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ombus. 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 trapez |
oid 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 10.21 |
a 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 trapezoid 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 equal 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 these 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 perpendicular 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 rect |
angles. 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 Zoe? 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 end |
points 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, O |
Q 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 Sphere 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 of 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 mid |
point 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 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. 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. Proving 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 |
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 k |
c 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 k |
d 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 sim |
ilitude 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 triangle, 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 14 |
365C12.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 prove 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. 14365 |
C12.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. 14 |
365C12.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 isos |
celes 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 of 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. |
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