text stringlengths 235 3.08k |
|---|
pgs 7/10/07 8:56 AM Page 518 518 Ratio, Proportion, and Similarity x s x When a diagonal of a square is drawn, the square is separated into two isosceles right triangles. We can express the length of a leg of the isosceles right triangle in terms of the length of the hypotenuse or the length of the hypotenuse in terms of the length of a leg. Let s be the length of the hypotenuse of an isosceles right triangle and x be the length of each leg. Use the Pythagorean Theorem to set up two equalities. Solve one for x and the other for s: Solve for x: a2 1 b2 5 c2 x2 1 x2 5 s2 2x2 5 s2 x2 5 s2 2 Solve for s: c2 5 a2 1 b2 s2 5 x2 1 x2 s2 5 2x2 " s2 2 s2 2? 2 2 $ $ 2 2 s The 30-60-Degree Right Triangle An altitude drawn to any side of an equilateral triangle bisects the base and separates the triangle into two congruent right triangles. Since the measure of each angle of an equilateral triangle is 60°, the measure of one of the acute angles of a right triangle formed by the altitude is 60° and the measure of the other acute angle is 30°. Each of the congruent right triangles formed by drawing an altitude to a side of an equilateral triangle is called a 30-60-degree right triangle. If two triangles are 30-60-degree right triangles, then they are similar by AA. In the diagram, ABC is an equilateral triangle with s the length of each side and h the length of an altitude. Then s is the length of the hypotenuse of the 30-60-degree triangle and h CD'AB, is the length of a leg. In the diagram, s AC s, AD, and CD h. 2 a2 b2 c2 C 30° s h A 2 s 2 B s2 4 h2 s2 h2 h2 4 4s2 3 4s2 3 h " 2 s 60° s 2 A D B 14365C12.pgs 7/10/07 8:56 AM Page 519 EXAMPLE 3 Solution Pythagorean Theorem 519 In right triangle ABC, the length of |
the hypotenuse,, is 6 centimeters and the length of one leg is 3 cenAB timeters. Find the length of the other leg. B a2 1 b2 5 c2 a2 1 (3)2 5 (6)2 a2 1 9 5 36 a2 5 27 a 5 27 5 " 6 cm 3 Answer C 3 cm A 9? 3 5 3 " " Note: The measure of one leg is one-half the measure of the hypotenuse and the length of the other leg is times the length of the hypotenuse. Therefore, this triangle is a 30-60-degree right triangle. We can use a calculator to verify this. " 2 3 Recall that tan A = find the measure of A. opp adj 5 BC AC 5 3 3 3 5 " 3 ". Use a graphing calculator to ENTER: 2nd TAN1 2nd ¯ 3 ENTER The calculator will return 60 as mA. Therefore, mB 30. Exercises Writing About Mathematics 1. Ira said that if the lengths of the sides of an obtuse triangle ABC are a, b, and c with c opposite the obtuse angle, then c2 a2 b2. Do you agree with Ira? Explain why or why not. (Hint: Make use of the altitude from one of the acute angles.) 2. Sean said that if the measures of the diagonals of a parallelogram are 6 and 8 and the mea- sure of one side of the parallelogram is 5 then the parallelogram is a rhombus. Do you agree with Sean? Explain why or why not. Developing Skills In 3–8, in each case the lengths of three sides of a triangle are given. Tell whether each triangle is a right triangle. 3. 6, 8, 10 4. 7, 8, 12 5. 5, 7, 8 6. 15, 36, 39 7. 14, 48, 50 8. 2, 2, 4 3 " 14365C12.pgs 7/10/07 8:56 AM Page 520 520 Ratio, Proportion, and Similarity 9. Find, to the nearest tenth of a centimeter, the length of a diagonal of a square if the mea- sure of one side is 8.0 centimeters. 10. Find the length of the side of a rhombus whose diagonals measure 40 centimeters and 96 centimeters. 11. The length of a side of a |
rhombus is 10 centimeters and the length of one diagonal is 120 millimeters. Find the length of the other diagonal. 12. The length of each side of a rhombus is 13 feet. If the length of the shorter diagonal is 10 feet, find the length of the longer diagonal. 13. Find the length of the diagonal of a rectangle whose sides measure 24 feet by 20 feet. 14. The diagonal of a square measures 12 feet. a. What is the exact measure of a side of the square? b. What is the area of the square? 15. What is the slant height of a cone whose height is 36 centimeters and whose radius is 15 centimeters? 16. One side of a rectangle is 9 feet longer than an adjacent side. The length of the diagonal is 45 feet. Find the dimensions of the rectangle. 17. One leg of a right triangle is 1 foot longer than the other leg. The hypotenuse is 9 feet longer than the shorter leg. Find the length of the sides of the triangle. Applying Skills 18. Marvin wants to determine the edges of a rectangular garden that is to be 10 feet by 24 feet. He has no way of determining the measure of an angle but he can determine lengths very accurately. He takes a piece of cord that is 60 feet long and makes a mark at 10 feet and at 34 feet from one end. Explain how the cord can help him to make sure that his garden is a rectangle. 19. A plot of land is in the shape of an isosceles trapezoid. The lengths of the parallel sides are 109 feet and 95 feet. The length of each of the other two sides is 25 feet. What is the area of the plot of land? 20. From a piece of cardboard, Shanti cut a semicircle with a radius of 10 inches. Then she used tape to join one half of the diameter along which the cardboard had been cut to the other half, forming a cone. What is the height of the cone that Shanti made? 21. The lengths of two adjacent sides of a parallelogram are 21 feet and 28 feet. If the length of a diagonal of the parallelogram is 35 feet, show that the parallelogram is a rectangle. 22. The lengths of the diagonals of a parallelogram are 140 centimeters and 48 centimeters. The length of one side of the parallelogram is 74 centimeters. Show that the parallelogram is a rhombus. 14365C12. |
pgs 7/10/07 8:56 AM Page 521 The Distance Formula 521 23. A young tree is braced by wires that are 9 feet long and fastened at a point on the trunk of the tree 5 feet from the ground. Find to the nearest tenth of a foot how far from the foot of the tree the wires should be fastened to the ground in order to be sure that the tree will be perpendicular to the ground. 24. The length of one side of an equilateral triangle is 12 feet. What is the distance from the centroid of the triangle to a side? (Express the answer in simplest radical form.) 12-10 THE DISTANCE FORMULA When two points in the coordinate plane are on the same vertical line, they have the same x-coordinate and the distance between them is the absolute value of the difference of their y-coordinates. In the diagram, the coordinates of A are (4, 8) and the coordinates of C are (4, 2). y A(4, 8) CA 5 8 2 2 5 6 When two points in the coordinate plane are on the same horizontal line, they have the same y-coordinate and the distance between them is the absolute value of the difference of their x-coordinates. In the diagram, the coordinates of B are (1, 2) and the coordinates of C are (4, 2). CB 5 1 2 4 5 3 1 O B(1, 2) 1 C(4, 2) x In ABC, C is a right angle and AB is the hypotenuse of a right triangle. Using the Pythagorean Theorem, we can find AB: AB2 5 CA2 1 CB2 AB2 5 62 1 32 AB2 5 36 1 9 AB 5 45 " AB 5 3 " 5 This example suggests a method that can be used to find a formula for the length of any segment in the coordinate plane. 14365C12.pgs 7/10/07 8:56 AM Page 522 522 Ratio, Proportion, and Similarity Let B(x1, y1) and A(x2, y2) be any two points in the coordinate plane. From A draw a vertical line and from B draw a horizontal line. Let the intersection of these two lines be C. The coordinates of C are (x2, y1). Let AB c, CB a x2, and CA b y2. Then, y1 x1 y |
O A(x2, y2) c b a B(x1, y1) C(x2, y1) x c2 a2 b2 x1 c2 x2 c (x2 2 x1)2 1 (y2 2 y1)2 2 y2 y1 2 " This result is called the distance formula. If the endpoints of a line segment in the coordinate plane are B(x1, y1) and A(x2, y2), then: AB " (x2 2 x1)2 1 (y2 2 y1)2 EXAMPLE 1 The coordinates of the vertices of quadrilateral ABCD are A(1, 3), B(6, 4), C(5, 3), and D(2, 4). y D a. Prove that ABCD is a rhombus. b. Prove that ABCD is not a square. 1 1 O A C x B Solution a. AB 5 " (6 2 (21))2 1 (24 2 (23))2 BC 5 5 5 5 " " " (7)2 1 (21)2 49 1 1 50 5 5 5 (5 2 6)2 1 (3 2 (24))2 (21)2 1 (7)2 " " 1 1 49 " 50 " (22 2 5)2 1 (4 2 3)2 DA 5 (27)2 1 (1)2 " " CD 5 5 5 5 49 1 1 " 50 " (21 2 (22))2 1 (23 2 4)2 (1)2 1 (27)2 " " 1 1 49 " 50 " 5 5 5 14365C12.pgs 7/10/07 8:56 AM Page 523 The Distance Formula 523 The lengths of the sides of the quadrilateral are equal. Therefore, the quadrilateral is a rhombus. b. If a rhombus is a square, then it has a right angle. METHOD 1 AC 5 5 5 5 " " " " (5 2 (21))2 1 (3 2 (23))2 If B is a right angle, then (6)2 1 (6)2 36 1 36 72 AC2 5 AB2 1 BC2 2 1 2 5? 50 72 A " B B 72 2 50 1 50 ✘ A " 50 2 B A " Therefore, ABC is not a right triangle, B is not a right angle |
and ABCD is not a square. METHOD 2 slope of AB 5 24 2 (23) 6 2 (21) slope of BC 5 3 2 (24) 5 2 6 5 21 7 5 21 7 5 7 21 5 27 AB The slope of Therefore, AB rhombus is not a square. is not equal to the negative reciprocal of the slope of. BC, B is not a right angle and the is not perpendicular to BC EXAMPLE 2 Prove that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices using the distance formula. y B(0, 2b) Proof We will use a coordinate proof. The triangle can be placed at any convenient position. Let right triangle ABC have vertices A(2a, 0), B(0, 2b), and C(0, 0). Let M be the midpoint of the hypotenuse. The coordinates of M, the midpoint of AB AB, are, 0 1 2b 2 2a 1 0 2 A 5 (a, b) B M(a, b) O C(0, 0) A(2a, 0) x 14365C12.pgs 7/10/07 8:56 AM Page 524 524 Ratio, Proportion, and Similarity y B(0, 2b) Then, since M is the midpoint of formula: AB, AM BM, and using the distance M(a, b) AM 5 BM 5 (a 2 0)2 1 (b 2 2b)2 " CM 5 " (a 2 0)2 1 (b 2 0)2 x C(0, 0) A(2a, 0) O 5 5 a2 1 (2b)2 " a2 1 b2 " 5 a2 1 b2 " Therefore, the midpoint of the hypotenuse is equidistant from the vertices of the triangle. EXAMPLE 3 Prove that the medians to the base angles of an isosceles triangle are congruent. Given: Isosceles ABC with vertices A(2a, 0), B(0, 2b), C(2a, 0). Let M be the midpoint of AB. and N be the midpoint of BC y B(0, 2b) M N Prove: CM > AN Proof The coordinates of M are The coordinates of N are A(2a, 0) O C |
(2a, 0) x 22a 1 0 2, 0 1 2b 2 A The length of B CM 5 (2a, b). is 2a 1 0 2, 0 1 2b 2 A The length of B AN is 5 (a, b). (2a 2 2a)2 1 (b 2 0)2 (22a 2 a)2 1 (0 2 b)2 " 5 (23a)2 1 b2 " " 5 (23a)2 1 (2b)2 " 5 9a2 1 b2 " CM AN; therefore, CM > AN. 5 9a2 1 b2 " Exercises Writing About Mathematics 1. Can the distance formula be used to find the length of a line segment when the endpoints of the segment are on the same horizontal line or on the same vertical line? Justify your answer. 2. Explain why x2 x1 2 (x2 x1)2. 14365C12.pgs 7/10/07 8:56 AM Page 525 The Distance Formula 525 Developing Skills In 3–10, the coordinates of the endpoints of in simplest form. AB are given. In each case, find the exact value of AB 3. A(1, 2), B(4, 6) 5. A(3, 2), B(5, 4) 7. A(1, 2), B(3, 4) 9. A(6, 2), B(1, 3) 4. A(–1, 6), B(4, 6) 6. A(0, 2), B(3, 1) 8. A(–5, 2), B(1, 6) 10. A(–3, 3), B(3, 3) 11. The coordinates of A are (0, 4) and the x-coordinate of B is 5. What is the y-coordinate of B if AB 13? (Two answers are possible.) 12. The coordinates of M are (2, 1) and the y-coordinate of N is 5. What is the x-coordinate of N if MN = 3? (Two answers are possible.) 5 " 13. The vertices of a quadrilateral are A(0, 2), B(5, 2), C(8, 2), D(3, 2). Prove that the quadrilateral is a rhombus using the distance formula. 14 |
. The vertices of a triangle are P(1, 1), Q(7, 1), and R(3, 3). a. Show that PQR is an isosceles triangle. b. Show that PQR is a right triangle using the Pythagorean Theorem. c. Show that the midpoint of the hypotenuse is equidistant from the vertices. 15. The vertices of a triangle are L(1, 1), M(7, 3), and N(2, 2). a. Show that LMN is a scalene triangle. b. Show that LMN is a right triangle using the Pythagorean Theorem. c. Show that the midpoint of MN is equidistant from the vertices. 16. The vertices of DEF are D(2, 3), E(5, 0), and F(2, 3). Show that 17. The coordinates of the vertices of BAT are B(2, 7), A(2, 1), and T(11, 4). DE > FE. a. Find the equation of the line that is the altitude from B to AT. b. Find the coordinates of D, the foot of the perpendicular or the foot of the altitude from part a. c. Find the length of the altitude BD. 18. The coordinates of the vertices of EDF are E(2, 0), D(4, 0), and F a. Find ED, DF, and FE. b. Is EDF equilateral? Justify your answer. 1, 3 3 ". B A 19. The vertices of ABC are A(1, 1), B(4, 1), and C(2, 4). a. Find the coordinates of the vertices of ABC, the image of ABC under the trans- formation D2. b. Show that distance is not preserved under the dilation. 14365C12.pgs 7/10/07 8:56 AM Page 526 526 Ratio, Proportion, and Similarity c. Show that ABC ABC using SSS. d. Use part c to show that the angle measures of ABC are preserved under the dilation. 20. The vertices of quadrilateral ABCD are A(2, 0), B(3, 1), C(4, 1), and D(3, 2). a. Show that ABCD is a parallelogram using the distance formula. b. Find the coordinates of the |
vertices of quadrilateral ABCD, the image of ABCD under the transformation D3. c. Show that ABCD is a parallelogram using the distance formula. d. Use part c to show that the images of the parallel segments of ABCD are also parallel under the dilation. 21. The vertices of quadrilateral ABCD are A(2, 2), B(2, 0), C(3, 3), and D(1, 1). Use the distance formula to prove that ABCD is a parallelogram but not a rhombus. 22. The vertices of ABC are A(0, 2), B(4, 6), and C(2, 4). Prove that ABC is an isosceles right triangle using the Pythagorean Theorem. Applying Skills 23. Use the distance formula to prove that (a, 0), (0, b) and (a, 0) are the vertices of an isosce- les triangle. 24. The vertices of square EFGH are E(0, 0), F(a, 0), G(a, a), and H(0, a). Prove that the diagonals of a square, EG and FH, are congruent and perpendicular. 25. The vertices of quadrilateral ABCD are A(0, 0), B(b, c), C(b a, c), and D(a, 0). Prove that if a2 b2 c2 then ABCD is a rhombus. 26. Prove the midpoint formula using the distance formula. Let P and Q have coordinates (x1, y1) and (x2, y2), respectively. Let M have coordinates point of if and only if PM MQ. PQ A x1 1 x2 2 y1 1 y2 2,. M is the mid- B a. Find PM. b. Find MQ. c. Show that PM MQ. 27. The vertices of WX are W(w, y) and X(x, z). a. What are the coordinates of b. Show, using the distance formula, that WX is k times the length of WX. under the dilation Dk?, the image of WX WrXr 14365C12.pgs 8/2/07 5:57 PM Page 527 CHAPTER SUMMARY Definitions to Know a • The ratio |
of two numbers, a and b, where b is not zero, is the number. b • A proportion is an equation that states that two ratios are equal. Chapter Summary 527 • In the proportion, the first and fourth terms, a and d, are the extremes of the proportion, and the second and third terms, b and c, are the means. b 5 c a d • If the means of a proportion are equal, the mean proportional is one of the means. • Two line segments are divided proportionally when the ratio of the lengths of the parts of one segment is equal to the ratio of the lengths of the parts of the other. • Two polygons are similar if there is a one-to-one correspondence between their vertices such that: 1. All pairs of corresponding angles are congruent. 2. The ratios of the lengths of all pairs of corresponding sides are equal. • The ratio of similitude of two similar polygons is the ratio of the lengths of corresponding sides. • A dilation of k is a transformation of the plane such that: 1. The image of point O, the center of dilation, is O. 2. When k is positive and the image of P is P, then h OP and h OPr are the same ray and OP kOP. 3. When k is negative and the image of P is P, then h OP and h OPr are opposite rays and OP kOP. • The projection of a point on a line is the foot of the perpendicular drawn from that point to the line. • The projection of a segment on a line, when the segment is not perpendicular to the line, is the segment whose endpoints are the projections of the endpoints of the given line segment on the line. • A Pythagorean triple is a set of three integers that can be the lengths of the sides of a right triangle. Postulates 12.1 Any geometric figure is similar to itself. (Reflexive property) 12.2 A similarity between two geometric figures may be expressed in either order. (Symmetric property) 12.3 Two geometric figures similar to the same geometric figure are similar to each other. (Transitive property) 12.4 For any given triangle there exist a similar triangle with any given ratio of similitude. 14365C12.pgs 7/10/07 8:56 AM Page 528 528 Ratio, Proportion, and Similarity Theorems 12.1 In a proportion, |
the product of the means is equal to the product of the extremes. In a proportion, the means may be interchanged. 12.1a 12.1b In a proportion, the extremes may be interchanged. 12.1c If the products of two pairs of factors are equal, the factors of one pair can be the means and the factors of the other the extremes of a proportion. 12.3 12.4 12.2 A line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side. Two line segments are divided proportionally if and only if the ratio of the length of a part of one segment to the length of the whole is equal to the ratio of the corresponding lengths of the other segment. Two triangles are similar if two angles of one triangle are congruent to two corresponding angles of the other. (AA) Two triangles are similar if the three ratios of corresponding sides are equal. (SSS) Two triangles are similar if the ratios of two pairs of corresponding sides are equal and the corresponding angles included between these sides are congruent. (SAS) 12.5 12.6 12.7 A line is parallel to one side of a triangle and intersects the other two sides if and only if the points of intersection divide the sides proportionally. 12.8 Under a dilation, angle measure is preserved. 12.9 Under a dilation, midpoint is preserved. 12.10 Under a dilation, collinearity is preserved. 12.11 12.12 If two triangles are similar, the lengths of corresponding altitudes have the same ratio as the lengths of any two corresponding sides. If two triangles are similar, the lengths of corresponding medians have the same ratio as the lengths of any two corresponding sides. If two triangles are similar, the lengths of corresponding angle bisectors have the same ratio as the lengths of any two corresponding sides. 12.14 Any two medians of a triangle intersect in a point that divides each 12.13 median in the ratio 2 : 1. 12.15 The medians of a triangle are concurrent. 12.16 The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to each other and to the original triangle. 12.16a The length of each leg of a right triangle is the mean proportional between the length of the projection of that leg on the hypotenuse and the length of the |
hypotenuse. 12.16b The length of the altitude to the hypotenuse of a right triangle is the mean proportional between the lengths of the projections of the legs on the hypotenuse. 12.17 A triangle is a right triangle if and only if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides. 14365C12.pgs 7/10/07 8:56 AM Page 529 Formulas In the coordinate plane, under a dilation of k with the center at the origin: Review Exercises 529 P(x, y) → P(kx, ky) or Dk(x, y) (kx, ky) If x is the length of a leg of an isosceles right triangle and s is the length of the hypotenuse, then: x s and s 2 " 2 x 2 " If s is the length of a side of an equilateral triangle and h is the length of an altitude then: If the endpoints of a line segment in the coordinate plane are B(x1, y1) and A(x2, y2), then: h 3 2 s " AB " (x2 2 x1)2 1 (y2 2 y1)2 VOCABULARY 12-1 Similar • Ratio of two numbers • Proportion • Extremes • Means • Mean proportional • Geometric mean 12-2 Midsegment theorem • Line segments divided proportionally 12-3 Similar polygons • Ratio of similitude • Constant of proportionality 12-4 Postulate of similarity • AA triangle similarity • SSS similarity theorem • SAS similarity theorem 12-5 Dilation • Enlargement • Contraction • Constant of dilation, k 12-7 Centroid 12-8 Projection of a point on a line • Projection of a segment on a line 12-9 Pythagorean Theorem • Pythagorean triple • 45-45-degree right triangle • 30-60-degree right triangle 12-10 Distance formula • Foot of an altitude REVIEW EXERCISES 1. Two triangles are similar. The lengths of the sides of the smaller triangle are 5, 6, and 9. The perimeter of the larger triangle is 50. What are the lengths of the sides of the larger triangle? 2. The measure of one angle of right ABC is 67° and the measure of an angle of right LMN is 23° |
. Are the triangles similar? Justify your answer. 14365C12.pgs 7/10/07 8:56 AM Page 530 530 Ratio, Proportion, and Similarity 3. A line parallel to side AB of ABC intersects AC at E and BC at F. If EC 12, AC 20, and AB 15, find EF. 4. A line intersects side AC FC 5, BC 15, prove that EFC ABC. of ABC at E and BC at F. If EC 4, AC 12, 5. The altitude to the hypotenuse of a right triangle divides the hypotenuse into two segments. If the length of the altitude is 12 and the length of the longer segment is 18, what is the length of the shorter segment? 6. In LMN, L is a right angle, b. Find MN. a. Find LP. LP is an altitude, MP 8, and PN 32. c. Find ML. d. Find NL. 7. The length of a side of an equilateral triangle is 18 centimeters. Find, to the nearest tenth of a centimeter, the length of the altitude of the triangle. 8. The length of the altitude to the base of an isosceles triangle is 10.0 cen- timeters and the length of the base is 14.0 centimeters. Find, to the nearest tenth of a centimeter, the length of each of the legs. 9. The coordinates of the endpoints of PQ are P(2, 7) and Q(8, 1). a. Find the coordinates of the endpoints of PrQr under the composition D2 + rx-axis b. What is the ratio. PQ PrQr? c. What are the coordinates of M, the midpoint of d. What are the coordinates of M, the image of M under PQ? D2 + rx-axis? e. What are the coordinates of N, the midpoint of f. Is M the midpoint of? Justify your answer. PrQr PrQr? 10. If g AB g CD and AD and BC intersect at E, prove that ABE DCE. 11. A line intersects AC at E and at F. If ABC EFC, g AB g EF. prove that BC 12. Find the length of the altitude to the bases of isosceles trapezoid KLMN if KL 20 cm, MN 38 cm, and KN 15 cm. |
13. Find the length of a side of a rhombus if the measures of the diagonals of the rhombus are 30 inches and 40 inches. 14365C12.pgs 7/10/07 8:56 AM Page 531 Review Exercises 531 14. The length of a side of a rhombus is 26.0 centimeters and the length of one diagonal is 28.0 centimeters. Find to the nearest tenth the length of the other diagonal. 15. The coordinates of the vertices of RST are R(4, 1), S(3, 2), and T(2, 1). a. Find the length of each side of the triangle in simplest radical form. b. Prove that the triangle is a right triangle. 16. The vertices of ABC are A(0, 0), B(4, 3), and C(0, 5). a. Prove that ABC is isosceles. b. The median to BC is AD. Find the coordinates of D. c. Find AD and DB. d. Prove that is the altitude to 17. The vertices of ABC are A(2, 1), B(2, 1), and C(0, 3). AD BC using the Pythagorean Theorem. a. Find the coordinates of ABC, the image of ABC under D3. b. Find, in radical form, the lengths of the sides of ABC and of ABC. c. Prove that ABC ABC. d. Find the coordinates of P, the centroid of ABC. CP PM 5 2 1 e. Let M be the midpoint of f. Find the coordinates of P, the centroid of ABC. g. Is P the image of P under D3? h. Let M be the midpoint of ArBr. Prove that. Prove that AB. CrP PMr 5 2 1. 18. A right circular cone is cut by a plane through a diameter of the base and the vertex of the cone. If the diameter of the base is 20 centimeters and the height of the cone is 24 centimeters, what is the slant height of the cone? Exploration A line parallel to the shorter sides of a rectangle can divide the rectangle into a square and a smaller rectangle. If the smaller rectangle is similar to the given rectangle, then both rectangles are called golden rectangles and the ratio of the lengths of their sides is called the golden ratio. The golden ratio is 1 1 A : 2 |
. 5 " B 14365C12.pgs 7/10/07 8:56 AM Page 532 532 Ratio, Proportion, and Similarity Follow the steps to construct a golden rectangle. You may use compass and straightedge or geometry software. STEP 1. Draw square ABCD. STEP 2. Construct E, the midpoint of AB. STEP 3. Draw the ray h AB. STEP 4. With E as the center and radius EC, draw an arc that h AB. Call this intersects point F. STEP 5. Draw the ray h DC. D C G A E 1 B F STEP 6. Construct the line perpendicular to of this line with be point G. h DC h AB through F. Let the intersection a. Let AB BC 2 and EB 1. Find EC EF, AF AE EF, and BF EF EB. Express each length as an irrational number in simplest radical form. b. Show that AFGD and FGCB are golden rectangles by showing that they are similar, that is, that AD 5 FG AF BF. c. Repeat steps 1 through 6 using a different square. Let AB x. Complete parts a and b. Do you obtain the same ratio? d. Research the golden rectangle and share your findings with your classmates. CUMULATIVE REVIEW Chapters 1–12 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The length and width of a rectangle are 16 centimeters and 12 centimeters. What is the length of a diagonal of the rectangle? (3) 25 cm (2) 20 cm (1) 4 cm (4) 4 cm 7 " 2. The measure of an angle is 12 degrees more than twice the measure of its supplement. What is the measure of the angle? (2) 56 (1) 26 (3) 64 (4) 124 3. What is the slope of the line through A(2, 8) and B(4, 1)? (1) 23 2 (2) 3 2 (3) 22 3 (4) 2 3 14365C12.pgs 7/10/07 8:56 AM Page 533 Cumulative Review 533 4. The measures of the opposite angles of a parallelogram are represented by 2x 34 and 3x 12. Find the value of x. (1) 22 (2) 46 (3) 78 (4) 126 5. Which of the following do not determine a plane? |
(1) three lines each perpendicular to the other two (2) two parallel lines (4) a line and a point not on it (3) two intersecting lines 6. Which of the following cannot be the measures of the sides of a triangle? (1) 5, 7, 8 (2) 3, 8, 9 (3) 5, 7, 7 (4) 2, 6, 8 7. Under a rotation of 90° about the origin, the image of the point whose coordinates are (3, 2) is the point whose coordinates are (1) (2, 3) (3) (3, 2) (2) (2, 3) (4) (2, 3) 8. If a conditional statement is true, which of the following must be true? (1) converse (2) inverse 9. In the figure, side AB of ABC is extended through B to D. If mCBD 105 and mA 53, what is the measure of C? (1) 22 (2) 52 (3) 75 (4) 158 (3) contrapositive (4) biconditional C A B D 10. A parallelogram with one right angle must be (1) a square. (2) a rectangle. (3) a rhombus. (4) a trapezoid. Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Given: ABCD with AB > CD, E not on › ‹, and ABCD BE > CE. Prove: AE > DE. 12. Given: perpendicular to plane p at R, points A and B in plane p, and RS RA > RB. l s _ _ l e _ _ Prove: SA > SB _ _ l s _ _ l e 14365C12.pgs 7/10/07 8:56 AM Page 534 534 Ratio, Proportion, and Similarity Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. The radius of the base of |
a right circular cone is one-half the slant height of the cone. The radius of the base is 2.50 feet. a. Find the lateral area of the cone to the nearest tenth of a square foot. b. Find the volume of the cone to the nearest tenth of a cubic foot. 14. The coordinates of the vertices of ABC are A(1, 0), B(4, 0), and C(2, 6). a. Write an equation of the line that contains the altitude from C to AB. b. Write an equation of the line that contains the altitude from B to AC. c. What are the coordinates of D, the intersection of the altitudes from C and from B? d. Write an equation for g AD. e. Is g AD perpendicular to BC, that is, does g AD contain the altitude from A to BC? Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. In the diagram, ABCD is a rectangle and ADE is an isosceles triangle, a median to. If with AD BC of AE > DE EF of ADE, is extended to intersect at G, prove that G is the midpoint BC. E D F A C G B 16. The coordinates of the vertices of ABC are A(2, 5), B(3, 1), and C(6, 4). a. Find the coordinates of ABC, the image of ABC under the com- position ry 5 x b. Show that ry 5 x + rx-axis. + rx-axis(x,y) R90°(x, y). 14365C13.pgs 7/12/07 3:56 PM Page 535 GEOMETRY OF THE CIRCLE Early geometers in many parts of the world knew that, for all circles, the ratio of the circumference of a circle to its diameter was a constant. Today, we write C p d 5 p to represent this constant. Euclid established that the ratio of the area of a circle to the square of its diame-, but early geometers did not use the symbol A d2 5 k ter was also a constant, that is,. How do these constants, p and k, relate |
to one another? Archimedes (287–212 B.C.) proposed that the area of a circle was equal to the area of a right triangle whose legs have lengths equal to the radius, r, and the circumference, C, of a circle. Thus A rC. He used indirect proof and the areas of inscribed and circumscribed polygons to prove his conjecture and to prove 1 2 310 71, p,3 1 7 that. Since this inequality can be written as 3.140845... 3.142857..., Archimedes’ approximation was correct to two decimal places. p Use Archimedes’ formula for the area of a circle and d 2r to show that and the facts that A and that 4k = pr2 C d 5 p.p CHAPTER 13 CHAPTER TABLE OF CONTENTS 13-1 Arcs and Angles 13-2 Arcs and Chords 13-3 Inscribed Angles and Their Measures 13-4 Tangents and Secants 13-5 Angles Formed by Tangents, Chords, and Secants 13-6 Measures of Tangent Segments, Chords, and Secant Segments 13-7 Circles in the Coordinate Plane 13-8 Tangents and Secants in the Coordinate Plane Chapter Summary Vocabulary Review Exercises Cumulative Review 535 14365C13.pgs 7/12/07 3:56 PM Page 536 536 Geometry of the Circle 13-1 ARCS AND ANGLES In Chapter 11, we defined a sphere and found that the intersection of a plane and a sphere was a circle. In this chapter, we will prove some important relationships involving the measures of angles and line segments associated with circles. Recall the definition of a circle. DEFINITION A circle is the set of all points in a plane that are equidistant from a fixed point of the plane called the center of the circle. If the center of a circle is point O, the circle is called circle O, written in symbols as (O. B A radius of a circle (plural, radii) is a line segment from the center of the circle to any point of the circle. The term radius is used to mean both the line segment and the length of the line segment. If A, B, and C are points of circle O, then are radii of the circle. Since the definition of a circle states that all points of the circle are equidistant |
from its center, OA OB OC. Thus, because equal line segments are congruent. We can state what we have just proved as a theorem. OA > OB > OC, and, OA OB OC O A C Theorem 13.1 All radii of the same circle are congruent. A circle separates a plane into three sets of points. If we let the length of the radius of circle O be r, then: • Point C is on the circle if OC r. • Point D is outside the circle if OD r. • Point E is inside the circle if OE r. C r O E D The interior of a circle is the set of all points whose distance from the cen- ter of the circle is less than the length of the radius of the circle. The exterior of a circle is the set of all points whose distance from the cen- ter of the circle is greater than the length of the radius of the circle. Central Angles Recall that an angle is the union of two rays having a common endpoint and that the common endpoint is called the vertex of the angle. DEFINITION A central angle of a circle is an angle whose vertex is the center of the circle. 14365C13.pgs 7/12/07 3:56 PM Page 537 Arcs and Angles 537 In the diagram, LOM and MOR are central angles because the vertex of each angle is point O, the center of the circle. L O M R Types of Arcs An arc of a circle is the part of the circle between two points on the circle. In the diagram, A, B, C, and D are points on circle O and AOB intersects the circle at two distinct points, A and B, separating the circle into two arcs Minor arc ( D ABX ) O A Major arc ( D )ACBX 1. If mAOB 180, points A and B and the points of the circle in the inte- rior of AOB make up minor arc AB, written as ABX. 2. Points A and B and the points of the circle not in the interior of AOB make up major arc AB. A major arc is usually named by three points: the two endpoints and any other point on the major arc. Thus, the major arc with endpoints A and B is written as ACBX or 3. If mAOC 180, points A and C separate circle O into two equal parts, ABC |
X each of which is called a semicircle. In the diagram above, name two different semicircles. and. ADBX ADCX Degree Measure of an Arc An arc of a circle is called an intercepted arc, or an arc intercepted by an angle, if each endpoint of the arc is on a different ray of the angle and the other points of the arc are in the interior of the angle. DEFINITION The degree measure of an arc is equal to the measure of the central angle that intercepts the arc. 14365C13.pgs 7/12/07 3:56 PM Page 538 538 Geometry of the Circle 80 G F 80 O E mGFEX 5 180 In circle O, GOE is a straight angle, mGOE 180, and mFOG 80.. Also, since Therefore, the degree measure of is 80°, written as FGX, mFEX 5 180 2 80 5 100 and mFGX 5 80 mFEGX 5 100 1 180 5 280 Therefore, mGFX 1 mFEGX 5 80 1 280 5 360 1. The degree measure of a semicircle is 180. Thus: 2. The degree measure of a major arc is equal to 360 minus the degree mea- sure of the minor arc having the same endpoints. Do not confuse the degree measure of an arc with the length of an arc. The degree measure of every circle is 360 but the circumference of a circle is 2p times the radius of the circle. Example: in circle O, OA 1 cm, and in circle O, OA 1.5 cm. In both circles, the degree measure of the circle is 360° but the circumference of circle O is 2p centimeters, and the circumference of circle O is 3p centimeters. O O A A Congruent Circles, Congruent Arcs, and Arc Addition DEFINITION Congruent circles are circles with congruent radii. If OC > OrCr, then circles O and O are congruent. C O C O DEFINITION Congruent arcs are arcs of the same circle or of congruent circles that are equal in measure. 14365C13.pgs 7/31/07 1:15 PM Page 539 Arcs and Angles 539 If mCDX 5 mCrDrX and O is not congruent to circle O, then and 60, then CDX have the same degree measure. (O |
> (Or CsDsX CDX > CrDrX is not congruent to. However, if circle CDX CsDsX even if D 60 C O C O 60 D 60 C D O Postulate 13.1 Arc Addition Postulate BCX and mABCX BCX ABX If point and no other points in common, then ABX m. m are two arcs of the same circle having a common endand ABCXBCX ABX The arc that is the sum of two arcs may be a minor arc, a major arc, or semimBCX 5 40, circle. For example, A, B, C, and D are points of circle O, mABX 5 90, and h OB and 1. Minor arc: Also, BCX h are opposite rays. OD ABX ACX mACX 5 mABX 1 mBCX 5 90 1 40 5 130 2. Major arc: BCDXABX ABDX mABDX 5 mABX 1 mBCDX = Also, 40 B C 90 O A 5 90 1 180 5 270 D h 3. Semicircle: Since OB BCX BCDXCDX Thus, and h OD are opposite rays, BOD is a straight angle. mBCX 1 mCDX 5 mBCDX 5 180, a semicircle. Also,. 14365C13.pgs 7/31/07 1:15 PM Page 540 540 Geometry of the Circle Theorem 13.2a In a circle or in congruent circles, if central angles are congruent, then their intercepted arcs are congruent Given Circle O circle O, AOB COD, and AOB AOB. ABX > ArBrX ABX > CDX and. B B A C A O O Prove Proof D It is given that AOB COD and AOB AOB. Therefore, mAOB mCOD mAOB because congruent angles have equal measures. Then since the degree measure of an arc is equal to the degree meamABX 5 mCDX 5 mArBrX. It is sure of the central angle that intercepts that arc, also given that circle O and circle O are congruent circles. Congruent arcs are defined as arcs of the same circle or of congruent circles that are equal in measure. Therefore, since their measures are equal, ABX > ArBrX ABX > CDX |
and. The converse of this theorem can be proved by using the same definitions and postulates. Theorem 13.2b In a circle or in congruent circles, central angles are congruent if their intercepted arcs are congruent. Theorems 13.2a and 13.2b can be written as a biconditional. Theorem 13.2 In a circle or in congruent circles, central angles are congruent if and only if their intercepted arcs are congruent. EXAMPLE 1 h OA Let and h OB a. mBOC b. be opposite rays and mAOC 75. Find: mBACX mACX mABX mBCX d. e. c. Solution a. m/BOC 5 m/AOB 2 m/AOC 5 180 2 75 5 105 A 75° C O B D 14541C13.pgs 1/25/08 3:48 PM Page 541 b. m c. m d. m mAOC 75 mBOC 105 ACX BCX ABX mAOB 180 mBACX 5 mBDAX 1 mACX e. 5 180 1 75 5 255 Arcs and Angles 541 mBACX 5 360 2 mBCX or 5 360 2 105 5 255 Answers a. 105 b. 75 c. 105 d. 180 e. 255 Exercises Writing About Mathematics 1. Kay said that if two lines intersect at the center of a circle, then they intercept two pairs of congruent arcs. Do you agree with Kay? Justify your answer. 2. Four points on a circle separate the circle into four congruent arcs: it true that g AC g'BD? Justify your answer. CDXBCXABX,,, and DAX. Is Developing Skills In 3–7, find the measure of the central angle that intercepts an arc with the given degree measure. 3. 35 4. 48 5. 90 6. 140 7. 180 In 8–12, find the measure of the arc intercepted by a central angle with the given measure. 8. 60 9. 75 10. 100 11. 120 12. 170 In 13–22, the endpoints of measure. 15. 13. mBOC mBCX mDAX mBCDX 19. 21. mAOC 17. AOC are on circle O, mAOB 89, and mCOD 42. |
Find each mABX 14. 16. mDOA 18. mBOD mDABX 22. mADCX 20. C 89° O 42° D B A 14541C13.pgs 1/25/08 3:48 PM Page 542 542 Geometry of the Circle 23. In 23–32, P, Q, S, and R are points on circle O, mPOQ 100, mQOS 110, and mSOR 35. Find each measure. mPQX mSRX mRPX 27. 29. mQOR mSRPX mQSX 24. 26. mROP mPQSX mQSRX mRPQX 110° 100° 35° 25. 28. 32. 31. 30. Q O R P S Applying Skills 33. Given: Circle O with ABX > CDX. Prove: AOB COD A O B D C 34. Given: and AB are on circle O. CD intersect at O, and the endpoints of Prove: AC > BD 35. In circle O, AOB'COD. Find mACX 36. Points A, B, C, and D lie on circle O, and a square. Hands-On Activity and mADCX AC'BD. AB and CD A C O D B at O. Prove that quadrilateral ABCD is For this activity, you may use compass, protractor, and straightedge, or geometry software. 1. Draw circle O with a radius of 2 inches and circle O with a radius of 3 inches. 2. Draw points A and B on the circle O so that mArBrX 60. so that 3. Show that AOB AOB. mABX 60 and points A and B on circle O 14365C13.pgs 7/12/07 3:56 PM Page 543 Arcs and Chords 543 13-2 ARCS AND CHORDS DEFINITION A chord of a circle is a line segment whose endpoints are points of the circle. A diameter of a circle is a chord that has the center of the circle as one of its points. AB and, are chords of circle O. In the diagram, Since O is a point of is a diameter. Since OA and OC are the lengths of the radius of circle O, OA OC. and O is the midpoint of AOC If the length of the |
radius of circle O is r, and the AOC AOC AOC length of the diameter is d, then d 5 AOC 5 OA 1 OC 5 r 1 r 5 2r d 2r That is: ABX, and major The endpoints of a chord are points on a circle and, therefore, determine two arcs of a circle, a minor arc and, central AOB, a major arc. In the diagram, chord are all determined by points A minor and B. We proved in the previous section that in a circle, congruent central angles intercept congruent arcs. Now we can prove that in a circle, congruent central angles have congruent chords and that congruent arcs have congruent chords. ABX AB C O A B A B O Theorem 13.3a In a circle or in congruent circles, congruent central angles have congruent chords. Given (O > (Or and COD AOB AOB A B A B Prove CD > AB > ArBr C O D O 14541C13.pgs 1/25/08 3:48 PM Page 544 544 Geometry of the Circle Proof We will show that COD AOB AOB by SAS. A B A B It is given that AOB COD and AOB AOB. Therefore, AOB COD AOB by the ArOr transitive property of congruence. Since the radii of congruent circles, these segments are all congruent: D, AO DO BO CO O,,, C O, and BrOr are DO > CO > AO > BO > ArOr > BrOr Therefore, by SAS, COD AOB AOB. Since corresponding parts of congruent triangles are congruent, CD > AB > ArBr. The converse of this theorem is also true. Theorem 13.3b In a circle or in congruent circles, congruent chords have congruent central angles. Given (O > (Or and CD > AB > ArBr Prove COD AOB AOB C D A O A B O Strategy This theorem can be proved in a manner similar to Theorem 13.3a: prove that COD AOB AOB by SSS. B The proof of Theorem 13.3b is left to the student. (See exercise 23.) Theorems 13.3a and 13.3b can be stated as |
a biconditional. Theorem 13.3 In a circle or in congruent circles, two chords are congruent if and only if their central angles are congruent. Since central angles and their intercepted arcs have equal degree measures, we can also prove the following theorems. Theorem 13.4a In a circle or in congruent circles, congruent arcs have congruent chords. Given (O > (Or and CDX > ABX > ArBrX Prove CD > AB > ArBr B B A O O A C D 14365C13.pgs 7/31/07 1:43 PM Page 545 Arcs and Chords 545 Proof First draw line segments from O to A, B, C, D, A, and B. Congruent arcs have congruent central angles. Therefore, COD AOB AOB. In a circle or in congruent circles, two chords are congruent if and only if their central angles are congruent. Therefore, CD > AB > ArBr. The converse of this theorem is also true. Theorem 13.4b In a circle or in congruent circles, congruent chords have congruent arcs. Given Prove CD > AB > ArBr and (O > (Or CDX > ABX > ArBrX Strategy First draw line segments from O to A, B, C, D, A and B. B B A O O A C D Prove COD AOB AOB by SSS. Then use congruent central angles to prove congruent arcs. The proof of Theorem 13.4b is left to the student. (See exercise 24.) Theorems 13.4a and 13.4b can be stated as a biconditional. Theorem 13.4 In a circle or in congruent circles, two chords are congruent if and only if their arcs are congruent. EXAMPLE 1 mABX In circle O, mBCX a. Find 35, mBOC 110, and mCDX. and AOD is a diameter. Solution a. b. Explain why AB = CD. mBCX 5 m/BOC 5 110 mCDX 5 180 2 mABX 2 mBCX 5 180 2 35 2 110 5 35 A 35 B O 110 D C b. In a circle, arcs with equal measure are congruent. |
Therefore, since mABX mCDX their arcs are congruent. Therefore, 35 and ABX 35, CDX AB > CD and AB = CD.. In a circle, chords are congruent if 14365C13.pgs 7/12/07 3:56 PM Page 546 546 Geometry of the Circle Chords Equidistant from the Center of a Circle We defined the distance from a point to a line as the length of the perpendicular from the point to the line. The perpendicular is the shortest line segment that can be drawn from a point to a line. These facts can be used to prove the following theorem. Theorem 13.5 A diameter perpendicular to a chord bisects the chord and its arcs. Given Diameter of circle O, chord, and COD ACX > BCX AB ADX > BDX. Prove AE > BE,, and AB'CD at E. A B C E O D Proof Statements Reasons and. OB 1. Two points determine a line. 1. Draw OA AB'CD 2. 3. AEO and BEO are right angles. 4. OA > OB OE > OE 5. 6. AOE BOE 7. AE > BE 8. AOE BOE ACX > BCX 9. 10. AOD is the supplement of AOE. BOD is the supplement of BOE. 11. AOD BOD ADX > BDX 12. 2. Given. 3. Perpendicular lines intersect to form right angles. 4. Radii of a circle are congruent. 5. Reflexive property of congruence. 6. HL. 7. Corresponding parts of congruent triangles are congruent. 8. Corresponding parts of congruent triangles are congruent. 9. In a circle, congruent central angles have congruent arcs. 10. If two angles form a linear pair, then they are supplementary. 11. Supplements of congruent angles are congruent. 12. In a circle, congruent central angles have congruent arcs. 14365C13.pgs 7/31/07 1:16 PM Page 547 Since a diameter is a segment of a line, the following corollary is also true: Arcs and Chords 547 Corollary 13.5a A line through the center of a circle that is perpendicular to a chord bisects the chord and its |
arcs. An apothem of a circle is a perpendicular line segment from the center of a circle to the midpoint of a chord. The term apothem also refers to the length of the segment. In the diagram, E is the midpoint of chord in circle O, AB'CD, or OE, is the apothem., and AB OE A C E O D Theorem 13.6 The perpendicular bisector of the chord of a circle contains the center of the circle. Given Circle O, and chord AB with midpoint M and perpendicular bisector k. Prove Point O is a point on k. Proof In the diagram, M is the midpoint of chord AB in circle O. A M O k B B Then, AM = MB and AO = OB (since these are radii). Points O and M are each equidistant from the endpoints of AB. Two points that are each equidis- tant from the endpoints of a line segment determine the perpendicular bisec- tor of the line segment. Therefore, g OM Through a point on a line there is only one perpendicular line. Thus, is the perpendicular bisector of g OM AB. and k are the same line, and O is on k, the perpendicular bisector of AB. Theorem 13.7a If two chords of a circle are congruent, then they are equidistant from the center of the circle. Given Circle O with AB > CD, OE'AB, and OF'CD. Prove OE > OF Proof A line through the center of a circle that is perpendicular to a chord bisects the chord and its arcs. Therefore, g OE and g OF bisect the congru- ent chords EB > FD OB > OD AB and CD. Since halves of congruent segments are congruent, OB. Draw. Therefore, OBE ODF by HL, and. Since and and OD OD OB OE > OF. are radii of the same circle, A C E O F B D 14365C13.pgs 7/12/07 3:56 PM Page 548 548 Geometry of the Circle The converse of this theorem is also true. Theorem 13.7b If two chords of a circle are equidistant from the center of the circle, then the chords are congruent. Given Circle O with OE'AB, OF'CD, and OE > OF. Prove AB > |
CD are radii of the same. Therefore, OBE ODF by HL, and and OD. A line through the center of a circle that is perpendicular to a OB OD. Since and OB OB > OD Proof Draw circle, EB > FD chord bisects the chord. Thus, gruent segments are congruent, AE > EB and AB > CD. CF > FD. Since doubles of con- A C E O F B D We can state theorems 13.7a and 13.7b as a biconditional. Theorem 13.7 Two chords are equidistant from the center of a circle if and only if the chords are congruent. What if two chords are not equidistant from the center of a circle? Which is the longer chord? We know that a diameter contains the center of the circle and is the longest chord of the circle. This suggests that the shorter chord is farther from the center of the circle. (1) Let and AB AB CD. CD be two chords of circle O and (2) Draw OE'AB and OF'CD. A E C O F B D (3) A line through the center of a circle that is perpendicular to the chord bisects the chord. Therefore, 1 2CD 5 FD. (4) The distance from a point to a line is the length of the perpendicular 1 2AB 5 EB and from the point to the line. Therefore, OE is the distance from the center, and OF is the distance from the center of the circle of the circle to. to CD AB Recall that the squares of equal quantities are equal, that the positive square roots of equal quantities are equal, and that when an inequality is multiplied by a negative number, the inequality is reversed. 14365C13.pgs 7/12/07 3:56 PM Page 549 (5) Since AB CD: Arcs and Chords 549 AB, CD 2AB, 1 1 2CD EB, FD EB2, FD2 2EB2. 2FD2 (6) Since OBE and ODF are right triangles and OB = OD: OB2 5 OD2 OE2 1 EB2 5 OF2 1 FD2 (7) When equal quantities are added to both sides of an inequality, the order of the inequality remains the same. Therefore, adding the equal quantities from step 6 to the inequality in step 5 gives: OE2 1 EB2 2 EB2. OF2 1 FD |
2 2 FD2 OE2. OF2 OE. OF Therefore, the shorter chord is farther from the center of the circle. We have just proved the following theorem: Theorem 13.8 In a circle, if the lengths of two chords are unequal, then the shorter chord is farther from the center. EXAMPLE 2 In circle O, mABX 90 and OA 6. a. Prove that AOB is a right triangle. b. Find AB. c. Find OC, the apothem to. AB Solution a. If mABX 90, then mAOB 90 because the measure of an arc is equal to the measure of the central angle that intercepts the arc. Since AOB is a right angle, AOB is a right triangle. A 6 O 90° C B 14365C13.pgs 7/12/07 3:56 PM Page 550 550 Geometry of the Circle A 6 O 90° C B b. Use the Pythagorean Theorem for right AOB. Since OA and OB are radii, OB OA 6. AB2 OA2 OB2 AB2 62 62 AB2 36 36 AB2 72 72 5 36 " ", AB OC'AB 2 5 6 2 " and bisects AB. Therefore, AB " c. Since OC is the apothem to. In right OCA, 2 OC2 1 AC2 AC 5 3 " 5 OA2 2 5 62 OC2 1 2 3 " B A 18 5 36 OC2 1 5 18 OC2 " Note: In the example, since AOB is an isosceles right triangle, mAOB is 45. Therefore AOC is also an isosceles right triangle and OC AC. " " OC 5 18 5 2 5 3 2 9 " Polygons Inscribed in a Circle If all of the vertices of a polygon are points of a circle, then the polygon is said to be inscribed in the circle. We can also say that the circle is circumscribed about the polygon. In the diagram: A 1. Polygon ABCD is inscribed in circle O. B 2. Circle O is circumscribed about polygon ABCD. D O C In an earlier chapter we proved that the perpendicular bisectors of the sides of a triangle meet at a point and that that point is equidistant from the verg tices of the triangle. In the diagram, PN are the perpendicular bis |
ectors of the sides of ABC. Every point on the perpendicular bisectors of a line segment is equidistant from the endpoints of the line segment. Therefore, PA PB PC and A, B, and C are points on a circle with center at P, that is, any triangle can be inscribed in a circle. g, PL g PM, and C P L M B N A 14365C13.pgs 7/12/07 3:56 PM Page 551 EXAMPLE 3 Prove that any rectangle can be inscribed in a circle. Proof Let ABCD be any rectangle. The diagonals of a rectangle Arcs and Chords 551 D AC > BD and AC BD. Since a rec- BD AC AE EC and BD BE ED. Halves of equal are congruent, so tangle is a parallelogram, the diagonals of a rectangle intersect at E, then and bisect each other. If 1 2 quantities are equal. Therefore, AE EC BE ED and the vertices of the rectangle are equidistant from E. Let E be the center of a circle with radius AE. The vertices of ABCD are on the circle and ABCD is inscribed in the circle. AC 1 2 A E C B Exercises Writing About Mathematics 1. Daniela said that if a chord is 3 inches from the center of a circle that has a radius of 5 inches, then a 3-4-5 right triangle is formed by the chord, its apothem, and a radius. Additionally, the length of the chord is 4 inches. Do you agree with Daniela? Explain why or why not. 2. Two angles that have the same measure are always congruent. Are two arcs that have the same measure always congruent? Explain why or why not. Developing Skills In 3–7, find the length of the radius of a circle whose diameter has the given measure. 3. 6 in. 4. 9 cm 5. 3 ft 6. 24 mm 7. " 24 cm In 8–12, find the length of the diameter of a circle whose radius has the given measure. 8. 5 in. 9. 12 ft 10. 7 cm 11. 6.2 mm 12. yd 5 " 13. In circle O, AOB is a diameter, AB 3x 13, and AO 2x 5. Find the length of the radius and of the diameter of the circle. In 14–21, DCOE |
is a diameter of circle O, AB is a chord of the circle, and OD'AB at C. 14. If AB 8 and OC 3, find OB. 16. If OC 20 and OB 25, find AB. 15. If AB 48 and OC 7, find OB. 17. If OC 12 and OB 18, find AB. 14365C13.pgs 7/12/07 3:56 PM Page 552 552 Geometry of the Circle 18. If AB 18 and OB 15, find OC. 19. If AB 20 and OB 15, find OC. 20. If mAOB 90, and AB 30 OB and DE., find 2 " 21. If mAOB 60, and AB 30, find OB and OC. 22. In circle O, chord LM is 3 centimeters from the center and chord RS is 5 centimeters from the center. Which is the longer chord? Applying Skills 23. Prove Theorem 13.3b, “In a circle or in congruent circles, congruent chords have congruent central angles.” 24. Prove Theorem 13.4b, “In a circle or in congruent circles, congruent chords have congruent arcs.” 25. Diameter AOB bisects chord CD of circle O intersects chord CD and is perpendicular to chord at E and bisects CD. CDX at B. Prove that AOB 26. The radius of a spherical ball is 13 centimeters. A piece that has a plane surface is cut off of the ball at a distance of 12 centimeters from the center of the ball. What is the radius of the circular faces of the cut pieces? 27. Triangle ABC is inscribed in circle O. The distance from the center of the circle to AB is greater than the distance from the center of the circle to ter of the circle to Which is the largest angle of ABC? Justify your answer. BC is greater than the distance from the center of the circle to AC. BC, and the distance from the cen- 13-3 INSCRIBED ANGLES AND THEIR MEASURES B In the diagram, ABC is an angle formed by two chords that have a common endpoint on the circle. A DEFINITION C An inscribed angle of a circle is an angle whose vertex is on the circle and whose sides contain chords of the circle. We can use the fact that the measure of a central angle is equal to the measure |
of its arc to find the relationship between ABC and the measure of its arc,.ACX 14365C13.pgs 7/12/07 3:56 PM Page 553 CASE 1 One of the sides of the inscribed angle contains a diameter of the circle. Inscribed Angles and Their Measures 553 B x x O 2x A 2x C OA Consider first an inscribed angle, ABC, with a diameter of circle O.. Then AOB is an isosceles triangle and mOAB mOBA x. Draw Angle AOC is an exterior angle and mAOC x x 2x. Since AOC is a central angle, mAOC. Therefore, mABC x mACX 5 2x BC 1. 2mACX We have shown that when one of the sides of an inscribed angle contains a diameter of the circle, the measure of the inscribed angle is equal to one-half the measure of its intercepted arc. Is this true for angles whose sides do not contain the center of the circle? B CASE 2 The center of the circle is in the interior of the angle. Let ABC be an inscribed angle in which the center of the circle is in the A O C interior of the angle. Draw BOD, a diameter of the circle. Then: mABD 1 2mADX and mDBC 1 2mDCX D Therefore: m/ABC 5 m/ABD 1 m/DBC 2mDCX 2mADX 1 1 5 1 2(mADX 1 mDCX) 5 1 2mACX 5 1 B O D A C CASE 3 The center of the circle is not in the interior of the angle. Let ABC be an inscribed angle in which the center of the circle is not in the interior of the angle. Draw BOD 1 2mADX, a diameter of the circle. Then: and mDBC 1 2mDCX mABD Therefore: m/ABC 5 m/ABD 2 m/DBC 2mDCX 2mADX 2 1 2(mADX 2 mDCX) 2mACX 5 1 5 1 5 1 These three possible positions of the sides of the circle with respect to the center the circle prove the following theorem: Theorem 13.9 The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc. There are two statements that can be derived from this theorem. 14365C13. |
pgs 7/12/07 3:56 PM Page 554 554 Geometry of the Circle Corollary 13.9a An angle inscribed in a semicircle is a right angle. Proof: In the diagram, ABC is inscribed in semicircle semicircle whose degree measure is 180°. Therefore, is a diameter of circle O, and is a. Also ADCX ABCX AOC mABC 2 mADCX 5 1 5 1 2(180) 5 90 B A C O D Since any triangle can be inscribed in a circle, the hypotenuse of a triangle can be the diameter of a circle with the midpoint of the hypotenuse the center of the circle. Corollary 13.9b If two inscribed angles of a circle intercept the same arc, then they are congruent. Proof: In the diagram, ABC and ADC are inscribed angles and each angle intercepts ACX and mADC. Since ABC and ADC have equal measures, they are congruent.. Therefore, mABC 2mACX 2mACX 1 1 EXAMPLE Triangle ABC is inscribed in circle O, mB 70, and mACX a. b. mA c. mC d. mABX B A D C mBCX 100. Find: Solution a. If the measure of an inscribed angle is one-half the measure of its intercepted arc, then the measure of the intercepted arc is twice the measure of the inscribed angle. A 2mACX m/B 5 1 2m/B 5 mACX 2(70) 5 mACX 140 5 mACX B 70 100 C 14365C13.pgs 7/12/07 3:56 PM Page 555 Inscribed Angles and Their Measures 555 c. m/C 5 180 2 (m/A 1 m/B) 5 180 2 (50 1 70) 5 180 2 120 5 60 b. m/A 5 1 2mBCX 2(100) 5 1 5 50 d. mABX 5 2m/C 5 2(60) 5 120 Answers a. 140° b. 50° c. 60° d. 120° SUMMARY Type of Angle Degree Measure Example Central Angle The measure of a central angle is equal to the measure of its intercepted arc. Inscribed Angle The measure of an inscribed angle is equal to one-half the measure of its intercepted arc. B A A 1 m1 mABX B 1 |
m/1 5 1 2m ABX Exercises Writing About Mathematics 1. Explain how you could use Corollary 13.9a to construct a right triangle with two given line segments as the hypotenuse and one leg. 14365C13.pgs 7/12/07 3:56 PM Page 556 556 Geometry of the Circle 2. In circle O, ABC is an inscribed angle and mACX 50. In circle O, PQR is an inscribed 50. Is ABC PQR if the circles are not congruent circles? Justify mPRX angle and your answer. Developing Skills In 3–7, B is a point on circle O not on 3. 88 4. 72 ACX 5. 170 ACX, an arc of circle O. Find mABC for each given m. 6. 200 7. 280 In 8–12, B is a point on circle O not on, an arc of circle O. Find m ACX for each given mABC. 8. 12 9. 45 11. 95 12. 125 ACX 10. 60 13. Triangle ABC is inscribed in a circle, mA 80 and 88. Find: mACX mBCX a. b. mB c. mC mABX d. mBACX e. 14. Triangle DEF is inscribed in a circle, DE > EF, and B mEFX 100. Find: a. mD mDEX b. c. mF d. mE mDFX e. A 80 88 C D A E 100 F B AC In 15–17, chords and 15. If mB 42 and mAEB 104, find: mBCX mADX a. mA BD b. c. intersect at E in circle O. d. mD e. mC E D C 16. If AB DC and mB 40, find: 17. If a. mD mADX mDCX a. mADX b. mABX b. mA mBCX c. c. mB 100, 110, and 18. Triangle ABC is inscribed in a circle and mBCX mABX mCAX b. a. c. d. mA e. mDEC 96, find: mBCX d. mAEB e. mC mABX mBCX : d. mA mCAX e. mB : 2 : 3 : 7. Find: |
f. mC 14365C13.pgs 7/12/07 3:57 PM Page 557 19. Triangle RST is inscribed in a circle and mSTX mTRX mRSX b. a. c. Inscribed Angles and Their Measures 557 mRSX 5 mSTX 5 mTRX. Find: d. mR e. mS f. mT Applying Skills 20. In circle O, LM and RS intersect at P. a. Prove that LPR SPM. b. If LP 15 cm, RP 12 cm, and SP 10 cm, find MP. 21. Triangle ABC is inscribed in a circle. If mABX 100 and mBCX isosceles. 22. Parallelogram ABCD is inscribed in a circle. mABCX mADCX. = a. Explain why mABCX mADCX. and b. Find c. Explain why parallelogram ABCD must be a rectangle. R P M L O S 130, prove that ABC is A D O B C D G F E Ex. 23 Ex. 24 Ex. 25 23. Triangle DEF is inscribed in a circle and G is any point not on mDEX 1 mEFX 5 mFGDX, show that DEF is a right triangle. DEFX. If 24. In circle O, ASA. AOC and BOD are diameters. If AB > CD, prove that ABC DCB by 25. Chords ABX 26. Prove that a trapezoid inscribed in a circle is isosceles. of circle O intersect at E. If and BD AC CDX, prove that ABC DCB. and major are arcs of circle O and AB CD. Prove that ADCX and major ADCX are arcs of circle O and ADX > BCX. Prove C B O 27. Minor ABCX ADX > BCX ABCX AB CD. 28. Minor that. 29. In circle O, AOC and BOD are diameters. Prove that AB CD. D A 14365C13.pgs 7/12/07 3:57 PM Page 558 558 Geometry of the Circle 30. Points A, B, C, D, E, and F are on circle O, AB CD EF,. ABCD and CDEF are trapezoids. Prove that CB ED and CAX > BDX > DFX > ECX. 31. |
Quadrilateral ABCD is inscribed in circle O, and. Prove that ABCD is not a parallelogram. CDX to ABX is not congruent 13-4 TANGENTS AND SECANTS In the diagram, line p has no points in common with the circle. Line m has one point in common with the circle. Line m is said to be tangent to the circle. Line k has two points in common with the circle. Line k is said to be a secant of the circle. P O A DEFINITION A tangent to a circle is a line in the plane of the circle that intersects the circle in one and only one point. DEFINITION A secant of a circle is a line that intersects the circle in two points. Let us begin by assuming that at every point on a circle, there exists exactly one tangent line. We can state this as a postulate. Postulate 13.2 At a given point on a given circle, one and only one line can be drawn that is tangent to the circle. OP Let P be any point on circle O and be a radius to that point. If line m containing points P and Q is perpendicular to, then OQ OP because the perpendicular is the shortest distance from a point to a line. Therefore, every point on the line except P is outside of circle O and line m must be tangent to the circle. This establishes the truth of the following theorem. OP Q m P O 14365C13.pgs 7/12/07 3:57 PM Page 559 Tangents and Secants 559 Theorem 13.10a If a line is perpendicular to a radius at a point on the circle, then the line is tangent to the circle. The converse of this theorem is also true. Theorem 13.10b If a line is tangent to a circle, then it is perpendicular to a radius at a point on the circle. Given Line m is tangent to circle O at P. Prove Line m is perpendicular to OP. Proof We can use an indirect proof. m b P O OP at P since, at a given point on a given line, one and only one line can Assume that m is not perpendicular to. OP Then there is some line b that is perpendicular to be drawn perpendicular to the given line. Then by Theorem 13.10a, b is a tangent to circle O at P. But this contradicts the postulate |
that states that at a given point on a circle, one and only one tangent can be drawn. Therefore, our assumption is false and its negation must be true. Line m is perpendicular to OP. We can state Theorems 13.10a and 13.10b as a biconditional. Theorem 13.10 A line is tangent to a circle if and only if it is perpendicular to a radius at its point of intersection with the circle. Common Tangents DEFINITION A common tangent is a line that is tangent to each of two circles. In the diagram, g AB is tangent to circle O at A and to circle O at B. Tangent is said to be a common internal tangent because the tangent intersects the line segment joining the centers of the circles. g AB A O O B 14365C13.pgs 7/12/07 3:57 PM Page 560 560 Geometry of the Circle In the diagram, g is tangent to cirCD cle P at C and to circle P at D. Tangent g is said to be a common external CD tangent because the tangent does not intersect the line segment joining the centers of the circles. P C P D The diagrams below show that two circles can have four, three, two, one, or no common tangents. 4 common tangents 3 common tangents 2 common tangents 1 common tangent No common tangents Two circles are said to be tangent to each other if they are tangent to the is tangent to circle O and to same line at the same point. In the diagram, circle O at T. Circles O and O are tangent externally because every point of one of the circles, except the point of tangency, is an external point of the other circle. g ST g MN In the diagram, is tangent to circle P and to circle P at M. Circles P and P are tangent internally because every point of one of the circles, except the point of tangency, is an internal point of the other circle. T O O S EXAMPLE Given: Circles O and O with a common internal, tangent to circle O at A and OOr tangent, circle O at B, and C the intersection of and g AB g AB. Prove: AC BC 5 OC OrC 14365C13.pgs 7/12/07 3:57 PM Page 561 Proof We will use similar triangles to prove |
the segments proportional. Tangents and Secants 561 g AB is tangent to circle O at A and circle O at B. A line tangent to a Line circle is perpendicular to a radius drawn to the point of tangency. Since perpendicular lines intersect to form right angles and all right angle are congruent, OAC OBC. Also, OCA OCB because vertical angles are congruent. Therefore, OCA OCB by AA. The lengths of corresponding sides of similar triangles are proportional. Therefore,. BC 5 OC AC OrC EXAMPLE 2 Circle O is tangent to g AB at A, O is tangent to g AB at B, and OOr intersects g AB at C. a. Prove that. b. If AC 8, AB 12, and OA 9, find OB. AC BC 5 OA OrB Solution a. We know that OAB OBA because they are right angles and that OCA OCB because they are vertical angles. Therefore, BC 5 OA AC OCA OCB by AA and OrB. b. AB 5 AC 1 BC 12 5 8 1 BC 4 5 BC BC 5 OA AC OrB 4 5 9 8 OrB 8OrB 5 36 OrB 5 36 8 5 9 2 Answer Tangent Segments DEFINITION A tangent segment is a segment of a tangent line, one of whose endpoints is the point of tangency. In the diagram, PQ segments of the tangents and g PQ PR and O from P. are tangent g PR to circle P Q O R Theorem 13.11 Tangent segments drawn to a circle from an external point are congruent. 14365C13.pgs 7/12/07 3:57 PM Page 562 562 Geometry of the Circle Given g PQ tangent to circle O at Q and g PR tangent to circle O at R. P Q O R Prove PQ > PR Proof Draw OQ, RP OP OR and, and. Since are OR. Since are tangent to the circle at Q and and OQ OQ > OR both radii of the same circle, QP R, OQP and ORP are both right angles, and OPQ and OPR are right is the hypotenuse of both OPQ and OPR. Therefore, triangles. Then OPQ OPR by HL. Corresponding parts of congruent triangles are congruent, so P |
Q > PR OP. The following corollaries are also true. Corollary 13.11a If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents. Given g PQ tangent to circle O at Q and g PR tangent to circle O at R. Prove h PO bisects RPQ. Strategy Use the proof of Theorem 13.11 to show that angles OPQ and RPO are congruent. P Q O R Corollary 13.11b If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle whose vertex is the center of the circle and whose rays are the two radii drawn to the points of tangency. Given g PQ tangent to circle O at Q and g PR tan- gent to circle O at R. Prove h OP bisects QOR. Strategy Use the proof of Theorem 13.11 to show that angles QOP and ROP are congruent. P Q O R 14365C13.pgs 7/12/07 3:57 PM Page 563 Tangents and Secants 563 The proofs of Corollaries 13.11a and 13.11b are left to the student. (See exercises 15 and 16.) A Polygon Circumscribed About a Circle A polygon is circumscribed about a circle if each side of the polygon is tangent to the circle. When a polygon is circumscribed about a circle, we also say that the circle is inscribed in the polygon. For example, in the diais tangent to gram, AB circle O at F, is tangent to circle O at H. Therefore, ABCD is circumscribed about circle O and circle O is inscribed in quadrilateral ABCD. is tangent to circle O at G, is tangent to circle O at E, DA CD BC If ABC is circumscribed about circle O, then we know that are the, OA bisectors of the angles of ABC, and O is the point at which the angle bisectors of the angles of a triangle intersect., and OB OC EXAMPLE 3 C BC CA, and are tangent to circle O at D, E, and, AB F, respectively. If AF 6, BE 7, |
and CE 5, find the perimeter of ABC. Solution Tangent segments drawn to a circle from an exter- nal point are congruent. AD AF 6 BD BE 7 CF CE 5 Therefore AB 5 AD 1 BD 5 6 1 7 5 13 BC 5 BE 1 CE 5 7 1 5 5 12 CA 5 CF 1 AF 5 5 1 6 5 11 Perimeter 5 AB 1 BC 1 CA 5 13 1 12 1 11 5 36 Answer 14365C13.pgs 7/12/07 3:57 PM Page 564 564 Geometry of the Circle EXAMPLE 4 Point P is a point on a line that is tangent to circle O at R, P is 12.0 centimeters from the center of the circle, and the length of the tangent segment from P is 8.0 centimeters. a. Find the exact length of the radius of the circle. b. Find the length of the radius to the nearest tenth. Solution a. P is 12 centimeters from the center of circle O; OP 12. R 8 cm P The length of the tangent segment is 8 centimeters; RP 8. A line tangent to a circle is perpendicular to the radius drawn to the point of tangency; OPR is a right triangle. O 12 cm RP2 OR2 OP2 82 OP2 122 64 OP2 144 OP2 80 b. Use a calculator to evaluate OP 80 5 16 " " 5 5 4 " 5 " 4. 5 " ENTER: 4 2nd ¯ 5 ENTER DISPLAY: 8.94427191 To the nearest tenth, OP 8.9. Answers a. 4 5 " cm b. 8.9 cm Exercises Writing About Mathematics 1. Line l is tangent to circle O at A and line m is tangent to circle O at B. If AOB is a diame- ter, does l intersect m? Justify your answer. 2. Explain the difference between a polygon inscribed in a circle and a circle inscribed in a polygon. 14365C13.pgs 7/12/07 3:57 PM Page 565 Tangents and Secants 565 Developing Skills In 3 and 4, ABC is circumscribed about circle O and D, E, and F are points of tangency. 3. If AD 5, EB 5, and CF 10, find the lengths of the sides of the triangle and show that the triangle is isosceles. 4. If AF 10, CE 20, and BD 30, |
find the lengths of the sides of the triangle and show that the triangle is a right triangle. A F D PQ In 5–11, T and R. is tangent to circle O at P, SQ is tangent to circle O at S, and OQ intersects circle O at C E B P 5. If OP 15 and PQ 20, find: a. OQ b. SQ c. TQ 6. If OQ 25 and PQ 24, find: a. OP b. RT c. RQ 7. If OP 10 and OQ 26, find: a. PQ b. RQ c. TQ 8. If OP 6 and TQ 13, find: a. OQ b. PQ c. SQ 9. If OS 9 and RQ 32, find: a. OQ b. SQ c. PQ 10. If PQ 3x, SQ 5x 8, and OS x 1, find: a. PQ b. SQ c. OS d. OQ 11. If SQ 2x, OS 2x 2, and OQ 3x 1, find: a. x b. SQ c. OS d. OQ 12. The sides of ABC are tangent to a circle at D, E, and F. If DB 4, BC 7, and the perimeter of the triangle is 30, find. BE b. EC c. CF d. AF e. AC f. AB 13. Line g RP is tangent to circle O at P and the circle at M, the midpoint of OR OR intersects. If RP 3.00 cm, A F C find the length of the radius of the circle: a. in radical form b. to the nearest hundredth O P 14. Points E, F, G, and H are the points of tangency to circle O of EFX FGX, AB BC is 70°, and DA is 50°. Find:, respectively. The measure of, and CD GHX of a. mEOF b. mFOG f. mEAO g. mEAH k. the sum of the measures of the angles of quadrilateral ABCD c. mGOH d. mHOE i. mGCF h. mFBE is 80°, of e. mAOE j. mHDG M, R E O B F GC A H D 14365C13.pgs 7/12/07 3 |
:57 PM Page 566 566 Geometry of the Circle Applying Skills 15. Prove Corollary 13.11a, “If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents.” 16. Prove Corollary 13.11b, “If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle whose vertex is the center of the circle and whose rays are the two radii drawn to the points of tangency.” g PQ are tangent to circle O at Q and R. 17. Lines and Q P g PR a. Prove that PQR PRQ. OP intersecting RQ at S and prove that QS RS b. Draw and OP'QR. c. If OP 10, SQ 4 and OS SP, find OS and SP. O R 18. Tangents AC and BC to circle O are perpendicular to each other at C. Prove: AC > AO a. b. OC " c. AOBC is a square. 2OA 19. Isosceles ABC is circumscribed about circle O. The points of tangency of the legs, AB and, are D and F, and the point of tangency of the base, 20. Line is a common external tangent to circle. BC AC point of g AB g O and circle O. AB and to circle O at B, and OA OB. g AB a. Prove that g OOr. is tangent to circle O at A is not parallel to g OOr and g AB. b. Let C be the intersection of Prove that OAC OBC. BC, is E. Prove that E is the mid- C A O B O c. If, and BC 12, find AC, AB, OC, OC, and OO. OA OrB 5 2 3 g AB is tangent to circle O at A and to circle O at B, and is a common internal tangent to circles O and O. 21. Line g AB OA OB. The intersection of a. Prove that OC OC. b. Prove that AC BC. OOr and g AB is C. O A C B O 14365C13.pgs 7/12/ |
07 3:57 PM Page 567 Angles Formed by Tangents, Chords, and Secants 567 Hands-On Activity Consider any regular polygon. Construct the angle bisectors of each interior angle. Since the interior angles are all congruent, the angles formed are all congruent. Since the sides of the regular polygon are all congruent, congruent isosceles triangles are formed by ASA. Any two adjacent triangles share a common leg. Therefore, they all share the same vertex. Since the legs of the triangles formed are all congruent, the vertex is equidistant from the vertices of the regular polygon. This common vertex is the center of the regular polygon. In this Hands-On Activity, we will use the center of a regular polygon to inscribe a circle in the polygon. a. Using geometry software or compass, protractor, and straightedge, construct a square, a regular pentagon, and a regular hexagon. For each figure: (1) Construct the center P of the regular polygon. (The center is the intersection of the angle bisectors of a regular polygon.) (2) Construct an apothem or perpendicular from P to one of the sides of regular polygon. (3) Construct a circle with center P and radius equal to the length of the apothem. b. Prove that the circles constructed in part a are inscribed inside of the polygon. Prove: (1) The apothems of each polygon are all congruent. (2) The foot of each apothem is on the circle. (3) The sides of the regular polygon are tangent to the circle. c. Let r be the distance from the center to a vertex of the regular polygon. Since the center is equidistant from each vertex, it is possible to circumscribe a circle about the polygon with radius r. Let a be the length of an apothem and s be the length of a side of the regular polygon. How is the radius, r, of the circumscribed circle related to the radius, a, of the inscribed circle? 13-5 ANGLES FORMED BY TANGENTS, CHORDS, AND SECANTS Angles Formed by a Tangent and a Chord g AB AC CD is a diameter. When is tangent to circle O at A, is In the diagram, AD is drawn, |
a chord, and ADC is a right angle because it is an angle inscribed in a semicircle, and ACD is the complement of CAD., BAC is a right angle, and DAB is the Also, complement of CAD. Therefore, since complements of the same angle are congruent, ACD DAB. We can conclude that since mACD =, then mDAB = 2mADX CA'AB 1 1. 2mADX A O B C D 14365C13.pgs 7/12/07 3:57 PM Page 568 568 Geometry of the Circle We can state what we have just proved on page 567 as a theorem. Theorem 13.12 The measure of an angle formed by a tangent and a chord that intersect at the point of tangency is equal to one-half the measure of the intercepted arc. Angles Formed by Two Intersecting Chords D O B E A C We can find how the measures of other angles and their intercepted arcs are intersect in the related. For example, in the diagram, two chords AB is drawn. Angle AED is an exterior angle of DEB. interior of circle O and DB Therefore, and CD mAED mBDE mDBE 2mDAX 2mBCX 1 1 2(mBCX 1 mDAX ) 1 1 BCX Notice that is the arc intercepted by BEC and is the arc intercepted by AED, the angle vertical to BEC. We can state this relationship as a theorem. DAX Theorem 13.13 The measure of an angle formed by two chords intersecting within a circle is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Angles Formed by Tangents and Secants We have shown how the measures of angles whose vertices are on the circle or within the circle are related to the measures of their intercepted arcs. Now we want to show how angles formed by two tangents, a tangent and a secant, or two secants, all of which have vertices outside the circle, are related to the measures of the intercepted arcs. A Tangent Intersecting a Secant In the diagram, g PTQ and g PRS is a tangent to circle O at R is a secant that intersects the circle at T is drawn. Then SRQ is an and at Q. Chord exterior angle of PRQ. RQ |
S Q R O T P 14365C13.pgs 7/12/07 3:57 PM Page 569 Angles Formed by Tangents, Chords, and Secants 569 mRQP mP mSRQ mP mSRQ mRQP mP mP 2mRQX 2mRTX 2(mRQX 2 mRTX) 1 1 1 Two Intersecting Secants In the diagram, g PTR that intersects the circle at R and T, and is a secant to circle O g PQS is a secant to circle O that intersects the circle is drawn. Then RQS at Q and S. Chord is an exterior angle of RQP. RQ mPRQ mP mRQS T R O Q S mP mRQS mPRQ mP mP 2mRSX 2mQTX 2(mRSX 2 mQTX ) 1 1 1 Two Intersecting Tangents In the diagram, tangent to the circle at Q, and T is a point on major g PRS is tangent to circle O at R, g is PQ RQX. is drawn. Then SRQ is an exterior angle of RQ Chord RQP. R S O Q P P mPQR mP mSRQ T mP mSRQ mPQR mP mP 2mRTQX 2mRQX 2(mRTQX 2 mRQX ) 1 1 1 For each pair of lines, a tangent and a secant, two secants, and two tangents, the steps necessary to prove the following theorem have been given: Theorem 13.14 The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs. 14365C13.pgs 7/12/07 3:57 PM Page 570 570 Geometry of the Circle EXAMPLE 1 A tangent and a secant are drawn to circle O from point P. The tangent intersects the circle at Q and mQRX : mRSX : mSQX 5 3 : 5 : 7 the secant at R and S. If, find: Q P O R mQRX a. d. mQRS mRSX |
b. e. mRQP mSQX c. f. mP S Solution Let mQRX 3x, mRSX 5x, and mSQX 7x. 3x 1 5x 1 7x 5 360 15x 5 360 x 5 24 mQRX 5 3x a. mRSX 5 5x b. 5 3(24) 5 72 5 5(24) 5 120 c. mSQX 5 7x 7(24) 5 168 d. m/QRS 5 1 5 1 2mSQX 2(168) e. m/RQP 5 1 5 1 2mQRX 2(72) f. 2(mSQX 2 mQRX ) m/P 5 1 5 1 2(168 2 72) 5 84 5 36 5 48 Answers a. 72° b. 120° c. 168° d. 84° e. 36° f. 48° Note: m/QRS 5 m/RQP 1 m/P 5 36 1 48 5 84 EXAMPLE 2 Two tangent segments, RP R. If mR is 70, find the measure of the minor arc into which the circle is divided. and RQ PQX, are drawn to circle O from an external point and of the major arc PSQX 14365C13.pgs 7/12/07 3:57 PM Page 571 Angles Formed by Tangents, Chords, and Secants 571 P R Q S Solution The sum of the minor arc and the major arc with the same endpoints is 360. Let x = mPQX. Then 360 x = mPSQX. mR 70 1 2(mPSQX 2 mPQX) 1 2(360 2 x 2 x) 1 2(360 2 2x) 70 70 180 x x 110 360 x 360 110 250 Answer mPQX 110 and mPSQX 250 SUMMARY Type of Angle Degree Measure Example Formed by a Tangent and a Chord The measure of an angle formed by a tangent and a chord that intersect at the point of tangency is equal to one-half the measure of the intercepted arc. Formed by Two Intersecting Chords The measure of an angle formed by two intersecting chords is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. A 1 B m |
/1 5 1 2mABX 2 1 A B D C m/1 5 1 m/2 5 1 2(mABX 1 mCDX) 2(mABX 1 mCDX) (Continued) 14365C13.pgs 7/12/07 3:57 PM Page 572 572 Geometry of the Circle SUMMARY (Continued) Type of Angle Degree Measure Example Formed by Tangents and Secants The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs/1 5 1 m/2 5 1 m/3 5 1 2(mABX 2 mACX) 2(mABX 2 mCDX) 2(mACBX 2 mABX) Exercises Writing About Mathematics 1. Nina said that a radius drawn to the point at which a secant intersects a circle cannot be perpendicular to the secant. Do you agree with Nina? Explain why or why not. 2. Two chords intersect at the center of a circle forming four central angles. Aaron said that the measure of one of these angles is one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Do you agree with Aaron? Explain why or why not. Developing Skills g PQS and 3. If In 3–8, secants mSTX mSTX mSTX 4. If 5. If 160 and 100 and 170 and 6. If mP 40 and 7. If mP 60 and 8. If mP 25 and g PRT mQRX mQRX mQRX mQRX mQRX mSTX intersect at P. 90, find mP. 40, find mP. 110, find mP. 86, find 50, find 110, find mSTX. mSTX..mQRX P Q R S T 14365C13.pgs 7/12/07 3:57 PM Page 573 Angles Formed by Tangents, Chords, and Secants 573 In 9–14, tangent g QP g PRT intersect at P. 70, find mP. 30, find mP. 120 and 170 and and secant mQRX mQRX mRTX 50 and mP 40, |
find 70 and 9. If 10. If 11. If mQTX mQTX mQRX mQRX mQRX 14. If mP 30 and 13. If 12. If 60 and mP 35, find mQRX 120, find mP. mQTX. mQTX. mQTX. 120, find Q P R T g RP and g QP intersect at P and S is on major arc QRX. 160, find mP. 80, find mP. 260, find mP. 210, find mP. 16. If 15. If In 15–20, tangents mRQX mRQX mRSQX mRSQX mRSQX 5 2mRQX 20. If mP 45, find 17. If 18. If 19. If, find mP. mRQX and mRSQX. P Q R S In 21–26, chords AB and intersect at E in the interior of a circle. mACX mDAX mACX mACX mACX mBCX 21. If 22. If 23. If 24. If 25. If 26. If CD mBDX mBCX mBDX mBDX 30 and 80, find mAEC. 180 and 100 find mAED. 25 and 20 and 45, find mDEB. 60, find mAED. mBDX. mDAX. 30 and mAEC 50, find 80 and mAEC 30, find CA E D B 27. In the diagram, g PA and g PB are tangent to circle O at A and and chord AC intersect at E, mCBX 125 and B. Diameter BD mP 55. Find: ABX a. m d. mDEC ADX b. m e. mPBD CDX c. m f. mPAC g. Show that BD is perpendicular to AC and bisects.AC P 55° A E O D C B 125° 14365C13.pgs 7/12/07 3:57 PM Page 574 574 Geometry of the Circle and secant segment are chords. If mP 45 and PBC are drawn to cir- AB 28. Tangent segment PA cle O and and mACX : mABX 5 5 : 2 ACX a. m d. mPAB AC, find: BCX b. m e. |
mCAB c. mACB f. mPAC Applying Skills 29. Tangent g PC intersects circle O at C, chord g AB CP, diameter COD intersects AB at E, and diameter AOF is extended to P. a. Prove that OPC OAE. mADX b. If mOAE 30, find, and mP. mCFX, mFBX, mBDX, mACX, A g ABC 30. Tangent intersects circle O at B, › ‹ AFOD secant intersects the circle at › ‹ CGOE intersects F and D, and secant the circle at G and E. If mEFBX 5 mDGBX, prove that AOC is A F E A O P 45 isosceles. 31. Segments BP AP and are tangent to circle O at A and B, respectively, and mAOB 120. Prove that ABP is equilateral. 32. Secant g ABC intersects a circle at A and B. Chord BD Prove that mCBD is drawn. 1 2mBDX. A B O P A B C O D 14365C13.pgs 7/12/07 3:57 PM Page 575 Measures of Tangent Segments, Chords, and Secant Segments 575 13-6 MEASURES OF TANGENT SEGMENTS, CHORDS, AND SECANT SEGMENTS Segments Formed by Two Intersecting Chords We have been proving theorems to establish the relationship between the measures of angles of a circle and the measures of the intercepted arcs. Now we will study the measures of tangent segments, secant segments, and chords. To do this, we will use what we know about similar triangles. Theorem 13.15 If two chords intersect within a circle, the product of the measures of the segments of one chord is equal to the product of the measures of the segments of the other. Given Chords circle O. AB and CD intersect at E in the interior of Prove (AE)(EB) (CE)(ED) Proof Statements Reasons A D E O C B AD and 1. Draw 2. A C and D B. CB 1. Two points determine a line. 2. Inscribed angles of a circle that inter- cept the same arc are congruent. 3. ADE CBE 3. AA. 4. CE 5 ED |
AE EB 5. (AE)(EB) (CE)(ED) 4. The lengths of the corresponding sides of similar triangles are in proportion. 5. In a proportion, the product of the means is equal to the product of the extremes. Segments Formed by a Tangent Intersecting a Secant Do similar relationships exist for tangent segments and secant segments? In the diagram, tangent segment is drawn to circle O, and secant segment PBC PA intersects the circle at B and C. C A P O B 14365C13.pgs 7/12/07 3:57 PM Page 576 576 Geometry of the Circle A O P B C PB, the part of the secant segment that is outside the circle, the We will call are drawn, C external segment of the secant. When chords PAB because the measure of each is one-half the measure of the intercepted. Also P P by the reflexive property. Therefore, BPA APC arc, by AA. The length of the corresponding sides of similar triangles are in proportion. Therefore: ABX and AB AC PA 5 PA PB PC and (PA)2 (PC)(PB) We can write what we have just proved as a theorem: Theorem 13.16 If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external segment. Note that both means of the proportion are PA, the length of the tangent segment. Therefore, we can say that the length of the tangent segment is the mean proportional between the lengths of the secant and its external segment. Theorem 13.16 can be stated in another way. Theorem 13.16 If a tangent and a secant are drawn to a circle from an external point, then the length of the tangent segment is the mean proportional between the lengths of the secant segment and its external segment. Segments Formed by Intersecting Secants What is the relationship of the lengths of two secants drawn to a circle from an external point? Let be two secant segments drawn to a circle as shown in the diagram. Draw a tangent segment to the circle from A. Since ADE ABC and AF A B D F C E AF 2 (AC)(AB) and AF 2 (AE)(AD), then (AC)(AB) (AE)(AD) |
Note: This relationship could also have been proved by showing that ABE ADC. 14365C13.pgs 7/12/07 3:57 PM Page 577 Measures of Tangent Segments, Chords, and Secant Segments 577 We can state this as a theorem: Theorem 13.17 If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. EXAMPLE 1 and PCD, and Two secant segments, PAB a tangent segment,, are drawn to a cirPE cle from an external point P. If PB 9 cm, PD 12 cm, and the external segment of is 1 centimeter longer than the exterPAB nal segment of, find: a. PA b. PC c. PE PCD P A O C E B D Solution Let x PC and x 1 PA. a. PA x 1 4 (PB)(PA) 5 (PD)(PC) 9PA 5 12PC 9(x 1 1) 5 12x 9x 1 9 5 12x 9 5 3x 3 5 x b. PC x 3 c. (PE)2 5 (PB)(PA) (PE)2 5 (9)(4) (PE)2 5 36 PE 5 6 (Use the positive square root.) Answers a. PA 4 cm b. PC 3 cm c. PE 6 cm EXAMPLE 2 In a circle, chords find ST and TR. PQ and RS intersect at T. If PT 2, TQ 10, and SR 9, P R 2 O T 10 Q 9 S 14365C13.pgs 7/12/07 3:57 PM Page 578 578 Geometry of the Circle Solution Since ST TR SR 9, let ST x and TR 9 x. P R 2 O T 10 Q 9 S (PT)(TQ) 5 (RT)(TS) (2)(10) 5 x(9 2 x) 20 5 9x 2 x2 x2 2 9x 1 20 5 0 (x 2 5)(x 2 4 Answer ST 5 and TR 4, or ST 4 and TR 5 EXAMPLE 3 Find the length of a chord that is 20 centimeters from the center of a circle if the length of the radius of the circle is 25 centimeters. Solution Draw diameter COD perpendicular to chord AB at E. Then |
OE is the distance from the center of the circle to the chord. OE 20 DE 5 OD 1 OE 5 25 1 20 5 45 CE 5 OC 2 OE 5 25 2 20 5 5 B 25 c m 5 cm C E A 20 cm O 25 cm D A diameter perpendicular to a chord bisects the chord. Therefore, AE EB. Let AE EB x. (AE)(EB) 5 (DE)(CE) (x)(x) 5 (45)(5) x2 5 225 x 5 15 (Use the positive square root.) Therefore, AB 5 AE 1 EB 5 x 1 x 5 30 cm Answer 14365C13.pgs 7/12/07 3:57 PM Page 579 Measures of Tangent Segments, Chords, and Secant Segments 579 SUMMARY Type of Segment Length Example Formed by Two Intersecting Chords If two chords intersect, the product of the measures of the segments of one chord is equal to the product of the measures of of the segments of the other. A D E O C B Formed by a Tangent Intersecting a Secant Formed by Two Intersecting Secants If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external segment. If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. (AE)(EB) 5 (CE)(ED) A O C P B (PA)2 5 (PC)(PB) P A C B O D (PB)(PA) 5 (PD)(PC) Exercises Writing About Mathematics 1. The length of chord AB in circle O is 24. Vanessa said that any chord of circle O that inter- AB sects the lengths of the segments is 144. Do you agree with Vanessa? Justify your answer. at its midpoint, M, is separated by M into two segments such that the product of 2. Secants answer. ABP and CDP are drawn to circle O. If AP CP, is BP DP? Justify your 14365C13.pgs 7/12/07 3:57 PM Page 580 580 Geometry of the Circle C E A D Developing Skills |
In 3–14, chords AB and intersect at E. CD 3. If CE 12, ED 2, and AE 3, find EB. 4. If CE 16, ED 3, and AE 8, find EB. 5. If AE 20, EB 5, and CE 10, find ED. 6. If AE 14, EB 3, and ED 6, find CE. 7. If CE 10, ED 4, and AE 5, find EB. 8. If CE 56, ED 14, and AE EB, find EB. 9. If CE 12, ED 2, and AE is 2 more than EB, find EB. 10. If CE 16, ED 12, and AE is 3 times EB, find EB. 11. If CE 8, ED 5, and AE is 6 more than EB, find EB. 12. If CE 9, ED 9, and AE is 24 less than EB, find EB. 13. If CE 24, ED 5, and AB 26, find AE and EB. 14. If AE 7, EB 4, and CD 16, find CE and ED. B In 15–22, g AF is tangent to circle O at F and secant ABC intersects circle O at B and C. 15. If AF 8 and AB 4, find AC. 16. If AB 3 and AC 12, find AF. 17. If AF 6 and AC 9, find AB. 18. If AB 4 and BC 12, find AF. 19. If AF 12 and BC is 3 times AB, find AC, AB, and BC. 20. If AF 10 and AC is 4 times AB, find AC, AB, and BC. 21. If AF 8 and CB 12, find AC, AB, and BC. 22. If AF 15 and CB 16, find AC, AB, and BC. A F B O C intersect at A outside the circle. h ABC h ADE In 23–30, secants and 23. If AB 8, AC 25, and AD 10, find AE. 24. If AB 6, AC 18, and AD 9, find AE. 25. If AD 12, AE 20, and AB 8, find AC. 26. If AD 9, AE 21, and AC is 5 times AB, find AB and AC. 27. If AB 3, AD 2, and DE 10, find AC. C E B D A 14365C13.pgs 7/12/07 3:57 PM Page 581 Circles |
in the Coordinate Plane 581 28. If AB 4, BC 12, and AD DE, find AE. 29. If AB 2, BC 7, and DE 3, find AD and AE. 30. If AB 6, BC 8, and DE 5, find AD and AE. 31. In a circle, diameter AB is extended through B to P and tangent segment PC is drawn. If BP 6 and PC 9, what is the measure of the diameter of the circle? 13-7 CIRCLES IN THE COORDINATE PLANE In the diagram, a circle with center at the origin and a radius with a length of 5 units is drawn in the coordinate plane. The points (5, 0), (0, 5), (5, 0) and (0, 5) are points on the circle. What other points are on the circle and what is the equation of the circle? Let P(x, y) be any other point on the circle. From P, draw a vertical line segment to the x-axis. Let this be point Q. Then OPQ is a right triangle with OQ x, PQ y, and OP 5. We can use the Pythagorean Theorem to write an equation for the circle: y (0, 5) P(x, y) (5, 0) 1 1O Q(x, 0) (5, 0) x (0, 5) OQ2 1 PQ2 5 OP2 y2 5 52 1 x2 The points (3, 4), (4, 3), (3, 4), (4, 3), (3, 4), (4, 3), (3, 4), and (4, 3) appear to be points on the circle and all make the equation x2 y2 52 true. The points (5, 0), (0, 5), (5, 0), and (0, 5) also make the equation true, as do points such as " If we replace 5 by the length of any radius, r, the equation of a circle whose 1, " and 22, 21 24 B A B A. center is at the origin is: x2 y2 r 2 How does the equation change if the center is not at the origin? For example, what is the equation of a circle whose center, C, is at (2, 4) and whose radius has a length of 5 units? The points (7, 4), (3, 4 |
), (2, 9), and (2, 1) are each 5 units from (2, 4) and are therefore points on the circle. Let P(x, y) be any other point on the circle. From P, draw a vertical line and from C, a horizontal line. (3, 4) y (2, 9) P(x, y) C (2, 4) Q(x, 4) 1 O 1 (2, 1) (7, 4) x 14365C13.pgs 7/12/07 3:57 PM Page 582 582 Geometry of the Circle Let the intersection of these two lines be Q. Then CPQ is a right triangle with: CQ x 2 PQ y 4 CP 5 We can use the Pythagorean Theorem to write an equation for the circle. CQ2 1 PQ2 (x 2 2)2 1 (y 2 4)2 5 52 5 CP2 The points (5, 8), (6, 7), (1, 8), (2, 7), (1, 0), (2, 1) (5, 0), and (6, 1) appear to be points on the circle and all make (x 2)2 (y 4)2 52 true. The points (7, 4), (3, 4), (2, 9), and (2, 1) also make the equation true, as do points whose coordinates are not integers. We can write a general equation for a circle with center at C(h, k) and radius r. Let P(x, y) be any point on the circle. From P draw a vertical line and from C draw a horizontal line. Let the intersection of these two lines be Q. Then CPQ is a right triangle with: CQ x h PQ y k CP r y P(x, y) r C(h, k) Q(x, k) O x We can use the Pythagorean Theorem to write an equation for the circle. CQ2 1 PQ2 5 CP2 (x 2 h)2 1 (y 2 k)2 5 r2 In general, the center-radius equation of a circle with radius r and center (h, k) is (x h)2 (y k)2 r2 AB A circle whose diameter has endpoints at A(3, 1) and B(5, 1) is shown at the right. The |
center of the circle, C, is the midpoint of the diameter. Recall that the coordinates of the midpoint of the segment whose endpoints are (a, b) and (c, d) are 23) 2 B A. The coordinates of C are, 21 1 (21) 2 B 5 (1, 1). y O 1 1 C A x B The length of the radius is the distance from C to any point on the circle. The distance between two points on the same vertical line, that is, with the same x-coordinates, is the absolute value of the difference of the y-coordinates. The length of the radius is the distance from C(1, 1) to A(3, 1). The length of the radius is 1 (3) 4. 14365C13.pgs 7/31/07 1:43 PM Page 583 Circles in the Coordinate Plane 583 In the equation of a circle with center at (h, k) and radius r, we use (x h)2 (y k)2 r2. For this circle with center at (1, 1) and radius 4, h 1, k 1, and r 4. The equation of the circle is: (x 1)2 (y (1))2 42 or (x 1)2 (y 1)2 16 The equation of a circle is a rule for a set of ordered pairs, that is, for a relation. For the circle (x 1)2 (y 1)2 16, (1, 5) and (1, 3) are two ordered pairs of the relation. Since these two ordered pairs have the same first element, this relation is not a function. EXAMPLE 1 a. Write an equation of a circle with center at (3, 2) and radius of length 7. b. What are the coordinates of the endpoints of the horizontal diameter? Solution a. The center of the circle is (h, k) (3, 2). The radius is r 7. The general form of the equation of a circle is (x h)2 (y k)2 r 2. The equation of the given circle is: (x 3)2 (y (2))2 72 (x 3)2 (y 2)2 49 or b. METHOD 1 If this circle were centered at the origin, then the endpoints of the horizontal diameter would be (7, 0) and (7, 0). However, |
the circle is centered at (3, 2). Shift these endpoints using the translation (7, 0) → (7 3, 0 2) (4, 2) (7, 0) → (7 3, 0 2) (10, 2) T3, 22 : y O 1 1 (0, 0) (3, 2) x METHOD 2 Since the center of the circle is (3, 2), the y-coordinates of the endpoints are both 2. Substitute y 2 into the equation and solve for x: (x 2 3)2 1 (y 1 2)2 5 49 (x 2 3)2 1 (22 1 2)2 5 49 x2 2 6x 1 9 1 0 5 49 x2 2 6x 2 40 5 0 (x 2 10)(x 1 4) 5 0 x 5 10 z x 5 24 The coordinates of the endpoints are (10, 2) and (4, 2). Answers a. (x 3)2 (y 2)2 49 b. (10, 2) and (4, 2) 14365C13.pgs 7/12/07 3:57 PM Page 584 584 Geometry of the Circle EXAMPLE 2 The equation of a circle is (x 1)2 (y 5)2 36. a. What are the coordinates of the center of the circle? b. What is the length of the radius of the circle? c. What are the coordinates of two points on the circle? Solution Compare the equation (x 1)2 (y 5)2 36 to the general form of the equation of a circle: (x h)2 (y k)2 r2 Therefore, h 1, k 5, r2 36, and r 6. a. The coordinates of the center are (1, 5). b. The length of the radius is 6. c. Points 7 units from (1, 5) on the same horizontal line are (8, 5) and (6, 5). Points 7 units from (1, 5) on the same vertical line are (1, 12) and (1, 2). Answers a. (1, 5) b. 6 c. (8, 5) and (6, 5) or (1, 12) and (1, 2) EXAMPLE 3 The equation of a circle is x2 y2 50. What is the length of the radius of the circle? Solution Compare the given |
equation to x2 y2 r2. r2 5 50 r 5 6 50 " Since a length is always positive, r 5. Answer r 5 6 r 5 65 " 25 2 " 2 " 2 " Exercises Writing About Mathematics 1. Cabel said that for every circle in the coordinate plane, there is always a diameter that is a vertical line segment and one that is a horizontal line segment. Do you agree with Cabel? Justify your answer. 2. Is 3x2 3y2 12 the equation of a circle? Explain why or why not. 14365C13.pgs 7/12/07 3:57 PM Page 585 Circles in the Coordinate Plane 585 Developing Skills In 3–8, write an equation of each circle that has the given point as center and the given value of r as the length of the radius. 3. (0, 0), r 3 6. (4, 2), r 10 5. (2, 0), r 6 8. (3, 3), r 2 4. (1, 3), r 5 7. (6, 0), r 9 In 9–16, write an equation of each circle that has a diameter with the given endpoints. 9. (2, 0) and (2, 0) 11. (2, 5) and (2, 13) 13. (5, 12) and (5, 12) 15. (7, 3) and (9, 10) In 17–22, write an equation of each circle. 17. y 18. y 10. (0, 4) and (0, 4) 12. (5, 3) and (3, 3) 14. (5, 9) and (7, 7) 16. (2, 2) and (18, 4) 19. y O –1 1 1 O 1 x 1 O 1 20. 21. O x –1 –1 x 22 In 23–28, find the center of each circle and graph each circle. 23. (x 2 2)2 1 (y 1 5)2 5 4 24. 25. 26. (x 1 4)2 1 (y 2 4) 5 36 y 2 1)2 5 25 y 1 3 4 B 2 5 81 25 2 1 A A A 14365C13.pgs 7/12/07 3:57 PM Page 586 586 Geometry of the Circle 27. 2x2 1 2y2 |
5 18 5(x 2 1)2 1 5(y 2 1)2 5 245 28. 29. Point C(2, 3) is the center of a circle and A(3, 9) is a point on the circle. Write an equa- tion of the circle. 30. Does the point (4, 4) lie on the circle whose center is at the origin and whose radius is Justify your answer. 32 "? 31. Is x2 4x 4 y2 2y 1 25 the equation of a circle? Explain why or why not. Applying Skills 32. In the figure on the right, the points A(2, 6), B(4, 0), and C(4, 0) appear to lie on a circle. y A(2, 6) a. Find the equation of the perpendicular bisector of. AB b. Find the equation of the perpendicular bisector of. BC. c. Find the equation of the perpendicular bisector of AC d. Find the circumcenter of ABC, the point of intersec- tion of the perpendicular bisectors. e. From what you know about perpendicular bisectors, why is the circumcenter equidistant from the vertices of ABC? f. Do the points A, B, C lie on a circle? Explain. 1 O C(4, 0) 1 x B(4, 0) 33. In the figure on the right, the circle with center at C(3, 1) appears to be inscribed in PQR with vertices P(1, 2), Q(3, 12), and R(7, 2). a. If the equations of the angle bisectors of PQR are 6x 8y 10, x 3, and 3x 4y 13, is C the incenter of PQR? b. From what you know about angle bisectors, why is the incenter equidistant from the sides of PQR? c. If S(3, 2) is a point on the circle, is the circle inscribed in PQR? Justify your answer. d. Write the equation of the circle. P(1, 2) y 1 O 1 R(7, 2) x C (3, 1) Q(3, 12) 14365C13.pgs 7/12/07 3:57 PM Page 587 34. In the figure on the right, the circle is circumscribed about ABC with |
vertices A(1, 3), B(5, 1), and C(5, 3). Find the equation of the circle. Justify your answer algebraically. Circles in the Coordinate Plane 587 A(1, 3) y B(5, 1) C(5, 3) 1 O 1 x 35. Bill Bekebrede wants to build a circular pond in his garden. The garden is in the shape of an equilateral triangle. The length of the altitude to one side of the triangle is 18 feet. To plan the pond, Bill made a scale drawing on graph paper, letting one vertex of the equilateral triangle OAB be O(0, 0) and another vertex be A(2s, 0). Therefore, the length of a side of the triangle is 2s. Bill knows that an inscribed circle has its center at the intersection of the angle bisectors of the triangle. Bill also knows that the altitude, median, and angle bisector from any vertex of an equilateral triangle are the same line. a. What is the exact length, in feet, of a side of the garden? b. In terms of s, what are the coordinates of B, the third vertex of the triangle? c. What are the coordinates of C, the intersection of the altitudes and of the angle bisectors of the triangle? d. What is the exact distance, in feet, from C to the sides of the garden? e. What should be the radius of the largest possible pond? 36. The director of the town park is planning walking paths within the park. One is to be a circular path with a radius of 1,300 feet. Two straight paths are to be perpendicular to each other. One of these straight paths is to be a diameter of the circle. The other is a chord of the circle. The two straight paths intersect 800 feet from the circle. Draw a model of the paths on graph paper letting 1 unit 100 feet. Place the center of the circle at (13, 13) and draw the diameter as a horizontal line and the chord as a vertical line. a. What is the equation of the circle? b. What are all the possible coordinates of the points at which the straight paths intersect the circular path? c. What are all the possible coordinates of the point at which the straight paths intersect? d. What are the lengths of the segments into which the point of intersection separates the straight paths? 14365C13.pgs 7/ |
12/07 3:57 PM Page 588 588 Geometry of the Circle 13-8 TANGENTS AND SECANTS IN THE COORDINATE PLANE Tangents in the Coordinate Plane The circle with center at the origin and radius 5 is shown on the graph. Let l be a line tangent to the circle at A(3, 4). Therefore, since a tangent is perpendicular to the radius drawn to the point of tangency. The slope of l is the negative reciprocal of the slope of l'OA. OA y P A(3, 4) 1 O 1 l x slope of OA Therefore, the slope of. We can use the slope of l and the point A(3, 4) to write the equation of l. l 5 23 4 y 2 4 x 2 3 5 23 4 4(y 2 4) 5 23(x 2 3) 4y 2 16 5 23x 1 9 3x 1 4y 5 25 The point P(1, 7) makes the equation true and is therefore a point on the tangent line 3x 4y 25. Secants in the Coordinate Plane A secant intersects a circle in two points. We can use an algebraic solution of a pair of equations to show that a given line is a secant. The equation of a circle with radius 10 and center at the origin is x2 y2 100. The equation of a line in the plane is x y 2. Is the line a secant of the circle? 14365C13.pgs 7/12/07 3:57 PM Page 589 Tangents and Secants in the Coordinate Plane 589 How to Proceed (1) Solve the pair of equations algebraically: (2) Solve the linear equation for y in terms of x: (3) Substitute the resulting expression for y in the equation of the circle: (4) Square the binomial: (5) Write the equation in standard form: (6) Divide by the common factor, 2: (7) Factor the quadratic equation: (8) Set each factor equal to zero: (9) Solve each equation for x: (10) For each value of x find the corresponding value of y: x2 1 y2 5 100 x 1 y 5 2 y 2 x x2 1 (2 2 x)2 5 100 x2 1 4 2 4x 1 x2 5 100 2x2 2 4x 2 96 |
5 0 x2 2 2x 2 48 5 0 (x 8)(x 6 26 26) y 5 8 y (–6, 8) 1 1O x (8, –6) The common solutions are (8, 6) and (6, 8). The line intersects the circle in two points and is therefore a secant. In the diagram, the circle is drawn with its center at the origin and radius 10. The line y 2 x is drawn with a y-intercept of 2 and a slope of 1. The line intersects the circle at (8, 6) and (6, 8). 14365C13.pgs 7/12/07 3:57 PM Page 590 590 Geometry of the Circle EXAMPLE 1 Find the coordinates of the points at which the line y 2x 1 intersects a circle with center at (0, 1) and radius of length. 20 " Solution In the equation (x h)2 (y k)2 r2, let h 0, k 1, and r = equation of the circle is: 20 ". The (x 0)2 (y (1))2 20 A " 2 B or x2 (y 1)2 20. Find the common solution of x2 (y 1)2 20 and y 2x 1. (1) The linear equation is solved for y in terms of x. Substitute, in the equation of the circle, the expression for y and simplify the result. (2) Square the monomial: (3) Write the equation in standard form: (4) Divide by the common factor, 5: x2 1 (y 1 1)2 5 20 x2 1 (2x 2 1 1 1)2 5 20 x2 1 (2x)2 5 20 x2 1 4x2 5 20 5x2 2 20 5 0 x2 2 4 5 0 (5) Factor the left side of the equation: (x 2 2)(x 1 2) 5 0 (6) Set each factor equal to zero: x 2 0 x 2 0 (7) Solve each equation for x: x 2 x 2 (8) For each value of x find the corresponding value of y: y 5 2x 2 1 y 5 2(2) 2 1 y 5 3 y 5 2x 2 1 y 5 2(22) 2 1 y 5 25 Answer The coordinates of the points of intersection are (2, 3) and (2, |
5). EXAMPLE 2 The line x + y 2 intersects the circle x2 y2 100 at A(8, 6) and B(6, 8). The line y 10 is tangent to the circle at C(0, 10). a. Find the coordinates of P, the point of intersection of the secant x + y 2 and the tangent y 10. b. Show that PC 2 (PA)(PB). 14365C13.pgs 7/12/07 3:57 PM Page 591 Tangents and Secants in the Coordinate Plane 591 (–8, 10) (–6, 8) y 1 1O x (8, –6) Solution a. Use substitution to find the intersection: If x y 2 and y 10, then x 10 2 and x 8. The coordinates of P are (8, 10). b. Use the distance formula, d = lengths of PC, PA, and PB. " (x2 2 x1)2 1 (y2 2 y1)2, to find the PA 5 " (28 2 8)2 1 (10 2 (26))2 PB 5 (28 2 (26))2 1 (10 2 8)2 " 5 256 1 256 " 5 " 5 16 256 " PC 5 " (28 2 0)2 1 (10 2 10)2 64 1 0 5 " 5 8 Then: PC2 5 82 5 64 and Therefore, PC 2 (PA)(PB). (PA)(PB) 5 (16 2)(2 2) " " 5 (32)(2) 5 64 14365C13.pgs 7/12/07 3:57 PM Page 592 592 Geometry of the Circle Exercises Writing About Mathematics 1. Ron said that if the x-coordinate of the center of a circle is equal to the length of the radius of the circle, then the y-axis is tangent to the circle. Do you agree with Ron? Explain why or why not. g AB g AB intersects a circle with center at C. The slope of tangent to the circle? Explain your answer. is m and the slope of m. Is 2. At A, g AB CA is Developing Skills In 3–14: a. Find the coordinates of the points of intersection of the circle and the line. b. Is the line a secant or a tangent to the circle? 3. x2 y2 36 y 6 6. x |
2 y2 10 y 3x 9. x2 y2 25 y = x 1 12. x2 y2 50 x y 10 4. x2 y2 100 x y 14 7. x2 y2 9 y x 3 10. x2 y2 20 x y 6 13. x2 y2 8 x y 4 5. x2 y2 25 x y 7 8. x2 y2 8 x y 11. x2 y2 18 y x 6 14. x2 (y 2)2 4 y x 4 In 15–18, write an equation of the line tangent to the given circle at the given point. 15. x2 y2 9 at (0, 3) 16. x2 y2 16 at (4, 0) 17. x2 y2 8 at (2, 2) 18. x2 y2 20 at (4, 2) Applying Skills 19. a. Write an equation of the secant that intersects x2 y2 25 at A(3, 4) and B(0, 5). b. Write an equation of the secant that intersects x2 y2 25 at D(0, 5) and E(0, 5). c. Find the coordinates of P, the intersection of d. Show that (PA)(PB) (PD)(PE). g AB and g DE. 20. a. Write an equation of the secant that intersects x2 y2 100 at A(6, 8) and B(8, 6). b. Write an equation of the tangent to x2 y2 100 at D(0, 10). c. Find the coordinates of P, the intersection of d. Show that (PA)(PB) (PD)2. g AB and the tangent line at D. 14365C13.pgs 8/2/07 6:00 PM Page 593 Chapter Summary 593 21. a. Write an equation of the tangent to x2 y2 18 at A(3, 3). b. Write an equation of the tangent to x2 y2 18 at B(3, 3). c. Find the point P at which the tangent to x2 y2 18 at A intersects the tangent to x2 y2 18 at B. d. Show that PA = PB. 22. Show that the line whose equation is x 2y 10 is tangent to the circle whose equation is x2 y2 20. 23 |
. a. Show that the points A(1, 7) and B(5, 7) lie on a circle whose radius is 5 and whose cen- ter is at (2, 3). b. What is the distance from the center of the circle to the chord AB? 24. Triangle ABC has vertices A(7, 10), B(2, 2), and C(2, 10). a. Find the coordinates of the points where the circle with equation (x 1)2 (y 7)2 9 intersects the sides of the triangle. b. Show that the sides of the triangle are tangent to the circle. c. Is the circle inscribed in the triangle? Explain. CHAPTER SUMMARY Definitions to Know • A circle is the set of all points in a plane that are equidistant from a fixed point of the plane called the center of the circle. • A radius of a circle (plural, radii) is a line segment from the center of the circle to any point of the circle. • A central angle of a circle is an angle whose vertex is the center of the circle. • An arc of a circle is the part of the circle between two points on the circle. • An arc of a circle is called an intercepted arc, or an arc intercepted by an angle, if each endpoint of the arc is on a different ray of the angle and the other points of the arc are in the interior of the angle. • The degree measure of an arc is equal to the measure of the central angle that intercepts the arc. • Congruent circles are circles with congruent radii. • Congruent arcs are arcs of the same circle or of congruent circles that are equal in measure. • A chord of a circle is a line segment whose endpoints are points of the circle. • A diameter of a circle is a chord that has the center of the circle as one of its points. • An inscribed angle of a circle is an angle whose vertex is on the circle and whose sides contain chords of the circle. 14365C13.pgs 8/2/07 6:00 PM Page 594 594 Geometry of the Circle • A tangent to a circle is a line in the plane of the circle that intersects the circle in one and only one point. • A secant of a circle is a line that intersects the circle in two points. • A common tangent is a line that is tang |
ent to each of two circles. • A tangent segment is a segment of a tangent line, one of whose endpoints is the point of tangency. Postulates 13.1 and ABX If point and no other points in common, then ABX m ABX. (Arc Addition Postulate) are two arcs of the same circle having a common endand ABCXBCX BCX BCX ABCX m m 13.2 At a given point on a given circle, one and only one line can be drawn that is tangent to the circle. Theorems and Corollaries 13.1 All radii of the same circle are congruent. 13.2 In a circle or in congruent circles, central angles are congruent if and only if their intercepted arcs are congruent. In a circle or in congruent circles, two chords are congruent if and only if their central angles are congruent. In a circle or in congruent circles, two chords are congruent if and only if their arcs are congruent. 13.3 13.4 13.5 A diameter perpendicular to a chord bisects the chord and its arcs. 13.5a A line through the center of a circle that is perpendicular to a chord 13.6 13.7 13.8 13.9 bisects the chord and its arcs. The perpendicular bisector of the chord of a circle contains the center of the circle. Two chords are equidistant from the center of a circle if and only if the chords are congruent. In a circle, if the lengths of two chords are unequal, the shorter chord is farther from the center. The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc. 13.9a An angle inscribed in a semicircle is a right angle. 13.9b If two inscribed angles of a circle intercept the same arc, then they are congruent. 13.10 A line is tangent to a circle if and only if it is perpendicular to a radius at its point of intersection with the circle. 13.11 Tangent segments drawn to a circle from an external point are congru- ent. 13.11a If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents. 13.11b If two tang |
ents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle whose vertex is the center of the circle and whose rays are the two radii drawn to the points of tangency. 14365C13.pgs 8/7/07 10:51 AM Page 595 Chapter Summary 595 13.12 The measure of an angle formed by a tangent and a chord that intersect at the point of tangency is equal to one-half the measure of the intercepted arc. 13.13 The measure of an angle formed by two chords intersecting within a circle is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. 13.16 13.15 13.14 The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs. If two chords intersect within a circle, the product of the measures of the segments of one chord is equal to the product of the measures of the segments of the other. If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external segment. If a tangent and a secant are drawn to a circle from an external point, then the length of the tangent segment is the mean proportional between the lengths of the secant segment and its external segment. If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. 13.17 13.16 Formulas Type of Angle Central Angle Degree Measure Example The measure of a central angle is equal to the measure of its intercepted arc. Inscribed Angle The measure of an inscribed angle is equal to one-half the measure of its intercepted arc. B A A 1 m1 mABX B 1 m/1 5 1 2m ABX (Continued) 14365C13.pgs 8/7/07 10:51 AM Page 596 596 Geometry of the Circle Formulas (Continued) Type of Angle Degree Measure Example Formed by a Tangent and a Chord The measure of an angle |
formed by a tangent and a chord that intersect at the point of tangency is equal to one-half the measure of the intercepted arc. Formed by Two Intersecting Chords The measure of an angle formed by two intersecting chords is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. Formed by Tangents and Secants The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs. A 1 B m/1 5 1 2mABX 2 1 A B D C m/1 5 1 m/2 5 1 2(mABX 1 mCDX) 2(mABX 1 mCDX/1 5 1 m/2 5 1 m/3 5 1 2(mABX 2 mACX) 2(mABX 2 mCDX) 2(mACBX 2 mABX) 14365C13.pgs 8/2/07 6:00 PM Page 597 Formulas (Continued) Vocabulary 597 Type of Segment Length Example Formed by Two Intersecting Chords If two chords intersect, the product of the measures of the segments of one chord is equal to the product of the measures of the segments of the other. A D E O C B Formed by a Tangent Intersecting a Secant Formed by Two Intersecting Secants If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external segment. If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. (AE)(EB) 5 (CE)(ED) A O C P B (PA)2 5 (PC)(PB) P A C B O D (PB)(PA) 5 (PD)(PC) The equation of a circle with radius r and center (h, k) is (x h)2 (y k)2 r 2 VOCABULARY 13-1 Circle • Center • Radius • Interior of a circle • Exterior of a circle • Central |
angle of a circle • Arc of a circle • Minor arc • Major arc • Semicircle • Intercepted arc • Degree measure of an arc • Congruent circles • Congruent arcs 13-2 Chord • Diameter • Apothem • Inscribed polygon • Circumscribed circle 13-3 Inscribed angle 14365C13.pgs 7/12/07 3:57 PM Page 598 598 Geometry of the Circle 13-4 Tangent to a circle • Secant of a circle • Common tangent • Common internal tangent • Common external tangent • Tangent externally • Tangent internally • Tangent segment • Circumscribed polygon • Inscribed circle • Center of a regular polygon 13-6 External segment 13-7 Center-radius equation of a circle REVIEW EXERCISES In 1–6, PA to circle O. Chords E. is a tangent and and AC PBC BD is a secant intersect at 1. If mABX mBCX 80, 100, find: mCDX a. mPAC 120, and P D A E O B C b. mCBD c. mAPC d. mDEC e. mAED a. 2. If mC 50, mDBC 55, and mPAC 100, find: d. mP mABX 3. If mCEB 80, ADX CDX c. mBEC b. m mBCX mABX CDX c. mCBD b. m 4. If AP 12 and PC 24, find PB and BC. 5. If PB 5 and BC 15, find AP. 6. If AC 11, DE 2, EB 12, and AE EC, find AE and EC. 120, and 70, find: d. mP a. m BCX e. m e. mPAC PA and secant seg- 7. Tangent segment PBC ment are drawn to circle O. If PB 8 and BC 10, PA is equal to (1) 12 (3) 80 (2) 4 5 " (4) 144 8. The equation of a circle with center at (2, 4) and radius of length 3 is (1) (x 2)2 (y 4)2 9 (2) (x 2)2 (y 4)2 9 (3) (x 2)2 (y 4)2 9 (2) (x 2)2 (y 4 |
)2 9 A P O B C 9. Two tangents that intersect at P intercept a major arc of 240° on the circle. What is the measure of P? 10. A chord that is 24 centimeters long is 9 centimeters from the center of a circle. What is the measure of the radius of the circle? 14365C13.pgs 8/2/07 6:01 PM Page 599 Review Exercises 599 B O C P O A D 11. Two circles, O and O, are tangent externally at P, OP 5, and OP 3. Segment is tangent to circle O at B and to circle O at intersects circle O at D C, and P, and circle O at P. If AD 2, find AB and AC. AOrO ABC 12. Isosceles ABC is inscribed in a circle. If the measure of the vertex angle, A, is 20 degrees less than twice the measure of each of the base angles, CAX find the measures of BCXABX, 13. Prove that a trapezoid inscribed in a circle is isosceles., and. 14. In circle O, chords and CD are parallel and intersects AD a. Prove that ABE and CDE are isosceles tri- BC AB at E. angles. ACX > BDX b. Prove that c. Prove that ABE CDE.. B E D A O C 15. Prove that if A, B, C, and D separate a circle into four congruent arcs, then quadrilateral ABCD is a square. 16. Prove, using a circumscribed circle, that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices of the triangle. 17. Secant segments PAB and drawn to circle O. If prove that from the center of the circle. PAB and CD AB PCD are equidistant PCD are, B F A O E C D P 18. An equilateral triangle is inscribed in a circle whose radius measures 12 centimeters. How far from the center of the circle is the centroid of the triangle? Exploration A regular polygon can be constructed by constructing the congruent isosceles triangles into which it can be divided. The measure of each base angle of the isosceles triangles is one-half the measure of an interior angle of the polygon. However, the interior angles of many regular |
polygons are not angles that can be constructed using compass and straightedge. For example, a regular polygon with nine sides has angles that measure 140 degrees. Each of the nine isosceles triangles of this polygon has base angles of 70 degrees which cannot be constructed with straightedge and compass. 14365C13.pgs 7/12/07 3:57 PM Page 600 600 Geometry of the Circle In this exploration, we will construct a regular triangle (equilateral triangle), a regular hexagon, a regular quadrilateral (a square), a regular octagon, and a regular dodecagon (a polygon with 12 sides) inscribed in a circle. a. Explain how a compass and a straightedge can be used to construct an equi- lateral triangle. Prove that your construction is valid. b. Explain how the construction in part a can be used to construct a regular hexagon. Prove that your construction is valid. c. Explain how a square, that is, a regular quadrilateral, can be inscribed in a circle using only a compass and a straightedge. (Hint: What is true about the diagonals of a square?) Prove that your construction is valid. d. Bisect the arcs determined by the chords that are sides of the square from the construction in part c. Join the endpoints of the chords that are formed to draw a regular octagon. Prove that this construction is valid. e. A regular octagon can also be constructed by constructing eight isosceles triangles. The interior angles of a regular octagon measure 135 degrees. Bisect a right angle to construct an angle of 45 degrees. The complement of that angle is an angle of 135 degrees. Bisect this angle to construct the base angle of the isosceles triangles needed to construct a regular octagon. f. Explain how a regular hexagon can be inscribed in a circle using only a compass and a straightedge. (Hint: Recall how a regular polygon can be divided into congruent isosceles triangles.) g. Bisect the arcs determined by the chords that are sides of the hexagon from part f to draw a regular dodecagon. h. A regular dodecagon can also be constructed by constructing twelve isosceles triangles. The interior angles of a regular dodecagon measure 150 degrees. Bisect a 60-degree angle to construct an angle of 30 degrees. The complement of that angle is an angle of |
150 degrees. Bisect this angle to construct the base angle of the isosceles triangles needed to construct a regular dodecagon. CUMULATIVE REVIEW Chapters 1–13 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The measure of A is 12 degrees more than twice the measure of its com- plement. The measure of A is (1) 26 (2) 39 (3) 64 (4) 124 14365C13.pgs 7/12/07 3:57 PM Page 601 Cumulative Review 601 2. The coordinates of the midpoint of a line segment with endpoints at (4, 9) (2) (3, 3) (3) (2, 24) (4) (6, 6) 3. What is the slope of a line that is perpendicular to the line whose equation and (2, 15) are (1) (1, 12) is 2x y 8? (1) 2 (2) 2 (3) 21 2 (4) 1 2 4. The altitude to the hypotenuse of a right triangle separates the hypotenuse into segments of length 6 and 12. The measure of the altitude is (4) (1) 18 6 (3) (2. The diagonals of a quadrilateral bisect each other. The quadrilateral can- not be a (1) trapezoid (2) rectangle (3) rhombus (4) square 6. Two triangles, ABC and DEF, are similar. If AB 12, DE 18, and the perimeter of ABC is 36, then the perimeter of DEF is (1) 24 (2) 42 (3) 54 (4) 162 7. Which of the following do not always lie in the same plane? (1) two points (2) three points (3) two lines (4) a line and a point not on the line 8. At A, the measure of an exterior angle of ABC is 110 degrees. If the measure of B is 45 degrees, what is the measure of C? (1) 55 (3) 70 (2) 65 (4) 135 9. Under the composition rx-axis + T2,3, what are the coordinates of the image of A(3, 5)? (1) (5, 2) (2) (5, 2) (3) (5, 8) (4 |
) (1, 2) 10. In the diagram, and g AB g AB g CD g CD at E and g EF at F. If intersects mAEF is represented by 3x and mCFE is represented by 2x 20, what is the value of x? (1) 4 (2) 12 (3) 32 (4) 96 A C F E B D 14365C13.pgs 7/12/07 3:57 PM Page 602 602 Geometry of the Circle Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The coordinates of the vertices of ABC are A(8, 0), B(4, 4), and C(0, 4). a. Find the coordinates of D, the midpoint of AB. b. Find the coordinates of E, the midpoint of BC. DE AC c. Is d. Are ABC and DBE similar triangles? Justify your answer.? Justify your answer. 12. ABCD is a quadrilateral, AC > BD E. Prove that ABCD is a rectangle. and AC and BD bisect each other at Part III Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. In the diagram, sect at E in plane p, g g'CD EF FA > FC.. If g AB and g EF EA > EC inter- g CD g'AB, prove that, and C F p E B D A 14. The length of the hypotenuse of a right triangle is 2 more than the length of the longer leg. The length of the shorter leg is 7 less than the length of the longer leg. Find the lengths of the sides of the right triangle. Part IV Answer all questions in this part. Each correct answer will receive 6 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 14365C13.pgs 7/12/ |
07 3:57 PM Page 603 Cumulative Review 603 15. a. Find the coordinates of A, the image of A(5, 2) under the composition R90 + ry 5 x. b. What single transformation is equivalent to R90 + ry 5 x? c. Is R90 + ry 5 x a direct isometry? Justify your answer. ADC 16. In the diagram, D is a point on such that AD : DC 1 : 3, and E is a point on BEC a. Show that AC : DC BC : EC. b. Prove that ABC DEC. such that BE : EC 1 : 3. A D B E C 14365C14.pgs 7/10/07 9:59 AM Page 604 CHAPTER 14 CHAPTER TABLE OF CONTENTS 14-1 Constructing Parallel Lines 14-2 The Meaning of Locus 14-3 Five Fundamental Loci 14-4 Points at a Fixed Distance in Coordinate Geometry 14-5 Equidistant Lines in Coordinate Geometry 14-6 Points Equidistant from a Point and a Line Chapter Summary Vocabulary Review Exercises Cumulative Review 604 LOCUS AND CONSTRUCTION Classical Greek construction problems limit the solution of the problem to the use of two instruments: the straightedge and the compass.There are three construction problems that have challenged mathematicians through the centuries and have been proved impossible: the duplication of the cube the trisection of an angle the squaring of the circle The duplication of the cube requires that a cube be constructed that is equal in volume to twice that of a given cube. The origin of this problem has many versions. For example, it is said to stem from an attempt at Delos to appease the god Apollo by doubling the size of the altar dedicated to Apollo. The trisection of an angle, separating the angle into three congruent parts using only a straightedge and compass, has intrigued mathematicians through the ages. The squaring of the circle means constructing a square equal in area to the area of a circle. This is equivalent to constructing a line segment whose length is equal to times the radius of the circle. p Although solutions to these problems have been presented using other instruments, solutions using only straightedge and compass have been proven to be impossible.! 14365C14.pgs 7/10/07 9:59 AM Page 605 Constructing Parallel Lines 605 14-1 CONSTRUCTING PARALLEL LINES In |
Chapter 5, we developed procedures to construct the following lines and rays: 1. a line segment congruent to a given line segment 2. an angle congruent to a given angle 3. the bisector of a given line segment 4. the bisector of a given angle 5. a line perpendicular to a given line through a given point on the line 6. a line perpendicular to a given line through a given point not on the line Then, in Chapter 9, we constructed parallel lines using the theorem that if two coplanar lines are each perpendicular to the same line, then they are parallel. Now we want to use the construction of congruent angles to construct par- allel lines. Two lines cut by a transversal are parallel if and only if the corresponding angles are congruent. For example, g CD in the diagram, the transversal intersects at G g at H. If EGB GHD, then CD and. Therefore, we can construct parallel lines by constructing congruent corresponding angles. g AB g AB g EF F H D B C A G E Construction 7 Construct a Line Parallel to a Given Line at a Given Point. Given g AB and point P not on g AB Construct A line through P that is parallel to g AB STEP 1. Through P, draw any line intersecting at R. Let S be g AB any point on the ray opposite h PR. STEP 2. At P, construct SPD PRB. Draw opposite ray of h PD h PC, forming, the g.CD Continued 14365C14.pgs 7/10/07 9:59 AM Page 606 606 Locus and Construction Construction 7 Construct a Line Parallel to a Given Line at a Given Point. (continued) Conclusion g AB CPD g Proof Corresponding angles, SPD and PRB, are congruent. Therefore, g AB CPD g. This construction can be used to construct the points and lines that satisfy other conditions. EXAMPLE 1 Given: g AB and PQ Construct: g CD g AB at a distance 2PQ from g AB. A B P Q Construction The distance from a point to a line is the length of the perpendicular from the point to the line. Therefore, we must locate points at a distance of 2PQ from a point on g AB at which to draw a parallel line. F C F C D F P RQ 1. Extend. PQ Locate point R on h such that PQ QR |
PQ, making PR 2PQ EF 3. On, locate point C at a distance PR from E. 2. Choose any point g AB E on construct the line h EF perpendicular. At E, g AB. to 4. At C, construct g CD perpendicular to g EF and therefore parallel to g.AB Conclusion g CD is parallel to g AB at a distance 2PQ from g.AB 14365C14.pgs 7/10/07 9:59 AM Page 607 Constructing Parallel Lines 607 Exercises Writing About Mathematics 1. In the example, every point on g CD is at a fixed distance, 2PQ, from g AB. Explain how you know that this is true. g CD g AB 2. Two lines, and, are parallel. A third line, g EF, is perpendicular to g AB at H. The perpendicular bisector of GH is the set of all points equidistant from Explain how you know that this is true. at G and to g AB and g CD g. CD Developing Skills In 3–9, complete each required construction using a compass and straightedge, or geometry software. Draw each given figure and do the required constructions. Draw a separate figure for each construction. Enlarge the given figure for convenience. P l m 3. Given: Parallel lines l and m and point P. Construct: a. a line through P that is parallel to l. b. a line that is parallel to l and to m and is equidistant from l and m. c. a line n that is parallel to l and to m such that l is equidistant from m and n. d. Is the line that you constructed in a parallel to m? Justify your answer. 4. Given: Line segments of length a and b and A Construct: a. a rectangle whose length is a and whose width is b. b. a square such that the length of a side is a. c. a parallelogram that has sides with measures a and b and an angle congruent to A. d. a rhombus that has sides with measure a and an angle congruent to A. 5. Given: AB a b A a. Divide AB into four congruent parts. A B b. Construct a circle whose radius is AB. c. Construct a circle whose diameter is AB. 14365C14.pgs 7/10/07 9:59 AM Page 608 608 Locus and |
Construction 6. Given: ABC Construct: a. a line parallel to g AB at C. b. the median to AC by first constructing the midpoint of AC. c. the median to BC by first constructing the midpoint of BC. C AC and the median to BC intersect at P, can A B d. If the median to the median to point of AB? Justify your answer. AB be drawn without first locating the mid- 7. Given: Obtuse triangle ABC Construct: a. the perpendicular bisector of. AC b. the perpendicular bisector of. BC c. M, the midpoint of AB. C A B AC and the intersect at P, d. If the perpendicular bisector of BC perpendicular bisector of what two points determine the perpendicular bisector of 8. Given: ABC Construct:? Justify your answer. AB a. the altitude to. AC b. the altitude to. BC c. If the altitude to AC and the altitude to BC at P, what two points determine the altitude to Justify your answer. intersect? AB 9. Given: ABC Construct: a. the angle bisector from B to. AC b. the angle bisector from A to c. If the angle bisector of B and the angle bisector of A intersect at P, what two points determine the? Justify your answer. angle bisector from C to. BC AB C C A A B B 10. a. Draw any line segment,. Draw any ray, AB h AC h AC, forming BAC. h AC such that DE 2AD. b. Choose any point, D, on. Construct DE on EB c. Draw d. Prove that AD : DE = AF : FB 1 : 2. and construct a line through D parallel to EB and intersecting h AB at F. 14365C14.pgs 7/10/07 9:59 AM Page 609 11. a. Draw an angle, LPR. h PL b. Choose point N on that PS : SQ 3 : 5.. Construct Q on h PL such that PQ 8PN. Locate S on PQ such The Meaning of Locus 609 QR c. Draw and construct a line through S parallel to 3 d. Prove that PST PQR with a constant of proportionality of. 8 QR and intersecting 14-2 THE MEANING OF LOCUS In a construction, the opening between the pencil and the point of the compass is a fixed distance |
, the length of the radius of a circle. The point of the compass determines a fixed point, point O in the diagram. If the length of the radius remains unchanged, all of the points in the plane that can be drawn by the compass form a circle, and any points that cannot be drawn by the compass do not lie on the circle. Thus, the circle is the set of all points at a fixed distance from a fixed point. This set is called a locus. h PR at T. O DEFINITION A locus is the set of all points that satisfy a given condition or set of conditions. The example of the circle given above helps us to understand what the def- inition means. Every definition can be written as a biconditional: p: A point is on the locus. q: A point satisfies the given conditions. 1. (p → q): If a point is on the locus, then the point satisfies the given conditions. All points on the circle are at a given fixed distance from the center. 2. (p → q): If a point is not on the locus, then the point does not satisfy the given conditions. Any point that is not on the circle is not at the given distance from the center. Recall that the statement (p → q) is the inverse of the statement (p → q). A locus is correct when both statements are true: the conditional and its inverse. We can restate the definition of locus in biconditional form: A point P is a point of the locus if and only if P satisfies the given conditions of the locus. 14365C14.pgs 7/10/07 9:59 AM Page 610 610 Locus and Construction Discovering a Locus Procedure To discover a probable locus: 1. Make a diagram that contains the fixed lines or points that are given. 2. Decide what condition must be satisfied and locate one point that meets the given condition. 3. Locate several other points that satisfy the given condition.These points should be sufficiently close together to develop the shape or the nature of the locus. 4. Use the points to draw a line or smooth curve that appears to be the locus. 5. Describe in words the geometric figure that appears to be the locus. Note: In this chapter, we will assume that all given points, segments, rays, lines, and circles lie in the same plane and the desired locus lies in that |
plane also. EXAMPLE 1 What is the locus of points equidistant from the endpoints of a given line segment? Solution Apply the steps of the procedure for discovering a probable locus. Make a diagram: is the given AB line segment. 2. Decide the condition to be satisfied: P is to be equidistant from A and B. Use a compass opened to any convenient radius to locate one such point, P. 3. Locate several other points equidistant from A and B, using a different opening of the compass for each point. 4. Through these points, draw the straight line that appears to be the locus. 5. Describe the locus in words: The locus is a straight line that is the perpendicular bisector of the given line segment. Answer 14365C14.pgs 7/10/07 9:59 AM Page 611 The Meaning of Locus 611 Note that in earlier chapters, we proved two theorems that justify these results: • If a point is equidistant from the endpoints of a line segment, then it lies on the perpendicular bisector of the segment. • If a point lies on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the segment. EXAMPLE 2 Construct the locus of points in the interior of an angle equidistant from the rays that form the sides of the given angle. Construction Corollaries 9.13b and 9.15a together state: A point is equidistant from the sides of an angle if and only if it lies on the bisector of the angle. Therefore, the required locus is the bisector of the angle. 1. Make a diagram: ABC is the given angle. A B C 2. Decide the condition to be satisfied: P is to be equidistant from h BC, the rays that are the sides of ABC. Construct the angle bisector. h BA and Use a compass to draw an arc with center B that intersects h at S. Then, with the compass open to a convenient radius, draw arcs BC from R and S that intersect in the interior of ABC. Call the intersection P. PR PS. at R and h BA A P C R B S 3. Through points P and B, draw the ray that is the locus. A P C R B S 14365C14.pgs 7/10/07 9:59 AM Page 612 6 |
12 Locus and Construction Exercises Writing About Mathematics 1. Are all of the points that are equidistant from the endpoints of a line segment that is 8 cen- timeters long 4 centimeters from the endpoints? Explain your answer. 2. What line segment do we measure to find the distance from a point to a line or to a ray? Developing Skills 3. What is the locus of points that are 10 centimeters from a given point? 4. What is the locus of points equidistant from two points, A and B, that are 8 meters apart? 5. What is the locus of points equidistant from two parallel lines 8 meters apart? 6. What is the locus of points 4 inches away from a given line, g? AB 7. What is the locus of points 3 inches from each of two parallel lines that are 6 inches apart? 8. What is the locus of points that are equidistant from two opposite sides of a square? 9. What is the locus of points that are equidistant from the vertices of two opposite angles of a square? 10. What is the locus of points that are equidistant from the four vertices of a square? (A locus can consist of a single point or no points.) 11. What is the locus of points in the interior of a circle whose radius measures 3 inches if the points are 2 inches from the circle? 12. What is the locus of points in the exterior of a circle whose radius measures 3 inches if the points are 2 inches from the circle? 13. What is the locus of points 2 inches from a circle whose radius measures 3 inches? 14. Circle O has a radius of length r, and it is given that r m. a. What is the locus of points in the exterior of circle O and at a distance m from the circle? b. What is the locus of points in the interior of circle O and at a distance m from the circle? c. What is the locus of points at a distance m from circle O? 15. Concentric circles have the same center. What is the locus of points equidistant from two concentric circles whose radii measure 10 centimeters and 18 centimeters? 16. A series of isosceles triangles are drawn, each of which has a fixed segment, AB, as its base. What is the locus of the vertices of the vertex angles of all such |
isosceles triangles? 17. Triangle ABC is drawn with a fixed base, AB, and an altitude to AB whose measure is 3 feet. What is the locus of points that can indicate vertex C in all such triangles? 14365C14.pgs 7/10/07 9:59 AM Page 613 Five Fundamental Loci 613 Applying Skills 18. What is the locus of the tip of the hour hand of a clock during a 12-hour period? 19. What is the locus of the center of a train wheel that is moving along a straight, level track? 20. What is the locus of the path of a car that is being driven down a street equidistant from the two opposite parallel curbs? 21. A dog is tied to a stake by a rope 6 meters long. What is the boundary of the surface over which he can move? 22. A boy walks through an open field that is bounded on two sides by straight, intersecting roads. He walks so that he is always equidistant from the two intersecting roads. Determine his path. 23. There are two stationary floats on a lake. A girl swims so that she is always equidistant from both floats. Determine her path. 24. A dime is rolled along a horizontal line so that the dime always touches the line. What is the locus of the center of the dime? 25. The outer edge of circular track is 40 feet from a central point. The track is 10 feet wide. What is path of a runner who runs on the track, 2 feet from the inner edge of the track? 14-3 FIVE FUNDAMENTAL LOCI There are five fundamental loci, each based on a different set of conditions. In each of the following, a condition is stated, and the locus that fits the condition is described in words and drawn below. Each of these loci has been shown as a construction in preceding sections. 1. Equidistant from two points: Find points equidistant from points A and B. Locus: The locus of points equidistant from two fixed points is the perpendicular bisector of the segment determined by the two points. A L ocus B 2. Equidistant from two intersecting lines: Find points equidistant from the intersecting lines g AB and g. CD Locus: The locus of points equidistant from two intersecting lines is a pair of lines that bisect |
the angles formed by the intersecting lines. A D Locus s u c o L C B 14365C14.pgs 7/10/07 9:59 AM Page 614 614 Locus and Construction 3. Equidistant from two parallel lines: Find points equidistant from the paral- lel lines g AB and g. CD Locus: The locus of points equidistant from two parallel lines is a third line, parallel to the given lines and midway between them. At a fixed distance from a line: Find points that are at distance d from the line g. AB Locus: The locus of points at a fixed distance from a line is a pair of lines, each parallel to the given line and at the fixed distance from the given line. Locus d A d Locus B 5. At a fixed distance from a point: Find points that are at a distance d from the fixed point A. Locus: The locus of points at a fixed distance from a fixed point is a circle whose center is the fixed point and whose radius is the fixed distance. L ocus d A These loci are often combined to find a point or set of points that satisfy two or more conditions. The resulting set is called a compound locus. Procedure To find the locus of points that satisfy two conditions: 1. Determine the locus of points that satisfy the first condition. Sketch a diagram showing these points. 2. Determine the locus of points that satisfy the second condition. Sketch these points on the diagram drawn in step 1. 3. Locate the points, if any exist, that are common to both loci. Steps 2 and 3 can be repeated if the locus must satisfy three or more conditions. EXAMPLE 1 Quadrilateral ABCD is a parallelogram. What is the locus of points equidistant from g AB and g CD and also equidistant from g AB and g?BC 14365C14.pgs 7/10/07 9:59 AM Page 615 Solution Follow the procedure for finding a compound locus. Five Fundamental Loci 615 (1) Since ABCD is a parallelogram, g CD g. The locus of points AB equidistant from two parallel lines is a third line parallel to the given lines and midway between them. In the g AB is equidistant from figure, g EF and g. CD (2) The lines g AB and g BC are |
intersecting lines. The locus of points equidistant from intersecting lines is a pair of lines that bisect the angles formed by the g JK g GH are equidistant from given lines. In the figure, g AB g EF (3) The point P at which and g BC. and intersects g EF g GH and the point Q at which intersects are equidistant from g JK and also equidistant from g AB g AB and and g CD g BC. Answer P and Q Note that only point P is equidistant from the three segments that are sides of the parallelogram, but both P and Q are equidistant from the lines of which these three sides are segments. Exercises Writing About Mathematics 1. If PQRS is a square, are the points that are equidistant from g PQ and g RS also equidistant from P and S? Explain your answer. 2. Show that the two lines that are equidistant from two intersecting lines are perpendicular to each other. Developing Skills In 3–10, sketch and describe each required locus. 3. The locus of points equidistant from two points that are 4 centimeters apart. 14365C14.pgs 7/10/07 9:59 AM Page 616 616 Locus and Construction 4. The locus of points that are 6 inches from the midpoint of a segment that is 1 foot long. 5. The locus of points equidistant from the endpoints of the base of an isosceles triangle. 6. The locus of points equidistant from the legs of an isosceles triangle. 7. The locus of points equidistant from the diagonals of a square. 8. The locus of points equidistant from the lines that contain the bases of a trapezoid. 9. The locus of points 4 centimeters from the midpoint of the base of an isosceles triangle if the base is 8 centimeters long. 10. The locus of points that are 6 centimeters from the altitude to the base of an isosceles trian- gle if the measure of the base is 12 centimeters 11. a. Sketch the locus of points equidistant from two parallel lines that are 4 centimeters apart. b. On the diagram drawn in a, place point P on one of the given parallel lines. Sketch the locus of points that are 3 centimeters from P. c |
. How many points are equidistant from the two parallel lines and 3 centimeters from P? 12. a. Construct the locus of points equidistant from the endpoints of a line segment. b. Construct the locus of points at a distance AB from M, the midpoint of 1 2(AB). AB c. How many points are equidistant from A and B and at a distance 1 2(AB) from the mid- point of AB? d. Draw line segments joining A and B to the points described in c to form a polygon. What kind of a polygon was formed? Explain your answer. 14-4 POINTS AT A FIXED DISTANCE IN COORDINATE GEOMETRY We know that the locus of points at a fixed distance from a given point is a circle whose radius is the fixed distance. In the coordinate plane, the locus of points r units from (h, k) is the circle whose equation is (x h)2 (y k)2 r2. For example, the equation of the locus of points 10 units from (2, 3) is: " (x 2)2 (y (3))2 ( 10)2 " or (x 2)2 (y 3)2 10 y 1 O 1 x (2, 3) We also know that the locus of points at a fixed distance from a given line is a pair of lines parallel to the given line. 14365C14.pgs 7/10/07 9:59 AM Page 617 Points at a Fixed Distance in Coordinate Geometry 617 For example, to write the equations of the locus of points 3 units from the horizontal line y 2, we need to write the equations of two horizontal lines, one 3 units above the line y 2 and the other 3 units below the line y 2. The equations of the locus are y 5 and y 1. To write the equations of the locus of points 3 units from the vertical line x 2, we need to write the equations of two vertical lines, one 3 units to the right of the line x 2 and the other 3 units to the left of the line x 2. The equations of the locus are x 5 and x 1. From these examples, we can infer the following 1O y 2 x 1 x 1 1O The locus of points d units from the horizontal line y a is the pair of lines y a d and y a |
d. The locus of points d units from the vertical line x a is the pair of lines x a d and x a d. EXAMPLE 1 What is the equation of the locus of points at a distance of (0, 1)? 20 " units from Solution The locus of points at a fixed distance from a point is the circle with the given point as center and the given distance as radius. The equation of the locus is (x 0)2 (y 1)2 20 A " 2 B or x2 (y 1)2 20 Answer EXAMPLE 2 What are the coordinates of the points on the line y 21 2x 1 at a distance of 20 " from (0, 1)? Solution From Example 1, we know that the set of all points at a distance of 20 from (0, 1) lie on the circle whose equation is x2 (y 1)2 20. Therefore, the points on the line y 1 at a distance of sections of the circle and the line. from (0, 1) are the inter- 21 2x " 20 " 14365C14.pgs 7/10/07 10:00 AM Page 618 618 Locus and Construction METHOD 1 Solve the system of equations graphically. Sketch the graphs and read the coordinates from the graph. y (4, 3) x2 ( y 1 )2 METHOD 2 Solve the system of equations algebraically. Substitute the value of y from the linear equation in the equation of the circle. 2 0 x (4, 1) y 1 x 1 2 x2 1 A x2 1 (y 2 1)2 5 20 2 5 20 21 2x 1 1 2 1 B x2 1 1 4x2 5 20 5 4x2 5 20 x2 5 16 x 5 64 Answer The points are (4, 1) and (4, 3). If x 4: y 5 21 2(4) 1 1 If x 4: y 5 21 2(24) 1 1 y 5 21 y 5 3 Exercises Writing About Mathematics 1. In Example 2, is the line y 21 2x 1 a secant of the circle x2 (y 1)2 20? Justify your answer. g PA and g PB 2. are tangent to circle O. Martin said that point O is in the locus of points equidistant from g PA and g PB. Do you agree with Martin? Explain why or why not. Develop |
ing Skills In 3–8, write an equation of the locus of points at the given distance d from the given point P. 3. d 4, P(0, 0) 6. d 7, P(1, 1) 4. d 1, P(1, 0) 7. d 10, P(3, 1) " 5. d 3, P(0, 2) 8. d 18, P(3, 5) " In 9–12, find the equations of the locus of points at the given distance d from the given line. 9. d 5, x 7 10. d 1, x 1 11. d 4, y 2 12. d 6, y 7 14365C14.pgs 7/10/07 10:00 AM Page 619 Equidistant Lines in Coordinate Geometry 619 In 13–16, find the coordinates of the points at the given distance from the given point and on the given line. 13. 5 units from (0, 0) on y x 1 15. 10 units from (0, 1) on y x 3 14. 13 units from (0, 0) on y x 7 16. 10 " units from (1, 1) on y x 2 In 17–22, write the equation(s) or coordinates and sketch each locus. 17. a. The locus of points that are 3 units from y 4. b. The locus of points that are 1 unit from x 2. c. The locus of points that are 3 units from y 4 and 1 unit from x 2. 18. a. The locus of points that are 3 units from (2, 2). b. The locus of points that are 3 units from y 1. c. The locus of points that are 3 units from (2, 2) and 3 units from y 1. 19. a. The locus of points that are 5 units from the origin. b. The locus of points that are 3 units from the x-axis. c. The locus of points that are 5 units from the origin and 3 units from the x-axis. 20. a. The locus of points that are 10 units from the origin. b. The locus of points that are 8 units from the y-axis. c. The locus of points that are 10 units from the origin and 8 units from the y-axis. 21. a. The loc |
us of points that are 2 units from (x 4)2 y2 16. b. The locus of points that are 2 units from the y-axis. c. The locus of points that are 2 units from (x 4)2 y2 16 and 2 units from the y-axis. 22. a. The locus of points that are 3 units from (x 1)2 (y 5)2 4. 5 2 3 b. The locus of points that are units from x. 2 3 c. The locus of points that are 3 units from (x 1)2 (y 5)2 4 and units from x. 2 5 2 14-5 EQUIDISTANT LINES IN COORDINATE GEOMETRY Equidistant from Two Points The locus of points equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points. For example, the locus of points equidistant from A(2, 1) and B(8, 5) is a line perpendicular to at its midpoint. AB 14365C14.pgs 7/10/07 10:00 AM Page 620 620 Locus and Construction midpoint of AB 5 2 1 8 2 A 5 (5, 2) 5 1 (21) 2, B y slope of AB 5 5 2 (21) 8 2 2 5 6 6 5 1 AB Therefore, the slope of a line perpendicuis 1. The perpendicular bisector of is the line through (5, 2) with slope 1. lar to AB The equation of this line is: y = B(8, 5) x 7 (5, 2) 1 O 1 A(2, 1) x y 2 2 x 2 5 5 21 y 2 2 5 2x 1 5 y 5 2x 1 7 EXAMPLE 1 Describe and write an equation for the locus of points equidistant from A(2, 5) and B(6, 1). M(2, 3) B(6, 1) x y 1 A(2, 5) 1 O y = 2x 1 Solution (1) Find the midpoint, M, of AB : M A 22 1 6 2, 5 1 1 2 (2) Find the slope of AB (2, 3) B : slope of AB 5 5 2 1 22 2 6 5 4 28 5 21 2 (3) The slope of a line perpendicular to AB is 2. (4 |
) Write an equation of the line through (2, 3) with slope 2 2x 2 4 y 5 2x 2 1 Answer The locus of points equidistant from A(2, 5) and B(6, 1) is the perpendicular. The equation of the locus is y 2x 1. bisector of AB 14365C14.pgs 7/10/07 10:00 AM Page 621 Equidistant Lines in Coordinate Geometry 621 Equidistant from Two Parallel Lines The locus of points equidistant from two parallel lines is a line parallel to the two lines and midway between them. For example, the locus of points equidistant from the vertical lines x 2 and x 6 is a vertical line midway between them. Since the given lines intersect the x-axis at (2, 0) and (6, 0), the line midway between them intersects the x-axis at (2, 0) and has the equation x 2. EXAMPLE 2 Write an equation of the locus of points equidistant from the parallel lines y 3x 2 and y 3x 6. Solution The locus is a line parallel to the given lines and midway between them. The slope of the locus is 3, the slope of the given lines. The y-intercept of the locus, b, is the average of the y-intercepts of the given lines. b 2 1 (26) 2 5 24 2 5 22 The equation of the locus is y 3x 2. Answer Note that in Example 2, we have used the midpoint of the y-intercepts of the given lines as the y-intercept of the locus. In Exercise 21, you will prove that the midpoint of the segment at which the two given parallel lines intercept the y-axis is the point at which the line equidistant from the given lines intersects the y-axis. Equidistant from Two Intersecting Lines The locus of points equidistant from two intersecting lines is a pair of lines that are perpendicular to each other and bisect the angles at which the given lines intersect. We will consider two special cases. 1. The locus of points equidistant from the axes The x-axis and the y-axis intersect at the origin to form right angles. Therefore, the lines that bisect the angles between the axes will also go through the origin and will form angles measuring 45° with the axes |
. One bisector will have a positive slope and one will have a negative slope. y O 45° B(a, a) A(a, 0) x B(a, a) 14365C14.pgs 7/10/07 10:00 AM Page 622 622 Locus and Construction y O 45° B(a, a) A(a, 0) x B(a, a) Let A(a, 0) be a point on the x-axis and B be a point on the bisector with a positive slope such that is perpendicular to the x-axis. The triangle formed by A, B, and the origin O is a 45-45 right triangle. Since 45-45 right triangles are isosceles, OA AB |a|, and the coordinates of B are (a, a). The line through (a, a) and the origin is y x. AB Similarly, if B is a point on the bisector with a negative slope, then the coor- dinates of B are (a, a). The line through (a, a) and the origin is y x. We have shown that the lines y x and y x are the locus of points equidistant from the axes. These lines are perpendicular to each other since their slopes, 1 and 1, are negative reciprocals. The locus of points in the coordinate plane equidistant from the axes is the pair of lines y x and y x. 2. The locus of points equidistant from two lines with slopes m and m that intersect at the origin y B y mx y mx A(a, ma) Let O(0, 0) and A(a, ma) be any two points on y mx and B(0, ma) be a point on the y-axis. Under a reflection in the yaxis, the image of O(0, 0) is O(0, 0) and the image of A(a, ma) is A(a, ma). The points O and A are on the line y mx. Therefore, the image of the line y mx is the line y mx since collinearity is preserved under a line reflection. Also, mAOB mAOB since angle measure is preserved under a line reflection. Therefore, the y-axis bisects the angle between the lines y mx and y mx. In a similar way, it |
can be shown that the x-axis bisects the other pair of angles between y mx and y mx. Therefore, the y-axis, together with the x-axis, is the locus of points equidistant from the lines y mx and y mx. A(a, ma) O x The locus of points in the coordinate plane equidistant from two lines with slopes m and m that intersect at the origin is the pair of lines y 0 and x 0, that is, the x-axis and the y-axis. Exercises Writing About Mathematics 1. In the coordinate plane, are the x-axis and the y-axis the locus of points equidistant from the intersecting lines y x and y x? Justify your answer. 14365C14.pgs 7/10/07 10:00 AM Page 623 Equidistant Lines in Coordinate Geometry 623 2. Ryan said that if the locus of points equidistant from y x 2 and y x 10 is y x 6, then the distance from y x 6 to y x 2 and to y x 10 is 4. Do you agree with Ryan? Justify your answer. Developing Skills In 3–8, find the equation of the locus of points equidistant from each given pair of points. 3. (1, 1) and (9, 1) 6. (0, 6) and (4, 2) 4. (3, 1) and (3, 3) 7. (2, 2) and (0, 2) 5. (0, 2) and (2, 0) 8. (4, 5) and (2, 1) In 9–14, find the equation of the locus of points equidistant from each given pair of parallel lines 9. x 1 and x 7 11. y x 3 and y x 9 13. y 2x 1 and y 2x 5 15. Find the coordinates of the locus of points equidistant from (2, 3) and (4, 3), and 3 units 10. y 0 and y 6 12. y x 2 and y x 6 14. 2x y 7 and y 2x 9 from (1, 3). 16. Find the coordinates of the locus of points equidistant from (2, 5) and (2, 3), and 4 units from (0, 1). Applying Skills 17. a. Find |
the equation of the locus of points equidistant from (3, 1) and (5, 5). b. Prove that the point (2, 6) is equidistant from the points (3, 1) and (5, 5) by showing that it lies on the line whose equation you wrote in a. c. Prove that the point (2, 6) is equidistant from (3, 1) and (5, 5) by using the distance for- mula. 18. a. Find the equation of the locus of points equidistant from the parallel lines y x 3 and y x 5. b. Show that point P(3, 2) is equidistant from the given parallel lines by showing that it lies on the line whose equation you wrote in a. c. Find the equation of the line that is perpendicular to the given parallel lines through point P(3, 2). d. Find the coordinates of point A at which the line whose equation you wrote in c inter- sects y x 3. e. Find the coordinates of point B at which the line whose equation you wrote in c inter- sects y x 5. f. Use the distance formula to show that PA PB, that is, that P is equidistant from y x 3 and y x 5. 19. Show that the locus of points equidistant from the line y x 1 and the line y x 1 is the y-axis and the line y 1. 14365C14.pgs 7/31/07 1:13 PM Page 624 624 Locus and Construction 20. Show that the locus of points equidistant from the line y 3x 2 and the line y 3x 2 is the y-axis and the line y 2. 21. Prove that the midpoint of the segment at which two given parallel lines intercept the y-axis is the point at which the line equidistant from the given lines intersects the y-axis. That is, if the y-intercept of the first line is b, and the y-intercept of the second line is c, then the y-intercept of the line equidistant from them is. b 1 c 2 a. Find the coordinates of A, the point where the first line intersects the y-axis. b. Find the coordinates of A, the point where the second line intersects the y-axis. c. Show that B d. Let |
B and B be points on the lines such that M is is the midpoint of M 0, A. AAr (0, bc) point on and lines. Show that ABM ABM. BBr BBr is perpendicular to both e. Using part d, explain why M is the point at which the line equidistant from the given lines intersects the y-axis. 14-6 POINTS EQUIDISTANT FROM A POINT AND A LINE We have seen that a straight line is the locus of points equidistant from two points or from two parallel lines. What is the locus of points equidistant from a given point and a given line? Consider a fixed horizontal g AB line and a fixed point F above the line. The point Pm, that is, the midpoint of the vertical line from F Pn Pn F g AB Locus to, is on the locus of points equidistant from the point and the line. Let Pn be any other point on the locus. As we move to the right or to the left from Pm along the distance from distance from F to Pn is along a slant line. The locus of points is a curve. to Pn continues to be the length of a vertical line, but the g, AB g AB Pm A Pn Pn Pn B 14365C14.pgs 7/10/07 10:00 AM Page 625 Points Equidistant from a Point and a Line 625 Consider a point and a horizontal line that are at a distance d from the origin. • Let F(0, d) be the point and l be the line. The equation of l is y d. • Let P(x, y) be any point equidistant from F and l. • Let M(x, d) be the point at which a vertical line from P intersects l. y F(0, d) O P(x, y) M(x, d) x l The distance from P to M is equal to the distance from P to F. PM 5 PF y 2 (2d) 5 (x 2 0)2 1 (y 2 d)2 " (y 1 d)2 5 x2 1 (y 2 d)2 y2 1 2dy 1 d2 5 x2 1 y2 2 2dy 1 d2 4dy 5 x2 1 4 For instance, if d, that is, if the |
given point is and the given line is y 5 21, then the equation of the locus is y x2. Recall that y x2 is the equa4 tion of a parabola whose turning point is the origin and whose axis of symmetry is the y-axis. 0, 1 4 B A Recall that under the translation, the image of 4dy x2 is Th, k 4d(y k) (x h)2. For example, if the fixed point is F(2, 1) and the fixed line is y 1, then (h, k) 5 (2, 0). d 1 and the turning point of the parabola is (2, 0), so Therefore, the equation of the parabola is 4(1)(y 0) (x 2)2 or 4y x2 4x 4. This equation can also be written as y 1 4x2 2 x 1 1. For any horizontal line and any point not on the line, the equation of the locus of points equidistant from the point and the line is a parabola whose equation can be written as y ax2 bx c. If the given point is above the line, the coefficient a is positive and the parabola opens upward. If the given point is below the line, the coefficient a is negative and the parabola opens downward. The axis of symmetry is a vertical line whose equation is x. Since the turning point is on the axis of symmetry, its x-coordinate is 2b 2a.2b 2a 14365C14.pgs 7/10/07 10:00 AM Page 626 626 Locus and Construction EXAMPLE 1 a. Draw the graph of y x2 4x 1 from x 1 to x 5. b. Write the coordinates of the turning point. c. Write an equation of the axis of symmetry. d. What are the coordinates of the fixed point and the fixed line for this parabola? Solution a. (1) Make a table of values using integral values of x from x 1 to x 5. (2) Plot the points whose coordinates are given in the table and draw a smooth curve through them. x 1 0 1 2 3 4 5 x2 4x 12 1 16 16 1 25 20. From the graph or from the table, the coordinates of the turning point appear to be (2, 5). We can verify this algebraically: x 5 2b 2a |
2(24) 2(1) 5 5 2 y 5 x2 2 4x 2 1 5 (2)2 2 4(2) 2 1 5 25 c. The axis of symmetry is the vertical line through the turning point, x 2. d. Note that the turning point of the parabola is (2, 5). When the turning point of the parabola y x2 has been moved 2 units to the right and 5 units down, the equation becomes the equation of the graph that we drew: y 2 (25) 5 (x 2 2)2 y 1 5 5 x2 2 4x 1 4 y 5 x2 2 4x 2 1 The turning point of the parabola is the midpoint of the perpendicular segment from the fixed point to the fixed line. Since the coefficient of y in the equation of the parabola is 1, 4d 1 or. The parabola opens d 5 1 4 14365C14.pgs 7/10/07 10:00 AM Page 627 Points Equidistant from a Point and a Line 627 upward so the fixed point is unit above the turning point and the fixed line is unit below the turning point. The coordinates of the fixed point are and the equation of the fixed line is y. 251 4 1 4 2, 243 4 B A 1 4 EXAMPLE 2 a. Draw the graph of y x2 2x 8 from x 4 to x 2. b. Write the coordinates of the turning point. c. Write an equation of the axis of symmetry. Solution a. (1) Make a table of values using integral values of x from x 4 to x 2. (2) Plot the points whose coordinates are given in the table and draw a smooth curve through them. x 4 3 2 1 0 1 2 x2 2x 8 16. From the graph or from the table, the coordinates of the turning point appear to be (1, 9). We can verify this algebraically: x 5 2b 2a 2(22) 2(21) 5 5 21 y 5 2x2 2 2x 1 8 5 2(21)2 2 2(21) 1 8 5 9 c. The axis of symmetry is the vertical line through the turning point, x 1. Here the parabola y x2 has been reflected in the x-axis so that the equation becomes y x2. Then that parabola has been moved 1 unit to the left and 9 units up so |
that the equation becomes (y 9) (x (1))2 or y 9 x2 2x 1, which can be written as y x2 2x 8 or y x2 2x 8. 14365C14.pgs 7/10/07 10:00 AM Page 628 628 Locus and Construction EXAMPLE 3 A parabola is equidistant from a given point and a line. How does the turning point of the parabola relate to the given point and line? Solution The x-coordinate of the turning point is the same as the x-coordinate of the given point and is halfway between the given point and the line. EXAMPLE 4 Solve the following system of equations graphically and check: y x2 2x 1 y x 3 Solution (1) Make a table of values using at least three integral values of x that are less than that of the turning point and three that are greater. The x-coordinate of the turning point is: 2b 2a 5 2(22) 2(1) 5 2 2 5 1 (2) Plot the points whose coordinates are given in the table and draw a smooth curve through them. x 2 1 0 1 2 3 4 x2 2x 16 3) On the same set of axes, sketch the graph of y x 3. 21 1. Move 1 unit down The y-intercept is 3. Start at the point (0, 3). The slope is 1 or and 1 unit to the right to find a second point of the line. From this point, again move 1 unit down and 1 unit to the right to find a third point. Draw a line through these three points2, 1) x (1, 4) y = x 3 1 O 1 14365C14.pgs 7/10/07 10:00 AM Page 629 Points Equidistant from a Point and a Line 629 (4) Read the coordinates of the points of intersection from the graph. The common solutions are (1, 4) and (2, 1). Answer (1, 4) and (2, 1) or x 1, y 4 and x 2, y 1 Exercises Writing About Mathematics 1. Luis said that the solutions to the equation x2 2x 8 0 are the x-coordinates of the points at which the graph of y x2 2x 8 intersects the y-axis. Do you agree with Luis? Explain why or why not. |
2. Amanda said that if the turning point of a parabola is (1, 0), then the x-axis is tangent to the parabola. Do you agree with Amanda? Explain why or why not. Developing Skills In 3–8, find the coordinates of the turning point and the equation of the axis of symmetry of each parabola. 3. y x2 6x 1 6. y x2 2x 5 4. y x2 2x 3 7. y x2 8x 4 5. y x2 4x 1 8. y x2 5x 2 In 9–16, find the common solution of each system of equations graphically and check your solution. 9. y x2 2x 2 y x 2 11. y x2 4x 3 y x 1 13. y x2 4x 2 y 2x 3 15. y x2 6x 5 y 7 2x 10. y x2 1 x y 1 12. y x2 2x 3 y 1 x 14. y x2 2x 2 y 2x 2 16. y 2x x2 y 2x 4 In 17–20 write the equation of the parabola that is the locus of points equidistant from each given point and line. F y 5 21 0, 1 17. 4 B 4 19. F(0, 2) and y 2 and A F 0, 21 18. 4 B 20. F(3, 3) and y 3 and y 5 1 4 A 14365C14.pgs 8/2/07 6:02 PM Page 630 630 Locus and Construction Hands-On Activity If the graph of an equation is moved h units in the horizontal direction and k units in the vertical direction, then x is replaced by x h and y is replaced by y k in the given equation. 1. The turning point of the parabola y ax2 is (0, 0). If the parabola y ax2 is moved so that the coordinates of the turning point are (3, 5), what is the equation of the parabola? 2. If the parabola y ax2 is moved so that the coordinates of the turning point are (h, k), what is the equation of the parabola? CHAPTER SUMMARY Loci • A locus of points is the set of all points, and only those points, that satisfy a given condition. • The locus of points equidistant from two |
fixed points that are the end- points of a segment is the perpendicular bisector of the segment. • The locus of points equidistant from two intersecting lines is a pair of lines that bisect the angles formed by the intersecting lines. • The locus of points equidistant from two parallel lines is a third line, paral- lel to the given lines and midway between them. • The locus of points at a fixed distance from a line is a pair of lines, each parallel to the given line and at the fixed distance from the given line. • The locus of points at a fixed distance from a fixed point is a circle whose center is the fixed point and whose radius is the fixed distance. • The locus of points equidistant from a fixed point and a line is a parabola. Loci in the Coordinate Plane • The locus of points in the coordinate plane r units from (h, k) is the circle whose equation is (x h)2 (y k)2 r2. • The locus of points d units from the horizontal line y a is the pair of lines y a d and y a d. • The locus of points d units from the vertical line x a is the pair of lines x a d and x a d. • The locus of points equidistant from A(a, c) and B(b, d) is a line perpen- dicular to AB at its midpoint • The locus of points equidistant from the axes is the pair of lines y x and y x. • The locus of points equidistant from the lines y mx and y mx is the pair of lines y 0 and x 0, that is, the x-axis and the y-axis. • The locus of points equidistant from (h, k d) and y k d is the parabola whose equation is 4d(y k) (x h)2. 14365C14.pgs 7/10/07 10:00 AM Page 631 Review Exercises 631 VOCABULARY 14-2 Locus • Concentric circles 14-3 Compound locus REVIEW EXERCISES 1. Construct: a. a right angle. b. an angle whose measure is 45°. c. parallelogram ABCD with AB a, BC b and mB 45. a b 2. Draw PQ. Construct |
S on PQ such that PS : SQ 2 : 3. In 3–6, sketch and describe each locus. 3. Equidistant from two points that are 6 centimeters apart. 4. Four centimeters from A and equidistant from A and B, the endpoints of a line segment that is 6 centimeters long. 5. Equidistant from the endpoints of a line segment that is 6 centimeters long and 2 centimeters from the midpoint of the segment. 6. Equidistant from parallel lines that are 5 inches apart and 4 inches from a point on one of the given lines. In 7–12, sketch the locus of points on graph paper and write the equation or equations of the locus. 7. 3 units from (1, 2). 8. 2 units from (2, 4) and 2 units from x 2. 9. Equidistant from x 1 and x 5. 10. Equidistant from y 2 and y 8. 11. Equidistant from y 2x and y 2x. 12. Equidistant from (1, 3) and (3, 1). 13. Find the coordinates of the points on the line x y 7 that are 5 units from the origin. 14. a. Draw the graph of y x2 4x 1. b. On the same set of axes, draw the graph of y 2x 2. c. What are the coordinates of the points of intersection of the graphs drawn in a and b? 14365C14.pgs 8/2/07 6:02 PM Page 632 632 Locus and Construction In 15–18, solve each system of equations graphically. 15. x2 y2 25 y x 1 17. y x2 2x 1 y x 1 x y 6 x 2 18. y x2 4x 2 16. (x 2)2 (y 1)2 4 19. A field is rectangular in shape and measures 80 feet by 120 feet. How many points are equidistant from any three sides of the field? (Hint: Sketch the locus of points equidistant from each pair of sides of the field.) 20. The coordinates of the vertices of isosceles trapezoid ABCD are A(0, 0), B(6, 0), C(4, 4), and D(2, 4). What are the coordinates of the locus of points equidistant from the vertices of the trapezoid? ( |
Hint: Sketch the locus of points equidistant from each pair of vertices.) Exploration An ellipse is the locus of points such that the sum of the distances from two fixed points F1 and F2 is a constant, k. Use the following procedure to create an ellipse. You will need a piece of string, a piece of thick cardboard, and two thumbtacks. STEP 1. Place two thumbtacks in the cardboard separated by a distance that is less than the length of the string. Call the thumbtacks F1 and F2. Attach one end of the string to F1 and the other to F2. The length of the string represents the sum of the distances from a point on the locus to the fixed points. F1 F2 P STEP 2. Place your pencil in the loop of string and pull the string taut to locate some point P. Keeping the string taut, slowly trace your pencil around the fixed points until you have created a closed figure. a. Prove that the closed figure you created is an ellipse. (Hint: Let k be the length of the string.) b. If F1 and F2 move closer and closer together, how is the shape of the ellipse affected? A hyperbola is the locus of points such that the difference of the distances from fixed points F1 and F2 is a constant, k. c. Explain why a hyperbola is not a closed figure. P F1 PF2 PF1 = k F2 14365C14.pgs 7/10/07 10:00 AM Page 633 CUMULATIVE REVIEW Part I Cumulative Review 633 Chapters 1–14 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following has a midpoint? g AB (1) (2) AB h AB (3) (4) ABC 2. What is the inverse of the statement “If two lines segments are congruent, then they have equal measures?” (1) If two line segments are not congruent, then they do not have equal measures. (2) If two line segments do not have equal measures, then they are not congruent. (3) If two line segments have equal measures, then they are congruent. (4) Two line segments are not congruent if they have unequal measures. |
AD 3. In triangle ABC, is the altitude to BC then which of the following may be false? (1) AB AC (2) BD CD (3) AB AD (4) mB mC. If D is the midpoint of BC, 4. If two angles of a triangle are congruent and complementary, then the tri- angle is (1) isosceles and right (2) scalene and right (3) isosceles and obtuse (4) scalene and acute 5. The equation of a line that is perpendicular to the line x 3y 6 is (1) y 3x 1 (2) y 3x 1 1 21 (3) y 3x 1 (4) y 1 3x 6. Under a line reflection, which of the following is not preserved? (1) angle measure (2) collinearity (3) orientation (4) midpoint 7. Which of the following is not sufficient to prove that quadrilateral ABCD is a parallelogram? AB > CD and (1) AB > CD and (2) BC > DA AB CD (3) (4) AC AC BD bisect each other and ⊥ BD 8. A prism with bases that are equilateral triangles has a height of 12.0 cen- timeters. If the length of each side of a base is 8.00 centimeters, what is the number of square centimeters in the lateral area of the prism? (1) 48.0 (2) 96.0 (3) 288 (4) 384 14365C14.pgs 7/10/07 10:00 AM Page 634 634 Locus and Construction 9. The altitude to the hypotenuse of a right triangle divides the hypotenuse into segments of lengths 4 and 45. The length of the shorter leg is 10. Triangle ABC is inscribed in circle O. If m (1) 6 5 " (2) 14 is the measure of ABC? (2) 55 (1) 45 (3) 2,009 110 and m (4) BCX 2,041 " 90, what " ABX (3) 80 (4) 110 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit 11. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.