tangent and a Chord The measure of an angle formed by a tangent and a chord that intersect at the |
point of tangency is equal to one-half the measure of the intercepted arc. Formed by Two Intersecting Chords The measure of an angle formed by two intersecting chords is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. A 1 B m/1 5 1 2mABX 2 1 A B D C m/1 5 1 m/2 5 1 2(mABX 1 mCDX) 2(mABX 1 mCDX) (Continued) 14365C13.pgs 7/12/07 3:57 PM Page 572 572 Geometry of the Circle SUMMARY (Continued) Type of Angle Degree Measure Example Formed by Tangents and Secants The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs/1 5 1 m/2 5 1 m/3 5 1 2(mABX 2 mACX) 2(mABX 2 mCDX) 2(mACBX 2 mABX) Exercises Writing About Mathematics 1. Nina said that a radius drawn to the point at which a secant intersects a circle cannot be perpendicular to the secant. Do you agree with Nina? Explain why or why not. 2. Two chords intersect at the center of a circle forming four central angles. Aaron said that the measure of one of these angles is one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Do you agree with Aaron? Explain why or why not. Developing Skills g PQS and 3. If In 3–8, secants mSTX mSTX mSTX 4. If 5. If 160 and 100 and 170 and 6. If mP 40 and 7. If mP 60 and 8. If mP 25 and g PRT mQRX mQRX mQRX mQRX mQRX mSTX intersect at P. 90, find mP. 40, find mP. 110, find mP. 86, find 50, find 110, find mSTX . mSTX . .mQRX P Q R S T 14365C13.pgs 7/12/07 3:57 PM Page 573 Angles Formed by Tangents, Chords, and Secants 573 In 9–14, tangent g QP g PRT intersect at P. 70, find mP. 30, find mP. 120 and 170 and and secant mQRX mQRX mRTX 50 and mP 40, find 70 and 9. If 10. If 11. If mQTX mQTX mQRX mQRX mQRX 14. If mP 30 and 13. If 12. If 60 and mP 35, find mQRX 120, find mP. mQTX . mQTX . mQTX . 120, find Q P R T g RP and g QP intersect at P and S is on major arc QRX . 160, find mP. 80, find mP. 260, find mP. 210, find mP. 16. If 15. If In 15–20, tangents mRQX mRQX mRSQX mRSQX mRSQX 5 2mRQX 20. If mP 45, find 17. If 18. If 19. If , find mP. mRQX and mRSQX . P Q R S In 21–26, chords AB and intersect at E in the interior of a circle. mACX mDAX mACX mACX mACX mBCX 21. If 22. If 23. If 24. If 25. If 26. If CD mBDX mBCX mBDX mBDX 30 and 80, find mAEC. 180 and 100 find mAED. 25 and 20 and 45, find mDEB. 60, find mAED. mBDX . mDAX . 30 and mAEC 50, find 80 and mAEC 30, find CA E D B 27. In the diagram, g PA and g PB are tangent to circle O at A and and chord AC intersect at E, mCBX 125 and B. Diameter BD mP 55. Find: ABX a. m d. mDEC ADX b. m e. mPBD CDX c. m f. mPAC g. Show that BD is perpendicular to AC and bisects .AC P 55° A E O D C B 125° 14365C13.pgs 7/12/07 3:57 PM Page 574 574 Geometry of the Circle and secant segment are chords. If mP 45 and PBC are drawn to cir- AB 28. Tangent segment PA cle O and and mACX : mABX 5 5 : 2 ACX a. m d. mPAB AC , find: BCX b. m e. mCAB c. mACB f. mPAC Applying Skills 29. Tangent g PC intersects circle O at C, chord g AB CP , diameter COD intersects AB at E, and diameter AOF is extended to P. a. Prove that OPC OAE. mADX b. If mOAE 30, find , and mP. mCFX , mFBX , mBDX , mACX , A g ABC 30. Tangent intersects circle O at B, › ‹ AFOD secant intersects the circle at › ‹ CGOE intersects F and D, and secant the circle at G and E. If mEFBX 5 mDGBX , prove that AOC is A F E A O P 45 isosceles. 31. Segments BP AP and are tangent to circle O at A and B, respectively, and mAOB 120. Prove that ABP is equilateral. 32. Secant g ABC intersects a circle at A and B. Chord BD Prove that mCBD is drawn. 1 2mBDX . A B O P A B C O D 14365C13.pgs 7/12/07 3:57 PM Page 575 Measures of Tangent Segments, Chords, and Secant Segments 575 13-6 MEASURES OF TANGENT SEGMENTS, CHORDS, AND SECANT SEGMENTS Segments Formed by Two Intersecting Chords We have been proving theorems to establish the relationship between the measures of angles of a circle and the measures of the intercepted arcs. Now we will study the measures of tangent segments, secant segments, and chords. To do this, we will use what we know about similar triangles. Theorem 13.15 If two chords intersect within a circle, the product of the measures of the segments of one chord is equal to the product of the measures of the segments of the other. Given Chords circle O. AB and CD intersect at E in the interior of Prove (AE)(EB) (CE)(ED) Proof Statements Reasons A D E O C B AD and 1. Draw 2. A C and D B . CB 1. Two points determine a line. 2. Inscribed angles of a circle that inter- cept the same arc are congruent. 3. ADE CBE 3. AA. 4. CE 5 ED AE EB 5. (AE)(EB) (CE)(ED) 4. The lengths of the corresponding sides of similar triangles are in proportion. 5. In a proportion, the product of the means is equal to the product of the extremes. Segments Formed by a Tangent Intersecting a Secant Do similar relationships exist for tangent segments and secant segments? In the diagram, tangent segment is drawn to circle O, and secant segment PBC PA intersects the circle at B and C. C A P O B 14365C13.pgs 7/12/07 3:57 PM Page 576 576 Geometry of the Circle A O P B C PB , the part of the secant segment that is outside the circle, the We will call are drawn, C external segment of the secant. When chords PAB because the measure of each is one-half the measure of the intercepted . Also P P by the reflexive property. Therefore, BPA APC arc, by AA. The length of the corresponding sides of similar triangles are in proportion. Therefore: ABX and AB AC PA 5 PA PB PC and (PA)2 (PC)(PB) We can write what we have just proved as a theorem: Theorem 13.16 If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external segment. Note that both means of the proportion are PA, the length of the tangent segment. Therefore, we can say that the length of the tangent segment is the mean proportional between the lengths of the secant and its external segment. Theorem 13.16 can be stated in another way. Theorem 13.16 If a tangent and a secant are drawn to a circle from an external point, then the length of the tangent segment is the mean proportional between the lengths of the secant segment and its external segment. Segments Formed by Intersecting Secants What is the relationship of the lengths of two secants drawn to a circle from an external point? Let be two secant segments drawn to a circle as shown in the diagram. Draw a tangent segment to the circle from A. Since ADE ABC and AF A B D F C E AF 2 (AC)(AB) and AF 2 (AE)(AD), then (AC)(AB) (AE)(AD) Note: This relationship could also have been proved by showing that ABE ADC. 14365C13.pgs 7/12/07 3:57 PM Page 577 Measures of Tangent Segments, Chords, and Secant Segments 577 We can state this as a theorem: Theorem 13.17 If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. EXAMPLE 1 and PCD , and Two secant segments, PAB a tangent segment, , are drawn to a cirPE cle from an external point P. If PB 9 cm, PD 12 cm, and the external segment of is 1 centimeter longer than the exterPAB nal segment of , find: a. PA b. PC c. PE PCD P A O C E B D Solution Let x PC and x 1 PA. a. PA x 1 4 (PB)(PA) 5 (PD)(PC) 9PA 5 12PC 9(x 1 1) 5 12x 9x 1 9 5 12x 9 5 3x 3 5 x b. PC x 3 c. (PE)2 5 (PB)(PA) (PE)2 5 (9)(4) (PE)2 5 36 PE 5 6 (Use the positive square root.) Answers a. PA 4 cm b. PC 3 cm c. PE 6 cm EXAMPLE 2 In a circle, chords find ST and TR. PQ and RS intersect at T. If PT 2, TQ 10, and SR 9, P R 2 O T 10 Q 9 S 14365C13.pgs 7/12/07 3:57 PM Page 578 578 Geometry of the Circle Solution Since ST TR SR 9, let ST x and TR 9 x. P R 2 O T 10 Q 9 S (PT)(TQ) 5 (RT)(TS) (2)(10) 5 x(9 2 x) 20 5 9x 2 x2 x2 2 9x 1 20 5 0 (x 2 5)(x 2 4 Answer ST 5 and TR 4, or ST 4 and TR 5 EXAMPLE 3 Find the length of a chord that is 20 centimeters from the center of a circle if the length of the radius of the circle is 25 centimeters. Solution Draw diameter COD perpendicular to chord AB at E. Then OE is the distance from the center of the circle to the chord. OE 20 DE 5 OD 1 OE 5 25 1 20 5 45 CE 5 OC 2 OE 5 25 2 20 5 5 B 25 c m 5 cm C E A 20 cm O 25 cm D A diameter perpendicular to a chord bisects the chord. Therefore, AE EB. Let AE EB x. (AE)(EB) 5 (DE)(CE) (x)(x) 5 (45)(5) x2 5 225 x 5 15 (Use the positive square root.) Therefore, AB 5 AE 1 EB 5 x 1 x 5 30 cm Answer 14365C13.pgs 7/12/07 3:57 PM Page 579 Measures of Tangent Segments, Chords, and Secant Segments 579 SUMMARY Type of Segment Length Example Formed by Two Intersecting Chords If two chords intersect, the product of the measures of the segments of one chord is equal to the product of the measures of of the segments of the other. A D E O C B Formed by a Tangent Intersecting a Secant Formed by Two Intersecting Secants If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external segment. If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. (AE)(EB) 5 (CE)(ED) A O C P B (PA)2 5 (PC)(PB) P A C B O D (PB)(PA) 5 (PD)(PC) Exercises Writing About Mathematics 1. The length of chord AB in circle O is 24. Vanessa said that any chord of circle O that inter- AB sects the lengths of the segments is 144. Do you agree with Vanessa? Justify your answer. at its midpoint, M, is separated by M into two segments such that the product of 2. Secants answer. ABP and CDP are drawn to circle O. If AP CP, is BP |
DP? Justify your 14365C13.pgs 7/12/07 3:57 PM Page 580 580 Geometry of the Circle C E A D Developing Skills In 3–14, chords AB and intersect at E. CD 3. If CE 12, ED 2, and AE 3, find EB. 4. If CE 16, ED 3, and AE 8, find EB. 5. If AE 20, EB 5, and CE 10, find ED. 6. If AE 14, EB 3, and ED 6, find CE. 7. If CE 10, ED 4, and AE 5, find EB. 8. If CE 56, ED 14, and AE EB, find EB. 9. If CE 12, ED 2, and AE is 2 more than EB, find EB. 10. If CE 16, ED 12, and AE is 3 times EB, find EB. 11. If CE 8, ED 5, and AE is 6 more than EB, find EB. 12. If CE 9, ED 9, and AE is 24 less than EB, find EB. 13. If CE 24, ED 5, and AB 26, find AE and EB. 14. If AE 7, EB 4, and CD 16, find CE and ED. B In 15–22, g AF is tangent to circle O at F and secant ABC intersects circle O at B and C. 15. If AF 8 and AB 4, find AC. 16. If AB 3 and AC 12, find AF. 17. If AF 6 and AC 9, find AB. 18. If AB 4 and BC 12, find AF. 19. If AF 12 and BC is 3 times AB, find AC, AB, and BC. 20. If AF 10 and AC is 4 times AB, find AC, AB, and BC. 21. If AF 8 and CB 12, find AC, AB, and BC. 22. If AF 15 and CB 16, find AC, AB, and BC. A F B O C intersect at A outside the circle. h ABC h ADE In 23–30, secants and 23. If AB 8, AC 25, and AD 10, find AE. 24. If AB 6, AC 18, and AD 9, find AE. 25. If AD 12, AE 20, and AB 8, find AC. 26. If AD 9, AE 21, and AC is 5 times AB, find AB and AC. 27. If AB 3, AD 2, and DE 10, find AC. C E B D A 14365C13.pgs 7/12/07 3:57 PM Page 581 Circles in the Coordinate Plane 581 28. If AB 4, BC 12, and AD DE, find AE. 29. If AB 2, BC 7, and DE 3, find AD and AE. 30. If AB 6, BC 8, and DE 5, find AD and AE. 31. In a circle, diameter AB is extended through B to P and tangent segment PC is drawn. If BP 6 and PC 9, what is the measure of the diameter of the circle? 13-7 CIRCLES IN THE COORDINATE PLANE In the diagram, a circle with center at the origin and a radius with a length of 5 units is drawn in the coordinate plane. The points (5, 0), (0, 5), (5, 0) and (0, 5) are points on the circle. What other points are on the circle and what is the equation of the circle? Let P(x, y) be any other point on the circle. From P, draw a vertical line segment to the x-axis. Let this be point Q. Then OPQ is a right triangle with OQ x, PQ y, and OP 5. We can use the Pythagorean Theorem to write an equation for the circle: y (0, 5) P(x, y) (5, 0) 1 1O Q(x, 0) (5, 0) x (0, 5) OQ2 1 PQ2 5 OP2 y2 5 52 1 x2 The points (3, 4), (4, 3), (3, 4), (4, 3), (3, 4), (4, 3), (3, 4), and (4, 3) appear to be points on the circle and all make the equation x2 y2 52 true. The points (5, 0), (0, 5), (5, 0), and (0, 5) also make the equation true, as do points such as " If we replace 5 by the length of any radius, r, the equation of a circle whose 1, " and 22, 21 24 B A B A . center is at the origin is: x2 y2 r 2 How does the equation change if the center is not at the origin? For example, what is the equation of a circle whose center, C, is at (2, 4) and whose radius has a length of 5 units? The points (7, 4), (3, 4), (2, 9), and (2, 1) are each 5 units from (2, 4) and are therefore points on the circle. Let P(x, y) be any other point on the circle. From P, draw a vertical line and from C, a horizontal line. (3, 4) y (2, 9) P(x, y) C (2, 4) Q(x, 4) 1 O 1 (2, 1) (7, 4) x 14365C13.pgs 7/12/07 3:57 PM Page 582 582 Geometry of the Circle Let the intersection of these two lines be Q. Then CPQ is a right triangle with: CQ x 2 PQ y 4 CP 5 We can use the Pythagorean Theorem to write an equation for the circle. CQ2 1 PQ2 (x 2 2)2 1 (y 2 4)2 5 52 5 CP2 The points (5, 8), (6, 7), (1, 8), (2, 7), (1, 0), (2, 1) (5, 0), and (6, 1) appear to be points on the circle and all make (x 2)2 (y 4)2 52 true. The points (7, 4), (3, 4), (2, 9), and (2, 1) also make the equation true, as do points whose coordinates are not integers. We can write a general equation for a circle with center at C(h, k) and radius r. Let P(x, y) be any point on the circle. From P draw a vertical line and from C draw a horizontal line. Let the intersection of these two lines be Q. Then CPQ is a right triangle with: CQ x h PQ y k CP r y P(x, y) r C(h, k) Q(x, k) O x We can use the Pythagorean Theorem to write an equation for the circle. CQ2 1 PQ2 5 CP2 (x 2 h)2 1 (y 2 k)2 5 r2 In general, the center-radius equation of a circle with radius r and center (h, k) is (x h)2 (y k)2 r2 AB A circle whose diameter has endpoints at A(3, 1) and B(5, 1) is shown at the right. The center of the circle, C, is the midpoint of the diameter. Recall that the coordinates of the midpoint of the segment whose endpoints are (a, b) and (c, d) are 23) 2 B A . The coordinates of C are , 21 1 (21) 2 B 5 (1, 1). y O 1 1 C A x B The length of the radius is the distance from C to any point on the circle. The distance between two points on the same vertical line, that is, with the same x-coordinates, is the absolute value of the difference of the y-coordinates. The length of the radius is the distance from C(1, 1) to A(3, 1). The length of the radius is 1 (3) 4. 14365C13.pgs 7/31/07 1:43 PM Page 583 Circles in the Coordinate Plane 583 In the equation of a circle with center at (h, k) and radius r, we use (x h)2 (y k)2 r2. For this circle with center at (1, 1) and radius 4, h 1, k 1, and r 4. The equation of the circle is: (x 1)2 (y (1))2 42 or (x 1)2 (y 1)2 16 The equation of a circle is a rule for a set of ordered pairs, that is, for a relation. For the circle (x 1)2 (y 1)2 16, (1, 5) and (1, 3) are two ordered pairs of the relation. Since these two ordered pairs have the same first element, this relation is not a function. EXAMPLE 1 a. Write an equation of a circle with center at (3, 2) and radius of length 7. b. What are the coordinates of the endpoints of the horizontal diameter? Solution a. The center of the circle is (h, k) (3, 2). The radius is r 7. The general form of the equation of a circle is (x h)2 (y k)2 r 2. The equation of the given circle is: (x 3)2 (y (2))2 72 (x 3)2 (y 2)2 49 or b. METHOD 1 If this circle were centered at the origin, then the endpoints of the horizontal diameter would be (7, 0) and (7, 0). However, the circle is centered at (3, 2). Shift these endpoints using the translation (7, 0) → (7 3, 0 2) (4, 2) (7, 0) → (7 3, 0 2) (10, 2) T3, 22 : y O 1 1 (0, 0) (3, 2) x METHOD 2 Since the center of the circle is (3, 2), the y-coordinates of the endpoints are both 2. Substitute y 2 into the equation and solve for x: (x 2 3)2 1 (y 1 2)2 5 49 (x 2 3)2 1 (22 1 2)2 5 49 x2 2 6x 1 9 1 0 5 49 x2 2 6x 2 40 5 0 (x 2 10)(x 1 4) 5 0 x 5 10 z x 5 24 The coordinates of the endpoints are (10, 2) and (4, 2). Answers a. (x 3)2 (y 2)2 49 b. (10, 2) and (4, 2) 14365C13.pgs 7/12/07 3:57 PM Page 584 584 Geometry of the Circle EXAMPLE 2 The equation of a circle is (x 1)2 (y 5)2 36. a. What are the coordinates of the center of the circle? b. What is the length of the radius of the circle? c. What are the coordinates of two points on the circle? Solution Compare the equation (x 1)2 (y 5)2 36 to the general form of the equation of a circle: (x h)2 (y k)2 r2 Therefore, h 1, k 5, r2 36, and r 6. a. The coordinates of the center are (1, 5). b. The length of the radius is 6. c. Points 7 units from (1, 5) on the same horizontal line are (8, 5) and (6, 5). Points 7 units from (1, 5) on the same vertical line are (1, 12) and (1, 2). Answers a. (1, 5) b. 6 c. (8, 5) and (6, 5) or (1, 12) and (1, 2) EXAMPLE 3 The equation of a circle is x2 y2 50. What is the length of the radius of the circle? Solution Compare the given equation to x2 y2 r2. r2 5 50 r 5 6 50 " Since a length is always positive, r 5 . Answer r 5 6 r 5 65 " 25 2 " 2 " 2 " Exercises Writing About Mathematics 1. Cabel said that for every circle in the coordinate plane, there is always a diameter that is a vertical line segment and one that is a horizontal line segment. Do you agree with Cabel? Justify your answer. 2. Is 3x2 3y2 12 the equation of a circle? Explain why or why not. 14365C13.pgs 7/12/07 3:57 PM Page 585 Circles in the Coordinate Plane 585 Developing Skills In 3–8, write an equation of each circle that has the given point as center and the given value of r as the length of the radius. 3. (0, 0), r 3 6. (4, 2), r 10 5. (2, 0), r 6 8. (3, 3), r 2 4. (1, 3), r 5 7. (6, 0), r 9 In 9–16, write an equation of each circle that has a diameter with the given endpoints. 9. (2, 0) and (2, 0) 11. (2, 5) and (2, 13) 13. (5, 12) and (5, 12) 15. (7, 3) and (9, 10) In 17–22, write an equation of each circle. 17. y 18. y 10. (0, 4) and (0, 4) 12. (5, 3) and (3, 3) 14. (5, 9) and (7, 7) 16. (2, 2) and (18, 4) 19. y O –1 1 1 O 1 x 1 O 1 20. 21. O x –1 –1 x 22 In 23–28, find the center of each circle and graph each circle. 23. (x 2 2)2 1 (y 1 5)2 5 4 24. 25. 26. (x 1 4)2 1 (y 2 4) 5 36 y 2 1)2 5 25 y 1 3 4 B 2 5 81 25 2 1 A A A 14365C13.pgs 7/12/07 3:57 PM Page 586 586 Geometry of the Circle 27. 2x2 1 2y2 5 18 5(x 2 1)2 1 5(y 2 1)2 5 245 28. 29. Point C(2, 3) is the center of a circle and A(3, 9) is a point on the circle. Write an equa- tion of the circle. 30. Does the point (4, 4) lie on the circle whose center is at the origin and whose radius is Justify your answer. 32 " ? 31. Is x2 4x 4 y2 2y 1 25 the equation of a circle? Explain why or why not. Applying Skills 32. In the figure on the right, the points A(2, 6), B(4, 0), and C(4, 0) appear to lie on a circle. y A(2, 6) a. Find the equation of the perpendicular bisector of . AB b. Find the equation of the perpendicular bisector of . BC . c. Find the equation of the perpendicular bisector of AC d. Find the circumcenter of ABC, the point of intersec- tion of the perpendicular bisectors. e. From what you know about perpendicular bisectors, why is the circumcenter equidistant from the vertices of ABC? f. Do the points A, B, C lie on a circle? Explain. 1 O C(4, 0) 1 x B(4, 0) 33. In the figure on the right, the circle with center at C(3, 1) appears to be inscribed in PQR with vertices P(1, 2), Q(3, 12), and R(7, 2). a. If the equations of the |
angle bisectors of PQR are 6x 8y 10, x 3, and 3x 4y 13, is C the incenter of PQR? b. From what you know about angle bisectors, why is the incenter equidistant from the sides of PQR? c. If S(3, 2) is a point on the circle, is the circle inscribed in PQR? Justify your answer. d. Write the equation of the circle. P(1, 2) y 1 O 1 R(7, 2) x C (3, 1) Q(3, 12) 14365C13.pgs 7/12/07 3:57 PM Page 587 34. In the figure on the right, the circle is circumscribed about ABC with vertices A(1, 3), B(5, 1), and C(5, 3). Find the equation of the circle. Justify your answer algebraically. Circles in the Coordinate Plane 587 A(1, 3) y B(5, 1) C(5, 3) 1 O 1 x 35. Bill Bekebrede wants to build a circular pond in his garden. The garden is in the shape of an equilateral triangle. The length of the altitude to one side of the triangle is 18 feet. To plan the pond, Bill made a scale drawing on graph paper, letting one vertex of the equilateral triangle OAB be O(0, 0) and another vertex be A(2s, 0). Therefore, the length of a side of the triangle is 2s. Bill knows that an inscribed circle has its center at the intersection of the angle bisectors of the triangle. Bill also knows that the altitude, median, and angle bisector from any vertex of an equilateral triangle are the same line. a. What is the exact length, in feet, of a side of the garden? b. In terms of s, what are the coordinates of B, the third vertex of the triangle? c. What are the coordinates of C, the intersection of the altitudes and of the angle bisectors of the triangle? d. What is the exact distance, in feet, from C to the sides of the garden? e. What should be the radius of the largest possible pond? 36. The director of the town park is planning walking paths within the park. One is to be a circular path with a radius of 1,300 feet. Two straight paths are to be perpendicular to each other. One of these straight paths is to be a diameter of the circle. The other is a chord of the circle. The two straight paths intersect 800 feet from the circle. Draw a model of the paths on graph paper letting 1 unit 100 feet. Place the center of the circle at (13, 13) and draw the diameter as a horizontal line and the chord as a vertical line. a. What is the equation of the circle? b. What are all the possible coordinates of the points at which the straight paths intersect the circular path? c. What are all the possible coordinates of the point at which the straight paths intersect? d. What are the lengths of the segments into which the point of intersection separates the straight paths? 14365C13.pgs 7/12/07 3:57 PM Page 588 588 Geometry of the Circle 13-8 TANGENTS AND SECANTS IN THE COORDINATE PLANE Tangents in the Coordinate Plane The circle with center at the origin and radius 5 is shown on the graph. Let l be a line tangent to the circle at A(3, 4). Therefore, since a tangent is perpendicular to the radius drawn to the point of tangency. The slope of l is the negative reciprocal of the slope of l ' OA . OA y P A(3, 4) 1 O 1 l x slope of OA Therefore, the slope of . We can use the slope of l and the point A(3, 4) to write the equation of l. l 5 23 4 y 2 4 x 2 3 5 23 4 4(y 2 4) 5 23(x 2 3) 4y 2 16 5 23x 1 9 3x 1 4y 5 25 The point P(1, 7) makes the equation true and is therefore a point on the tangent line 3x 4y 25. Secants in the Coordinate Plane A secant intersects a circle in two points. We can use an algebraic solution of a pair of equations to show that a given line is a secant. The equation of a circle with radius 10 and center at the origin is x2 y2 100. The equation of a line in the plane is x y 2. Is the line a secant of the circle? 14365C13.pgs 7/12/07 3:57 PM Page 589 Tangents and Secants in the Coordinate Plane 589 How to Proceed (1) Solve the pair of equations algebraically: (2) Solve the linear equation for y in terms of x: (3) Substitute the resulting expression for y in the equation of the circle: (4) Square the binomial: (5) Write the equation in standard form: (6) Divide by the common factor, 2: (7) Factor the quadratic equation: (8) Set each factor equal to zero: (9) Solve each equation for x: (10) For each value of x find the corresponding value of y: x2 1 y2 5 100 x 1 y 5 2 y 2 x x2 1 (2 2 x)2 5 100 x2 1 4 2 4x 1 x2 5 100 2x2 2 4x 2 96 5 0 x2 2 2x 2 48 5 0 (x 8)(x 6 26 26) y 5 8 y (–6, 8) 1 1O x (8, –6) The common solutions are (8, 6) and (6, 8). The line intersects the circle in two points and is therefore a secant. In the diagram, the circle is drawn with its center at the origin and radius 10. The line y 2 x is drawn with a y-intercept of 2 and a slope of 1. The line intersects the circle at (8, 6) and (6, 8). 14365C13.pgs 7/12/07 3:57 PM Page 590 590 Geometry of the Circle EXAMPLE 1 Find the coordinates of the points at which the line y 2x 1 intersects a circle with center at (0, 1) and radius of length . 20 " Solution In the equation (x h)2 (y k)2 r2, let h 0, k 1, and r = equation of the circle is: 20 " . The (x 0)2 (y (1))2 20 A " 2 B or x2 (y 1)2 20. Find the common solution of x2 (y 1)2 20 and y 2x 1. (1) The linear equation is solved for y in terms of x. Substitute, in the equation of the circle, the expression for y and simplify the result. (2) Square the monomial: (3) Write the equation in standard form: (4) Divide by the common factor, 5: x2 1 (y 1 1)2 5 20 x2 1 (2x 2 1 1 1)2 5 20 x2 1 (2x)2 5 20 x2 1 4x2 5 20 5x2 2 20 5 0 x2 2 4 5 0 (5) Factor the left side of the equation: (x 2 2)(x 1 2) 5 0 (6) Set each factor equal to zero: x 2 0 x 2 0 (7) Solve each equation for x: x 2 x 2 (8) For each value of x find the corresponding value of y: y 5 2x 2 1 y 5 2(2) 2 1 y 5 3 y 5 2x 2 1 y 5 2(22) 2 1 y 5 25 Answer The coordinates of the points of intersection are (2, 3) and (2, 5). EXAMPLE 2 The line x + y 2 intersects the circle x2 y2 100 at A(8, 6) and B(6, 8). The line y 10 is tangent to the circle at C(0, 10). a. Find the coordinates of P, the point of intersection of the secant x + y 2 and the tangent y 10. b. Show that PC 2 (PA)(PB). 14365C13.pgs 7/12/07 3:57 PM Page 591 Tangents and Secants in the Coordinate Plane 591 (–8, 10) (–6, 8) y 1 1O x (8, –6) Solution a. Use substitution to find the intersection: If x y 2 and y 10, then x 10 2 and x 8. The coordinates of P are (8, 10). b. Use the distance formula, d = lengths of PC, PA, and PB. " (x2 2 x1)2 1 (y2 2 y1)2 , to find the PA 5 " (28 2 8)2 1 (10 2 (26))2 PB 5 (28 2 (26))2 1 (10 2 8)2 " 5 256 1 256 " 5 " 5 16 256 " PC 5 " (28 2 0)2 1 (10 2 10)2 64 1 0 5 " 5 8 Then: PC2 5 82 5 64 and Therefore, PC 2 (PA)(PB). (PA)(PB) 5 (16 2)(2 2) " " 5 (32)(2) 5 64 14365C13.pgs 7/12/07 3:57 PM Page 592 592 Geometry of the Circle Exercises Writing About Mathematics 1. Ron said that if the x-coordinate of the center of a circle is equal to the length of the radius of the circle, then the y-axis is tangent to the circle. Do you agree with Ron? Explain why or why not. g AB g AB intersects a circle with center at C. The slope of tangent to the circle? Explain your answer. is m and the slope of m. Is 2. At A, g AB CA is Developing Skills In 3–14: a. Find the coordinates of the points of intersection of the circle and the line. b. Is the line a secant or a tangent to the circle? 3. x2 y2 36 y 6 6. x2 y2 10 y 3x 9. x2 y2 25 y = x 1 12. x2 y2 50 x y 10 4. x2 y2 100 x y 14 7. x2 y2 9 y x 3 10. x2 y2 20 x y 6 13. x2 y2 8 x y 4 5. x2 y2 25 x y 7 8. x2 y2 8 x y 11. x2 y2 18 y x 6 14. x2 (y 2)2 4 y x 4 In 15–18, write an equation of the line tangent to the given circle at the given point. 15. x2 y2 9 at (0, 3) 16. x2 y2 16 at (4, 0) 17. x2 y2 8 at (2, 2) 18. x2 y2 20 at (4, 2) Applying Skills 19. a. Write an equation of the secant that intersects x2 y2 25 at A(3, 4) and B(0, 5). b. Write an equation of the secant that intersects x2 y2 25 at D(0, 5) and E(0, 5). c. Find the coordinates of P, the intersection of d. Show that (PA)(PB) (PD)(PE). g AB and g DE . 20. a. Write an equation of the secant that intersects x2 y2 100 at A(6, 8) and B(8, 6). b. Write an equation of the tangent to x2 y2 100 at D(0, 10). c. Find the coordinates of P, the intersection of d. Show that (PA)(PB) (PD)2. g AB and the tangent line at D. 14365C13.pgs 8/2/07 6:00 PM Page 593 Chapter Summary 593 21. a. Write an equation of the tangent to x2 y2 18 at A(3, 3). b. Write an equation of the tangent to x2 y2 18 at B(3, 3). c. Find the point P at which the tangent to x2 y2 18 at A intersects the tangent to x2 y2 18 at B. d. Show that PA = PB. 22. Show that the line whose equation is x 2y 10 is tangent to the circle whose equation is x2 y2 20. 23. a. Show that the points A(1, 7) and B(5, 7) lie on a circle whose radius is 5 and whose cen- ter is at (2, 3). b. What is the distance from the center of the circle to the chord AB ? 24. Triangle ABC has vertices A(7, 10), B(2, 2), and C(2, 10). a. Find the coordinates of the points where the circle with equation (x 1)2 (y 7)2 9 intersects the sides of the triangle. b. Show that the sides of the triangle are tangent to the circle. c. Is the circle inscribed in the triangle? Explain. CHAPTER SUMMARY Definitions to Know • A circle is the set of all points in a plane that are equidistant from a fixed point of the plane called the center of the circle. • A radius of a circle (plural, radii) is a line segment from the center of the circle to any point of the circle. • A central angle of a circle is an angle whose vertex is the center of the circle. • An arc of a circle is the part of the circle between two points on the circle. • An arc of a circle is called an intercepted arc, or an arc intercepted by an angle, if each endpoint of the arc is on a different ray of the angle and the other points of the arc are in the interior of the angle. • The degree measure of an arc is equal to the measure of the central angle that intercepts the arc. • Congruent circles are circles with congruent radii. • Congruent arcs are arcs of the same circle or of congruent circles that are equal in measure. • A chord of a circle is a line segment w |
hose endpoints are points of the circle. • A diameter of a circle is a chord that has the center of the circle as one of its points. • An inscribed angle of a circle is an angle whose vertex is on the circle and whose sides contain chords of the circle. 14365C13.pgs 8/2/07 6:00 PM Page 594 594 Geometry of the Circle • A tangent to a circle is a line in the plane of the circle that intersects the circle in one and only one point. • A secant of a circle is a line that intersects the circle in two points. • A common tangent is a line that is tangent to each of two circles. • A tangent segment is a segment of a tangent line, one of whose endpoints is the point of tangency. Postulates 13.1 and ABX If point and no other points in common, then ABX m ABX . (Arc Addition Postulate) are two arcs of the same circle having a common endand ABCXBCX BCX BCX ABCX m m 13.2 At a given point on a given circle, one and only one line can be drawn that is tangent to the circle. Theorems and Corollaries 13.1 All radii of the same circle are congruent. 13.2 In a circle or in congruent circles, central angles are congruent if and only if their intercepted arcs are congruent. In a circle or in congruent circles, two chords are congruent if and only if their central angles are congruent. In a circle or in congruent circles, two chords are congruent if and only if their arcs are congruent. 13.3 13.4 13.5 A diameter perpendicular to a chord bisects the chord and its arcs. 13.5a A line through the center of a circle that is perpendicular to a chord 13.6 13.7 13.8 13.9 bisects the chord and its arcs. The perpendicular bisector of the chord of a circle contains the center of the circle. Two chords are equidistant from the center of a circle if and only if the chords are congruent. In a circle, if the lengths of two chords are unequal, the shorter chord is farther from the center. The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc. 13.9a An angle inscribed in a semicircle is a right angle. 13.9b If two inscribed angles of a circle intercept the same arc, then they are congruent. 13.10 A line is tangent to a circle if and only if it is perpendicular to a radius at its point of intersection with the circle. 13.11 Tangent segments drawn to a circle from an external point are congru- ent. 13.11a If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents. 13.11b If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle whose vertex is the center of the circle and whose rays are the two radii drawn to the points of tangency. 14365C13.pgs 8/7/07 10:51 AM Page 595 Chapter Summary 595 13.12 The measure of an angle formed by a tangent and a chord that intersect at the point of tangency is equal to one-half the measure of the intercepted arc. 13.13 The measure of an angle formed by two chords intersecting within a circle is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. 13.16 13.15 13.14 The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs. If two chords intersect within a circle, the product of the measures of the segments of one chord is equal to the product of the measures of the segments of the other. If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external segment. If a tangent and a secant are drawn to a circle from an external point, then the length of the tangent segment is the mean proportional between the lengths of the secant segment and its external segment. If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. 13.17 13.16 Formulas Type of Angle Central Angle Degree Measure Example The measure of a central angle is equal to the measure of its intercepted arc. Inscribed Angle The measure of an inscribed angle is equal to one-half the measure of its intercepted arc. B A A 1 m1 mABX B 1 m/1 5 1 2m ABX (Continued) 14365C13.pgs 8/7/07 10:51 AM Page 596 596 Geometry of the Circle Formulas (Continued) Type of Angle Degree Measure Example Formed by a Tangent and a Chord The measure of an angle formed by a tangent and a chord that intersect at the point of tangency is equal to one-half the measure of the intercepted arc. Formed by Two Intersecting Chords The measure of an angle formed by two intersecting chords is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Formed by Tangents and Secants The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs. A 1 B m/1 5 1 2mABX 2 1 A B D C m/1 5 1 m/2 5 1 2(mABX 1 mCDX) 2(mABX 1 mCDX/1 5 1 m/2 5 1 m/3 5 1 2(mABX 2 mACX) 2(mABX 2 mCDX) 2(mACBX 2 mABX) 14365C13.pgs 8/2/07 6:00 PM Page 597 Formulas (Continued) Vocabulary 597 Type of Segment Length Example Formed by Two Intersecting Chords If two chords intersect, the product of the measures of the segments of one chord is equal to the product of the measures of the segments of the other. A D E O C B Formed by a Tangent Intersecting a Secant Formed by Two Intersecting Secants If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external segment. If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. (AE)(EB) 5 (CE)(ED) A O C P B (PA)2 5 (PC)(PB) P A C B O D (PB)(PA) 5 (PD)(PC) The equation of a circle with radius r and center (h, k) is (x h)2 (y k)2 r 2 VOCABULARY 13-1 Circle • Center • Radius • Interior of a circle • Exterior of a circle • Central angle of a circle • Arc of a circle • Minor arc • Major arc • Semicircle • Intercepted arc • Degree measure of an arc • Congruent circles • Congruent arcs 13-2 Chord • Diameter • Apothem • Inscribed polygon • Circumscribed circle 13-3 Inscribed angle 14365C13.pgs 7/12/07 3:57 PM Page 598 598 Geometry of the Circle 13-4 Tangent to a circle • Secant of a circle • Common tangent • Common internal tangent • Common external tangent • Tangent externally • Tangent internally • Tangent segment • Circumscribed polygon • Inscribed circle • Center of a regular polygon 13-6 External segment 13-7 Center-radius equation of a circle REVIEW EXERCISES In 1–6, PA to circle O. Chords E. is a tangent and and AC PBC BD is a secant intersect at 1. If mABX mBCX 80, 100, find: mCDX a. mPAC 120, and P D A E O B C b. mCBD c. mAPC d. mDEC e. mAED a. 2. If mC 50, mDBC 55, and mPAC 100, find: d. mP mABX 3. If mCEB 80, ADX CDX c. mBEC b. m mBCX mABX CDX c. mCBD b. m 4. If AP 12 and PC 24, find PB and BC. 5. If PB 5 and BC 15, find AP. 6. If AC 11, DE 2, EB 12, and AE EC, find AE and EC. 120, and 70, find: d. mP a. m BCX e. m e. mPAC PA and secant seg- 7. Tangent segment PBC ment are drawn to circle O. If PB 8 and BC 10, PA is equal to (1) 12 (3) 80 (2) 4 5 " (4) 144 8. The equation of a circle with center at (2, 4) and radius of length 3 is (1) (x 2)2 (y 4)2 9 (2) (x 2)2 (y 4)2 9 (3) (x 2)2 (y 4)2 9 (2) (x 2)2 (y 4)2 9 A P O B C 9. Two tangents that intersect at P intercept a major arc of 240° on the circle. What is the measure of P? 10. A chord that is 24 centimeters long is 9 centimeters from the center of a circle. What is the measure of the radius of the circle? 14365C13.pgs 8/2/07 6:01 PM Page 599 Review Exercises 599 B O C P O A D 11. Two circles, O and O, are tangent externally at P, OP 5, and OP 3. Segment is tangent to circle O at B and to circle O at intersects circle O at D C, and P, and circle O at P. If AD 2, find AB and AC. AOrO ABC 12. Isosceles ABC is inscribed in a circle. If the measure of the vertex angle, A, is 20 degrees less than twice the measure of each of the base angles, CAX find the measures of BCXABX , 13. Prove that a trapezoid inscribed in a circle is isosceles. , and . 14. In circle O, chords and CD are parallel and intersects AD a. Prove that ABE and CDE are isosceles tri- BC AB at E. angles. ACX > BDX b. Prove that c. Prove that ABE CDE. . B E D A O C 15. Prove that if A, B, C, and D separate a circle into four congruent arcs, then quadrilateral ABCD is a square. 16. Prove, using a circumscribed circle, that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices of the triangle. 17. Secant segments PAB and drawn to circle O. If prove that from the center of the circle. PAB and CD AB PCD are equidistant PCD are , B F A O E C D P 18. An equilateral triangle is inscribed in a circle whose radius measures 12 centimeters. How far from the center of the circle is the centroid of the triangle? Exploration A regular polygon can be constructed by constructing the congruent isosceles triangles into which it can be divided. The measure of each base angle of the isosceles triangles is one-half the measure of an interior angle of the polygon. However, the interior angles of many regular polygons are not angles that can be constructed using compass and straightedge. For example, a regular polygon with nine sides has angles that measure 140 degrees. Each of the nine isosceles triangles of this polygon has base a |
ngles of 70 degrees which cannot be constructed with straightedge and compass. 14365C13.pgs 7/12/07 3:57 PM Page 600 600 Geometry of the Circle In this exploration, we will construct a regular triangle (equilateral triangle), a regular hexagon, a regular quadrilateral (a square), a regular octagon, and a regular dodecagon (a polygon with 12 sides) inscribed in a circle. a. Explain how a compass and a straightedge can be used to construct an equi- lateral triangle. Prove that your construction is valid. b. Explain how the construction in part a can be used to construct a regular hexagon. Prove that your construction is valid. c. Explain how a square, that is, a regular quadrilateral, can be inscribed in a circle using only a compass and a straightedge. (Hint: What is true about the diagonals of a square?) Prove that your construction is valid. d. Bisect the arcs determined by the chords that are sides of the square from the construction in part c. Join the endpoints of the chords that are formed to draw a regular octagon. Prove that this construction is valid. e. A regular octagon can also be constructed by constructing eight isosceles triangles. The interior angles of a regular octagon measure 135 degrees. Bisect a right angle to construct an angle of 45 degrees. The complement of that angle is an angle of 135 degrees. Bisect this angle to construct the base angle of the isosceles triangles needed to construct a regular octagon. f. Explain how a regular hexagon can be inscribed in a circle using only a compass and a straightedge. (Hint: Recall how a regular polygon can be divided into congruent isosceles triangles.) g. Bisect the arcs determined by the chords that are sides of the hexagon from part f to draw a regular dodecagon. h. A regular dodecagon can also be constructed by constructing twelve isosceles triangles. The interior angles of a regular dodecagon measure 150 degrees. Bisect a 60-degree angle to construct an angle of 30 degrees. The complement of that angle is an angle of 150 degrees. Bisect this angle to construct the base angle of the isosceles triangles needed to construct a regular dodecagon. CUMULATIVE REVIEW Chapters 1–13 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The measure of A is 12 degrees more than twice the measure of its com- plement. The measure of A is (1) 26 (2) 39 (3) 64 (4) 124 14365C13.pgs 7/12/07 3:57 PM Page 601 Cumulative Review 601 2. The coordinates of the midpoint of a line segment with endpoints at (4, 9) (2) (3, 3) (3) (2, 24) (4) (6, 6) 3. What is the slope of a line that is perpendicular to the line whose equation and (2, 15) are (1) (1, 12) is 2x y 8? (1) 2 (2) 2 (3) 21 2 (4) 1 2 4. The altitude to the hypotenuse of a right triangle separates the hypotenuse into segments of length 6 and 12. The measure of the altitude is (4) (1) 18 6 (3) (2. The diagonals of a quadrilateral bisect each other. The quadrilateral can- not be a (1) trapezoid (2) rectangle (3) rhombus (4) square 6. Two triangles, ABC and DEF, are similar. If AB 12, DE 18, and the perimeter of ABC is 36, then the perimeter of DEF is (1) 24 (2) 42 (3) 54 (4) 162 7. Which of the following do not always lie in the same plane? (1) two points (2) three points (3) two lines (4) a line and a point not on the line 8. At A, the measure of an exterior angle of ABC is 110 degrees. If the measure of B is 45 degrees, what is the measure of C? (1) 55 (3) 70 (2) 65 (4) 135 9. Under the composition rx-axis + T2,3 , what are the coordinates of the image of A(3, 5)? (1) (5, 2) (2) (5, 2) (3) (5, 8) (4) (1, 2) 10. In the diagram, and g AB g AB g CD g CD at E and g EF at F. If intersects mAEF is represented by 3x and mCFE is represented by 2x 20, what is the value of x? (1) 4 (2) 12 (3) 32 (4) 96 A C F E B D 14365C13.pgs 7/12/07 3:57 PM Page 602 602 Geometry of the Circle Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The coordinates of the vertices of ABC are A(8, 0), B(4, 4), and C(0, 4). a. Find the coordinates of D, the midpoint of AB . b. Find the coordinates of E, the midpoint of BC . DE AC c. Is d. Are ABC and DBE similar triangles? Justify your answer. ? Justify your answer. 12. ABCD is a quadrilateral, AC > BD E. Prove that ABCD is a rectangle. and AC and BD bisect each other at Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. In the diagram, sect at E in plane p, g g ' CD EF FA > FC . . If g AB and g EF EA > EC inter- g CD g ' AB , prove that , and C F p E B D A 14. The length of the hypotenuse of a right triangle is 2 more than the length of the longer leg. The length of the shorter leg is 7 less than the length of the longer leg. Find the lengths of the sides of the right triangle. Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 14365C13.pgs 7/12/07 3:57 PM Page 603 Cumulative Review 603 15. a. Find the coordinates of A, the image of A(5, 2) under the composition R90 + ry 5 x . b. What single transformation is equivalent to R90 + ry 5 x ? c. Is R90 + ry 5 x a direct isometry? Justify your answer. ADC 16. In the diagram, D is a point on such that AD : DC 1 : 3, and E is a point on BEC a. Show that AC : DC BC : EC. b. Prove that ABC DEC. such that BE : EC 1 : 3. A D B E C 14365C14.pgs 7/10/07 9:59 AM Page 604 CHAPTER 14 CHAPTER TABLE OF CONTENTS 14-1 Constructing Parallel Lines 14-2 The Meaning of Locus 14-3 Five Fundamental Loci 14-4 Points at a Fixed Distance in Coordinate Geometry 14-5 Equidistant Lines in Coordinate Geometry 14-6 Points Equidistant from a Point and a Line Chapter Summary Vocabulary Review Exercises Cumulative Review 604 LOCUS AND CONSTRUCTION Classical Greek construction problems limit the solution of the problem to the use of two instruments: the straightedge and the compass.There are three construction problems that have challenged mathematicians through the centuries and have been proved impossible: the duplication of the cube the trisection of an angle the squaring of the circle The duplication of the cube requires that a cube be constructed that is equal in volume to twice that of a given cube. The origin of this problem has many versions. For example, it is said to stem from an attempt at Delos to appease the god Apollo by doubling the size of the altar dedicated to Apollo. The trisection of an angle, separating the angle into three congruent parts using only a straightedge and compass, has intrigued mathematicians through the ages. The squaring of the circle means constructing a square equal in area to the area of a circle. This is equivalent to constructing a line segment whose length is equal to times the radius of the circle. p Although solutions to these problems have been presented using other instruments, solutions using only straightedge and compass have been proven to be impossible. ! 14365C14.pgs 7/10/07 9:59 AM Page 605 Constructing Parallel Lines 605 14-1 CONSTRUCTING PARALLEL LINES In Chapter 5, we developed procedures to construct the following lines and rays: 1. a line segment congruent to a given line segment 2. an angle congruent to a given angle 3. the bisector of a given line segment 4. the bisector of a given angle 5. a line perpendicular to a given line through a given point on the line 6. a line perpendicular to a given line through a given point not on the line Then, in Chapter 9, we constructed parallel lines using the theorem that if two coplanar lines are each perpendicular to the same line, then they are parallel. Now we want to use the construction of congruent angles to construct par- allel lines. Two lines cut by a transversal are parallel if and only if the corresponding angles are congruent. For example, g CD in the diagram, the transversal intersects at G g at H. If EGB GHD, then CD and . Therefore, we can construct parallel lines by constructing congruent corresponding angles. g AB g AB g EF F H D B C A G E Construction 7 Construct a Line Parallel to a Given Line at a Given Point. Given g AB and point P not on g AB Construct A line through P that is parallel to g AB STEP 1. Through P, draw any line intersecting at R. Let S be g AB any point on the ray opposite h PR . STEP 2. At P, construct SPD PRB. Draw opposite ray of h PD h PC , forming , the g .CD Continued 14365C14.pgs 7/10/07 9:59 AM Page 606 606 Locus and Construction Construction 7 Construct a Line Parallel to a Given Line at a Given Point. (continued) Conclusion g AB CPD g Proof Corresponding angles, SPD and PRB, are congruent. Therefore, g AB CPD g . This construction can be used to construct the points and lines that satisfy other conditions. EXAMPLE 1 Given: g AB and PQ Construct: g CD g AB at a distance 2PQ from g AB . A B P Q Construction The distance from a point to a line is the length of the perpendicular from the point to the line. Therefore, we must locate points at a distance of 2PQ from a point on g AB at which to draw a parallel line. F C F C D F P RQ 1. Extend . PQ Locate point R on h such that PQ QR PQ, making PR 2PQ EF 3. On , locate point C at a distance PR from E. 2. Choose any point g AB E on construct the line h EF perpendicular . At E, g AB . to 4. At C, construct g CD perpendicular to g EF and therefore parallel to g .AB Con |
clusion g CD is parallel to g AB at a distance 2PQ from g .AB 14365C14.pgs 7/10/07 9:59 AM Page 607 Constructing Parallel Lines 607 Exercises Writing About Mathematics 1. In the example, every point on g CD is at a fixed distance, 2PQ, from g AB . Explain how you know that this is true. g CD g AB 2. Two lines, and , are parallel. A third line, g EF , is perpendicular to g AB at H. The perpendicular bisector of GH is the set of all points equidistant from Explain how you know that this is true. at G and to g AB and g CD g . CD Developing Skills In 3–9, complete each required construction using a compass and straightedge, or geometry software. Draw each given figure and do the required constructions. Draw a separate figure for each construction. Enlarge the given figure for convenience. P l m 3. Given: Parallel lines l and m and point P. Construct: a. a line through P that is parallel to l. b. a line that is parallel to l and to m and is equidistant from l and m. c. a line n that is parallel to l and to m such that l is equidistant from m and n. d. Is the line that you constructed in a parallel to m? Justify your answer. 4. Given: Line segments of length a and b and A Construct: a. a rectangle whose length is a and whose width is b. b. a square such that the length of a side is a. c. a parallelogram that has sides with measures a and b and an angle congruent to A. d. a rhombus that has sides with measure a and an angle congruent to A. 5. Given: AB a b A a. Divide AB into four congruent parts. A B b. Construct a circle whose radius is AB. c. Construct a circle whose diameter is AB. 14365C14.pgs 7/10/07 9:59 AM Page 608 608 Locus and Construction 6. Given: ABC Construct: a. a line parallel to g AB at C. b. the median to AC by first constructing the midpoint of AC . c. the median to BC by first constructing the midpoint of BC . C AC and the median to BC intersect at P, can A B d. If the median to the median to point of AB ? Justify your answer. AB be drawn without first locating the mid- 7. Given: Obtuse triangle ABC Construct: a. the perpendicular bisector of . AC b. the perpendicular bisector of . BC c. M, the midpoint of AB . C A B AC and the intersect at P, d. If the perpendicular bisector of BC perpendicular bisector of what two points determine the perpendicular bisector of 8. Given: ABC Construct: ? Justify your answer. AB a. the altitude to . AC b. the altitude to . BC c. If the altitude to AC and the altitude to BC at P, what two points determine the altitude to Justify your answer. intersect ? AB 9. Given: ABC Construct: a. the angle bisector from B to . AC b. the angle bisector from A to c. If the angle bisector of B and the angle bisector of A intersect at P, what two points determine the ? Justify your answer. angle bisector from C to . BC AB C C A A B B 10. a. Draw any line segment, . Draw any ray, AB h AC h AC , forming BAC. h AC such that DE 2AD. b. Choose any point, D, on . Construct DE on EB c. Draw d. Prove that AD : DE = AF : FB 1 : 2. and construct a line through D parallel to EB and intersecting h AB at F. 14365C14.pgs 7/10/07 9:59 AM Page 609 11. a. Draw an angle, LPR. h PL b. Choose point N on that PS : SQ 3 : 5. . Construct Q on h PL such that PQ 8PN. Locate S on PQ such The Meaning of Locus 609 QR c. Draw and construct a line through S parallel to 3 d. Prove that PST PQR with a constant of proportionality of . 8 QR and intersecting 14-2 THE MEANING OF LOCUS In a construction, the opening between the pencil and the point of the compass is a fixed distance, the length of the radius of a circle. The point of the compass determines a fixed point, point O in the diagram. If the length of the radius remains unchanged, all of the points in the plane that can be drawn by the compass form a circle, and any points that cannot be drawn by the compass do not lie on the circle. Thus, the circle is the set of all points at a fixed distance from a fixed point. This set is called a locus. h PR at T. O DEFINITION A locus is the set of all points that satisfy a given condition or set of conditions. The example of the circle given above helps us to understand what the def- inition means. Every definition can be written as a biconditional: p: A point is on the locus. q: A point satisfies the given conditions. 1. (p → q): If a point is on the locus, then the point satisfies the given conditions. All points on the circle are at a given fixed distance from the center. 2. (p → q): If a point is not on the locus, then the point does not satisfy the given conditions. Any point that is not on the circle is not at the given distance from the center. Recall that the statement (p → q) is the inverse of the statement (p → q). A locus is correct when both statements are true: the conditional and its inverse. We can restate the definition of locus in biconditional form: A point P is a point of the locus if and only if P satisfies the given conditions of the locus. 14365C14.pgs 7/10/07 9:59 AM Page 610 610 Locus and Construction Discovering a Locus Procedure To discover a probable locus: 1. Make a diagram that contains the fixed lines or points that are given. 2. Decide what condition must be satisfied and locate one point that meets the given condition. 3. Locate several other points that satisfy the given condition.These points should be sufficiently close together to develop the shape or the nature of the locus. 4. Use the points to draw a line or smooth curve that appears to be the locus. 5. Describe in words the geometric figure that appears to be the locus. Note: In this chapter, we will assume that all given points, segments, rays, lines, and circles lie in the same plane and the desired locus lies in that plane also. EXAMPLE 1 What is the locus of points equidistant from the endpoints of a given line segment? Solution Apply the steps of the procedure for discovering a probable locus. Make a diagram: is the given AB line segment. 2. Decide the condition to be satisfied: P is to be equidistant from A and B. Use a compass opened to any convenient radius to locate one such point, P. 3. Locate several other points equidistant from A and B, using a different opening of the compass for each point. 4. Through these points, draw the straight line that appears to be the locus. 5. Describe the locus in words: The locus is a straight line that is the perpendicular bisector of the given line segment. Answer 14365C14.pgs 7/10/07 9:59 AM Page 611 The Meaning of Locus 611 Note that in earlier chapters, we proved two theorems that justify these results: • If a point is equidistant from the endpoints of a line segment, then it lies on the perpendicular bisector of the segment. • If a point lies on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the segment. EXAMPLE 2 Construct the locus of points in the interior of an angle equidistant from the rays that form the sides of the given angle. Construction Corollaries 9.13b and 9.15a together state: A point is equidistant from the sides of an angle if and only if it lies on the bisector of the angle. Therefore, the required locus is the bisector of the angle. 1. Make a diagram: ABC is the given angle. A B C 2. Decide the condition to be satisfied: P is to be equidistant from h BC , the rays that are the sides of ABC. Construct the angle bisector. h BA and Use a compass to draw an arc with center B that intersects h at S. Then, with the compass open to a convenient radius, draw arcs BC from R and S that intersect in the interior of ABC. Call the intersection P. PR PS. at R and h BA A P C R B S 3. Through points P and B, draw the ray that is the locus. A P C R B S 14365C14.pgs 7/10/07 9:59 AM Page 612 612 Locus and Construction Exercises Writing About Mathematics 1. Are all of the points that are equidistant from the endpoints of a line segment that is 8 cen- timeters long 4 centimeters from the endpoints? Explain your answer. 2. What line segment do we measure to find the distance from a point to a line or to a ray? Developing Skills 3. What is the locus of points that are 10 centimeters from a given point? 4. What is the locus of points equidistant from two points, A and B, that are 8 meters apart? 5. What is the locus of points equidistant from two parallel lines 8 meters apart? 6. What is the locus of points 4 inches away from a given line, g ? AB 7. What is the locus of points 3 inches from each of two parallel lines that are 6 inches apart? 8. What is the locus of points that are equidistant from two opposite sides of a square? 9. What is the locus of points that are equidistant from the vertices of two opposite angles of a square? 10. What is the locus of points that are equidistant from the four vertices of a square? (A locus can consist of a single point or no points.) 11. What is the locus of points in the interior of a circle whose radius measures 3 inches if the points are 2 inches from the circle? 12. What is the locus of points in the exterior of a circle whose radius measures 3 inches if the points are 2 inches from the circle? 13. What is the locus of points 2 inches from a circle whose radius measures 3 inches? 14. Circle O has a radius of length r, and it is given that r m. a. What is the locus of points in the exterior of circle O and at a distance m from the circle? b. What is the locus of points in the interior of circle O and at a distance m from the circle? c. What is the locus of points at a distance m from circle O? 15. Concentric circles have the same center. What is the locus of points equidistant from two concentric circles whose radii measure 10 centimeters and 18 centimeters? 16. A series of isosceles triangles are drawn, each of which has a fixed segment, AB , as its base. What is the locus of the vertices of the vertex angles of all such isosceles triangles? 17. Triangle ABC is drawn with a fixed base, AB , and an altitude to AB whose measure is 3 feet. What is the locus of points that can indicate vertex C in all |
such triangles? 14365C14.pgs 7/10/07 9:59 AM Page 613 Five Fundamental Loci 613 Applying Skills 18. What is the locus of the tip of the hour hand of a clock during a 12-hour period? 19. What is the locus of the center of a train wheel that is moving along a straight, level track? 20. What is the locus of the path of a car that is being driven down a street equidistant from the two opposite parallel curbs? 21. A dog is tied to a stake by a rope 6 meters long. What is the boundary of the surface over which he can move? 22. A boy walks through an open field that is bounded on two sides by straight, intersecting roads. He walks so that he is always equidistant from the two intersecting roads. Determine his path. 23. There are two stationary floats on a lake. A girl swims so that she is always equidistant from both floats. Determine her path. 24. A dime is rolled along a horizontal line so that the dime always touches the line. What is the locus of the center of the dime? 25. The outer edge of circular track is 40 feet from a central point. The track is 10 feet wide. What is path of a runner who runs on the track, 2 feet from the inner edge of the track? 14-3 FIVE FUNDAMENTAL LOCI There are five fundamental loci, each based on a different set of conditions. In each of the following, a condition is stated, and the locus that fits the condition is described in words and drawn below. Each of these loci has been shown as a construction in preceding sections. 1. Equidistant from two points: Find points equidistant from points A and B. Locus: The locus of points equidistant from two fixed points is the perpendicular bisector of the segment determined by the two points. A L ocus B 2. Equidistant from two intersecting lines: Find points equidistant from the intersecting lines g AB and g . CD Locus: The locus of points equidistant from two intersecting lines is a pair of lines that bisect the angles formed by the intersecting lines. A D Locus s u c o L C B 14365C14.pgs 7/10/07 9:59 AM Page 614 614 Locus and Construction 3. Equidistant from two parallel lines: Find points equidistant from the paral- lel lines g AB and g . CD Locus: The locus of points equidistant from two parallel lines is a third line, parallel to the given lines and midway between them. At a fixed distance from a line: Find points that are at distance d from the line g . AB Locus: The locus of points at a fixed distance from a line is a pair of lines, each parallel to the given line and at the fixed distance from the given line. Locus d A d Locus B 5. At a fixed distance from a point: Find points that are at a distance d from the fixed point A. Locus: The locus of points at a fixed distance from a fixed point is a circle whose center is the fixed point and whose radius is the fixed distance. L ocus d A These loci are often combined to find a point or set of points that satisfy two or more conditions. The resulting set is called a compound locus. Procedure To find the locus of points that satisfy two conditions: 1. Determine the locus of points that satisfy the first condition. Sketch a diagram showing these points. 2. Determine the locus of points that satisfy the second condition. Sketch these points on the diagram drawn in step 1. 3. Locate the points, if any exist, that are common to both loci. Steps 2 and 3 can be repeated if the locus must satisfy three or more conditions. EXAMPLE 1 Quadrilateral ABCD is a parallelogram. What is the locus of points equidistant from g AB and g CD and also equidistant from g AB and g ?BC 14365C14.pgs 7/10/07 9:59 AM Page 615 Solution Follow the procedure for finding a compound locus. Five Fundamental Loci 615 (1) Since ABCD is a parallelogram, g CD g . The locus of points AB equidistant from two parallel lines is a third line parallel to the given lines and midway between them. In the g AB is equidistant from figure, g EF and g . CD (2) The lines g AB and g BC are intersecting lines. The locus of points equidistant from intersecting lines is a pair of lines that bisect the angles formed by the g JK g GH are equidistant from given lines. In the figure, g AB g EF (3) The point P at which and g BC . and intersects g EF g GH and the point Q at which intersects are equidistant from g JK and also equidistant from g AB g AB and and g CD g BC . Answer P and Q Note that only point P is equidistant from the three segments that are sides of the parallelogram, but both P and Q are equidistant from the lines of which these three sides are segments. Exercises Writing About Mathematics 1. If PQRS is a square, are the points that are equidistant from g PQ and g RS also equidistant from P and S? Explain your answer. 2. Show that the two lines that are equidistant from two intersecting lines are perpendicular to each other. Developing Skills In 3–10, sketch and describe each required locus. 3. The locus of points equidistant from two points that are 4 centimeters apart. 14365C14.pgs 7/10/07 9:59 AM Page 616 616 Locus and Construction 4. The locus of points that are 6 inches from the midpoint of a segment that is 1 foot long. 5. The locus of points equidistant from the endpoints of the base of an isosceles triangle. 6. The locus of points equidistant from the legs of an isosceles triangle. 7. The locus of points equidistant from the diagonals of a square. 8. The locus of points equidistant from the lines that contain the bases of a trapezoid. 9. The locus of points 4 centimeters from the midpoint of the base of an isosceles triangle if the base is 8 centimeters long. 10. The locus of points that are 6 centimeters from the altitude to the base of an isosceles trian- gle if the measure of the base is 12 centimeters 11. a. Sketch the locus of points equidistant from two parallel lines that are 4 centimeters apart. b. On the diagram drawn in a, place point P on one of the given parallel lines. Sketch the locus of points that are 3 centimeters from P. c. How many points are equidistant from the two parallel lines and 3 centimeters from P? 12. a. Construct the locus of points equidistant from the endpoints of a line segment . b. Construct the locus of points at a distance AB from M, the midpoint of 1 2(AB) . AB c. How many points are equidistant from A and B and at a distance 1 2(AB) from the mid- point of AB ? d. Draw line segments joining A and B to the points described in c to form a polygon. What kind of a polygon was formed? Explain your answer. 14-4 POINTS AT A FIXED DISTANCE IN COORDINATE GEOMETRY We know that the locus of points at a fixed distance from a given point is a circle whose radius is the fixed distance. In the coordinate plane, the locus of points r units from (h, k) is the circle whose equation is (x h)2 (y k)2 r2. For example, the equation of the locus of points 10 units from (2, 3) is: " (x 2)2 (y (3))2 ( 10)2 " or (x 2)2 (y 3)2 10 y 1 O 1 x (2, 3) We also know that the locus of points at a fixed distance from a given line is a pair of lines parallel to the given line. 14365C14.pgs 7/10/07 9:59 AM Page 617 Points at a Fixed Distance in Coordinate Geometry 617 For example, to write the equations of the locus of points 3 units from the horizontal line y 2, we need to write the equations of two horizontal lines, one 3 units above the line y 2 and the other 3 units below the line y 2. The equations of the locus are y 5 and y 1. To write the equations of the locus of points 3 units from the vertical line x 2, we need to write the equations of two vertical lines, one 3 units to the right of the line x 2 and the other 3 units to the left of the line x 2. The equations of the locus are x 5 and x 1. From these examples, we can infer the following 1O y 2 x 1 x 1 1O The locus of points d units from the horizontal line y a is the pair of lines y a d and y a d. The locus of points d units from the vertical line x a is the pair of lines x a d and x a d. EXAMPLE 1 What is the equation of the locus of points at a distance of (0, 1)? 20 " units from Solution The locus of points at a fixed distance from a point is the circle with the given point as center and the given distance as radius. The equation of the locus is (x 0)2 (y 1)2 20 A " 2 B or x2 (y 1)2 20 Answer EXAMPLE 2 What are the coordinates of the points on the line y 21 2x 1 at a distance of 20 " from (0, 1)? Solution From Example 1, we know that the set of all points at a distance of 20 from (0, 1) lie on the circle whose equation is x2 (y 1)2 20. Therefore, the points on the line y 1 at a distance of sections of the circle and the line. from (0, 1) are the inter- 21 2x " 20 " 14365C14.pgs 7/10/07 10:00 AM Page 618 618 Locus and Construction METHOD 1 Solve the system of equations graphically. Sketch the graphs and read the coordinates from the graph. y (4, 3) x2 ( y 1 )2 METHOD 2 Solve the system of equations algebraically. Substitute the value of y from the linear equation in the equation of the circle. 2 0 x (4, 1) y 1 x 1 2 x2 1 A x2 1 (y 2 1)2 5 20 2 5 20 21 2x 1 1 2 1 B x2 1 1 4x2 5 20 5 4x2 5 20 x2 5 16 x 5 64 Answer The points are (4, 1) and (4, 3). If x 4: y 5 21 2(4) 1 1 If x 4: y 5 21 2(24) 1 1 y 5 21 y 5 3 Exercises Writing About Mathematics 1. In Example 2, is the line y 21 2x 1 a secant of the circle x2 (y 1)2 20? Justify your answer. g PA and g PB 2. are tangent to circle O. Martin said that point O is in the locus of points equidistant from g PA and g PB . Do you agree with Martin? Explain why or why not. Developing Skills In 3–8, write an equation of the locus of points at the given distance d from the given point P. 3. d 4, P(0, 0) 6. d 7, P(1, 1) 4. d 1, P(1, 0) 7. d 10 , P(3, 1) " 5. d 3, P(0, 2) 8. d 18 , P(3, 5) " In 9–12, find the equations of the locus of points at the given distance d from the given line. 9. d 5, x 7 10. d 1, x 1 11. d 4, y 2 12. d 6, y 7 14365C14.pgs 7/10/07 10:00 AM Page 619 Equidistant Lines in Coordinate Geometry 619 In 13–16, find the coordinates of the points at the given distance from the given point and on the given line. 13. 5 uni |
ts from (0, 0) on y x 1 15. 10 units from (0, 1) on y x 3 14. 13 units from (0, 0) on y x 7 16. 10 " units from (1, 1) on y x 2 In 17–22, write the equation(s) or coordinates and sketch each locus. 17. a. The locus of points that are 3 units from y 4. b. The locus of points that are 1 unit from x 2. c. The locus of points that are 3 units from y 4 and 1 unit from x 2. 18. a. The locus of points that are 3 units from (2, 2). b. The locus of points that are 3 units from y 1. c. The locus of points that are 3 units from (2, 2) and 3 units from y 1. 19. a. The locus of points that are 5 units from the origin. b. The locus of points that are 3 units from the x-axis. c. The locus of points that are 5 units from the origin and 3 units from the x-axis. 20. a. The locus of points that are 10 units from the origin. b. The locus of points that are 8 units from the y-axis. c. The locus of points that are 10 units from the origin and 8 units from the y-axis. 21. a. The locus of points that are 2 units from (x 4)2 y2 16. b. The locus of points that are 2 units from the y-axis. c. The locus of points that are 2 units from (x 4)2 y2 16 and 2 units from the y-axis. 22. a. The locus of points that are 3 units from (x 1)2 (y 5)2 4. 5 2 3 b. The locus of points that are units from x . 2 3 c. The locus of points that are 3 units from (x 1)2 (y 5)2 4 and units from x . 2 5 2 14-5 EQUIDISTANT LINES IN COORDINATE GEOMETRY Equidistant from Two Points The locus of points equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points. For example, the locus of points equidistant from A(2, 1) and B(8, 5) is a line perpendicular to at its midpoint. AB 14365C14.pgs 7/10/07 10:00 AM Page 620 620 Locus and Construction midpoint of AB 5 2 1 8 2 A 5 (5, 2) 5 1 (21) 2 , B y slope of AB 5 5 2 (21) 8 2 2 5 6 6 5 1 AB Therefore, the slope of a line perpendicuis 1. The perpendicular bisector of is the line through (5, 2) with slope 1. lar to AB The equation of this line is: y = B(8, 5) x 7 (5, 2) 1 O 1 A(2, 1) x y 2 2 x 2 5 5 21 y 2 2 5 2x 1 5 y 5 2x 1 7 EXAMPLE 1 Describe and write an equation for the locus of points equidistant from A(2, 5) and B(6, 1). M(2, 3) B(6, 1) x y 1 A(2, 5) 1 O y = 2x 1 Solution (1) Find the midpoint, M, of AB : M A 22 1 6 2 , 5 1 1 2 (2) Find the slope of AB (2, 3) B : slope of AB 5 5 2 1 22 2 6 5 4 28 5 21 2 (3) The slope of a line perpendicular to AB is 2. (4) Write an equation of the line through (2, 3) with slope 2 2x 2 4 y 5 2x 2 1 Answer The locus of points equidistant from A(2, 5) and B(6, 1) is the perpendicular . The equation of the locus is y 2x 1. bisector of AB 14365C14.pgs 7/10/07 10:00 AM Page 621 Equidistant Lines in Coordinate Geometry 621 Equidistant from Two Parallel Lines The locus of points equidistant from two parallel lines is a line parallel to the two lines and midway between them. For example, the locus of points equidistant from the vertical lines x 2 and x 6 is a vertical line midway between them. Since the given lines intersect the x-axis at (2, 0) and (6, 0), the line midway between them intersects the x-axis at (2, 0) and has the equation x 2. EXAMPLE 2 Write an equation of the locus of points equidistant from the parallel lines y 3x 2 and y 3x 6. Solution The locus is a line parallel to the given lines and midway between them. The slope of the locus is 3, the slope of the given lines. The y-intercept of the locus, b, is the average of the y-intercepts of the given lines. b 2 1 (26) 2 5 24 2 5 22 The equation of the locus is y 3x 2. Answer Note that in Example 2, we have used the midpoint of the y-intercepts of the given lines as the y-intercept of the locus. In Exercise 21, you will prove that the midpoint of the segment at which the two given parallel lines intercept the y-axis is the point at which the line equidistant from the given lines intersects the y-axis. Equidistant from Two Intersecting Lines The locus of points equidistant from two intersecting lines is a pair of lines that are perpendicular to each other and bisect the angles at which the given lines intersect. We will consider two special cases. 1. The locus of points equidistant from the axes The x-axis and the y-axis intersect at the origin to form right angles. Therefore, the lines that bisect the angles between the axes will also go through the origin and will form angles measuring 45° with the axes. One bisector will have a positive slope and one will have a negative slope. y O 45° B(a, a) A(a, 0) x B(a, a) 14365C14.pgs 7/10/07 10:00 AM Page 622 622 Locus and Construction y O 45° B(a, a) A(a, 0) x B(a, a) Let A(a, 0) be a point on the x-axis and B be a point on the bisector with a positive slope such that is perpendicular to the x-axis. The triangle formed by A, B, and the origin O is a 45-45 right triangle. Since 45-45 right triangles are isosceles, OA AB |a|, and the coordinates of B are (a, a). The line through (a, a) and the origin is y x. AB Similarly, if B is a point on the bisector with a negative slope, then the coor- dinates of B are (a, a). The line through (a, a) and the origin is y x. We have shown that the lines y x and y x are the locus of points equidistant from the axes. These lines are perpendicular to each other since their slopes, 1 and 1, are negative reciprocals. The locus of points in the coordinate plane equidistant from the axes is the pair of lines y x and y x. 2. The locus of points equidistant from two lines with slopes m and m that intersect at the origin y B y mx y mx A(a, ma) Let O(0, 0) and A(a, ma) be any two points on y mx and B(0, ma) be a point on the y-axis. Under a reflection in the yaxis, the image of O(0, 0) is O(0, 0) and the image of A(a, ma) is A(a, ma). The points O and A are on the line y mx. Therefore, the image of the line y mx is the line y mx since collinearity is preserved under a line reflection. Also, mAOB mAOB since angle measure is preserved under a line reflection. Therefore, the y-axis bisects the angle between the lines y mx and y mx. In a similar way, it can be shown that the x-axis bisects the other pair of angles between y mx and y mx. Therefore, the y-axis, together with the x-axis, is the locus of points equidistant from the lines y mx and y mx. A(a, ma) O x The locus of points in the coordinate plane equidistant from two lines with slopes m and m that intersect at the origin is the pair of lines y 0 and x 0, that is, the x-axis and the y-axis. Exercises Writing About Mathematics 1. In the coordinate plane, are the x-axis and the y-axis the locus of points equidistant from the intersecting lines y x and y x? Justify your answer. 14365C14.pgs 7/10/07 10:00 AM Page 623 Equidistant Lines in Coordinate Geometry 623 2. Ryan said that if the locus of points equidistant from y x 2 and y x 10 is y x 6, then the distance from y x 6 to y x 2 and to y x 10 is 4. Do you agree with Ryan? Justify your answer. Developing Skills In 3–8, find the equation of the locus of points equidistant from each given pair of points. 3. (1, 1) and (9, 1) 6. (0, 6) and (4, 2) 4. (3, 1) and (3, 3) 7. (2, 2) and (0, 2) 5. (0, 2) and (2, 0) 8. (4, 5) and (2, 1) In 9–14, find the equation of the locus of points equidistant from each given pair of parallel lines 9. x 1 and x 7 11. y x 3 and y x 9 13. y 2x 1 and y 2x 5 15. Find the coordinates of the locus of points equidistant from (2, 3) and (4, 3), and 3 units 10. y 0 and y 6 12. y x 2 and y x 6 14. 2x y 7 and y 2x 9 from (1, 3). 16. Find the coordinates of the locus of points equidistant from (2, 5) and (2, 3), and 4 units from (0, 1). Applying Skills 17. a. Find the equation of the locus of points equidistant from (3, 1) and (5, 5). b. Prove that the point (2, 6) is equidistant from the points (3, 1) and (5, 5) by showing that it lies on the line whose equation you wrote in a. c. Prove that the point (2, 6) is equidistant from (3, 1) and (5, 5) by using the distance for- mula. 18. a. Find the equation of the locus of points equidistant from the parallel lines y x 3 and y x 5. b. Show that point P(3, 2) is equidistant from the given parallel lines by showing that it lies on the line whose equation you wrote in a. c. Find the equation of the line that is perpendicular to the given parallel lines through point P(3, 2). d. Find the coordinates of point A at which the line whose equation you wrote in c inter- sects y x 3. e. Find the coordinates of point B at which the line whose equation you wrote in c inter- sects y x 5. f. Use the distance formula to show that PA PB, that is, that P is equidistant from y x 3 and y x 5. 19. Show that the locus of points equidistant from the line y x 1 and the line y x 1 is the y-axis and the line y 1. 14365C14.pgs 7/31/07 1:13 PM Page 624 624 Locus and Construction 20. Show that the locus of points equidistant from the line y 3x 2 and the line y 3x 2 is the y-axis and the line y 2. 21. Prove that the midpoint of the segment at which two given parallel lines intercept the y-axis is the point at which the line equidistant from the given lines intersects the y-axis. That is, if the y-intercept of the first line is b, and the y-intercept of the second line is c, then the y-intercept of the line equidistant from them is . b 1 c 2 a. Find the coordinates of A, the point where the first line intersects the y-axis. b. Find the coordinates of A, the point where the second line intersects the y-axis. c. Show that B d. Let B and B be points on the lines such that M is is the midpoint of M 0, A . AAr (0, bc) point on and lines. Show that ABM ABM. BBr BBr is perpendicular to both e. Using part d, explain why M is the point at which the line equidistant from the given lines intersects the y-axis. 14-6 POINTS EQUIDISTANT FROM A POINT AND A LINE We have seen that a straight line is the locus of points equidistant from two points or from two parallel lines. What is the locus of points equidistant from a given point and a given line? Consider a fixed horizontal g AB line and a fixed po |
int F above the line. The point Pm, that is, the midpoint of the vertical line from F Pn Pn F g AB Locus to , is on the locus of points equidistant from the point and the line. Let Pn be any other point on the locus. As we move to the right or to the left from Pm along the distance from distance from F to Pn is along a slant line. The locus of points is a curve. to Pn continues to be the length of a vertical line, but the g , AB g AB Pm A Pn Pn Pn B 14365C14.pgs 7/10/07 10:00 AM Page 625 Points Equidistant from a Point and a Line 625 Consider a point and a horizontal line that are at a distance d from the origin. • Let F(0, d) be the point and l be the line. The equation of l is y d. • Let P(x, y) be any point equidistant from F and l. • Let M(x, d) be the point at which a vertical line from P intersects l. y F(0, d) O P(x, y) M(x, d) x l The distance from P to M is equal to the distance from P to F. PM 5 PF y 2 (2d) 5 (x 2 0)2 1 (y 2 d)2 " (y 1 d)2 5 x2 1 (y 2 d)2 y2 1 2dy 1 d2 5 x2 1 y2 2 2dy 1 d2 4dy 5 x2 1 4 For instance, if d , that is, if the given point is and the given line is y 5 21 , then the equation of the locus is y x2. Recall that y x2 is the equa4 tion of a parabola whose turning point is the origin and whose axis of symmetry is the y-axis. 0, 1 4 B A Recall that under the translation , the image of 4dy x2 is Th, k 4d(y k) (x h)2. For example, if the fixed point is F(2, 1) and the fixed line is y 1, then (h, k) 5 (2, 0) . d 1 and the turning point of the parabola is (2, 0), so Therefore, the equation of the parabola is 4(1)(y 0) (x 2)2 or 4y x2 4x 4. This equation can also be written as y 1 4x2 2 x 1 1 . For any horizontal line and any point not on the line, the equation of the locus of points equidistant from the point and the line is a parabola whose equation can be written as y ax2 bx c. If the given point is above the line, the coefficient a is positive and the parabola opens upward. If the given point is below the line, the coefficient a is negative and the parabola opens downward. The axis of symmetry is a vertical line whose equation is x . Since the turning point is on the axis of symmetry, its x-coordinate is 2b 2a .2b 2a 14365C14.pgs 7/10/07 10:00 AM Page 626 626 Locus and Construction EXAMPLE 1 a. Draw the graph of y x2 4x 1 from x 1 to x 5. b. Write the coordinates of the turning point. c. Write an equation of the axis of symmetry. d. What are the coordinates of the fixed point and the fixed line for this parabola? Solution a. (1) Make a table of values using integral values of x from x 1 to x 5. (2) Plot the points whose coordinates are given in the table and draw a smooth curve through them. x 1 0 1 2 3 4 5 x2 4x 12 1 16 16 1 25 20 . From the graph or from the table, the coordinates of the turning point appear to be (2, 5). We can verify this algebraically: x 5 2b 2a 2(24) 2(1) 5 5 2 y 5 x2 2 4x 2 1 5 (2)2 2 4(2) 2 1 5 25 c. The axis of symmetry is the vertical line through the turning point, x 2. d. Note that the turning point of the parabola is (2, 5). When the turning point of the parabola y x2 has been moved 2 units to the right and 5 units down, the equation becomes the equation of the graph that we drew: y 2 (25) 5 (x 2 2)2 y 1 5 5 x2 2 4x 1 4 y 5 x2 2 4x 2 1 The turning point of the parabola is the midpoint of the perpendicular segment from the fixed point to the fixed line. Since the coefficient of y in the equation of the parabola is 1, 4d 1 or . The parabola opens d 5 1 4 14365C14.pgs 7/10/07 10:00 AM Page 627 Points Equidistant from a Point and a Line 627 upward so the fixed point is unit above the turning point and the fixed line is unit below the turning point. The coordinates of the fixed point are and the equation of the fixed line is y . 251 4 1 4 2, 243 4 B A 1 4 EXAMPLE 2 a. Draw the graph of y x2 2x 8 from x 4 to x 2. b. Write the coordinates of the turning point. c. Write an equation of the axis of symmetry. Solution a. (1) Make a table of values using integral values of x from x 4 to x 2. (2) Plot the points whose coordinates are given in the table and draw a smooth curve through them. x 4 3 2 1 0 1 2 x2 2x 8 16 . From the graph or from the table, the coordinates of the turning point appear to be (1, 9). We can verify this algebraically: x 5 2b 2a 2(22) 2(21) 5 5 21 y 5 2x2 2 2x 1 8 5 2(21)2 2 2(21) 1 8 5 9 c. The axis of symmetry is the vertical line through the turning point, x 1. Here the parabola y x2 has been reflected in the x-axis so that the equation becomes y x2. Then that parabola has been moved 1 unit to the left and 9 units up so that the equation becomes (y 9) (x (1))2 or y 9 x2 2x 1, which can be written as y x2 2x 8 or y x2 2x 8. 14365C14.pgs 7/10/07 10:00 AM Page 628 628 Locus and Construction EXAMPLE 3 A parabola is equidistant from a given point and a line. How does the turning point of the parabola relate to the given point and line? Solution The x-coordinate of the turning point is the same as the x-coordinate of the given point and is halfway between the given point and the line. EXAMPLE 4 Solve the following system of equations graphically and check: y x2 2x 1 y x 3 Solution (1) Make a table of values using at least three integral values of x that are less than that of the turning point and three that are greater. The x-coordinate of the turning point is: 2b 2a 5 2(22) 2(1) 5 2 2 5 1 (2) Plot the points whose coordinates are given in the table and draw a smooth curve through them. x 2 1 0 1 2 3 4 x2 2x 16 3) On the same set of axes, sketch the graph of y x 3. 21 1 . Move 1 unit down The y-intercept is 3. Start at the point (0, 3). The slope is 1 or and 1 unit to the right to find a second point of the line. From this point, again move 1 unit down and 1 unit to the right to find a third point. Draw a line through these three points2, 1) x (1, 4) y = x 3 1 O 1 14365C14.pgs 7/10/07 10:00 AM Page 629 Points Equidistant from a Point and a Line 629 (4) Read the coordinates of the points of intersection from the graph. The common solutions are (1, 4) and (2, 1). Answer (1, 4) and (2, 1) or x 1, y 4 and x 2, y 1 Exercises Writing About Mathematics 1. Luis said that the solutions to the equation x2 2x 8 0 are the x-coordinates of the points at which the graph of y x2 2x 8 intersects the y-axis. Do you agree with Luis? Explain why or why not. 2. Amanda said that if the turning point of a parabola is (1, 0), then the x-axis is tangent to the parabola. Do you agree with Amanda? Explain why or why not. Developing Skills In 3–8, find the coordinates of the turning point and the equation of the axis of symmetry of each parabola. 3. y x2 6x 1 6. y x2 2x 5 4. y x2 2x 3 7. y x2 8x 4 5. y x2 4x 1 8. y x2 5x 2 In 9–16, find the common solution of each system of equations graphically and check your solution. 9. y x2 2x 2 y x 2 11. y x2 4x 3 y x 1 13. y x2 4x 2 y 2x 3 15. y x2 6x 5 y 7 2x 10. y x2 1 x y 1 12. y x2 2x 3 y 1 x 14. y x2 2x 2 y 2x 2 16. y 2x x2 y 2x 4 In 17–20 write the equation of the parabola that is the locus of points equidistant from each given point and line. F y 5 21 0, 1 17. 4 B 4 19. F(0, 2) and y 2 and A F 0, 21 18. 4 B 20. F(3, 3) and y 3 and y 5 1 4 A 14365C14.pgs 8/2/07 6:02 PM Page 630 630 Locus and Construction Hands-On Activity If the graph of an equation is moved h units in the horizontal direction and k units in the vertical direction, then x is replaced by x h and y is replaced by y k in the given equation. 1. The turning point of the parabola y ax2 is (0, 0). If the parabola y ax2 is moved so that the coordinates of the turning point are (3, 5), what is the equation of the parabola? 2. If the parabola y ax2 is moved so that the coordinates of the turning point are (h, k), what is the equation of the parabola? CHAPTER SUMMARY Loci • A locus of points is the set of all points, and only those points, that satisfy a given condition. • The locus of points equidistant from two fixed points that are the end- points of a segment is the perpendicular bisector of the segment. • The locus of points equidistant from two intersecting lines is a pair of lines that bisect the angles formed by the intersecting lines. • The locus of points equidistant from two parallel lines is a third line, paral- lel to the given lines and midway between them. • The locus of points at a fixed distance from a line is a pair of lines, each parallel to the given line and at the fixed distance from the given line. • The locus of points at a fixed distance from a fixed point is a circle whose center is the fixed point and whose radius is the fixed distance. • The locus of points equidistant from a fixed point and a line is a parabola. Loci in the Coordinate Plane • The locus of points in the coordinate plane r units from (h, k) is the circle whose equation is (x h)2 (y k)2 r2. • The locus of points d units from the horizontal line y a is the pair of lines y a d and y a d. • The locus of points d units from the vertical line x a is the pair of lines x a d and x a d. • The locus of points equidistant from A(a, c) and B(b, d) is a line perpen- dicular to AB at its midpoint • The locus of points equidistant from the axes is the pair of lines y x and y x. • The locus of points equidistant from the lines y mx and y mx is the pair of lines y 0 and x 0, that is, the x-axis and the y-axis. • The locus of points equidistant from (h, k d) and y k d is the parabola whose equation is 4d(y k) (x h)2. 14365C14.pgs 7/10/07 10:00 AM Page 631 Review Exercises 631 VOCABULARY 14-2 Locus • Concentric circles 14-3 Compound locus REVIEW EXERCISES 1. Construct: a. a right angle. b. an angle whose measure is 45°. c. parallelogram ABCD with AB a, BC b and mB 45. a b 2. Draw PQ . Construct S on PQ such that PS : SQ 2 : 3. In 3–6, sketch and describe each locus. 3. Equidistant from two points that are 6 centimeters apart. 4. Four centimeters from A and equidistant from A and B, the endpoints of a line segment that is 6 centimeters long. 5. Equidistant from the endpoints of a line segment tha |
t is 6 centimeters long and 2 centimeters from the midpoint of the segment. 6. Equidistant from parallel lines that are 5 inches apart and 4 inches from a point on one of the given lines. In 7–12, sketch the locus of points on graph paper and write the equation or equations of the locus. 7. 3 units from (1, 2). 8. 2 units from (2, 4) and 2 units from x 2. 9. Equidistant from x 1 and x 5. 10. Equidistant from y 2 and y 8. 11. Equidistant from y 2x and y 2x. 12. Equidistant from (1, 3) and (3, 1). 13. Find the coordinates of the points on the line x y 7 that are 5 units from the origin. 14. a. Draw the graph of y x2 4x 1. b. On the same set of axes, draw the graph of y 2x 2. c. What are the coordinates of the points of intersection of the graphs drawn in a and b? 14365C14.pgs 8/2/07 6:02 PM Page 632 632 Locus and Construction In 15–18, solve each system of equations graphically. 15. x2 y2 25 y x 1 17. y x2 2x 1 y x 1 x y 6 x 2 18. y x2 4x 2 16. (x 2)2 (y 1)2 4 19. A field is rectangular in shape and measures 80 feet by 120 feet. How many points are equidistant from any three sides of the field? (Hint: Sketch the locus of points equidistant from each pair of sides of the field.) 20. The coordinates of the vertices of isosceles trapezoid ABCD are A(0, 0), B(6, 0), C(4, 4), and D(2, 4). What are the coordinates of the locus of points equidistant from the vertices of the trapezoid? (Hint: Sketch the locus of points equidistant from each pair of vertices.) Exploration An ellipse is the locus of points such that the sum of the distances from two fixed points F1 and F2 is a constant, k. Use the following procedure to create an ellipse. You will need a piece of string, a piece of thick cardboard, and two thumbtacks. STEP 1. Place two thumbtacks in the cardboard separated by a distance that is less than the length of the string. Call the thumbtacks F1 and F2. Attach one end of the string to F1 and the other to F2. The length of the string represents the sum of the distances from a point on the locus to the fixed points. F1 F2 P STEP 2. Place your pencil in the loop of string and pull the string taut to locate some point P. Keeping the string taut, slowly trace your pencil around the fixed points until you have created a closed figure. a. Prove that the closed figure you created is an ellipse. (Hint: Let k be the length of the string.) b. If F1 and F2 move closer and closer together, how is the shape of the ellipse affected? A hyperbola is the locus of points such that the difference of the distances from fixed points F1 and F2 is a constant, k. c. Explain why a hyperbola is not a closed figure. P F1 PF2 PF1 = k F2 14365C14.pgs 7/10/07 10:00 AM Page 633 CUMULATIVE REVIEW Part I Cumulative Review 633 Chapters 1–14 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following has a midpoint? g AB (1) (2) AB h AB (3) (4) ABC 2. What is the inverse of the statement “If two lines segments are congruent, then they have equal measures?” (1) If two line segments are not congruent, then they do not have equal measures. (2) If two line segments do not have equal measures, then they are not congruent. (3) If two line segments have equal measures, then they are congruent. (4) Two line segments are not congruent if they have unequal measures. AD 3. In triangle ABC, is the altitude to BC then which of the following may be false? (1) AB AC (2) BD CD (3) AB AD (4) mB mC . If D is the midpoint of BC , 4. If two angles of a triangle are congruent and complementary, then the tri- angle is (1) isosceles and right (2) scalene and right (3) isosceles and obtuse (4) scalene and acute 5. The equation of a line that is perpendicular to the line x 3y 6 is (1) y 3x 1 (2) y 3x 1 1 21 (3) y 3x 1 (4) y 1 3x 6. Under a line reflection, which of the following is not preserved? (1) angle measure (2) collinearity (3) orientation (4) midpoint 7. Which of the following is not sufficient to prove that quadrilateral ABCD is a parallelogram? AB > CD and (1) AB > CD and (2) BC > DA AB CD (3) (4) AC AC BD bisect each other and ⊥ BD 8. A prism with bases that are equilateral triangles has a height of 12.0 cen- timeters. If the length of each side of a base is 8.00 centimeters, what is the number of square centimeters in the lateral area of the prism? (1) 48.0 (2) 96.0 (3) 288 (4) 384 14365C14.pgs 7/10/07 10:00 AM Page 634 634 Locus and Construction 9. The altitude to the hypotenuse of a right triangle divides the hypotenuse into segments of lengths 4 and 45. The length of the shorter leg is 10. Triangle ABC is inscribed in circle O. If m (1) 6 5 " (2) 14 is the measure of ABC? (2) 55 (1) 45 (3) 2,009 110 and m (4) BCX 2,041 " 90, what " ABX (3) 80 (4) 110 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit 11. D is a point on side and E is a point on side . If AD 6, DB 9, and AC 20, find AE and EC. AB AC DE BC of ABC such that 12. A pile of gravel is in the form of a cone. The circumference of the pile of gravel is 75 feet and its height is 12 feet. How many cubic feet of gravel does the pile contain? Give your answer to the nearest hundred cubic feet. Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. a. On graph paper, sketch the graph of y x2 x 2. b. On the same set of axes, sketch the graph of y x 5. c. What are the common solutions of the equations y x2 x 2 and y x 5? 14. a. Show that the line whose equation is x y 4 intersects the circle whose equation is x2 y2 8 in exactly one point and is therefore tangent to the circle. b. Show that the radius to the point of tangency is perpendicular to the tangent. 14365C14.pgs 7/10/07 10:00 AM Page 635 Cumulative Review 635 Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. a. The figure below shows the construction of the perpendicular bisector, g DE , of segment AB . Identify all congruent segments and angles in the g DE is the perpen- construction and state the theorems that prove that dicular bisector of . AB E A M D B b. The figure below shows the construction of FDE congruent to CAB. Identify all congruent lines and angles in the construction and state the theorems that prove that FDE CAB. C B A F E D 16. a. Quadrilateral ABCD is inscribed in circle O, AB CD , and AC is a diameter. Prove that ABCD is a rectangle. b. In quadrilateral ABCD, diagonals AC and CDE are similar but not congruent, prove that the quadrilateral is a trapezoid. intersect at E. If ABE and BD c. Triangle ABC is equilateral. From D, the midpoint of AB , and to F, the midpoint of , a line BC segment is drawn to E, the midpoint of AC . Prove that DECF is a rhombus. 14365Index.pgs 7/13/07 10:19 AM Page 636 INDEX A AA triangle similarity (AA), 490. See also Angle-angle triangle similarity AAS triangle congruence, 352. See also Angle-angle-side triangle congruence Abscissa, 210 Absolute value, 7–8 Acute angle, 17 Acute triangle, 25, 105 Addition of angles, 20–21 associative property of, 4–5 closure property of, 4 commutative property of, 4 distributive property of multiplication over, 5 inequality postulates involving, 267–268 of line segments, 12–13 Addition postulate, 119–120, 267–268 Additive identity, 5 Additive inverses, 5 Adjacent angles, 145 Adjacent interior angle, 277 Adjacent sides, 380 Adjacent vertices, 368, 380 Algebra, Boolean, 34 Alternate exterior angles, 330 Alternate interior angles, 330 Altitude(s). See also Height of cone, 456 of cylinder, 453 of pyramid, 449 of rectangle, 389 of triangle, 175 concurrence of, 317–320 Analytic geometry, 209 Angle(s), 15 acute, 17, 105 addition of, 20–21 adjacent, 145 alternate exterior, 330 alternate interior, 330 base, 25 bisectors of, 20, 103, 137, 199 central, 536–537 classifying according to their measures, 17 classifying triangles according to, 25–26 complementary, 145 congruent, 19, 103 construction of, congruent to a given angle, 197 corresponding, 330 defined, 15 definitions involving pairs of, 145–146 dihedral, 424 exterior, of, 16, 330 polygon, 276–277, 369 triangle, 277–279 formed by tangent and chord, 567–568 formed by tangent and secant, 568–571 formed by two intersecting chords, 568 inequalities involving, for triangle, 281–284 inscribed, and their measures, 552–555 interior, of, 16, 330 polygon, 368–369 linear pair of, 148 measure of, 16 naming of, 16 obtuse, 17 plane, 424 postulates of, 135–138 proving theorems about, 144–145 right, 17 sides of, 15 straight, 16, 17 subtraction of, 20–21 sum of measures of, of triangle, 347–350 supplementary, 145–146, 148–149 theorems involving pairs of, 146–148 trisection of, 604 using congruent triangles to prove congruent, 178–179 vertex of, 15, 25 vertical, 149 Angle addition postulate, 120 Angle-angle-side triangle congruence, 352. See also AAS triangle congruence Angle-angle triangle similarity, 490. See also AA triangle similarity Angle bisector(s) concurrence of, of triangle, 364–365 of triangle, 176 Angle measure, preservation of under dilation, 497 under glide reflection, 244 under line reflection, 217 under point reflection, 228 under rotation about a fixed point, 239 under translation, 234 Angle-side-angle triangle congruence, 162. See also ASA triangle congruence Antecedent, 55 Apothem of circle, |
547 of polygon, 567 Arc(s) of circle, 537 congruent, 538 degree measure of, 537–538 intercepted, 537 major, 537 minor, 537 types of, 537 Arc addition postulate, 539 Archimedes, 535 Area of a polygon, 409–410 Argument, valid, 75, 109 Arrowheads, 2 ASA triangle congruence, 162. See also Angle-side-angle triangle congruence Associative property of addition, 4–5 of multiplication, 5 Axiom, 93, 109–110 Axis of symmetry, 219, 625 B Base(s) of cone, 456 of cylinder, 453 of polyhedron, 440 of rectangle, 389 of regular pyramid, 450 of trapezoid, 402 Base angles of isosceles trapezoid, 403 of isosceles triangle, 25 lower, 404 upper, 404 Basic constructions, 196–202 Baudha¯ yana, 474 Betweenness, 8 14365Index.pgs 7/13/07 10:19 AM Page 637 Biconditional(s), 69–73 applications of, 70–73 definitions as, 97–99 Bisector of angle, 20, 103, 137 of line segment, 12 Bolyai, János, 379 Boole, George, 34 Boolean algebra, 34 C Cartesian coordinates, 209 Cavalieri, Bonaventura, 419 Cavalieri’s Principle, 419 Center of circle, 460, 536 chords equidistant from, 546–550 of regular polygon, 450 of sphere, 459 Center-radius equation of a circle, 582 Centimeter, cubic, 446 Central angle of circle, 536–537 Centroid, 506 Chord(s) angles formed by, 567–568 of circle, 543 equidistant from center of circle, 546–550 segments formed by two intersecting, 575 Circle(s), 460 apothem of, 547 arc of, 537 center of, 460, 536 central angle of, 536–537 chords of, 543, 546–550 circumference of, 535 concentric, 612 congruent, 538 in coordinate plane, 581–584 defined, 536 diameter of, 543 exterior of, 536 inscribed angles of, 552–553 interior of, 536 polygons circumscribed about, 563–564 polygons inscribed in, 550–551 radius of, 536 secant of, 558 squaring of, 604 tangent to, 558 Circular cylinder, 453 surface area of, 454 volume of, 454 Circumcenter, 194 Circumference of circle, 535 Circumscribed circle, 550 Circumscribed polygon, 563 Closed sentence, 37 Closure property of addition, 4 of multiplication, 4 Collinear set of points, 7 Collinearity, preservation of under line reflection, 217 under translation, 234 Common external tangent, 560 Common internal tangent, 559 Common tangent, 559–561 Commutative property of addition, 4 of multiplication, 4 Compass, 196, 604 Complementary angles, 145 Complements, 145 Composition of transformations, 243, 251–252 Compound locus, 614 Compound sentence(s), 42, 53 conjunctions as, 42–46 Compound statement(s), 42, 53 conditionals as, 4253–57 disjunctions as, 48–50 Concave polygon, 368 Concentric circles, 612 Conclusion(s), 55 drawing, 80–83 Concurrence, 193 of altitudes of triangle, 317–320 of angle bisectors of triangle, 364–365 Conditional(s), 53–57, 60, 335 contrapositive of, 64–65 converse of, 63–64 false, 63, 65 hidden, 55–57, 98 inverse of, 61–62 parts of, 55 as they relate to proof, 138–139 true, 63, 64 truth value of, 54–55 Cone(s), 456–458 altitude of, 456 base of, 456 frustum of, 459 height of, 456 right circular, 456 slant height of, 456 surface area of, 457–458 volume of, 457–458 Congruence equivalence relations of, 156–157 of triangles, 134, 174–203 Congruent angles, 19, 103 Congruent arcs, 538 Congruent circles, 538 Congruent polygons, 154–155 corresponding parts of, 155 Congruent segments, 9 Congruent triangles, 155–156 in proving line segments congruent and angles congruent, 178–179 using two pairs of, 186 Conjecture, 95, 97 Conjunct, 42 Conjunction, 42–46 Consecutive angles of polygon, 368 of quadrilateral, 380 Consecutive sides of quadrilateral, 380 Consecutive vertices of polygon, 368 of quadrilateral, 380 Consequent, 55 Constant of dilation, k, 501 Constant of proportionality, 486 Index 637 Constructions of angle bisector, 199 of congruent angles, 197 of congruent segments, 196–197 locus in, 609–611 of midpoint, 198 of parallel lines, 605–606 of perpendicular bisector, 198 of perpendicular through a point not on the line, 201 of perpendicular through a point on the line, 200 Contraction, 496 Contrapositive, 64–65 Converse, 63–64 of Isosceles Triangle Theorem, 357–360 of Pythagorean Theorem, 516–517 Converse statement, 335 Convex polygon, 368 Coordinate geometry, 209, 290 equidistant lines in, 619–622 points at fixed distance in, 616–618 Coordinate of point, 4, 7 Coordinate plane, 210 circles in, 581–584 dilations in, 247–248 line reflections in, 222–225 locating points in, 210–211 parallel lines in, 342–344 point reflections in, 227–231 rotations in, 238–242 secants in, 588–591 tangents in, 588 translations in, 232–235 preservation of angle measure, 232 preservation of collinearity, 232 preservation of distance, 232 preservation of midpoint, 232 Coordinate proof, 313–314 of general theorems, 313 for special cases, 313 Coordinates, 210 of point in plane, 210–213 rectangular, 211 Coplanar lines, 329 Corollary, 182 Corresponding angles, 155, 330 Corresponding parts of congruent polygons, 155 Corresponding sides, 155 Counterexample, 95 Cube, 445 duplication of, 604 Cubic centimeter, 446 Cylinder, 453–455 altitude of, 453 bases of, 453 circular, 453 height of, 453 lateral surface of, 453 right circular, 453 D Decagon, 367 Deductive reasoning, 97, 100–103, 150–151 Definition(s), 7, 93 as biconditionals, 97–99 qualities of good, 7 14365Index.pgs 7/13/07 10:19 AM Page 638 638 Index Definition(s) cont. using, in proofs, 141–142 writing as conditionals, 98 Degree measure of angle, 16 of arc, 537–538 DeMorgan, Augustus, 34 DeMorgan’s Laws, 34 Descartes, René, 209, 290 Detachment, law of, 75, 101, 105 Diagonal of polygon, 368 of quadrilateral, 380 Diagram(s) tree, 42 using, in geometry, 26–27 Diameter, 543 Dihedral angle, 424 Dilation(s), 495–499 in coordinate plane, 247–248 defined, 247 preservation of angle measure under, 496–497 preservation of collinearity under, 498 preservation of midpoint under, 497 Direct isometry, 252–253 Direct proof, 105–106 Disjunct, 48 Disjunction, 48–50, 76 Disjunctive inference, law of, 76–78 Distance, between two parallel lines, 383 between two planes, 437 between two points, 7–8 from a point to a line, 20 preservation of, under glide reflection, 244 preservation of, under point reflection, 228 preservation of, under rotation about a fixed point, 239 preservation of, under translation, 234 Distance formula, 521–522, 522 Distance postulate, 136 Distributive property, 5 Divide and average method, 174 Division postulate, 124, 270–271 Dk, 247, 496 Domain, 36, 250 E Edge of polyhedron, 440 Elements (Euclid), 1, 262, 474 Endpoint, of ray, 15 Enlargement, 496 Epicureans, 262 Equality reflexive property of, 110–111 symmetric property of, 111 transitive property of, 111, 263–264 Equation(s) of line, 295–299 solving, with biconditionals, 70–71 Equiangular triangle, 25 Equidistant, 191 Equidistant lines in coordinate geometry, 619–622 Equilateral triangle(s), 24, 25, 181–183 properties of, 183 Equivalence relation, 111 of congruence, 156–157 of similarity, 487–488 Eratosthenes, 1 Euclid, 1, 93, 134, 262, 328, 379, 535 parallel postulate of, 328 Euclid Freed of Every Flaw, 379 Exclusive or, 51 Exterior angle(s), 330 alternate, 330 of polygon, 276–277, 369–371 of triangle, 277–279 Exterior angle inequality theorem, 277 Exterior angle theorem, 349 Exterior of angle, 16 Exterior of circle, 536 External segment, 576 Extremes, 476 F Face(s) of polyhedron, 440 Fermat, Pierre de, 209 Fixed point, 214 Foot of altitude, 525 Foot of perpendicular, 20 Formula(s) angles, central, 555 inscribed, 555 formed by tangents, chords, and secants, 571–572 of polygons, 369 area of a rectangle, 409 circle, 582 distance, 522 Heron’s, 174 lateral area, of cone, 457, 467 of cylinder, 454, 467 of prism, 442, 467 of pyramid, 467 midpoint, 304 point-slope, 297 segments formed by tangents, chords, and secants, 579 slope, 292, 297 surface area, of cone, 457 of cylinder, 454, 467 of prism, 442, 467 of pyramid, 467 of sphere, 462, 467 volume, of cone, 457, 467 of cylinder, 454, 467 of prism, 446, 467 of pyramid, 449, 467 of sphere, 462, 467 Foundations of Geometry (Hilbert), 93 45-45-degree right triangle, 517–518 Frustum of cone, 459 Function(s) defined, 250 transformations as, 250–254 G Galileo, 419 Generalization, 94 General quadrilateral, 380 Geometric constructions, 196. See also Constuctions Geometric inequalities, 262–285 basic inequality postulates, 263–265 inequalities involving lengths of the sides of a triangle, 273–274 inequalities involving sides and angles of a triangle, 281–284 inequality involving an exterior angle of a triangle, 276–279 inequality postulates involving addition and subtraction, 267–268 inequality postulates involving multiplication and division, 270–271 Geometric mean, 478 Geometry analytic, 209 coordinate, 209, 290 deductive reasoning, 100–103 defined, 2 definitions as biconditionals, 97–99 inductive reasoning, 94–97 non-Euclidean, 376 spherical, 32 proving statements in, 93–130 addition and subtraction postulates, 118–122 direct proofs, 105–108 indirect proofs, 105–108 multiplication and division postulates, 124–126 postulates, theorems, and proof, 109–115 substitution postulate, 115–117 solid, 420 using diagrams in, 26–27 using logic to form proof, 100–103 Glide reflection, 243–245 Graphing polygons, 212–213 Graphs, 4 Great circle of sphere, 460, 461 H Half-line, 14–15 Heath, Thomas L., 1 Height. See also Altitude of cone, 456 of cylinder, 453 of prism, 440 of pyramid, 449 Heron of Alexandria, 174 Heron’s formula, 174 Hexagon, 367 Hidden conditional, 55–57, 98 Hilbert, David, 93 HL triangle congruence theorem, 362–365 Hypotenuse, 26 Hypotenuse-leg triangle congruence theorem, 362–365. See also HL triangle congruence theorem Hypothesis, 55 I Identity additive, 5 14365Index.pgs 7/13/07 10:19 AM Page 639 multiplicative, 5 Identity property, 5 If p then q, 53 Image, 214, 215 Incenter, 364 Included angle, 24 Included side, 24 Inclusive or, 51 Incomplete sentences, 35 Indirect proof, 105–108, 283, 309, 331, 336, 425, 429, 431, 434, 436, 559 Inductive reasoning, 94–97 Inequality geometric, 262–28 |
5 involving exterior angle of triangle, 276–279 involving lengths of sides of triangle, 273–274 Lateral sides of prism, 440 Lateral surface of cylinder, 453 Law(s) DeMorgan’s, 34 of Detachment, 75, 101, 105 of Disjunctive Inference, 76 of logic, 35, 74–78 Leg(s) proving right triangles congruent by hypotenuse, 362–365 of right triangle, 26 of trapezoid, 402 Leibniz, Gottfried, 34, 290 Length of line segment, 9 Line(s), 1, 2, 7, 420–422 coplanar, 329 equation of, 295–299 equidistant, in coordinate geometry, involving sides and angles of triangle, 281–284 transitive property of, 264 Inequality postulate(s), 263, 265 619–622 number, 3–4 order of points on, 8 parallel, 421, 605–606 involving addition and subtraction, in coordinate plate, 342–344 267–268 involving multiplication and division, 270–271 relating whole quantity and its parts, 263 parallel to a plane, 433 perpendicular, 20, 100, 149 methods of proving, 193 planes and, 423–431 slopes of, 307–310 transitive property, 263–264 points equidistant from point and, Inscribed angle of circle, 552–553 measures of, 552–555 Inscribed circle, 563 Inscribed polygon, 550 Intercepted arc, 537 Interior angle(s), 330 alternate, 330 on the same side of the transversal, 330 of polygon, 368–369 Interior of angle, 16 Interior of circle, 536 Intersecting lines, equidistant from two, 621–622 624–629 postulates of, 135–138 skew, 421–422 slope of, 291–294 straight, 1, 2 Linear pair of angles, 148 Line of reflection, 214 Line reflection(s), 214–220 in coordinate plane, 222–225 preservation of angle measure under, 217 preservation of collinearity under, 217 preservation of distance under, 215 preservation of midpoint under, 217 Intersection of perpendicular bisectors Line segment(s), 9 of sides of triangle, 193–195 Inverse of a conditional, 61 Inverse property, 5 Inverses, 61–62 additive, 5 multiplicative, 5 Isometry, 244 direct, 252–253 opposite, 253 Isosceles quadrilateral, 379 Isosceles trapezoid(s) base angles of, 403, 404 properties of, 403 proving that quadrilateral is, 403–407 Isosceles triangle(s), 24, 25, 181–183, 451 base angle of, 25 parts of, 25 vertex angle of, 25 Isosceles triangle theorem, 181 converse of, 357–360 L Lateral area of prism, 442 Lateral edges of prism, 440, 441 addition of, 12–13 associated with triangles, 175–177 bisector of, 12 congruent, 9 construction of congruent segment, 196–197 construction of perpendicular bisector and midpoint, 198 divided proportionally, 482 formed by intersecting secants, 576–578 formed by tangent intersecting secant, 575–576 formed by two intersecting chords, 575 length or measure of, 9 on a line, projection of, 510 methods of proving perpendicular, 193 midpoint of, 11–12, 300–305 perpendicular bisector of, 191–195 postulates of, 135–138 proportions involving, 480–484 subtraction of, 12–13 tangent, 561–563 Index 639 using congruent triangles to prove congruent, 178–179 Line symmetry, 218–220 Lobachevsky, Nicolai, 379 Locus, 604–630 compound, 614 discovering, 610–611 equidistant from two intersecting lines, 613 equidistant from two parallel lines, 614 equidistant from two points, 613 fixed distance from line, 614 fixed distance from point, 614 meaning of, 609–611 Logic, 34–92 biconditionals in, 69–73 conditionals in, 53–57 conjunctions in, 42–46 contrapositive in, 64–65 converse in, 63–64 defined, 35 disjunctions in, 48–50 drawing conclusions in, 80–83 equivalents in, 65–67 in forming geometry proof, 100–103 inverse in, 61–62 law(s) of, 35, 74–78 detachment, 75 disjunctive inference, 76–78 negations in, 38 nonmathematical sentences and phrases in, 35–36 open sentences in, 36–37 sentences and their truth values in, 35 statements and symbols in, 37 symbols in, 38–39 two uses of the word or, 51 Logical equivalents, 65–67 Lower base angles of isosceles trapezoid, 404 M Major arc, 537 Mathematical sentences, 35 Mean(s), 476 geometric, 478 Mean proportional, 478–479, 512 Measure of angle, 16, 17, 552–555 formed by tangent intersecting a secant, 568 formed by two intersecting chords, 568 formed by two intersecting secants, 569 formed by two intersecting tangents, 569 of arc, 537 of central angle of a circle, 537 of inscribed angle of a circle, 553 of segments formed by intersecting chords, 575 intersecting secants, 576–577 tangent intersecting a secant, 575–576 Median of trapezoid, 405 14365Index.pgs 7/13/07 10:19 AM Page 640 640 Index Median cont. of triangle, 175–176 Metrica (Heron of Alexandria), 174 Midpoint of line segment, 11–12, 300–305 preservation of under dilation, 497 under glide reflection, 244 under line reflection, 217 under point reflection, 228 under rotation about a fixed point, 239 under translation, 234 Midpoint formula, 304 Midsegment, 346–347, 480 Midsegment theorem, 480 Minor arc, 537 Multiplication associative property of, 5 closure property of, 4 commutative property of, 4 distributive property of, 5 inequality postulates involving, 270–271 Multiplication postulate, 124, 270–271 Multiplication property of zero, 5 Multiplicative identity, 5 Multiplicative inverses, 5 N Negation, 38 Negative rotation, 239 Negative slope, 293 Newton, Isaac, 290 n-gon, 367 Noncollinear set of points, 7 Nonadjacent interior angle, 277 Nonmathematical sentences, phrases and, 35–36 No slope, 293 Number(s) properties of system, 110 rational, 3 real, 3 Number line, 3–4 distance between two points on the real, 8 Numerical operation, 4 O Obtuse angle, 17 Obtuse triangle, 25 Octagon, 367 One-to-one correspondence, 154 Open sentence, 36–37 Opposite angle, 24 Opposite isometry, 253 Opposite rays, 15 Opposite side, 24 Or exclusive, 51 inclusive, 51 Order of points on a line, 8–9 Ordered pair, 210 Ordinate, 210 Orientation, 252–254 Origin, 210 Orthocenter, 319 P Parabola, turning point of, 626 Paragraph proof, 101–102 Parallelepiped(s), 441, 442 rectangular, 442 volume of, 446 Parallel lines, 328–372, 421 constructing, 605–606 in coordinate plane, 342–344 defined, 329 equidistant from two, 621 methods of proving, 333 planes and, 433–439 properties of, 335–340 proving, 329–333 in space, 421 transversals and, 330–332 Parallelogram(s), 380–383 proof of quadrilateral as, 385–387 properties of, 383 Parallel planes, 438 Partition postulate, 118–119 Pentagon, 23, 367 Perpendicular bisector construction of, 198 of line segment, 191–195 Perpendicular Bisector Concurrence Theorem, 193 Perpendicular lines, 20, 100, 149 construction of through given point on line, 200 through point not on the given line, 201 methods of proving, 193 planes and, 423–431 slopes of, 307–310 Phrase(s), 35 nonmathematical sentences and, 35–36 Plane(s), 420–422. See also Coordinate plane coordinates of point in, 210–213 defined, 2 distance between two, 437 naming of, 2 parallel, 438 symmetry, 464 Plane angle, 424 Playfair, John, 328 Playfair’s postulate, 328 Point(s), 1, 2, 7, 420–422 collinear, 7 coordinate of, 4, 7 distance between two, 7–8 equidistant from two, 619–620 finding coordinates of, in plane, 211 fixed, 214 at fixed distance in coordinate geometry, 616–618 locating, in coordinate plane, 210–211 noncollinear set of, 7 order of, on a line, 8 points equidistant from line and, 624–629 projection of, on line, 510 Point reflection, 227–229 in coordinate plane, 229–231 properties of, 228–229 Point reflections in the coordinate plane, 247–248 preservation of angle measure under, 228 preservation of collinearity under, 228 preservation of distance under, 228 preservation of midpoint under, 228 Points equidistant from point and line, 624–629 Point-slope form of equation of line, 297 Point symmetry, 229 Polygon(s), 23, 367 areas of, 409–410 circumscribed about a circle, 563–564 concave, 368 convex, 368 diagonals of, 368 exterior angles of, 276–277, 369–371 graphing, 212–213 inscribed in circle, 550–551 interior angles of, 368–369 regular, 369 sides of, 23 similar, 486–488 Polyhedron(s), 440–442 bases of, 440 edges of, 440 faces of, 440 vertices of, 440 Positive rotation, 239 Positive slope, 292 Postulate(s), 93, 109–110 addition, 119–120, 267–268, 539 basic inequality, 263–265 of the coordinate plane, 210 division, 124, 270–271 Euclid’s parallel, 328 first, in proving statements, 110–113 of lines, line segments, and angles, 135–138 multiplication, 124, 270–271 partition, 118–119 Playfair’s, 328 powers, 124 relating a whole quantity and its parts, 263 roots, 125–126 of similarity, 490 substitution, 115–117, 264 subtraction, 121–122, 267–268 trichotomy, 264–265 using, in proofs, 141–142 Powers postulate, 124 Preimage, 214, 215 Premise, 55, 75 Prism, 440 altitude of, 440 height of, 440 lateral area of, 442 lateral edges of, 440, 441 lateral sides of, 440 right, 441 surface area of, 440–443 total surface area of, 442 volume of, 446–447 14365Index.pgs 7/13/07 10:19 AM Page 641 Projection of point on a line, 510 of segment on a line, 510 Proof(s), 100 conditional statements as they relate to, 138–139 by contradiction, 105 direct, 105–106 indirect, 106–108, 283, 309, 331, 336, 425, 429, 431, 434, 436, 559 paragraph, 101–102 transformational, 308 two-column, 101 using postulates and definitions in, 141–142 Proportion(s), 476–477 extremes of, 476 involving line segments, 480–484 in right triangle, 510–513 Proportionality, constant of, 486 Proving triangles congruent by AAS, 352–355 ASA, 161–163 SAS, 158–159 SSS, 165 HL, 362–365 similar by, 489–494 AA, 489–494 SSS, 489–494 SAS, 489–494 Pyramid(s), 449 altitude of, 449 height of, 449 properties of regular, 450–451 regular, 449 slant height of, 449 surface area of, 449 vertex of, 449 volume of, 449 Pythagoras, 1, 474 Pythagorean Theorem, 474, 515 converse of, 516–517 Pythagorean triple, 517 Q Quadrilateral(s), 23, 212, 379–412 consecutive angles of, 380 diagonal of, 380 general, 380 isosceles, 379 opposite angles of, 380 opposite sides of, 380 proof of as isosceles trapezoid, 403–407 as parallelogram, 385–387 as rectangle, 390 as rhombus, 394–396 as square, 399–401 Quarter turn, 240 R Radius of circle, 536 of sphere, 459 Range, 250 Ra |
tio defined, 475–476 of similitude, 486 Rational numbers, 3 Ray, 14–15 endpoint of, 15 Real number line, distance between two points on, 8 Real numbers, 3 properties of system, 4–5 Reasoning deductive, 97, 100–103, 150–151 inductive, 94–97 Rectangle(s), 389–391 altitude of, 389 angles of, 389 base of, 389 diagonals of, 389 proof of quadrilateral as, 390 properties of, 390 Rectangular coordinates, 211 Rectangular parallelepiped, 442 Rectangular solid, 442–443 Reflection(s) glide, 243–245 line, 214–220 line of, 214 in line y x, 224–225 point, 227–231 in x-axis, 223 in y-axis, 222–223 Reflexive property, 156, 487 of equality, 110–111 Regular polygon(s), 369 center of, 450 Regular pyramid(s), 449 lateral faces of, 451 lateral sides of, 451 properties of, 450–451 Remote interior angle, 277 Replacement set, 36 Rhombus, 393–396 diagonals of, 394 proof of quadrilateral as, 394–396 properties of, 394 Riemann, Georg, 379 Right angle, 17 Right circular cone, 456 Right circular conical surface, 456 Right circular cylinder, 453 Right prism, 441 Right triangle(s), 25, 26, 515 45-45 degree, 517–518 proportions in, 510–513 proving congruent by hypotenuse, leg, 362–365 Pythagorean theorem and, 515–517 30-60 degree, 518 rk, 217 RO, 230 Roots postulate, 125–126 Rotation in coordinate plane, 238–239, 240–242 defined, 238 negative, 239 positive, 239 preservation of distance under, about a fixed point, 239 Index 641 Rotational symmetry, 239–242 RP, 229 RP,d, 239 rx-axis, 223 ry-axis, 222 ryx, 224 S Saccheri, Girolamo, 379 SAS triangle congruence, 158. See also Side-angle-side triangle congruence SAS similarity theorem (SAS), 491–492. See also Side-angleside similarity theorem Scalene triangle, 24, 25, 98 Secant(s) angle formed by two intersecting, 568–571 of circle, 558 in coordinate plane, 588–591 external segment of, 576 segment formed by intersecting, 576–578 segment formed by tangent intersecting, 575–576 Segment. See Line segments Semicircle, 537 Semiperimeter, 174 Sentence closed, 37 compound, 42, 53 incomplete, 35 mathematical, 35 nonmathematical, 35–36 open, 36–37 truth value of, 35 Set, 2, 7 noncollinear, of points, 7 replacement, 36 solution, 36 truth, 36 Side(s) adjacent, 380 of angle, 15 classifying triangles according to, 24–25 consecutive, 380 corresponding, 155 inequalities involving, for triangle, 281–284 of polygon, 23 Side-angle-side similarity theorem, 491–492. See also SAS similarity theorem Side-angle-side triangle congruence, 158, See also SAS triangle congruence Side-side-side similarity theorem, 491. See also SSS similarity theorem Side-side-side triangle congruence, 165. See also SSS triangle congruence Side-side-side triangle similarity, 491. See also SSS triangle similarity Similar polygons, 486–488 14365Index.pgs 7/13/07 10:19 AM Page 642 642 Index Similarity equivalence relation of, 487–488 postulate of, 490 Similitude, ratio of, 486 Skew lines, 421–422 Slant height of cone, 456 of pyramid, 449 Slope of line, 291–294 negative, 293 of perpendicular lines, 307–310 positive, 292 undefined, 293 zero, 293 Slope-intercept form of an equation, 298 Solid geometry, 420 Solid, rectangular, 442–443 Solution set, 36 Sphere, 459–463 center of, 459 great circle of, 460 radius of, 459 surface area of, 462–463 volume of, 462–463 Square(s), 399–401 properties of, 399 proving that quadrilateral is, 399–401 Squaring of circle, 604 SSS similarity theorem (SSS), 491. See also, Side-side-side similarity theorem SSS triangle congruence, 165. See also Side-side-side triangle congruence Statement(s) compound, 42, 53 negation of, 38 postulates used in proving, 110–113 symbols and, 37 Straight angle, 16, 17 Straight line, 1, 2 Straightedge, 196, 604 Substitution postulate, 115–117, 264, 267–268 Subtraction of angles, 20–21 inequality postulates involving, 267–268 of line segments, 12–13 Subtraction postulate, 121–122 Sum of two angles, 20 Supplementary angles, 145–146, 148–149 Supplements, 146 Surface, 1 Surface area of circular cylinder, 454 of cone, 457–458 of prism, 440–443 of pyramid, 449 of sphere, 462–463 Symbols logic, 38–39 statements and, 37 Symmetric property, 156, 487 of equality, 111 Symmetry axis of, 219, 625 line, 218–220 line of, 219 point, 229 rotational, 239–242 translational, 235 Symmetry plane, 464 T Ta,b, 235 Tangent(s) angles formed by two intersecting, 567–568, 568–571 to a circle, 558 common, 559–561 common external, 560 in coordinate plane, 588 externally, 560 internally, 560 segments formed by, intersecting a secant, 575–576 Tangent segment, 561–563 Tangent to circle, 558 Terms, undefined, 1, 2 Tessellation, 259 Thales, 1 Theorem(s), 110 defined, 141 Exterior Angle, 349 Exterior Angle Inequality, 277 Hypotenuse-Leg Triangle Congruence, 362–365 involving pairs of angles, 146–148 Isosceles Triangle, 181 converse of, 357–360 Midsegment, 480 Perpendicular Bisector Concurrence, 193 proving, about angles, 144–145 Pythagorean, 474, 515 converse of, 515–517 Triangle Inequality, 273 Triangle Similarity, 490 30-60-degree right triangle, 518 Total surface area of prism, 442 Transformation(s), 215 composition of, 251–252 as functions, 250–254 dilation, 247–248, 495–499 glide reflection, 243–245 line reflection, 214–220 point reflection, 227–231 rotation, 238–239, 240–242 translation, 232–235 Transformational proof, 308 Transitive property, 156, 157, 263–264, 487, 488 of equality, 111 of inequality, 264 Translation(s) of a units in the horizontal direction and b units in the vertical direction, 233 in coordinate plane, 232–235 preservation of angle measure, 232 preservation of collinearity, 232 preservation of distance, 232 preservation of midpoint, 232 defined, 232 Translational symmetry, 235 Transversal(s) defined, 330 parallel lines and, 330–332 Trapezoid, 402–407 bases of, 402 legs of, 402 median of, 405 Tree diagram, 42 Triangle(s), 23 acute, 25, 105 altitude of, 175 angle bisector of, 176 centroid of, 506 classifying according to angles, 25–26 according to sides, 24–25 concurrence of altitudes of, 317–320 concurrence of angle bisectors of, 364–365 concurrence of medians of, 506–509 congruence of, 134, 155–156, 174–203 defined, 23 equiangular, 25 equilateral, 24, 25, 181–183 exterior angles of, 277–279 included sides and included angles of, 24 inequalities involving an exterior angle of, 276–279 involving lengths of sides of, 273–274 involving sides and angles of, 281–284 isosceles, 24, 25, 181–183, 451 line segments associated with, 175–177 median of, 175–176 obtuse, 25 opposite sides and opposite angles in, 24 proportional relations among segments related to, 502–505 proving congruent by AAS, 352–355 ASA, 161–163 SAS, 158–159 SSS, 165 HL, 362–365 proving similar by AA, 489–494 SSS, 489–494 SAS, 489–494 right, 25, 26, 515 45-45 degree, 517–518 proportions in, 510–513 Pythagorean theorem and, 515–517 30-60 degree, 518 scalene, 24, 25, 98 sum of measures of angles of, 347–350 Triangle Inequality Theorem, 273 Triangle Similarity Theorem, 490 Triangular prism, volume of, 447 14365Index.pgs 7/13/07 10:19 AM Page 643 Trichotomy postulate, 264–265 Trisection of angle, 604 Truth set, 36 Truth table, 42–44, 51 Truth value, 35, 37, 38 for conditional p → q, 54–55 sentences and, 35 al-Tusi, Nasir al-Din, 379 Two-column proof, 101 U Undefined term, 1, 2 Upper base angles, 404 of isosceles trapezoid, 404 V Valid argument, 75, 109 Value(s) absolute, 7–8 truth, 35, 37, 38, 54–55 Vertex of angle, 15 of cone, 456 of pyramid, 449 Vertex angle of isosceles triangle, 25 Vertical angles, 149 as congruent, 150–151 Vertices adjacent, 368, 380 consecutive, 380 of polyhedron, 440 Volume of circular cylinder, 454 of cone, 457–458 of prism, 446–447 of pyramid, 449 of sphere, 462–463 of triangular prism, 447 Index 643 X x-axis, reflection in, 223 x-coordinate, 210 x-intercept, 296 (x, y), 210 Y y-axis, reflection in, 222–223 y-coordinate, 210 y-intercept, 296 Z Zero, multiplication property of, 5 Zero slope, 293 14365Index.pgs 7/13/07 10:19 AM Page 644 |
This design is called a tessellation, or tiling, of regular hexagons. IMPROVING YOUR ALGEGRA SKILLS Algebraic Magic Squares I A magic square is an arrangement of numbers in a square grid. The numbers in every row, column, or diagonal add up to the same number. For example, in the magic square on the left, the sum of each row, column, and diagonal is 18. Complete the 5-by-5 magic square on the right. Use only the numbers in this list: 6, 7, 9, 13, 17, 21, 23, 24, 27, and 28. 5 4 9 10 6 2 3 8 7 20 29 22 19 12 10 25 18 11 8 14 26 30 15 16 12 CHAPTER 0 Geometric Art L E S S O N 0.4 Everything is an illusion, including this notion. STANISLAW J. LEC Op Art Op art, or optical art, is a form of abstract art that uses lines or geometric patterns to create a special visual effect. The contrasting dark and light regions sometimes appear to be in motion or to represent a change in surface, direction, and dimension. Victor Vasarely was one artist who transformed grids so that spheres seem to bulge from them. Recall the series Tsiga I, II, and III that appears in Lesson 0.1. Harlequin, shown at right, is a rare Vasarely work that includes a human form. Still, you can see Vasarely’s trademark sphere in the clown’s bulging belly In Hesitate, by contemporary op artist Bridget Riley (b 1931), what effect do the changing dots produce? In Harlequin, Victor Vasarely used curved lines and shading to create the form of a clown in motion. Op art is fun and easy to create. To create one kind of op art design, first make a design in outline. Next, draw horizontal or vertical lines, gradually varying the space between the lines to create an illusion of hills and valleys. Finally, color in alternating spaces. The Wavy Letter Step 1 Step 2 Step 3 LESSON 0.4 Op Art 13 To create the next design, first locate a point on each of the four sides of a square. Each point should be the same distance from a corner, as shown. Your compass is a good tool for measuring equal lengths. Connect these four points to create another square within the first. Repeat the process until the squares appear to converge on the center. Be careful that you don’t fall in! The Square Spiral Step 1 Step 2 Step 3 Step 4 Here are some other examples of op art. Square tunnel or top of pyramid? Amish quilt, tumbling block design Japanese Op Art, Hajime Juchi, Dover Publications Op art by Carmen Apodaca, geometry student You can create any of the designs on this page using just a compass and straightedge (and doing some careful coloring). Can you figure out how each of these op art designs was created? 14 CHAPTER 0 Geometric Art EXERCISES 1. What is the optical effect in each piece of art in this lesson? 2. Nature creates its own optical art. At first the black and white stripes of a zebra appear to work against it, standing out against the golden brown grasses of the African plain. However, the stripes do provide the zebras with very effective protection from predators. When and how? 3. Select one type of op art design from this lesson and create your own version of it. 4. Create an op art design that has reflectional symmetry, but not rotational symmetry. 5. Antoni Gaudí (1852–1926) designed the Bishop’s Palace in Astorga, Spain. List as many geometric shapes as you can recognize on the palace (flat, two-dimensional shapes such as rectangles as well as solid, threedimensional shapes such as cylinders). What type of symmetry do you see on the palace? Bishop’s Palace, Astorga, Spain IMPROVING YOUR REASONING SKILLS Bagels In the original computer game of bagels, a player determines a three-digit number (no digit repeated) by making educated guesses. After each guess, the computer gives a clue about the guess. Here are the clues. bagels: no digit is correct pico: one digit is correct but in the wrong position fermi: one digit is correct and in the correct position In each of the games below, a number of guesses have been made, with the clue for each guess shown to its right. From the given set of guesses and clues, determine the three-digit number. If there is more than one solution, find them all. Game 1 ? ? ? bagels pico pico pico fermi pico Game 2 ? ? ? bagels pico pico fermi fermi pico pico LESSON 0.4 Op Art 15 Knot Designs Knot designs are geometric designs that appear to weave or to interlace like a knot. Some of the earliest known designs are found in Celtic art from the northern regions of England and Scotland. In their carved stone designs, the artists imitated the rich geometric patterns of three-dimensional crafts such as weaving and basketry. The Book of Kells (8th and 9th centuries) is the most famous collection of Celtic knot designs. L E S S O N 0.5 In the old days, a love-sick sailor might send his sweetheart a length of fishline loosely tied in a love knot. If the knot was returned pulled tight it meant the passion was strong. But if the knot was returned untied— ah, matey, time to ship out. OLD SAILOR’S TALE Carved knot pattern from Nigeria Celtic knot design Today a very familiar knot design is the set of interconnected rings (shown at right) used as the logo for the Olympic Games. Here are the steps for creating two examples of knot designs. Look them over before you begin the exercises. Step 1 Step 2 Step 3 Step 4 You can use a similar approach to create a knot design with rings. Step 1 Step 2 Step 3 Step 4 16 CHAPTER 0 Geometric Art Here are some more examples of knot designs. Knot design by Scott Shanks, geometry student Tiger Tail, Diane Cassell, parent of geometry student Medieval Russian knot design Japanese knot design The last woodcut made by M. C. Escher is a knot design called Snakes. The rings and the snakes interlace, and the design has 3-fold rotational symmetry. Snakes, M. C. Escher, 1969/ ©2002 Cordon Art B. V.–Baarn–Holland. All rights reserved. LESSON 0.5 Knot Designs 17 EXERCISES 1. Name a culture or country whose art uses knot designs. You will need Construction tools for Exercise 3 2. Create a knot design of your own, using only straight lines on graph paper. 3. Create a knot design of your own with rotational symmetry, using a compass or a circle template. 4. Sketch five rings linked together so that you could separate all five by cutting open one ring. 5. The coat of arms of the Borromeo family, who lived during the Italian Renaissance (ca. 15th century), showed a very interesting knot design known as the Borromean Rings. In it, three rings are linked together so that if any one ring is removed the remaining two rings are no longer connected. Got that? Good. Sketch the Borromean Rings. 6. The Chokwe storytellers of northeastern Angola are called Akwa kuta sona (“those who know how to draw”). When they sit down to draw and to tell their stories, they clear the ground and set up a grid of points in the sand with their fingertips, as shown below left. Then they begin to tell a story and, at the same time, trace a finger through the sand to create a lusona design with one smooth, continuous motion. Try your hand at creating sona (plural of lusona). Begin with the correct number of dots. Then, in one motion, re-create one of the sona below. The initial dot grid is shown for the rat. Initial dot grid Rat Mbemba bird Scorpion 7. In Greek mythology, the Gordian knot was such a complicated knot that no one could undo it. Oracles claimed that whoever could undo the knot would become the ruler of Gordium. When Alexander the Great (356–323 B.C.E.) came upon the knot, he simply cut it with his sword and claimed he had fulfilled the prophecy, so the throne was his. The expression “cutting the Gordian knot” is still used today. What do you think it means? Science Mathematician DeWitt Sumners at Florida State University and biophysicist Sylvia Spenger at the University of California, Berkeley, have discovered that when a virus attacks DNA, it creates a knot on the DNA. 18 CHAPTER 0 Geometric Art 8. The square knot and granny knot are very similar but do very different things. Compare their symmetries. Use string to re-create the two knots and explain their differences. Square knot Granny knot 9. Cut a long strip of paper from a sheet of lined paper or graph paper. Tie the strip of paper snugly, but without wrinkles, into a simple knot. What shape does the knot create? Sketch your knot. SYMBOLIC ART Japanese artist Kunito Nagaoka (b 1940) uses geometry in his work. Nagaoka was born in Nagano, Japan, and was raised near the active volcano Asama. In Japan, he experienced earthquakes and typhoons as well as the human tragedies of Hiroshima and Nagasaki. In 1966, he moved to Berlin, Germany, a city rebuilt in concrete from the ruins of World War II. These experiences clearly influenced his work. You can find other examples of symbolic art at www.keymath.com/DG . Look at the etching shown here, or another piece of symbolic art. Write a paragraph describing what you think might have happened in the scene or what you think it might represent. What types of geometric figures do you find? Use geometric shapes in your own sketch or painting to evoke a feeling or to tell a story. Write a one- or two-page story related to your art. ISEKI/PY XVIII (1978), Kunito Nagaoka LESSON 0.5 Knot Designs 19 Islamic Tile Designs Islamic art is rich in geometric forms. Early Islamic, or Muslim, artists became familiar with geometry through the works of Euclid, Pythagoras, and other mathematicians of antiquity, and they used geometric patterns extensively in their art and architecture. L E S S O N 0.6 Patience with small details makes perfect a large work, like the universe. JALALUDDIN RUMI An exterior wall of the Dome of the Rock (660–750 C.E.) mosque in Jerusalem Islam forbids the representation of humans or animals in religious art. So, instead, the artists use intricate geometric patterns. One of the most striking examples of Islamic architecture is the Alhambra Palace, in Granada, Spain. Built over 600 years ago by Moors and Spaniards, the Alhambra is filled from floor to |
ceiling with marvelous geometric patterns. The designs you see on this page are but a few of the hundreds of intricate geometric patterns found in the tile work and the inlaid wood ceilings of buildings like the Alhambra and the Dome of the Rock. Carpets and hand-tooled bronze plates from the Islamic world also show geometric designs. The patterns often elaborate on basic grids of regular hexagons, equilateral triangles, or squares. These complex Islamic patterns were constructed with no more than a compass and a straightedge. Repeating patterns like these are called tessellations. You’ll learn more about tessellations in Chapter 7. Alcove in the Hall of Ambassadors, the Alhambra, a Moorish palace in Granada, Spain 20 CHAPTER 0 Geometric Art The two examples below show how to create one tile in a square-based and a hexagon-based design. The hexagon-based pattern is also a knot design. 8-pointed Star Step 1 Step 2 Step 3 Step 4 Step 5 Hexagon Tile Design Step 1 Step 2 Step 3 Step 4 Step 5 LESSON 0.6 Islamic Tile Designs 21 In Morocco, zillij, the art of using glazed tiles to form geometric patterns, is the most common practice for making mosaics. Zillij artists cut stars, octagons, and other shapes from clay tiles and place them upside down into the lines of their design. When the tiling is complete, artists pour concrete over the tiles to form a slab. When the concrete dries, they lift the whole mosaic, displaying the colors and connected shapes, and mount it against a fountain, palace, or other building. EXERCISES 1. Name two countries where you can find Islamic architecture. 2. What is the name of the famous palace in Granada, Spain, where you can find beautiful examples of tile patterns? You will need Construction tools for Exercises 5–7 3. Using tracing paper or transparency film, trace a few tiles from the 8-pointed star design. Notice that you can slide, or translate, the tracing in a straight line horizontally, vertically, and even diagonally to other positions so that the tracing will fit exactly onto the tiles again. What is the shortest translation distance you can find, in centimeters? 4. Notice that when you rotate your tracing from Exercise 3 about certain points in the tessellation, the tracing fits exactly onto the tiles again. Find two different points of rotation. (Put your pencil on the point and try rotating the tracing paper or transparency.) How many times in one rotation can you make the tiles match up again? 22 CHAPTER 0 Geometric Art Architecture After studying buildings in other Muslim countries, the architect of the Petronas Twin Towers, Cesar Pelli (b 1926), decided that geometric tiling patterns would be key to the design. For the floor plan, his team used a very traditional tile design, the 8-pointed star—two intersecting squares. To add space and connect the design to the traditional “arabesques,” the design team added arcs of circles between the eight points. 5. Currently the tallest buildings in the world are the Petronas Twin Towers in Kuala Lumpur, Malaysia. Notice that the floor plans of the towers have the shape of Islamic designs. Use your compass and straightedge to re-create the design of the base of the Petronas Twin Towers, shown at right. 6. Use your protractor and ruler to draw a square tile. Use your compass, straightedge, and eraser to modify and decorate it. See the example in this lesson for ideas, but yours can be different. Be creative! 7. Construct a regular hexagon tile and modify and decorate it. See the example in this lesson for ideas, but yours can be different. 8. Create a tessellation with one of the designs you made in Exercises 6 and 7. Trace or photocopy several copies and paste them together in a tile pattern. (You can also create your tessellation using geometry software and print out a copy.) Add finishing touches to your tessellation by adding, erasing, or whiting out lines as desired. If you want, see if you can interweave a knot design within your tessellation. Color your tessellation. PHOTO OR VIDEO SAFARI In Lesson 0.1, you saw a few examples of geometry and symmetry in nature and art. Now go out with your group and document examples of geometry in nature and art. Use a camera or video camera to take pictures of as many examples of geometry in nature and art as you can. Look for many different types of symmetry, and try to photograph art and crafts from many different cultures. Consider visiting museums and art galleries, but make sure it’s okay to take pictures when you visit. You might find examples in your home or in the homes of friends and neighbors. If you take photographs, write captions for them that describe the geometry and the types of symmetries you find. If you record video, record your commentary on the soundtrack. LESSON 0.6 Islamic Tile Designs 23 ● CHAPTER 11 REVIEW ● CHAPTER 0 REVIEW ● CHAPTER 0 REVIEW ● CHAPTER 0 REVIEW ● CHAPTER CHAPTER 0 R E V I E W In this chapter, you described the geometric shapes and symmetries you see in nature, in everyday objects, in art, and in architecture. You learned that geometry appears in many types of art—ancient and modern, from every culture—and you learned specific ways in which some cultures use geometry in their art. You also used a compass and straightedge to create your own works of geometric art. The end of a chapter is a good time to review and organize your work. Each chapter in this book will end with a review lesson. EXERCISES 1. List three cultures that use geometry in their art. You will need Construction tools for Exercises 4, 5, and 10 2. What is the optical effect of the op art design at right? 3. Name the basic tools of geometry you used in this chapter and describe their uses. 4. With a compass, draw a 12-petal daisy. 5. Construction With a compass and straightedge, construct a regular hexagon. 6. List three things in nature that have geometric shapes. Name their shapes. 7. Draw an original knot design. 8. Which of the wheels below have reflectional symmetry? Hot Blocks (1966–67), Edna Andrade How many lines of symmetry does each have? Wheel A Wheel B Wheel C Wheel D 9. Which of the wheels in Exercise 8 have only rotational symmetry? What kind of rotational symmetry does each have? 24 CHAPTER 0 Geometric Art ● CHAPTER 0 REVIEW ● CHAPTER 0 REVIEW ● CHAPTER 0 REVIEW ● CHAPTER 0 REVIEW ● CHAPTER 0 10. A mandala is a circular design arranged in rings that radiate from the center. (See the Cultural Connection below.) Use your compass and straightedge to create a mandala. Draw several circles using the same point as the center. Create a geometric design in the center circle, and decorate each ring with a symmetric geometric design. Color or decorate your mandala. Two examples are shown below. The first mandala uses daisy designs. The second mandala is a combination knot and Islamic design by Scott Shanks, geometry student. 11. Create your own personal mandala. You might include your name, cultural symbols, photos of friends and relatives, and symbols that have personal meaning for you. Color it. 12. Create one mandala that uses techniques from Islamic art, is a knot design, and also has optical effects. Cultural The word mandala comes from Sanskrit, the classical language of India, and means “circle” or “center.” Hindus use mandala designs for meditation. The Aztec calendar stone below left is an example of a mandala. In the center is the mask of the sun god. Notice the symbols are arranged symmetrically within each circle. The rose windows in many gothic cathedrals, like the one below right from the Chartres Cathedral in France, are also mandalas. Notice all the circles within circles, each one filled with a design or picture. CHAPTER 0 REVIEW 25 EW ● CHAPTER 0 REVIEW ● CHAPTER 0 REVIEW ● CHAPTER 0 REVIEW ● CHAPTER 0 REVIEW ● CHAPTE 13. Did you know that “flags” is the most widely read topic of the World Book Encyclopedia? Research answers to these questions. More information about flags is available at www.keymath.com/DG . a. Is the flag of Puerto Rico symmetric? Explain. b. Does the flag of Kenya have rotational symmetry? Explain. c. Name a country whose flag has both rotational and reflectional symmetry. Sketch the flag. Assessing What You’ve Learned KEEPING A PORTFOLIO This section suggests how you might review, organize, and communicate to others what you’ve learned. Whether you follow these suggestions or directions from your teacher, or use study strategies of your own, be sure to reflect on all you’ve learned. An essential part of learning is being able to show yourself and others how much you know and what you can do. Assessment isn’t limited to tests and quizzes. Assessment isn’t even limited to what your teacher sees or what makes up your grade. Every piece of art you make, and every project or exercise you complete, gives you a chance to demonstrate to somebody—yourself, at least—what you’re capable of. BEGIN A PORTFOLIO This chapter is primarily about art, so you might organize your work the way a professional artist does—in a portfolio. A portfolio is different from a notebook, both for an artist and for a geometry student. An artist’s notebook might contain everything from scratch work to practice sketches to random ideas jotted down. A portfolio is reserved for an artist’s most significant or best work. It’s his or her portfolio that an artist presents to the world to demonstrate what he or she is capable of doing. The portfolio can also show how an artist’s work has changed over time. Review all the work you’ve done so far and choose one or more examples of your best art projects to include in your portfolio. Write a paragraph or two about each piece, addressing these questions: What is the piece an example of ? Does this piece represent your best work? Why else did you choose it? What mathematics did you learn or apply in this piece? How would you improve the piece if you redid or revised it? Portfolios are an ongoing and ever-changing display of your work an |
d growth. As you finish each chapter, update your portfolio by adding new work. 26 CHAPTER 0 Geometric Art CHAPTER 1 Introducing Geometry Although I am absolutely without training or knowledge in the exact sciences, I often seem to have more in common with mathematicians than with my fellow artists. M. C. ESCHER Three Worlds, M. C. Escher, 1955 ©2002 Cordon Art B. V.–Baarn–Holland. All rights reserved In this chapter you will ● write your own definitions of many geometry terms and geometric figures ● start a notebook with a list of all the terms and their definitions ● develop very useful visual thinking skills L E S S O N 1.1 Nature’s Great Book is written in mathematical symbols. GALILEO GALILEI Building Blocks of Geometry Three building blocks of geometry are points, lines, and planes. A point is the most basic building block of geometry. It has no size. It has only location. You represent a point with a dot, and you name it with a capital letter. The point shown below is called P. A tiny seed is a physical model of a point. P Mathematical model of a point A line is a straight, continuous arrangement of infinitely many points. It has infinite length but no thickness. It extends forever in two directions. You name a line by giving the letter names of any two points on the line and by placing the line symbol above the letters, for example, AB or BA. B A A piece of spaghetti is a physical model of a line. A line, however, is longer, straighter, and thinner than any piece of spaghetti ever made. Mathematical model of a line A plane has length and width but no thickness. It is like a flat surface that extends infinitely along its length and width. You represent a plane with a four-sided figure, like a tilted piece of paper, drawn in perspective. Of course, this actually illustrates only part of a plane. You name a plane with a script capital letter, such as . A flat piece of rolledout dough is a model of a plane, but a plane is broader, wider, and thinner than any piece of dough you could roll. Mathematical model of a plane 28 CHAPTER 1 Introducing Geometry Investigation Mathematical Models Step 1 Step 2 Identify examples of points, lines, and planes in these pictures. Explain in your own words what point, line, and plane mean. It can be difficult to explain what points, lines, and planes are. Yet, you probably recognized several models of each in the investigation. Early mathematicians tried to define these terms. By permission of Johnny Hart and Creators Syndicate, Inc. LESSON 1.1 Building Blocks of Geometry 29 The ancient Greeks said, “A point is that which has no part. A line is breadthless length.” The Mohist philosophers of ancient China said, “The line is divided into parts, and that part which has no remaining part is a point.” Those definitions don’t help much, do they? A definition is a statement that clarifies or explains the meaning of a word or a phrase. However, it is impossible to define point, line, and plane without using words or phrases that themselves need definition. So these terms remain undefined. Yet, they are the basis for all of geometry. Using the undefined terms point, line, and plane, you can define all other geometry terms and geometric figures. Many are defined in this book, and others will be defined by you and your classmates. Keep a definition list in your notebook, and each time you encounter new geometry vocabulary, add the term to your list. Illustrate each definition with a simple sketch. Here are your first definitions. Begin your list and draw sketches for all definitions. Collinear means on the same line. A B C Points A, B, and C are collinear. Coplanar means on the same plane. D E F Points D, E, and F are coplanar. Name three balls that are collinear. Name three balls that are coplanar but not collinear. Name four balls that are not coplanar. B A C D F E G 30 CHAPTER 1 Introducing Geometry A line segment consists of two points called the endpoints of the segment and all the points between them that are collinear with the two points. Line segment B A Endpoints You can write line segment AB, using a segment symbol, as AB or BA. There are two ways to write the length of a segment. You can write AB 2 in., meaning the distance from A to B is 2 inches. You can also use an m for “measure” in front of the segment name, and write the distance as mAB 2 in. If no measurement units are used for the length of a segment, it is understood that the choice of units is not important, or is based on the length of the smallest square in the grid. Figure A A 2 B Figure B y M 4 5 4 N x AB 2 in., or mAB 2 in. mMN 5 units Two segments are congruent segments if and only if they have the same measure or length. The symbol for congruence is , and you say it as “is congruent to.” You use the equals symbol, , between equal numbers and the congruence symbol, , between congruent figures. C 3.2 cm 3.2 cm A D AC DC AC DC When drawing figures, you show congruent segments by making identical markings. These single marks mean these two segments are congruent to each other. B A C S P These double marks SP RQ, mean that and these triple marks mean that PQ SR. R Q The midpoint of a segment is the point on the segment that is the same distance from both endpoints. The midpoint bisects the segment, or divides the segment into two congruent segments. LESSON 1.1 Building Blocks of Geometry 31 EXAMPLE Study the diagrams below. a. Name each midpoint and the segment it bisects. b. Name all the congruent segments. Use the congruence symbol to write your answers. C 2 cm G E F 2 cm H D J K L M N P Solution Look carefully at the markings and apply the midpoint definition. a. CF FD, so F is the midpoint of CD; JK KL, so K is the midpoint of JL. b. CF FD, HJ HL, and JK KL. Even though EF and FG appear to have the same length, you cannot assume they are congruent without the markings. The same is true for MN and NP. Ray AB is the part of AB that contains point A and all the points on AB that are on the same side of point A as point B. Imagine cutting off all the points to the left of point A. Endpoint Ray AB A Y B A Y B In the figure above, AY and AB are two ways to name the same ray. Note that AB is not the same as BA! A ray begins at a point and extends infinitely in one direction. You need two letters to name a ray. The first letter is the endpoint of the ray, and the second letter is any other point that the ray passes through. AB BA B B A A Physical model of a ray: beams of light 32 CHAPTER 1 Introducing Geometry EXERCISES 1. Identify the models in the photos below for point, segment, plane, collinear points, and coplanar points. For Exercises 2–4, name each line in two different ways. 2. P T 3. A R T 4. y 5 M SA x 8 For Exercises 5–7, draw two points and label them. Then use a ruler to draw each line. Don’t forget to use arrowheads to show that it extends indefinitely. 5. AB 6. KL 7. DE with D(3, 0) and E(0, 3) For Exercises 8–10, name each line segment. A 8. C 9. y 5 P 10. R I Q x 10 T For Exercises 11 and 12, draw and label each line segment. 11. AB 12. RS with R(0, 3) and S(2, 11) For Exercises 13 and 14, use your ruler to find the length of each line segment to the nearest tenth of a centimeter. Write your answer in the form mAB ? . A C 13. 14. D B For Exercises 15–17, use your ruler to draw each segment as accurately as you can. Label each segment. 15. AB 4.5 cm 16. CD 3 in. 17. EF 24.8 cm LESSON 1.1 Building Blocks of Geometry 33 18. Name each midpoint and the segment it bisects 19. Draw two segments that have the same midpoint. Mark your drawing to show congruent segments. 20. Draw and mark a figure in which M is the midpoint of ST, SP PT, and T is the midpoint of PQ. For Exercises 21–23, name the ray in two different ways. 21. A B C 22. M N P 23. Z Y X For Exercises 24–26, draw and label each ray. 25. YX 24. AB 26. MN 27. Draw a plane containing four coplanar points A, B, C, and D, with exactly three collinear points A, B, and D. 28. Given two points A and B, there is only one segment that you can name: AB. With three collinear points A, B, and C, there are three different segments that you can name: AB, AC, and BC. With five collinear points A, B, C, D, and E, how many different segments can you name? For Exercises 29–31, draw axes onto graph paper and locate point A(4, 0) as shown. 29. Draw AB, where point B has coordinates (2, 6). 30. Draw OM with endpoint (0, 0) that goes through point M(2, 2). 31. Draw CD through points C(2, 1) and D(2, 3). y 5 –5 x A (4, 0) –5 Career Woodworkers use a tool called a plane to shave a rough wooden surface to create a perfectly smooth planar surface. The smooth board can then be made into a tabletop, a door, or a cabinet. Woodworking is a very precise process. Producing high-quality pieces requires an understanding of lines, planes, and angles as well as careful measurements. 34 CHAPTER 1 Introducing Geometry 32. If the signs of the coordinates of collinear points P(6, 2), Q(5, 2), and R(4, 6) are reversed, are the three new points still collinear? Draw a picture and explain why. 33. Draw a segment with midpoint N(3, 2). Label it PQ. 34. Copy triangle TRY shown at right. Use your ruler to find the midpoint A of side TR and the midpoint G of side TY. Draw AG. T R Y SPIRAL DESIGNS The circle design shown below is used in a variety of cultures to create mosaic decorations. The spiral design may have been inspired by patterns in nature. Notice that the seeds on the sunflower also spiral out from the center. Here are the steps to make the spirals. Step 1 Step 2 Step 3 Step 4 The more circles and radii you draw, the more detailed your design will be. Create and decorate your own spiral design. LESSON 1.1 Building Blocks of Geometry 35 ALGEBRA SKILLS 1 ● USING YOUR ALGEBRA SKILLS 1 ● USING YOUR ALGEBRA SKILLS 1 ● USING YO USING YOUR ALGEBRA SKILLS 1 Midpoint A midpoint is the point on a line segment that is the same distance from both endpoints. You can think of a midpoint as bei |
ng halfway between two locations. You know how to mark a midpoint. But when the position and location matter, such as in navigation and geography, you can use a coordinate grid and some algebra to find the exact location of the midpoint. You can calculate the coordinates of the midpoint of a segment on a coordinate grid using a formula. Coordinate Midpoint Property If x1, y1 then the coordinates of the midpoint are and x2, y2 are the coordinates of the endpoints of a segment, x1 x2, y1 2 2 y2 History Surveyors and mapmakers of ancient Egypt, China, Greece, and Rome used various coordinate systems to locate points. Egyptians made extensive use of square grids and used the first known rectangular coordinates at Saqqara around 2650 B.C.E. By the seventeenth century, the age of European exploration, the need for accurate maps and the development of easy-to-use algebraic symbols gave rise to modern coordinate geometry. Notice the lines of latitude and longitude in this seventeenth-century map. 36 CHAPTER 1 Introducing Geometry ALGEBRA SKILLS 1 ● USING YOUR ALGEBRA SKILLS 1 ● USING YOUR ALGEBRA SKILLS 1 ● USING YO EXAMPLE Solution Segment AB has endpoints (8, 5) and (3, 6). Find the coordinates of the midpoint of AB. y The midpoint is not on a grid intersection point, so we can use the coordinate midpoint property. 3) 2.5 x2 (8 x x1 2 2 14) 0.5 y2 (13 y y1 2 2 The midpoint of AB is (2.5, 0.5). A (–8, 5) (–2.5, –0.5) x B (3, –6) EXERCISES For Exercises 1–3, find the coordinates of the midpoint of the segment with each pair of endpoints. 1. (12, 7) and (6, 15) 2. (17, 8) and (1, 11) 3. (14, 7) and (3, 18) 4. One endpoint of a segment is (12, 8). The midpoint is (3, 18). Find the coordinates of the other endpoint. 5. A classmate tells you, “Finding the coordinates of a midpoint is easy. You just find the averages.” Is there any truth to it? Explain what you think your classmate means. 6. Find the two points on AB that divide the segment into three congruent parts. Point A has coordinates (0, 0) and point B has coordinates (9, 6). Explain your method. 7. Describe a way to find points that divide a segment into fourths. 8. In each figure below, imagine drawing the diagonals AC and BD. a. Find the midpoint of AC and the midpoint of BD in each figure. b. What do you notice about the midpoints? y A B B 10 5 B C (35.5, 10) C (22, 8.5) Figure 1 Figure 2 Figure 3 A D 5 C 10 D (16, 0.5) A 15 20 25 D (29.5, 1) 30 35 x USING YOUR ALGEBRA SKILLS 1 Midpoint 37 L E S S O N 1.2 Inspiration is needed in geometry, just as much as in poetry. ALEKSANDR PUSHKIN Poolroom Math People use angles every day. Plumbers measure the angle between connecting pipes to make a good fitting. Woodworkers adjust their saw blades to cut wood at just the correct angle. Air traffic controllers use angles to direct planes. And good pool players must know their angles to plan their shots. Is the angle between the two hands of the wristwatch smaller than the angle between the hands of the large clock? “Little Benji,” the wristwatch Big Ben at the Houses of Parliament in London, England 38 CHAPTER 1 Introducing Geometry You can use the terms that you defined in Lesson 1.1 to write a precise definition of angle. An angle is formed by two rays that share a common endpoint, provided that the two rays are noncollinear. In other words, the rays cannot lie on the same line. The common endpoint of the two rays is the vertex of the angle. The two rays are the sides of the angle. You can name the angle in the figure below angle TAP or angle PAT, or use the angle symbol and write TAP or PAT. Notice that the vertex must be the middle letter, and the first and last letters each name a point on a different ray. Since there are no other angles with vertex A, you can also simply call this A. T Vertex A Sides P Career In sports medicine, specialists may examine the healing rate of an injured joint by its angle of recovery. For example, a physician may assess how much physical therapy a patient needs by measuring the degree to which a patient can bend his or her ankle from the floor. EXAMPLE A Name all the angles in these drawings. Solution The angles are T, V, TUV, 1, TUR, XAY, YAZ, and XAZ. (Did you get them all?) Notice that 1 is a shorter way to name RUV Which angles in Example A seem big to you? Which seem small? The measure of an angle is the smallest amount of rotation about the vertex from one ray to the other, measured in degrees. According to this definition, the measure of an angle can be any value between 0° and 180°. The smallest amount of rotation PG to PS from is 136°. 136° P 224° G S The geometry tool you use to measure an angle is a protractor. Here’s how you use it. Step 1: Place the center mark of the protractor on the vertex. Step 2: Line up the 0-mark with one side of the angle. Step 3: Read the measure on the protractor scale. 0 30 40 5 0 6 150 140 television antenna is a physical model of an angle. Note that changing the length of the antenna doesn’t change the angle 90 100 110 12 80 70 1 0 0 90 0 1 3 0 60 Step 4: Be sure you read the scale that has the 0-mark you are using! The angle in the diagram measures 34° and not 146°. LESSON 1.2 Poolroom Math 39 To show the measure of an angle, use an m before the angle symbol. For example, mZAP 34° means the measure of ZAP is 34 degrees. EXAMPLE B Use your protractor to measure these angles as accurately as you can. Which ones measure more than 90°? Z 34° A P mZAP 34° 3 1 2 Solution Measuring to the nearest degree, you should get these approximate answers. (The symbol means “is approximately equal to.”) m3 92° 2 and 3 measure more than 90°. m1 16° m2 164° Two angles are congruent angles if and only if they have the same measure. You use identical markings to show that two angles in a figure are congruent. D G O These markings mean that DOG CAT and mDOG mCAT. A C T A ray is the angle bisector if it contains the vertex and divides the angle into two congruent angles. In the figure at right, CD bisects ACB so that ACD BCD. C B D A Science Earth takes 365.25 days to travel a full 360° around the Sun. That means that each day, Earth travels a little less than 1° in its orbit around the Sun. Meanwhile, Earth also completes one full rotation each day, making the Sun appear to rise and set. By how many degrees does the Sun’s position in the sky change every hour? 40 CHAPTER 1 Introducing Geometry EXAMPLE C Look for angle bisectors and congruent angles in the figures below. a. Name each angle bisector and the angle it bisects. b. Name all the congruent angles in the figure. Use the congruence symbol and name the angles so there is no confusion about which angle you mean. E 44° 43° F G H P 52° 52° Q S R M Y E N O Solution a. Use the angle bisector definition. SRP PRQ, so RP bisects SRQ. b. SRP PRQ and YMN OME. Investigation Virtual Pool Pocket billiards, or pool, is a game of angles. When a ball bounces off the pool table’s cushion, its path forms two angles with the edge of the cushion. The incoming angle is formed by the cushion and the path of the ball approaching the cushion. Outgoing angle Cushion Incoming angle The outgoing angle is formed by the cushion and the path of the ball leaving the cushion. As it turns out, the measure of the outgoing angle equals the measure of the incoming angle. Computer scientist Nesli O’Hare is also a professional pool player. LESSON 1.2 Poolroom Math 41 Use your protractor to study these shots Step 1 Step 2 Step 3 Step 4 Step 5 Use your protractor to find the measure of 1. Which is the correct outgoing angle? Which point—A or B—will the ball hit? Which point on the cushion—W, X, or Y—should the white ball hit so that the ray of the outgoing angle passes through the center of the 8-ball? Compare your results with your group members’ results. Does everyone agree? How would you hit the white ball against the cushion so that the ball passes over the same spot on the way back? How would you hit the ball so that it bounces off three different points on the cushions, without ever touching cushion CP? EXERCISES 1. Name each angle in three E T different ways. N F U 1 O 2 R For Exercises 2–4, draw and label each angle. 2. TAN 3. BIG 4. SML 5. For each figure at right, list the angles that you can name using only the vertex letter. S R T P Q 42 CHAPTER 1 Introducing Geometry B C D E F 6. Draw a figure that contains at least three angles and requires three letters to name each angle. For Exercises 7–14, find the measure of each angle. B C Y 0 30 40 5 0 6 150 140 90 100 110 12 80 70 1 0 0 90 0 1 3 0 60 . mAQB ? 8. mAQC ? 9. mXQA ? 10. mAQY ? 11. mZQY ? 12. mZQX ? 13. mCQB ? 14. mXQY ? For Exercises 15–19, use your protractor to find the measure of the angle to the nearest degree. 15. mMAC ? M C A 17. mS ? 18. mSON ? 19. mNOR ? 16. mIBM ? I N B M S O Q R 20. Which angle below has the greater measure, SML or BIG? Why? S M L I B G LESSON 1.2 Poolroom Math 43 For Exercises 21–23, use your protractor to draw angles with these measures. Label them. 21. mA 44° 22. mB 90° 23. mCDE 135° 24. Use your protractor to draw the angle bisector of A in Exercise 21 and the angle bisector of D in Exercise 23. Use markings to show that the two halves are congruent. 25. Copy triangle CAN shown at right. Use your protractor to find the angle bisector of A. Label the point where it crosses CN point Y. Use your ruler to find the midpoint of CN and label it D. Are D and Y the same point? C A For Exercises 26–28, draw a clock face with hands to show these times. 26. 3:30 27. 3:40 28. 3:15 29. Give an example of a time when the angle made by the hands of the clock will be greater than 90°. For Exercises 30–33, copy each figure and mark it with all the given information. 30. TH 6 mTHO 90° OH 8 H 31. RA SA mT mH RT SH T H N T O R A S 32. AT AG AI AN AGT ATG GI TN 33. BW TI WO IO WBT ITB BWO TIO For Exercises 34 and 35, write down what you know from the markings. Do not use your protractor or your ruler. 34. MI ? IC ? mM ? Y E |
K M C I 35. MEO ? SUE ? OU ? S U E M O 44 CHAPTER 1 Introducing Geometry For Exercises 36–38, do not use a protractor. Recall from Chapter 0 that a complete rotation around a point is 360°. Find the angle measures represented by each letter. 36. 37. 38. 15° 21° x 1 4 rotation y 41° 37° 1 2 rotation z 74° 135° 87° 39. If the 4-ball is hit as shown, will it go into the corner pocket? Find the path of the ball using only your protractor and straightedge. 40. What is the measure of the incoming angle? Which point will the ball pass through? Use your protractor to find out. D C B A 41. The principle you just learned for billiard balls is also true for a ray of light reflecting from a mirror. What you “see” in the mirror is actually light from an object bouncing off the mirror and traveling to your eye. Will you be able to see your shoes in this mirror? Copy the illustration and draw rays to show the light traveling from your shoes to the mirror and back to your eye. LESSON 1.2 Poolroom Math 45 Review 42. Use your ruler to draw a segment with length 12 cm. Then use your ruler to locate the midpoint. Label and mark the figure. 43. The balancing point of an object is called its center of gravity. Where is the center of gravity of a thin, rodlike piece of wire or tubing? Copy the thin wire shown below onto your paper. Mark the balance point or center of gravity. 44. Explain the difference between MS DG and MS DG. IMPROVING YOUR VISUAL THINKING SKILLS Coin Swap I and II 1. Arrange two dimes and two pennies on a grid of five squares, as shown. Your task is to switch the position of the two dimes and two pennies in exactly eight moves. A coin can slide into an empty square next to it, or it can jump over one coin into an empty space. Record your solution by drawing eight diagrams that show the moves. 2. Arrange three dimes and three pennies on a grid of seven squares, as shown. Follow the same rules as above to switch the position of the three dimes and three pennies in exactly 15 moves. Record your solution by listing in order which coin is moved. For example, your list might begin PDP. . . . 46 CHAPTER 1 Introducing Geometry L E S S O N 1.3 “When I use a word,” Humpty replied in a scornful tone,“it means just what I choose it to mean—neither more nor less.” “The question is,” said Alice,“whether you can make a word mean so many different things.” LEWIS CARROLL What’s a Widget? Good definitions are very important in geometry. In this lesson you will write your own geometry definitions. Which creatures in the last group are Widgets? A C B D Widgets Not Widgets Who are Widgets? You might have asked yourself, “What things do all the Widgets have in common, and what things do Widgets have that others do not have?” In other words, what characteristics make a Widget a Widget? They all have colorful bodies with nothing else inside; two tails—one like a crescent moon, the other like an eyeball. By observing what a Widget is and what a Widget isn’t, you identified the characteristics that distinguish a Widget from a non-Widget. Based on these characteristics, you should have selected A as the only Widget in the last group. This same process can help you write good definitions of geometric figures. This statement defines a protractor: “A protractor is a geometry tool used to measure angles.” First, you classify what it is (a geometry tool), then you say how it differs from other geometry tools (it is the one you use to measure angles). What should go in the blanks to define a square? A square is a that . Classify it. What is it? How does it differ from others? Once you’ve written a definition, you should test it. To do this, you look for a counterexample. That is, try to create a figure that fits your definition but isn’t what you’re trying to define. If you can come up with a counterexample for your definition, you don’t have a good definition. EXAMPLE A Everyone knows, “A square is a figure with four equal sides.” What’s wrong with this definition? a. Sketch a counterexample. (You can probably find more than one!) b. Write a better definition for a square. LESSON 1.3 What’s a Widget? 47 Solution You probably noticed that “figure” is not specific enough to classify a square, and that “four equal sides” does not specify how it differs from the first counterexample shown below. a. Three counterexamples are shown here, and you may have found others, too. b. One better definition is “A square is a 4-sided figure that has all sides congruent and all angles measuring 90 degrees.” Beginning Steps to Creating a Good Definition 1. Classify your term. What is it? (“A square is a 4-sided figure . . .”) 2. Differentiate your term. How does it differ from others in that class? (“. . . that has four congruent sides and four right angles.”) 3. Test your definition by looking for a counterexample. Ready to write a couple of definitions? First, here are two more types of markings that are very important in geometry. A restaurant counter example The same number of arrow marks indicates that lines are parallel. The symbol means “is parallel to.” A small square in the corner of an angle indicates that it measures 90°. The symbol means “is perpendicular to.” EXAMPLE B Define these terms: a. Parallel lines b. Perpendicular lines k i j i j k Solution Following these steps, classify and differentiate each term. Classify. Differentiate. a. Parallel lines are lines in the same plane that never meet. b. Perpendicular lines are lines that meet at 90 angles. Why do you need to say “in the same plane” for parallel lines but not for perpendicular lines? Sketch or demonstrate a counterexample to show the following definition is incomplete: “Parallel lines are lines that never meet.” (Two lines that do not intersect and are noncoplanar are skew lines.) 48 CHAPTER 1 Introducing Geometry Investigation Defining Angles Here are some examples and non-examples of special types of angles. Step 1 Step 2 Step 3 Step 4 Write a definition for each boldfaced term. Make sure your definitions highlight important differences. Trade definitions and test each other’s definitions by looking for counterexamples. If another group member finds a counterexample to one of your definitions, write a better definition. As a group, decide on the best definition for each term. As a class, agree on common definitions. Add these to your notebook. Draw and label a picture to illustrate each definition. Right Angle 90° 46° 105° Right angles Not right angles Acute Angle 89° 33° 91° 56° 100° Acute angles Not acute angles Notice the many congruent angles in this Navajo transitional Wedgeweave blanket. Are they right, acute, or obtuse angles? Obtuse Angle 101° 91° 129° 38° 89° Obtuse angles Not obtuse angles LESSON 1.3 What’s a Widget? 49 Pair of Vertical Angles 10 Pairs of vertical angles: Not pairs of vertical angles: 1 and 2 3 and 4 AED and BEC AEC and DEB Linear Pair of Angles C A E B D 1 2 3 4 Linear pairs of angles: 1 and 2 3 and 4 AED and AEC BED and DEA Pair of Complementary Angles m1 m2 90° 1 2 3 4 1 and 2 3 and 4 5 and6 7 and 8 9 and 10 1 2 30° 3 4 A B 150° 5 6 Not linear pairs of angles: 1 and 2 3 and 4 5 and 6 A and B m1 m2 90° G 40° 52° H 3 4 1 2 Pairs of complementary angles: 1 and 2 3 and 4 Not pairs of complementary angles: G and H 1 and 2 3 and 4 Pair of Supplementary Angles m3 m4 180° m4 m5 180 Pairs of supplementary angles: Not pairs of supplementary angles: 1 and 2 3 and 4 1, 2, and 3 4 and 5 What types of angles or angle pairs do you see in this magnified view of a computer chip? 50 CHAPTER 1 Introducing Geometry How did you do? Did you notice the difference between a supplementary pair and a linear pair? Did you make it clear which is which in your definitions? The more you practice writing geometry definitions, the better you will get at it. The design of this African Kente cloth contains examples of parallel and perpendicular lines, obtuse and acute angles, and complementary and supplementary angle pairs. To learn about the significance of Kente cloth designs, visit www.keymath.com/DG . EXERCISES For Exercises 1–8, draw and carefully label the figures. Use the appropriate marks to indicate right angles, parallel lines, congruent segments, and congruent angles. Use a protractor and a ruler when you need to. 1. Acute angle DOG with a measure of 45° 3. Obtuse angle BIG with angle bisector IE 5. PE AR 7. Complementary angles A and B with mA 40° 2. Right angle RTE 4. DG MS 6. Vertical angles ABC and DBE 8. Supplementary angles C and D with mD 40° 9. Which creatures in the last group below are Zoids? What makes a Zoid a Zoid? A C B D Zoids Not Zoids Who are Zoids? 10. What are the characteristics of a good definition? For Exercises 11–20, four of the statements are true. Make a sketch or demonstrate each true statement. For each false statement, draw a counterexample. 11. For every line segment there is exactly one midpoint. 12. For every angle there is exactly one angle bisector. 13. If two different lines intersect, then they intersect at one and only one point. 14. If two different circles intersect, then they intersect at one and only one point. LESSON 1.3 What’s a Widget? 51 15. Through a given point on a line there is one and only one line perpendicular to the given line. 16. In every triangle there is exactly one right angle. 17. Through a point not on a line, one and only one line can be constructed parallel to the given line. 18. If CA AT, then A is the midpoint of CT. 19. If mD 40° and mC 140°, then angles C and D are a linear pair. 20. If point A is not the midpoint of CT, then CA AT. 21. There is something wrong with this definition for a pair of vertical angles: “If AB and CD intersect at point P, then APC and BPD are a pair of vertical angles.” Sketch a counterexample to show why it is not correct. Can you add a phrase to correct it? y 5 S –5 T x 5 –5 R A B Mirror Review For Exercises 22 and 23, refer to the graph at right. 22. Find possible coordinates of a point |
P so that points P, T, and S are collinear. 23. Find possible coordinates of a point Q so that QR TS. 24. A partial mirror reflects some light and lets the rest of the light pass through. In the figure at right, half the light from point A passes through the partial mirror to point B. Copy the figure, then draw the outgoing angle for the light reflected from the mirror. What do you notice about the ray of reflected light and the ray of light that passes through? Science Albert Abraham Michelson (1852–1931) designed the Michelson Interferometer to find the wavelength of light. A modern version of the experiment uses a partial mirror to split a laser beam so that it travels in two different directions, and mirrors to recombine the separated beams. Partial mirror Mirrors Lens Laser 52 CHAPTER 1 Introducing Geometry 25. Find possible coordinates of points A, B, and C so that BAC is a right angle, BAT is an acute angle, ABS is an obtuse angle, and the points C, T, and R are collinear. 26. If D is the midpoint of AC and C is the midpoint of AB, and AD 3 cm, what is the length of AB? 27. If BD is the angle bisector of ABC, BE is the angle bisector of ABD, and mDBC 24°, what is mEBC? 28. Draw and label a figure that has two congruent segments and three congruent angles. Mark the congruent angles and congruent segments. y 6 S –12 –6 x T R 29. Show how three lines in a plane can intersect in no points, exactly one point, exactly two points, or exactly three points. 30. Show how it is possible for two triangles to intersect in one point, two points, three points, four points, five points, or six points, but not seven points. Show how they can intersect in infinitely many points. 31. Each pizza is cut into slices from the center. a. What fraction of the pizza is left? b. What fraction of the pizza is missing? c. If the pizza is cut into nine equal slices, how many degrees is each angle at the center of the pizza? 60° 120° IMPROVING YOUR VISUAL THINKING SKILLS Polyominoes In 1953, United States mathematician Solomon Golomb introduced polyominoes at the Harvard Mathematics Club, and they have been played with and enjoyed throughout the world ever since. Polyominoes are shapes made by connecting congruent squares. The squares are joined together side to side. (A complete side must touch a complete side.) Some of the smaller polyominoes are shown below. There is only one monomino and only one domino, but there are two trominoes, as shown. There are five tetrominoes— one is shown. Sketch the other four. Monomino Domino Trominoes Tetromino LESSON 1.3 What’s a Widget? 53 L E S S O N 1.4 There are two kinds of people in this world: those who divide everything into two groups, and those who don’t. KENNETH BOULDING Polygons A polygon is a closed figure in a plane, formed by connecting line segments endpoint to endpoint with each segment intersecting exactly two others. Each line segment is called a side of the polygon. Each endpoint where the sides meet is called a vertex of the polygon. Polygons Not polygons You classify a polygon by the number of sides it has. Familiar polygons have specific names, listed in this table. The ones without specific names are called n-sided polygons, or n-gons. For instance, you call a 25-sided polygon a 25-gon. Consecutive angles E Consecutive vertices A B D C Consecutive sides To name a polygon, list the vertices in consecutive order. You can name the pentagon above pentagon ABCDE. You can also call it DCBAE, but not BCAED. When the polygon is a triangle, you use the triangle symbol. For example, ABC means triangle ABC. A C E D A B B C Pentagon ABCDE ABC A diagonal of a polygon is a line segment that connects two nonconsecutive vertices. A polygon is convex if no diagonal is outside the polygon. A polygon is concave if at least one diagonal is outside the polygon. Sides Name 3 4 5 6 7 8 9 10 11 12 n Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon Nonagon Decagon Undecagon Dodecagon n-gon Diagonal E D F G Convex polygons: All diagonals are inside Concave polygons: One or more diagonals are outside 54 CHAPTER 1 Introducing Geometry Recall that two segments or two angles are congruent if and only if they have the same measures. Two polygons are congruent polygons if and only if they are exactly the same size and shape. “If and only if ” means that the statements work both ways. If polygons are congruent, then corresponding sides and angles are congruent. If corresponding sides and angles are congruent, then polygons are congruent. For example, if quadrilateral CAMP is congruent to quadrilateral SITE, then their four pairs of corresponding angles and four pairs of corresponding sides are also congruent. When you write a statement of congruence, always write the letters of the corresponding vertices in an order that shows the correspondences. How does the shape of the framework of this Marc Chagall (1887–1985) stained glass window support the various shapes of the design? P M E T C I A S CAMP SITE EXAMPLE Which polygon is congruent to ABCDE? ABCDE ? F A 100 130 130 E D B C K J Q 100 T P 130 L 130 S U G H N M W V Solution All corresponding sides and angles must be congruent, so polygon ABCDE polygon QLMNP. You could also say ABCDE QPNML, because all the congruent parts would still match. In an equilateral polygon, all the sides have equal length. In an equiangular polygon, all the angles have equal measure. A regular polygon is both equilateral and equiangular. Equiangular octagon Equilateral octagon Regular octagon LESSON 1.4 Polygons 55 EXERCISES For Exercises 1–8, classify each polygon. Assume that the sides of the chips and crackers are straight. 1. 5. 2. 6. 3. 7. 4. 8. For Exercises 9–11, draw an example of each polygon. 9. Quadrilateral 10. Dodecagon 11. Octagon For Exercises 12 and 13, give one possible name for each polygon. 12. R 13. F O 14. B F I E V R U C 15. Name a pair of consecutive angles and a pair of consecutive sides in the L O figure below. H A C Y T N 16. Draw a concave hexagon. How many diagonals does it have? The repeating pattern of squares and triangles creates a geometric tree in this quilt design by Diane Venters. What other polygons can you find in this quilt? 56 CHAPTER 1 Introducing Geometry 17. Name the diagonals of pentagon ABCDE. For Exercises 18 and 19, use the information given to name the triangle that is congruent to the first one. D 18. EAR ? A I 19. OLD ? E R T N 20. In the figure at right, THINK POWER. a. Find the missing measures. b. If mP 87° and mW 165°, which angles in THINK do you know? Write their measures. 21. If pentagon FIVER is congruent to pentagon PANCH 58 N 20 I L K 34 T 28 44 H then which side in pentagon FIVER is congruent to side PA? Which angle in pentagon PANCH is congruent to IVE? 22. Draw an equilateral concave pentagon. Then draw an equiangular concave pentagon. For Exercises 23–26, copy the given polygon and segment onto graph paper. Give the coordinates of the missing points. 23. CAR PET y 24. TUNA FISH y 6 R P C –5 x 5 E A –6 25. BLUE FISH y 6 E U L 6 x –6 H F B –6 6 A H F N x 6 –7 T U –6 26. RECT ANGL y R L 7 G –7 E T x 12 C LESSON 1.4 Polygons 57 y C (0, 5) B (4, 4) x A (0, 5) 27. Draw an equilateral octagon ABCDEFGH with A(5, 0), B(4,4), and C(0, 5) as three of its vertices. Is it regular? For Exercises 28–32, sketch and carefully label the figure. Mark the congruences. 28. Pentagon PENTA with PE EN 29. Hexagon NGAXEH with HEX EXA 30. Equiangular quadrilateral QUAD with QU QD 31. A hexagon with exactly one line of reflectional symmetry 32. Two different equilateral pentagons with perimeter 25 cm 33. Use your compass, protractor, and straightedge to draw a regular pentagon. 34. A rectangle with perimeter 198 cm is divided into five congruent rectangles as shown in the diagram at right. What is the perimeter of one of the five congruent rectangles? Review 35. Name a pair of complementary angles and a pair of vertical angles in the figure at right. 36. Draw AB, CD, and EF with AB CD and CD EF. 37. Draw a counterexample to show that this statement is false: “If a rectangle has perimeter 50 meters, then a pair of adjacent sides measures 10 meters and 15 meters.” A S 18° T 72° O R C E 38. Is it possible for four lines in a plane to intersect in exactly zero points? One point? Two points? Three points? Four points? Five points? Six points? Draw a figure to support each of your answers. IMPROVING YOUR VISUAL THINKING SKILLS Pentominoes I In Polyominoes I, you learned about shapes called polyominoes. Polyominoes with five squares are called pentominoes. There are 12 pentominoes. Can you find them all? One is shown at right. Use graph paper or square dot paper to sketch all 12. 58 CHAPTER 1 Introducing Geometry L E S S O N 1.5 The difference between the right word and the almost right word is the difference between lightning and the lightning bug. MARK TWAIN Triangles and Special Quadrilaterals You have learned to be careful with geometry definitions. It turns out that you also have to be careful with diagrams. When you look at a diagram, be careful not to assume too much from it. To assume something is to accept it as true without facts or proof. Things you can assume: You may assume that lines are straight, and if two lines intersect, they intersect at one point. Lightning You may assume that points on a line are collinear and that all points shown in a diagram are coplanar unless planes are drawn to show that they are noncoplanar. Not lightning Things you can’t assume: You may not assume that just because two lines or segments look parallel that they are parallel—they must be marked parallel! You may not assume that two lines are perpendicular just because they look perpendicular—they must be marked perpendicular! Pairs of angles, segments, or polygons are not necessarily congruent, unless they are marked with information that tells you they must be congruent! EXAMPLE In the diagrams below, which pairs of lines are perpendicular? Which |
pairs of lines are parallel? Which pair of triangles is congruent LESSON 1.5 Triangles and Special Quadrilaterals 59 Solution By studying the markings, you can tell that AB CD, JK JM, and STU XYZ. In this lesson you will write definitions that classify different kinds of triangles and special quadrilaterals, based on relationships among their sides and angles. Investigation Triangles and Special Quadrilaterals Write a good definition of each boldfaced term. Discuss your definitions with others in your group. Agree on a common set of definitions for your class and add them to your definition list. In your notebook, draw and label a figure to illustrate each definition. Right Triangle 75° 90° 26° 64° 15° 58° 87° 65° 91° 35° 24° Right triangles Not right triangles Acute Triangle 51° 63° 66° 40° 28° 80° 72° 85° 55° Acute triangles Obtuse Triangle 41° 112° 27° 118° 40° 22° 50° 27° 122° 31° 40° Not acute triangles 104° 130° 72° 68° 57° 88° 66° 26° 33° Obtuse triangles Not obtuse triangles What shape is the basis for the design on this Islamic textile from Uzbekistan? 60 CHAPTER 1 Introducing Geometry The Sol LeWitt (b 1928, United States) design inside this art museum uses triangles and quadrilaterals to create a painting the size of an entire room. Sol LeWitt, Wall Drawing #652— On three walls, continuous forms with color ink washes superimposed, color in wash. Collection: Indianapolis Museum of Art, Indianapolis, IN. September, 1990. Courtesy of the artist. Scalene Triangle 15 17 8 6 8 7 60° 28 20 11 6 5 11 8 4 Scalene triangles Not scalene triangles Equilateral Triangle 8 8 8 Equilateral triangles Not equilateral triangles Isosceles Triangle Isosceles triangles Not isosceles triangles LESSON 1.5 Triangles and Special Quadrilaterals 61 In an isosceles triangle, the angle between the two sides of equal length is called the vertex angle. The side opposite the vertex angle is called the base of the isoceles triangle. The two angles opposite the two sides of equal length are called the base angles of the isoceles triangle. Base Base angles Vertex angle Now let’s write definitions for quadrilaterals. Trapezoid Trapezoids Not trapezoids How does this ancient pyramid in Mexico use shapes for practical purposes and also for its overall attractiveness? At the Acoma Pueblo Dwellings in New Mexico, how do sunlight and shadows enhance the existing shape formations? How many shapes make up the overall triangular shapes of these pyramids at the Louvre in Paris? 62 CHAPTER 1 Introducing Geometry Kite Kites Not kites Recreation Today’s kite designers use lightweight plastics, synthetic fabrics, and complex shapes to sustain kites in the air longer than earlier kites that were made of wood, cloth, and had the basic “kite” shape. Many countries even hold annual kite festivals where contestants fly flat kites, box kites, and fighter kites. The design will determine the fastest and most durable kite in the festival. Parallelogram Rhombus Rectangle Parallelograms Not parallelograms Rhombuses Not rhombuses 90° 90° 90° 90° Rectangles Not rectangles LESSON 1.5 Triangles and Special Quadrilaterals 63 Square S P D R Q A C B Squares Not squares As you learned in the investigation, a square is not a square unless it has the proper markings. Keep this in mind as you work on the exercises. EXERCISES 1. Based on the marks, what can you assume to be true in each figure For Exercises 2–9, match the term on the left with its figure on the right. A. E. G. 2. Scalene right triangle 3. Isosceles right triangle 4. Isosceles obtuse triangle 5. Trapezoid 6. Rhombus 7. Rectangle 8. Kite 9. Parallelogram B. F. C. D. H. I. For Exercises 10–18, sketch and label the figure. Mark the figures. 10. Isosceles acute triangle ACT with AC CT 11. Scalene triangle SCL with angle bisector CM 12. Isosceles right triangle CAR with mCRA 90° 64 CHAPTER 1 Introducing Geometry 13. Trapezoid ZOID with ZO ID 14. Kite BENF with BE EN 15. Rhombus EQUL with diagonals EU and QL intersecting at A 16. Rectangle RGHT with diagonals RH and GT intersecting at I 17. Two different isosceles triangles with perimeter 4a b 18. Two noncongruent triangles, each with side 6 cm and an angle measuring 40° 19. Draw a hexagon with exactly two outside diagonals. 20. Draw a regular quadrilateral. What is another name for this shape? 21. Find the other two vertices of a square with one vertex (0, 0) and another vertex (4, 2). Can you find another answer? Austrian architect and artist Friedensreich Hundertwasser (1928–2000) designed the apartment house Hundertwasser-House (1986) with a square spiral staircase in its center. What other shapes do you see? For Exercises 22–24, use the graphs below. Can you find more than one answer? y 5 x –5 –5 R Y –5 y 5 –5 M –5 E y 5 C –5 x –5 L x –5 –5 22. Locate a point L so that LRY is an isosceles triangle. 23. Locate a point O so that MOE is an isosceles right triangle. 24. Locate a point R so that CRL is an isosceles right triangle. Review For Exercises 25–29, tell whether the statement is true or false. For each false statement, sketch a counterexample or explain why the statement is false. 25. An acute angle is an angle whose measure is less than 90°. 26. If two lines intersect to form a right angle, then the lines are perpendicular. 27. A diagonal is a line segment that connects any two vertices of a polygon. LESSON 1.5 Triangles and Special Quadrilaterals 65 28. A ray that divides the angle into two angles is the angle bisector. y 29. An obtuse triangle has exactly one angle whose measure is greater than 90°. (–3, 5) (2, 4) 30. Use the ordered pair rule (x, y) → (x 1, y 3) to relocate the (–4, 1) four vertices of the given quadrilateral. Connect the four new points to create a new quadrilateral. Do the two quadrilaterals appear congruent? Check your guess with tracing paper or patty paper. (1, 1) x 31. Suppose a set of thin rods is glued together into a triangle as shown. How would you place the triangular arrangement of rods onto the edge of a ruler so that they balance? Explain why. DRAWING THE IMPOSSIBLE You experienced some optical illusions with op art in Chapter 0. Some optical illusions are tricks—they at first appear to be drawings of real objects, but actually they are impossible to make, except on paper. Reproduce the two impossible objects by drawing them on full sheets of paper. Can you create an impossible object of your own? Try it. From WALTER WICK’S OPTICAL TRICKS. Published by Cartwheel Books, a division of Scholastic Inc. ©1998 by Walter Wick. Reprinted by permission. 66 CHAPTER 1 Introducing Geometry L E S S O N 1.6 I can never remember things I didn’t understand in the first place. AMY TAN Circles Unless you walked to school this morning, you arrived on a vehicle with circular wheels. A circle is the set of all points in a plane at a given distance (radius) from a given point (center) in the plane. You name a circle by its center. The circle on the bicycle wheel, with center O, is called circle O. When you see a dot at the center of a circle, you can assume that it represents the center point. A segment from the center to a point on the edge of the circle is called a radius. Its length is also called the radius. The diameter is a line segment containing the center, with its endpoints on the circle. The length of this segment is also called the diameter. By permission of Johnny Hart and Creators Syndicate, Inc. O LESSON 1.6 Circles 67 If two or more circles have the same radius, they are congruent circles. If two or more coplanar circles share the same center, they are concentric circles. Congruent circles Concentric circles An arc of a circle is two points on the circle and the continuous (unbroken) part of the circle between the two points. The two points are called the endpoints of the arc. A B or BA . You classify arcs into three types: semicircles, minor You write arc AB as AB arcs, and major arcs. A semicircle is an arc of a circle whose endpoints are the endpoints of a diameter. A minor arc is an arc of a circle that is smaller than a semicircle. A major arc is an arc of a circle that is larger than a semicircle. You can name minor arcs with the letters of the two endpoints. For semicircles and major arcs, you need three points to make clear which arc you mean—the first and last letters are the endpoints and the middle letter is any other point on the arc. P Minor arc AP Semicircle APD A O D Major arc PAD Arcs have a degree measure, just as angles do. A full circle has an arc measure of 360°, a semicircle has an arc measure of 180°, and so on. You find the arc measure by measuring the central angle, the angle with its vertex at the center of the circle, and sides passing through the endpoints of the arc. BC a minor arc mBC 45° B 45° C BDC a major arc mBDC 315° 315° D 68 CHAPTER 1 Introducing Geometry Investigation Defining Circle Terms Step 1 Write a good definition of each boldfaced term. Discuss your definitions with others in your group. Agree on a common set of definitions as a class and add them to your definition list. In your notebook, draw and label a figure to illustrate each definition. Chord Chords: AB, CD, EF, GH, and IJ Diameter C A E O F B D Not chords: PQ, RS, TU , and VW R T W P S Q V U Diameters: AB, CD, and EF Not diameters: PQ, RS, TU, and VW Tangent Tangents: AB, CD, and EF Not tangents: PQ, RS, TU, and VW Note: You can say AB is a tangent, or you can say AB is tangent to circle O. The point where the tangent touches the circle is called the point of tangency. Step 2 Step 3 Can a chord of a circle also be a diameter of the circle? Can it be a tangent? Explain why or why not. Can two circles be tangent to the same line at the same point? Draw a sketch and explain. LESSON 1.6 Circles 69 In each photo, find examples of the terms introduced in this lesson. Circular irrigation on a farm Japanese wood bridge CAD design of pistons in a car engine EXERCISES For Exercises 1–8, use the diagram at right. Points E, P, and |
C are collinear. 1. Name three chords. 2. Name one diameter. 3. Name five radii. 4. Name five minor arcs. 5. Name two semicircles. 6. Name two major arcs. 7. Name two tangents. 8. Name a point of tangency. 9. Name two types of vehicles that use wheels, two household appliances that use wheels, and two uses of the wheel in the world of entertainment. 10. In the figure at right, what is mPQ ? ? mPRQ 11. Use your compass and protractor to make an arc with measure 65°. Now make an arc with measure 215°. Label each arc with its measure 110° O R 70 CHAPTER 1 Introducing Geometry 12. Name two places or objects where concentric circles appear. Bring an example of a set of concentric circles to class tomorrow. You might look in a magazine for a photo or make a copy of a photo from a book (but not this book!). 13. Sketch two circles that appear to be concentric. Then use your compass to construct a pair of concentric circles. 14. Sketch circle P. Sketch a triangle inside circle P so that the three sides of the triangle are chords of the circle. This triangle is “inscribed” in the circle. Sketch another circle and label it Q. Sketch a triangle in the exterior of circle Q so that the three sides of the triangle are tangents of the circle. This triangle is “circumscribed” about the circle. 15. Use your compass to construct two circles with the same radius intersecting at two points. Label the centers P and Q. Label the points of intersection of the two circles A and B. Construct quadrilateral PAQB. What type of quadrilateral is it? 16. Do you remember the daisy construction from Chapter 0? Construct a circle with radius s. With the same compass setting, divide the circle into six congruent arcs. Construct the chords to form a regular hexagon inscribed in the circle. Construct radii to each of the six vertices. What type of triangles are formed? What is the ratio of the perimeter of the hexagon to the diameter of the circle? 17. Sketch the path made by the midpoint of a radius of a circle if the radius is rotated about the center. For Exercises 18–20, use the ordered pair rule shown to relocate the four points on the given circle. Can the four new points be connected to create a new circle? Does the new figure appear congruent to the original circle? 18. (x, y) → (x 1, y 2) y 19. (x, y) → (2x, 2y) 20. (x, y) → (2x, y5 x 5 –5 5 –5 –5 Review For Exercises 21–24, draw each kind of triangle or write “not possible” and explain why. Use your geometry tools to make your drawings as accurate as possible. 21. Isosceles right triangle 23. Scalene obtuse triangle 22. Scalene isosceles triangle 24. Isosceles obtuse triangle LESSON 1.6 Circles 71 For Exercises 25–33, sketch, label, and mark the figure. 25. Obtuse scalene triangle FAT with mFAT 100° 26. Trapezoid TRAP with TR AP and TRA a right angle 27. Two different (noncongruent) quadrilaterals with angles of 60°, 60°, 120°, and 120° 28. Equilateral right triangle 29. Right isosceles triangle RGT with RT GT and mRTG 90° 30. An equilateral triangle with perimeter 12a 6b 31. Two triangles that are not congruent, each with angles measuring 50° and 70° 32. Rhombus EQUI with perimeter 8p and mIEQ 55° 33. Kite KITE with TE 2EK and mTEK 120° IMPROVING YOUR REASONING SKILLS Checkerboard Puzzle 1. Four checkers—three red and one black—are arranged on the corner of a checkerboard, as shown at right. Any checker can jump any other checker. The checker that was jumped over is then removed. With exactly three horizontal or vertical jumps, remove all three red checkers, leaving the single black checker. Record your solution. 2. Now, with exactly seven horizontal or vertical jumps, remove all seven red checkers, leaving the single black checker. Record your solution 72 CHAPTER 1 Introducing Geometry L E S S O N 1.7 You can observe a lot just by watching. YOGI BERRA A Picture Is Worth a Thousand Words A picture is worth a thousand words! That expression certainly applies to geometry. A drawing of an object often conveys information more quickly than a long written description. People in many occupations use drawings and sketches to communicate ideas. Architects create blueprints. Composers create musical scores. Choreographers visualize and map out sequences of dance steps. Basketball coaches design plays. Interior designers—well, you get the picture. Visualization skills are extremely important in geometry. So far, you have visualized geometric situations in every lesson. To visualize a plane, you pictured a flat surface extending infinitely. In another lesson, you visualized the number of different ways that four lines can intersect. Can you picture what the hands of a clock look like when it is 3:30? 10 9 8 12 1 11 By drawing diagrams, you apply visual thinking to problem solving. Let’s look at some examples that show how to use visual thinking to solve word problems. EXAMPLE A Volumes 1 and 2 of a two-volume set of math books sit next to each other on a shelf. They sit in their proper order: Volume 1 on the left and Volume 2 on the right. Each front and back cover is 1 inch thick, and the 8 pages portion of each book is 1 inch thick. If a bookworm starts at the first page of Volume 1 and burrows all the way through to the last page of Volume 2, how far will she travel? Take a moment and try to solve the problem in your head. Bookplate for Albert Ernst Bosman, M. C. Escher, 1946 ©2002 Cordon Art B. V.–Baarn–Holland. All rights reserved. LESSON 1.7 A Picture Is Worth a Thousand Words 73 Solution Did you get 21 inches? It seems reasonable, doesn’t it? 4 Guess what? That’s not the answer. Let’s get organized. Reread the problem to identify what information you are given and what you are trying to find. First page of Volume 1 You are given the thickness of each cover, the thickness of the page portion, and the position of the books on the shelf. You are trying to find how far it is from the first page of Volume 1 to the last page of Volume 2. Draw a picture and locate the position of the pages referred to in the problem. Now “look” how easy it is to solve the problem. She traveled only 1 inch through the two covers! 4 Last page of Volume 2 EXAMPLE B In Reasonville, many streets are named after famous mathematicians. Streets that end in an “s” run east–west. All other streets might run either way. Wiles Street runs perpendicular to Germain Street. Fermat Street runs parallel to Germain Street. Which direction does Fermat Street run? Mathematics Sophie Germain (1776–1831), a French mathematician with no formal education, wrote a prize treatise, contributed to many theories, and worked extensively on Fermat’s Last Theorem. The French mathematician Pierre de Fermat (1601–1665) developed analytic geometry. His algebraic approach is what made his influence on geometry so strong. Andrew Wiles (b 1953), an English mathematician at Princeton University, began trying to prove Fermat’s Last Theorem when he was just 10 years old. In 1993, after spending most of his career working on the theorem, sometimes in complete isolation, he announced a proof of the problem. Solution Did you make a diagram? You can start your diagram with the first piece of information. Then you can add to the diagram as new information is added. Wiles Street ends in an “s,” so it runs east–west. You are trying to find the direction of Fermat Street. 74 CHAPTER 1 Introducing Geometry EXAMPLE C N N W Wiles E W Wiles E W Wiles E Germain Germain Fermat S S Initial diagram Improved diagram Final diagram The final diagram reveals the answer. Fermat Street runs north–south. Sometimes there is more than one point or even many points that satisfy a set of conditions. The set of points is called a locus of points. Let’s look at an example showing how to solve a locus problem. Harold, Dina, and Linda are standing on a flat, dry field reading their treasure map. Harold is standing at one of the features marked on the map, a gnarled tree stump, and Dina is standing atop a large black boulder. The map shows that the treasure is buried 60 meters from the tree stump and 40 meters from the large black boulder. Harold and Dina are standing 80 meters apart. What is the locus of points where the treasure might be buried? Solution Start by drawing a diagram based on the information given in the first two sentences, then add to the diagram as new information is added. Can you visualize all the points that are 60 meters from the tree stump? Mark them on your diagram. They should lie on a circle. The treasure is also 40 meters from the boulder. All the possible points lie in a circle around the boulder. The two possible spots where the treasure might be buried, or the locus of points, are the points where the two circles intersect. H 80 m D 60 m H D 60 m H Treasure here 40 m D Or here Initial diagram Improved diagram Final diagram LESSON 1.7 A Picture Is Worth a Thousand Words 75 EXERCISES 1. Surgeons, engineers, carpenters, plumbers, electricians, and furniture movers all rely on trained experience with visual thinking. Describe how one of these tradespeople or someone in another occupation uses visual thinking in his or her work. You will need Construction tools for Exercises 33 and 34 Now try your hand at some word problems. Read each problem carefully, determine what you are trying to find, and draw and label a diagram. Finally, solve the problem. 2. In the city of Rectangulus, all the streets running east–west are numbered and those streets running north–south are lettered. The even-numbered streets are one-way east and the odd-numbered streets are one-way west. All the vowel-lettered avenues are one-way north and the rest are two-way. Can a car traveling south on S Street make a legal left turn onto 14th Street? 3. Freddie the Frog is at the bottom of a 30-foot well. Each day he jumps up 3 feet, but then, during the night, he slides back down 2 feet. How many days will it take Freddie to get to the top and out? 4. Mary Ann is building a fence around the ou |
ter edge of a rectangular garden plot that measures 25 feet by 45 feet. She will set the posts 5 feet apart. How many posts will she need? 5. Midway through a 2000-meter race, a photo is taken of five runners. It shows Meg 20 meters behind Edith. Edith is 50 meters ahead of Wanda, who is 20 meters behind Olivia. Olivia is 40 meters behind Nadine. Who is ahead? In your diagram, use M for Meg, E for Edith, and so on. 6. Here is an exercise taken from Marilyn vos Savant’s Ask Marilyn® column in Parade magazine. It is a good example of a difficultsounding problem becoming clear once a diagram has been made. Try it. A 30-foot cable is suspended between the tops of two 20-foot poles on level ground. The lowest point of the cable is 5 feet above the ground. What is the distance between the two poles? 7. Points A and B lie in a plane. Sketch the locus of points in the plane that are equally distant from points A and B. Sketch the locus of points in space that are equally distant from points A and B. 8. Draw an angle. Label it A. Sketch the locus of points in the plane of angle A that are the same distance from the two sides of angle A. 9. Line AB lies in plane . Sketch the locus of points in plane that are 3 cm from AB. Sketch the locus of points in space that are 3 cm from AB. 76 CHAPTER 1 Introducing Geometry 10. Beth Mack and her dog Trouble are exploring in the woods east of Birnam Woods Road, which runs north-south. They begin walking in a zigzag pattern: 1 km south, 1 km west, 1 km south, 2 km west, 1 km south, 3 km west, and so on. They walk at the rate of 4 km/h. If they started 15 km east of Birnam Woods Road at 3:00 P.M., and the sun sets at 7:30 P.M., will they reach Birnam Woods Road before sunset? In geometry you will use visual thinking all the time. In Exercises 11 and 12 you will be asked to locate and recognize congruent geometric figures even if they are in different positions due to translations (slides), rotations (turns), or reflections (flips). 11. If trapezoid ABCD were rotated 90° counterclockwise about (0, 0), to what (x, y) location would points A, B, C, and D be relocated? 12. If CYN were reflected over the y-axis, to what location would points C, N, and Y be relocated? y D (1, 2) C (3, 2) A (0, 0) x B (5, 0) y N (0, 3) Y (4, 1) x C (–3, –1) 13. What was the ordered pair rule used to relocate the four vertices of ABCD to ABCD4 A A –4 14. Which lines are perpendicular? Which lines are parallel 15. Sketch the next two figures in the pattern below. If this pattern were to continue, what would be the perimeter of the eighth figure in the pattern? (Assume the length of each segment is 1 cm.) , , , . . . , LESSON 1.7 A Picture Is Worth a Thousand Words 77 16. Many of the geometric figures you have defined are closely related to one another. A diagram can help you see the relationships among them. At right is a concept map showing the relationships among members of the triangle family. This type of concept map is known as a tree diagram because the relationships are shown as branches of a tree. Copy and fill in the missing branches of the tree diagram for triangles. 17. At right is a concept map showing the relationships among some members of the parallelogram family. This type of concept map is known as a Venn diagram. Fill in the missing names. Acute Triangles Obtuse ? ? Scalene Isosceles Isosceles Isosceles Scalene Equilateral Parallelograms ? Squares ? A net is a two-dimensional pattern that you can cut and fold to form a three-dimensional figure. Another visual thinking skill you will need is the ability to visualize nets being folded into solid objects and geometric solids being unfolded into nets. The net below left can be folded into a cube and the net below right can be folded into a pyramid. Net for a cube Net for a square-based pyramid 18. Which net(s) will fold to make a cube? A. D. B. E. C. F. For Exercises 19–22, match the net with its geometric solid. 19. A. 20. B. 21. C. 22. D. 78 CHAPTER 1 Introducing Geometry Review For Exercises 23–32, write the words or the symbols that make the statement true. 23. The three undefined terms of geometry are ? , ? , and ? . 24. “Line AB” may be written using a symbol as ? . 25. “Arc AB” may be written using a symbol as ? . 26. The point where the two sides of an angle meet is the ? of the angle. 27. “Ray AB” may be written using a symbol as ? . 28. “Line AB is parallel to segment CD” is written in symbolic form as ? . 29. The geometry tool you use to measure an angle is a ? . 30. “Angle ABC” is written in symbolic form as ? . 31. The sentence “Segment AB is perpendicular to line CD” is written in symbolic form as ? . 32. The angle formed by a light ray coming into a mirror is ? the angle formed by a light ray leaving the mirror. William Thomas Williams, DO YOU THINK A IS B, acrylic on canvas, 1975–77, Fisk University Galleries, Nashville, Tennessee. 33. Use your compass to draw two congruent circles intersecting in exactly one point. How does the distance between the two centers compare with the radius? 34. Use your compass to construct two congruent circles so that each circle passes through the center of the other circle. Label the centers P and Q. Construct PQ connecting the centers. Label the points of intersection of the two circles A and B. Construct chord AB. What is the relationship between AB and PQ? IMPROVING YOUR VISUAL THINKING SKILLS Hexominoes Polyominoes with six squares are called hexominoes. There are 35 different hexominoes. There is 1 with a longest string of six squares; there are 3 with a longest string of five squares, 13 with a longest string of four squares, 17 with a longest string of three squares; and there is 1 with a longest string of two squares. Use graph paper to sketch the 35 hexominoes which are nets for a cube. Here is one hexomino that does fold into a cube. LESSON 1.7 A Picture Is Worth a Thousand Words 79 L E S S O N 1.8 When curiosity turns to serious matters, it’s called research. MARIE VON EBNERESCHENBACH Space Geometry Lesson 1.1 introduced you to point, line, and plane. Throughout this chapter you have used these terms to define a wide range of other geometric figures, from rays to polygons. You did most of your work on a single flat surface, a single plane. Some problems, however, required you to step out of a single plane to visualize geometry in space. In this lesson you will learn more about space geometry, or solid geometry. Space is the set of all points. Unlike lines and planes, space cannot be contained in a flat surface. Space is three-dimensional, or “3-D.” In an “edge view,” you see the front edge of a building as a vertical line, and the other edges as diagonal lines. Isometric dot paper helps you draw these lines, as you can see in the steps below. Let’s practice the visual thinking skill of presenting three-dimensional (3-D) objects in two-dimensional (2-D) drawings. The geometric solid you are probably most familiar with is a box, or rectangular prism. Below are steps for making a two-dimensional drawing of a rectangular prism. This type of drawing is called an isometric drawing. It shows three sides of an object in one view (an edge view). This method works best with isometric dot grid paper. After practicing, you will be able to draw the box without the aid of the dot grid. Step 1 Step 2 Step 3 Step 4 Use dashed lines for edges that you couldn’t see if the object were solid. 80 CHAPTER 1 Introducing Geometry The three-dimensional objects you will study include the six types of geometric solids shown below. Prism Pyramid Cylinder Cone Sphere Hemisphere The shapes of these solids are probably already familiar to you even if you are not familiar with their proper names. The ability to draw these geometric solids is an important visual thinking skill. Here are some drawing tips. Remember to use dashes for the hidden lines. Step 1 Step 2 Step 3 Step 4 Step 1 Step 2 Step 1 Step 2 Step 3 Step 4 LESSON 1.8 Space Geometry 81 Pyramid Cylinder Prism Cone Sphere Step 1 Step 2 Step 1 Step 2 Hemisphere Step 1 Step 2 Solid geometry also involves visualizing points and lines in space. In the following investigation, you will have to visualize relationships between geometric figures in a plane and in space. Investigation Space Geometry Step 1 Step 2 Make a sketch or use physical objects to demonstrate each statement in the list below. Work with your group to determine whether each statement is true or false. If the statement is false, draw a picture and explain why it is false. 1. Only one line can be drawn through two different points. 2. Only one plane can pass through one line and a point that is not on the line. 3. If two coplanar lines are both perpendicular to a third line in the same plane, then the two lines are parallel. 4. If two planes do not intersect, then they are parallel. 5. If two lines do not intersect, then they must be parallel. 6. If a line is perpendicular to two lines in a plane, but the line is not contained in the plane, then the line is perpendicular to the plane. 82 CHAPTER 1 Introducing Geometry EXERCISES For Exercises 1–6, draw each figure. Study the drawing tips provided on the previous page before you start. 1. Cylinder 2. Cone 3. Prism with a hexagonal base 4. Sphere 5. Pyramid with a heptagonal base 6. Hemisphere 7. The photo at right shows a prism-shaped building with a pyramid roof and a cylindrical porch. Draw a cylindrical building with a cone roof and a prismshaped porch. For Exercises 8 and 9, make a drawing to scale of each figure. Use isometric dot grid paper. Label each figure. (For example, in Exercise 8, draw the solid so that the dimensions measure 2 units by 3 units by 4 units, then label the figure with meters.) 8. A rectangular solid 2 m by 3 m by 4 m, sitting on its A police station, or koban, in Tokyo, Japan biggest face. 9. A rectangular solid 3 inches by 4 inches by 5 inches, resting on its smallest face. Draw lines on the three visible surfaces showing |
how you can divide the solid into cubic-inch boxes. How many such boxes will fit in the solid? For Exercises 10–12, use isometric dot grid paper to draw the figure shown. 10. 11. 12. For Exercises 13–15, sketch the three-dimensional figure formed by folding each net into a solid. Name the solid. 13. 14. 15. LESSON 1.8 Space Geometry 83 For Exercises 16 and 17, find the lengths x and y. (Every angle on each block is a right angle.) 16. 20 45 x 17 y 13 45 18 17. 3 8 y 2 2 7 x 2 In Exercises 18 and 19, each figure represents a two-dimensional figure with a wire attached. The three-dimensional solid formed by spinning the figure on the wire between your fingers is called a solid of revolution. Sketch the solid of revolution formed by each two-dimensional figure. 18. 19. When a solid is cut by a plane, the resulting two-dimensional figure is called a section. For Exercises 20 and 21, sketch the section formed when each solid is sliced by the plane, as shown. 20. 21. A real-life example of a “solid of revolution” is a clay pot on a potter’s wheel. Slicing a block of clay reveals a section of the solid. Here, the section is a rectangle. 84 CHAPTER 1 Introducing Geometry All of the statements in Exercises 22–29 are true except for two. Make a sketch to demonstrate each true statement. For each false statement, draw a sketch and explain why it is false. 22. Only one plane can pass through three noncollinear points. 23. If a line intersects a plane that does not contain the line, then the intersection is exactly one point. 24. If two lines are perpendicular to the same line, then they are parallel. 25. If two different planes intersect, then their intersection is a line. Physical models can help you visualize the intersections of lines and planes in space. Can you see examples of intersecting lines in this photo? Parallel lines? Planes? Points? 26. If a line and a plane have no points in common, then they are parallel. 27. If a plane intersects two parallel planes, then the lines of intersection are parallel. 28. If three random planes intersect (no two are parallel and all three do not share the same line), then they divide space into six parts. 29. If two lines are perpendicular to the same plane, then they are parallel to each other. Review 30. If the kite DIAN were rotated 90° clockwise about the origin, to what location would point A be relocated? 31. Use your ruler to measure the perimeter of WIM (in centimeters) and your protractor to measure the largest angle. 32. Use your geometry tools to draw a triangle with two sides of length 8 cm and length 13 cm and the angle between them measuring 120°. y N (0, 0) x D (–3, –1) I (–3, –3) A (–1, –3) I W M IMPROVING YOUR VISUAL THINKING SKILLS Equal Distances Here’s a real challenge: Show four points A, B, C, and D so that AB BC AC AD BD CD. LESSON 1.8 Space Geometry 85 Geometric Probability I You probably know what probability means. The probability, or likelihood, of a particular outcome is the ratio of the number of successful outcomes to the number of possible outcomes. So the probability of rolling a 4 on a 6-sided die is 1 . Or, you can name 6 an event that involves more than one outcome, like getting the total 4 on two 6-sided dice. Since each die can come up in six different ways, there are 6 6 or 36 combinations (count ’em!). You can get the total 4 with a 1 and a 3, a 3 and a 1, or a 2 and a 2. So the probability 1 3 of getting the total 4 is . or 1 2 3 6 Anyway, that’s the theory. Activity Chances Are In this activity you’ll see that you can apply probability theory to geometric figures. The Spinner After you’ve finished your homework and have eaten dinner, you play a game of chance using the spinner at right. Where the spinner lands determines how you’ll spend the evening. Sector A: Playing with your younger brother the whole evening Sector B: Half the evening playing with your younger brother and half the evening watching TV Sector C: Cleaning the birdcage, the hamster cage, and the aquarium the whole evening Sector D: Playing in a band in a friend’s garage the whole evening You will need ● a protractor ● a ruler 86 CHAPTER 1 Introducing Geometry Step 1 Step 2 What is the probability of landing in each sector? What is the probability that you’ll spend at least half the evening with your younger brother? What is the probability that you won’t spend any time with him? The Bridge A computer programmer who is trying to win money on a TV survival program builds a 120 ft rope bridge across a piranha-infested river 90 ft below. Step 3 Step 4 If the rope breaks where he is standing (a random point), but he is able to cling to one end of it, what is the probability that he’ll avoid getting wet (or worse)? Suppose the probability that the rope breaks at all is 1 . Also suppose that, as long 2 as he doesn’t fall more than 30 ft, the probability that he can climb back up is 3 . 4 What is the probability that he won’t fall at all? What is the probability that if he does, he’ll be able to climb back up? The Bus Stop Noriko arrives at the bus stop at a random time each day. Her bus stops there every 20 minutes between 3:00 and 4:30 P.M. Step 5 Step 6 Draw a number line to show stopping times. (Don’t worry about the length of time that the bus is actually stopped. Assume it is 0 minutes.) What is the probability that she will have to wait 5 minutes or more? 10 minutes or more? Hint: What line lengths represent possible waiting time? Step 7 If the bus stops for exactly 3 minutes, how do your answers to Step 5 change? Step 8 Step 9 List the geometric properties you needed in each of the three scenarios above and tell how your answers depended on them. How is geometric probability like the probability you’ve studied before? How is it different? Step 10 Create your own geometric probability problem. EXPLORATION Geometric Probability 1 87 ● CHAPTER 11 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER CHAPTER 1 R E V I E W It may seem that there’s a lot to memorize in this chapter. But having defined terms yourself, you’re more likely to remember and understand them. The key is to practice using these new terms and to be organized. Do the following exercises, then read Assessing What You’ve Learned for tips on staying organized. Whether you’ve been keeping a good list or not, go back now through each lesson in the chapter and double-check that you’ve completed each definition and that you understand it. For example, if someone mentions a geometry term to you, can you sketch it? If you are shown a geometric figure can you classify it? Compare your list of geometry terms with the lists of your group members. EXERCISES For Exercises 1–16, identify the statement as true or false. For each false statement, explain why it is false or sketch a counterexample. 1. The three basic building blocks of geometry are point, line, and plane. 2. “The ray through point P from point Q” is written in symbolic form as PQ. 3. “The length of segment PQ”can be written as PQ. 4. The vertex of angle PDQ is point P. 5. The symbol for perpendicular is . 6. A scalene triangle is a triangle with no two sides the same length. 7. An acute angle is an angle whose measure is more than 90°. 8. If AB intersects CD at point P, then APD and APC are a pair of vertical angles. 9. A diagonal is a line segment in a polygon connecting any two nonconsecutive vertices. 10. If two lines lie in the same plane and are perpendicular to the same line, then they are parallel. 11. If the sum of the measures of two angles is 180°, then the two angles are complementary. 12. A trapezoid is a quadrilateral having exactly one pair of parallel sides. 13. A polygon with ten sides is a decagon. 88 CHAPTER 1 Introducing Geometry A knowledge of parallel lines, planes, arcs, circles, and symmetry is necessary to build durable guitars that sound pleasing. ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 14. A square is a rectangle with all the sides equal in length. 15. A pentagon has five sides and six diagonals. 16. The largest chord of a circle is a diameter of the circle. For Exercises 17–25, match each term with its figure below, or write “no match.” 17. Isosceles acute triangle 18. Isosceles right triangle 19. Rhombus 20. Trapezoid 23. Concave polygon 21. Pyramid 24. Chord 22. Cylinder 25. Minor arc A. E. I. ? B. F. J. M. N. D. H. ? L. C. G. K. O. For Exercises 26–33, sketch, label, and mark each figure. 26. Kite KYTE with KY YT 27. Scalene triangle PTS with PS 3, ST 5, PT 7, and angle bisector SO 28. Hexagon REGINA with diagonal AG parallel to sides RE and NI 29. Trapezoid TRAP with AR and PT the nonparallel sides. Let E be the midpoint of PT and let Y be the midpoint of AR. Draw EY. 30. A triangle with exactly one line of reflectional symmetry 31. A circle with center at P, radii PA and PT, and chord TA creating a minor arc TA CHAPTER 1 REVIEW 89 EW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTE 32. A pair of concentric circles with the diameter AB of the inner circle perpendicular at B to a chord CD of the larger circle 33. A pyramid with a pentagonal base 34. Draw a rectangular prism 2 inches by 3 inches by 5 inches, resting on its largest face. Draw lines on the three visible faces, showing how the solid can be divided into 30 smaller cubes. 35. Use your protractor to draw a 125° angle. 36. Use your protractor, ruler, and compass to draw an isosceles triangle with a vertex angle having a measure of 40°. 37. Use your geomety tools to draw a regular octagon. 38. What is the measure of A? Use your protractor. For Exercises 39–42, find the lengths x and y. y x 39. 42. 40. 1 3 8 3 2 4 4 18 y 7 x 2 2 2 20 A 41. y x x x 10 12 2y y 8 x y 12 43. If D is the midpoint of AC, C is the midpoint of AB, and BD 12 cm, what is the length of AB? 44. If BD is the angle bisector of ABC and BE is the angle bisector of DBC, find mEBA if mDBE 32°? 45. Wha |
t is the measure of the angle formed by the hands of the clock at 2:30? 46. If the pizza is cut into 12 congruent pieces, how many degrees are in each central angle? 47. Make a concept map (a tree diagram or a Venn diagram) to organize these quadrilaterals: rhombus, rectangle, square, trapezoid. 90 CHAPTER 1 Introducing Geometry ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 9 in. 14 in. 48. The box at right is wrapped with two strips of ribbon, as shown. What is the minimum length of ribbon needed to decorate the box? 5 in. 49. At one point in a race, Rico was 15 ft behind Paul and 18 ft ahead of Joe. Joe was trailing George by 30 ft. Paul was ahead of George by how many ft? 50. A large aluminum ladder was resting vertically against the research shed at midnight when it began to slide down the side of the shed. A burglar was clinging to the ladder’s midpoint, holding a pencil flashlight that was visible in the dark. Witness Jill Seymour claimed to see the ladder slide. What did she see? That is, what was the path taken by the bulb of the flashlight? Draw a diagram showing the path. (Devise a physical test to check your visual thinking. You might try sliding a meterstick against a wall, or you might plot points on graph paper.) 51. Jiminey Cricket is caught in a windstorm. At 5:00 P.M. he is 500 cm away from his home. Each time he jumps toward home he leaps a distance of 50 cm, but before he regains strength to jump again he is blown back 40 cm. If it takes a full minute between jumps, how long will it take Jiminey to get home? 52. If the right triangle BAR were rotated 90° clockwise about point B, to what location would point A be relocated? 53. Sketch the three-dimensional figure formed by folding the net below into a solid. y R (2, 2) x A (–2, –1) B (2, –1) 54. Sketch the solid of revolution formed when you spin the two-dimensional figure about the line. 55. Sketch the section formed when the solid is sliced by the plane, as shown. CHAPTER 1 REVIEW 91 EW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTER 1 REVIEW ● CHAPTE 56. Use an isometric dot grid to sketch the 57. Sketch the figure shown with the red edge figure shown below. vertical and facing the viewer. Assessing What You’ve Learned ORGANIZE YOUR NOTEBOOK Is this textbook filling up with folded-up papers stuffed between pages? If so, that’s a bad sign! But it’s not too late to get organized. Keeping a well-organized notebook is one of the best habits you can develop to improve and assess your learning. You should have sections for your classwork, definition list, and homework exercises. There should be room to make corrections and to summarize what you learned and write down questions you still have. Many books include a definition list (sometimes called a glossary) in the back. This book makes you responsible for your own glossary, so it’s essential that, in addition to taking good notes, you keep a complete definition list that you can refer to. You started a definition list in Lesson 1.1. Get help from classmates or your teacher on any definition you don’t understand. As you progress through the course, your notebook will become more and more important. A good way to review a chapter is to read through the chapter and your notes and write a one-page summary of the chapter. And if you create a one-page summary for each chapter, the summaries will be very helpful to you when it comes time for midterms and final exams. You’ll find no better learning and study aid than a summary page for each chapter, and your definition list, kept in an organized notebook. UPDATE YOUR PORTFOLIO If you did the project in this chapter, document your work and add it to your portfolio. Choose one homework assignment that demonstrates your best work in terms of completeness, correctness, and neatness. Add it (or a copy of it) to your portfolio. 92 CHAPTER 1 Introducing Geometry CHAPTER 2 Reasoning in Geometry That which an artist makes is a mirror image of what he sees around him. M. C. ESCHER Hand with Reflecting Sphere (Self-Portrait in Spherical Mirror), M. C. Escher ©2002 Cordon Art B. V.–Baarn–Holland. All rights reserved In this chapter you will ● perform geometry investigations and make many discoveries by observing common features or patterns ● use your discoveries to solve problems through a process called inductive reasoning ● use inductive reasoning to discover patterns ● learn to use deductive reasoning ● learn about vertical angles and linear pairs ● make conjectures L E S S O N 2.1 We have to reinvent the wheel every once in a while, not because we need a lot of wheels; but because we need a lot of inventors. BRUCE JOYCE Inductive Reasoning As a child you learned by experimenting with the natural world around you. You learned how to walk, to talk, and to ride your first bicycle, all by trial and error. From experience you learned to turn a water faucet on with a counterclockwise motion and to turn it off with a clockwise motion. You achieved most of your learning by a process called inductive reasoning. It is the process of observing data, recognizing patterns, and making generalizations about those patterns. Geometry is rooted in inductive reasoning. In ancient Egypt and Babylonia, geometry began when people developed procedures for measurement after much experience and observation. Assessors and surveyors used these procedures to calculate land areas and to reestablish the boundaries of agricultural fields after floods. Engineers used the procedures to build canals, reservoirs, and the Great Pyramids. Throughout this course you will use inductive reasoning. You will perform investigations, observe similarities and patterns, and make many discoveries that you can use to solve problems. Language The word “geometry” means “measure of the earth” and was originally inspired by the ancient Egyptians. The ancient Egyptians devised a complex system of land surveying in order to reestablish land boundaries that were erased each spring by the annual flooding of the Nile River. Inductive reasoning guides scientists, investors, and business managers. All of these professionals use past experience to assess what is likely to happen in the future. When you use inductive reasoning to make a generalization, the generalization is called a conjecture. Consider the following example from science. 94 CHAPTER 2 Reasoning in Geometry EXAMPLE A Solution A scientist dips a platinum wire into a solution containing salt (sodium chloride), passes the wire over a flame, and observes that it produces an orange-yellow flame. She does this with many other solutions that contain salt, finding that they all produce an orange-yellow flame. Make a conjecture based on her findings. The scientist tested many other solutions containing salt, and found no counterexamples. You should conjecture: “If a solution contains sodium chloride, then in a flame test it produces an orange-yellow flame.” Platinum wire flame test Like scientists, mathematicians often use inductive reasoning to make discoveries. For example, a mathematician might use inductive reasoning to find patterns in a number sequence. Once he knows the pattern, he can find the next term. EXAMPLE B Consider the sequence 2, 4, 7, 11, . . . Make a conjecture about the rule for generating the sequence. Then find the next three terms. Solution Look at the numbers you add to get each term. The 1st term in the sequence is 2. You add 2 to find the 2nd term. Then you add 3 to find the 3rd term, and so on. +2 +3 +4 2, 4, 7, 11 You can conjecture that if the pattern continues, you always add the next counting number to get the next term. The next three terms in the sequence will be 16, 22, and 29. +5 +6 +7 11, 16, 22, 29 In the following investigation you will use inductive reasoning to recognize a pattern in a series of drawings and use it to find a term much farther out in a sequence. LESSON 2.1 Inductive Reasoning 95 Investigation Shape Shifters Look at the sequence of shapes below. Pay close attention to the patterns that occur in every other shape. Step 1 Step 2 Step 3 Step 4 Step 5 What patterns do you notice in the 1st, 3rd, and 5th shapes? What patterns do you notice in the 2nd, 4th, and 6th shapes? Draw the next two shapes in the sequence. Use the patterns you discovered to draw the 25th shape. Describe the 30th shape in the sequence. You do not have to draw it! Sometimes a conjecture is difficult to find because the data collected are unorganized or the observer is mistaking coincidence with cause and effect. Good use of inductive reasoning depends on the quantity and quality of data. Sometimes not enough information or data have been collected to make a proper conjecture. For example, if you are asked to find the next term in the pattern 3, 5, 7, you might conjecture that the next term is 9—the next odd number. Someone else might notice that the pattern is the consecutive odd primes and say that the next term is 11. If the pattern was 3, 5, 7, 11, 13, what would you be more likely to conjecture? EXERCISES 1. On his way to the local Hunting and Gathering Convention, caveperson Stony Grok picks up a rock, drops it into a lake, and notices that it sinks. He picks up a second rock, drops it into the lake, and notices that it also sinks. He does this five more times. Each time, the rock sinks straight to the bottom of the lake. Stony conjectures: “Ura nok seblu,” which translates to ? . What counterexample would Stony Grok need to find to disprove, or at least to refine, his conjecture? 96 CHAPTER 2 Reasoning in Geometry 2. Sean draws these geometric figures on paper. His sister Courtney measures each angle with a protractor. They add the measures of each pair of angles to form a conjecture. Write their conjecture. 90° 90° 150° 30° 45° 135° For Exercises 3–10, use inductive reasoning to find the next two terms in each sequence. 3. 1, 10, 100, 1000, ? , ? 5. 7, 3, 1, 5, 9, 13, ? , ? |
7. 1, 1, 2, 3, 5, 8, 13, ? , ? 9. 32, 30, 26, 20, 12, 2. 1, 3, 6, 10, 15, 21, ? , ? 8. 1, 4, 9, 16, 25, 36, ? , ? 10. 1, 2, 4, 8, 16, 32, ? , ? For Exercises 11–16, use inductive reasoning to draw the next shape in each picture pattern. 11. 13. 15. 12. 14. 16. Use the rule provided to generate the first five terms of the sequence in Exercise 17 and the next five terms of the sequence in Exercise 18. 17. 3n 2 18. 1, 3, 6, 10, . . . , n(n 2 1), . . . 19. Now it’s your turn. Generate the first five terms of a sequence. Give the sequence to a member of your family or to a friend and ask him or her to find the next two terms in the sequence. Can he or she find your pattern? 20. Write the first five terms of two different sequences in which 12 is the 3rd term. 21. Think of a situation in which you have used inductive reasoning. Write a paragraph describing what happened and explaining why you think it was inductive reasoning. LESSON 2.1 Inductive Reasoning 97 22. Look at the pattern in these pairs of equations. Decide if the conjecture is true. If it is not true, find a counterexample. 122 144 132 169 1032 10609 1122 12544 and and and and 212 441 312 961 3012 90601 2112 44521 Conjecture: If two numbers have the same digits in reverse order, then the squares of those numbers will have identical digits but in reverse order. Review 23. Sketch the section formed when the cone is sliced by the plane, as shown. 24. 25. 26. Sketch the three-dimensional figure formed by folding the net below into a solid. 27. Sketch the figure shown 28. Sketch the solid of below but with the red edge vertical and facing you. revolution formed when the two-dimensional figure is rotated about the line. For Exercises 29–38, write the word that makes the statement true. 29. Points are ? if they lie on the same line. 30. A triangle with two congruent sides is ? . 31. The geometry tool used to measure the size of an angle in degrees is called a(n) ? . 32. A(n) ? of a circle connects its center to a point on the circle. 33. A segment connecting any two non-adjacent vertices in a polygon is called a(n) ? . 34. A polygon with 12 sides is called a(n) ? . 35. A trapezoid has exactly one pair of ? sides. 98 CHAPTER 2 Reasoning in Geometry 36. A(n) ? polygon is both equiangular and equilateral. 37. If angles are complementary, then their measures add to ? . 38. If two lines intersect to form a right angle, then they are ? . For Exercises 39–42, sketch and label the figure. 39. Pentagon GIANT with diagonal AG parallel to side NT 40. A quadrilateral that has reflectional symmetry but not rotational symmetry 41. A prism with a hexagonal base 42. A counterexample to show that the following statement is false: The diagonals of a kite bisect the angles. IMPROVING YOUR REASONING SKILLS Puzzling Patterns These patterns are “different.” Your task is to find the next term. 1. 18, 46, 94, 63, 52, 61, ? 2. O, T, T, F, F, S, S, E, N, ? 3. 1, 4, 3, 16, 5, 36, 7, ? 4. 4, 8, 61, 221, 244, 884, ? 5. 6, 8, 5, 10, 3, 14, 1, ? 6. B, 0, C, 2, D, 0, E, 3, F, 3, G, ? 7. 2, 3, 6, 1, 8, 6, 8, 4, 8, 4, 8, 3, 2, 3, 2, 3, ? 8 Where do the X, Y, and Z go? LESSON 2.1 Inductive Reasoning 99 Deductive Reasoning Have you ever noticed that the days are longer in the summer? Or that mosquitoes appear after a summer rain? Over the years you have made conjectures, using inductive reasoning, based on patterns you have observed. When you make a conjecture, the process of discovery may not always help explain why the conjecture works. You need another kind of reasoning to help answer this question. Deductive reasoning is the process of showing that certain statements follow logically from agreed-upon assumptions and proven facts. When you use deductive reasoning, you try to reason in an orderly way to convince yourself or someone else that your conclusion is valid. If your initial statements are true, and you give a logical argument, then you have shown that your conclusion is true. For example, in a trial, lawyers use deductive arguments to show how the evidence that they present proves their case. A lawyer might make a very good argument. But first, the court must believe the evidence and accept it as true. L E S S O N 2.2 That’s the way things come clear. All of a sudden. And then you realize how obvious they’ve been all along. MADELEINE L’ENGLE The success of an attorney’s case depends on the jury accepting the evidence as true and following the steps in her deductive reasoning. You use deductive reasoning in algebra. When you provide a reason for each step in the process of solving an equation, you are using deductive reasoning. Here is an example. EXAMPLE A Solve the equation for x. Give a reason for each step in the process. Solution 3(2x 1) 2(2x 1) 7 42 5x 3(2x 1) 2(2x 1) 7 42 5x 5(2x 1) 7 42 5x 5(2x 1) 35 5x 10x 5 35 5x 10x 30 5x 15x 30 x 2 The original equation. Combining like terms. Subtraction property of equality. Distributive property. Subtraction property of equality. Addition property of equality. Division property of equality. 100 CHAPTER 2 Reasoning in Geometry The next example shows how to use both kinds of reasoning: inductive reasoning to discover the property and deductive reasoning to explain why it works. EXAMPLE B In each diagram, AC bisects obtuse angle BAD. Classify BAD, DAC, and CAB as acute, right, or obtuse. Then complete the conjecture. C A B C B A D mBAD 120° A mBAD 158° D B D C mBAD 92° Conjecture: If an obtuse angle is bisected, then the two newly formed congruent angles are ? . Justify your answers with a deductive argument. Solution In each diagram, BAD is obtuse because mBAD is greater than 90°. In each diagram, the angles formed by the bisector are acute because their measures— 60°, 79°, and 46°—are less than 90°. So one possible conjecture is Conjecture: If an obtuse angle is bisected, then the two newly formed congruent angles are acute. Why? According to our definition of an angle, every angle measure is less than 180°. So, using algebra, if m is the measure of an obtuse angle, then m 180°. When you bisect an angle, the two newly formed angles each measure half of the original angle, or m 1 m. If m 180°, then 1 1 (180), 2 2 2 so 1 m 90°. The two angles are each less 2 than 90°, so they are acute. m1_ 2 m m1_ 2 Science Here is an example of inductive reasoning, supported by deductive reasoning. El Niño is the warming of water in the tropical Pacific Ocean, which produces unusual weather conditions and storms worldwide. For centuries, farmers living in the Andes Mountains of South America have observed the stars in the Pleiades constellation to predict El Niño conditions. If the Pleiades look dim in June, they predict an El Niño year. What is the connection? Scientists have recently found that in an El Niño year, increased evaporation from the ocean produces high-altitude clouds that are invisible to the eye, but create a haze that makes stars more difficult to see. Therefore, the pattern that the Andean farmers knew about for centuries is now supported by a scientific explanation. To find out more about this story, go to www.keymath.com/DG . El Niño Conditions Increased convection Convective loop Equator Warm upper ocean layer Cold lower ocean layer 120°E 80°W LESSON 2.2 Deductive Reasoning 101 Inductive reasoning allows you to discover new ideas based on observed patterns. Deductive reasoning can help explain why your conjectures are true. Inductive and deductive reasoning work very well together. In this investigation you will use inductive reasoning to form a conjecture and deductive reasoning to explain why it’s true. Investigation Overlapping Segments In each segment, AB CD. Step 1 Step 2 Step 3 Step 4 25 cm 75 cm 25 cm 36 cm 80 cm 36 cm A B C D A B C D From the markings on each diagram, determine the lengths of AC and BD. What do you discover about these segments? Draw a new segment. Label it AD. Place your own points B and C on AD so that AB CD. A B C D Measure AC and BD. How do these lengths compare? Complete the conclusion of this conjecture: If AD has points A, B, C, and D in that order with AB CD, then ? . Now you will use deductive reasoning and algebra to explain why the conjecture from Step 4 is true. Step 5 Use deductive reasoning to convince your group that AC will always equal BD. Take turns explaining to each other. Write your argument algebraically. In the investigation you used both inductive and deductive reasoning to convince yourself of the overlapping segments property. You will use a similar process in the next lesson to discover and prove the overlapping angles property in Exercise 17. Good use of deductive reasoning depends on the quality of the argument. Just like the saying, “A chain is only as strong as its weakest link,” a deductive argument is only as good (or as true) as the statements used in the argument. A conclusion in a deductive argument is true only if all the statements in the argument are true. Also, the statements in your argument must clearly follow from each other. Did you use clear arguments in explaining the investigation steps? Did you point out that BC is part of both AC and BD? Did you point out that if you add the same amount to things that are equal the resulting sum must be equal? 102 CHAPTER 2 Reasoning in Geometry EXERCISES 1. When you use ? reasoning you are generalizing from careful observation that something is probably true. When you use ? reasoning you are establishing that, if a set of properties is accepted as true, something else must be true. 2. A and B are complementary. mA 25°. What is mB? What type of reasoning do you use, inductive or deductive reasoning, when solving this problem? 3. If the pattern continues, what are the next two terms? What type of reasoning do you use, inductive or deductive reasoning, when solving this problem? 1 4 9 16 4. DGT is isosceles with TD DG. If the perimeter of DGT is 756 cm and GT 240 cm, then DG ? . What type of reasoning do you use, inductive or deduc |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.