1. P 5 36 units The unknown side is a leg x x x x. 2. The unknown side is a hypotenuse.

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1 Chapter 7 Prerequisite Skills (p. 40). The triangle is an equilateral triangle.. The triangle is a right triangle.. The triangle is an acute triangle. 4. The triangle is an obtuse isosceles triangle. 5. Ï 45 5 Ï 9 p 5 5 Ï 5 6. ( Ï 7 ) 5 ( Ï 7 ) 5 9(7) Ï p Ï 5 5 Ï Ï 5 7 p Ï Ï p Ï 5 7 Ï ( 5) 5 4 5( 4) 5 6( 4) P 5 6 units Q 5 8 units R 5 9 units S 5 8 units T 5 8 units Sum of areas 5 P Q R S T 5 (6) 8 (9) units The side lengths of the squares are equal to the lengths of the two legs of Triangle P.. rea of square in Step 4 5 P Q R S T 5 44 units. The side length of this square is equal to the length of the hpotenuse of Triangle P.. The sum of the areas of the two squares in Step is equal to the area of the square in Step 4. So, the sum of the squares of the legs of a right triangle is equal to the square of the hpotenuse. 4. The legs of Triangle P are congruent, and the meet to form a right angle. The conjecture in Eercise is not true for all isosceles triangles. For eample: Lesson 7. Investigating Geometr ctivit 7. (p. 4) Step : R S Q Copright b McDougal Littell, a division of Houghton Mifflin Compan. Step 4: T P P R P T R S R P Q The conjecture is true for all right triangles. In the figures below, ou can see that 4T 5 C 4T. So, 5 C. T C T T 7. Guided Practice (pp. 4 46). The unknown side is a leg The unknown side is a hpotenuse Ï T T T T C T T Geometr Worked-Out Solution Ke 97

2 . (Length of ladder) 5 (Distance from house) (Height of ladder) Ï 565 ø.8 The length of the ladder is about.8 feet. 4. The Pthagorean Theorem is onl true for right triangles. 5. The base of each right triangle is (0) h h 99 5 h Ï 5 h rea 5 (base)(height) 5 (0)( Ï ) 5 45 Ï ø 49.5 The area of the triangle is about 49.5 square feet. 6. The base of each right triangle is 0 m h h h 4 5 h rea 5 (base)(height) 5 (0)(4) 5 40 The area of the triangle is 40 square meters. 7. Using the Pthagorean Theorem: Using a Pthagorean triple: common triple is, 4, 5. Multipl each number b. p p 5 5 p 5 5 So, the length of the hpotenuse is 5 inches. 8. Using the Pthagorean Theorem: Using a Pthagorean triple: a common triple is 7, 4, 5. Multipl each number b. 7 p p p 5 50 So, the length of the hpotenuse is 50 centimeters. 7. Eercises (pp ) Skill Practice. set of three positive integers a, b, and c that satisfies the equation c 5 a b is called a Pthagorean triple.. In order to use the Pthagorean Theorem, ou must have the lengths of two of the sides of a right triangle , , In the Pthagorean Theorem, b and c were substituted incorrectl. a b 5 c It is not algebraicall correct to simplif 7 4 as (7 4). The solution should be: h h h 4. ø h The height of the fire escape landing is about 4. feet h h h 9.4 ø h The height of the backboard frame is about 9.4 inches b b b.9 ø b The base of the frame is about.9 feet.. The base of each right triangle is (6) 5 8 meters h h 5 5 h 5 5 h rea 5 (base)(height) 5 (6)(5) 5 0 The area of the triangle is 0 square meters. Copright b McDougal Littell, a division of Houghton Mifflin Compan. 98 Geometr Worked-Out Solution Ke

3 Copright b McDougal Littell, a division of Houghton Mifflin Compan.. The base of each right triangle is () 5 6 feet h h 44 5 h 5 h rea 5 (base)(height) 5 ()() 5 9 The area of the triangle is 9 square feet.. The base of each right triangle is 6 centimeters h h 64 5 h 8 5 h rea 5 (base)(height) 5 ()(8) 5 48 The area of the triangle is 48 square centimeters. 4. common triple is 7, 4, 5. Multipl each length b. 7 p 5 4 p p 5 75 So, common triple is, 4, 5. Multipl each length b 0. p p p So, common triple is 8, 5, 7. Multipl each length b ; 8 p p p So, 5. c 5 a b c c c 5 89 c 5 7 The length of the hpotenuse is 7 inches and 5; common Pthagorean triple is 8, 5, 7. Multipl the lengths b. 8 p p p 5 5 The missing side is a leg with length of and 5; common Pthagorean triple is, 4, 5. Multipl the lengths b 5. p p p The missing side is a leg with length of and 96; common Pthagorean triple is 7, 4, 5. Multipl the lengths b 4. 7 p p p The missing side is a hpotenuse with length of and 48; common Pthagorean triple is 5,,. Multipl the lengths b 4. 5 p p p The missing side is a hpotenuse with length of and 85; common Pthagorean triple is 8, 5, 7. Multipl the lengths b 5. 8 p p p The missing side is a leg with length of and 75; common Pthagorean triple is 7, 4, 5. Multipl the lengths b. 7 p 5 4 p p 5 75 The missing side is a leg with length of Ï 5 5 Ï 6. Let h represent the height. 5 5 h h 6 5 h 4 5 h Ï 65 Geometr Worked-Out Solution Ke 99

4 7. ; b b 96 5 b 6 5 b rea 5 (base)(height) 5 (5)(6) 5 70 The area of the triangle is 70 square feet. 8. (4 4) 5 () ( 4) ( 6) or 5 6 ecause cannot be zero, the value of is u 6 0 t s 6 r 9 9 So, b Problem Solving. nd base ,00 ø 7. Home plate 90 ft ball thrown from home plate to second base must go about 7. feet ft r s r s 5 5 r 44 5 s 5 5 r 5 s t t 64 5 t 8 5 t u 5 s t Ï 0. b b 5 4 (4 b ) 5 5 b b 4 (4 b ) b b b b 5 7 8b b 5 7 Solve the sstem of equations. 8b b 5 7 b 5 5 8b 5 5 b 5 9 and b ft The balloon is 8 feet above the ground.. The hpotenuse is 65, because it must be the longest of the three sides. 4. a ø P ø ø 86.9 The perimeter of the field is about 86.9 feet. b. P Number of dogwoods to plant You will need about 9 dogwood seedlings. c. Number of dogwoods cost of dogwoods 5 total cost 9 5 total cost 8 5 total cost The trees will cost $8. Copright b McDougal Littell, a division of Houghton Mifflin Compan. 00 Geometr Worked-Out Solution Ke

5 5. a b. C C CE C CE 0 Ï 0 60 ø 60.8 Ï 0 0 ø Ï 0 60 ø 6. Ï 00 0 ø Ï 0 60 ø 67. Ï 90 0 ø Ï ø 7. Ï 80 0 ø Ï ø 78. Ï 70 0 ø Ï ø 84.9 Ï 60 0 ø Ï ø 9. Ï 50 0 ø Ï Ï Ï ø 08. Ï 0 0 ø Ï ø 6.6 Ï 0 0 ø Ï 0 60 ø 5. Ï 0 0 ø Ï 0 60 ø 4. Ï Statements. nc and ndef are right triangles.. c 5 a b, f 5 d e. Given. b 5 e, c 5 f. Given Reasons. Pthagorean Theorem 4. f 5 a e 4. Substitution Propert of Equalit 5. a e 5 d e 5. Substitution Propert of Equalit 6. a 5 d 6. Subtraction Propert of Equalit 7. a 5 d 7. Definition of square roots 8. C > EF 8. Definition of congruent segments 9. C > DFE 9. Right ngles Congruence Theorem 0. nc > ndef 0. SS Congruence Postulate b. The shortest distance that ou must travel is 50 feet. c. 8. Row 5 ft 5 ft Copright b McDougal Littell, a division of Houghton Mifflin Compan. 60 ft 00 ft C 0 ft 50 ft E 0 ft 6. Consider a right triangle, with coordinates (0, 0), (0, a), and (b, 0). (0, a) (0, 0) (b, 0) The distance from (0, a) to (b, 0) is d 5 Ï (b 0) (0 a) 5 Ï b a d 5 Ï a b d 5 a b. 7. Given: nc and ndef are right triangles; b 5 e, c 5 f Prove: nc > ndef b c D e f D Row 5 ft If the trees are staggered so that trees in row are halfwa between the trees in row, at the minimum distance of 5 feet, an equilateral triangle is formed with each leg having a length of 5 feet. isect the base to form two right triangles h 5 ft 5 ft h h h.5 ft.5 ft 4. ø h The minimum distance between the rows is about 4. feet. Mied Review 9. ( Ï 7 ) 5 Ï 7 p Ï (4 Ï ) 5 4 ( Ï ) 5 6 p (6 Ï 8 ) 5 (6) ( Ï 8 ) 5 6 p (8 Ï ) 5 (8) ( Ï ) 5 64 p feet, 6 feet, feet > 6 6 > > 9 > The length of the third side must be greater than feet and less than 9 feet. C a F d E Geometr Worked-Out Solution Ke 0

6 44. 5 inches, inches, inches 5 > 5 > > 6 6 > The length of the third side must be greater than 6 inches and less than 6 inches meters, meters, meters 4 > 4 > > 7 5 > The length of the third side must be greater than 7 meters and less than 5 meters. 46. inches, 7 inches, inches > 7 7 > > 5 9 > The length of the third side must be greater than 5 inches and less than 9 inches ards, 8 ards, ards 8 > > > 0 6 > The length of the third side must be greater than 0 ards and less than 6 ards meters, 9 meters, meters 7 > > > 66 > The length of the third side must be greater than meters and less than 66 meters The triangles are similar, because all of the ratios are equal. Triangle, Triangle The scale factor of Triangle to Triangle is Þ 9 Þ 0 The triangles are not similar, because the ratios are not equal. Lesson 7. Investigating Geometr ctivit 7. (p. 440). If a triangle is a right triangle, then the square of the length of the hpotenuse is equal to the sum of the squares of the lengths of the legs. If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle.. The converse of the Pthagorean Theorem is true because an obtuse or acute triangle will not produce the same equivalenc in the squares of the side lengths.. Let C be the largest angle in nc. If m C < 908, then < C C. If m C 5 908, then 5 C C. If m C > 908, then > C C. 4. If > C C, then the triangle is an obtuse triangle. 5. If < C C, then the triangle is an acute triangle. 6. If 5 C C, then the triangle is a right triangle. 7. Guided Practice (pp ) (4 Ï ) The triangle is a right triangle Þ The triangle is not a right triangle.. ( Ï 6 ) The triangle is a right triangle > 6 9 > 4 0 > The side lengths, 4, and 6 can form a triangle. c? a b 6? 4 6? > 5 The triangle is obtuse. 5. No, triangles with side lengths,, and 4 could not be used to verif that ou have perpendicular lines, because the side lengths do not form a right triangle. 4 Þ 7. Eercises (pp ) Skill Practice. The longest side of a right triangle is called a hpotenuse.. The side lengths of a triangle can be used to classif a triangle as acute, obtuse, or right b comparing the square of the length of the longest side to the sum of the squares of the lengths of the two other sides. If c 5 a b, the triangle is a right triangle. If c > a b, the triangle is an obtuse triangle. If c < a b, the triangle is an acute triangle The triangle is a right triangle. Copright b McDougal Littell, a division of Houghton Mifflin Compan. 0 Geometr Worked-Out Solution Ke

7 Copright b McDougal Littell, a division of Houghton Mifflin Compan Þ The triangle is not a right triangle. 5. ( Ï 5 ) Þ 40 The triangle is not a right triangle. 6. (4 Ï 9 ) Þ 96 The triangle is not a right triangle. 7. ( Ï 6 ) The triangle is a right triangle The triangle is a right triangle The triangle is a right triangle Þ 8 The triangle is not a right triangle The triangle is a right triangle.. ( Ï 4 ) The triangle is a right triangle.. (7 Ï 5 ) The triangle is a right triangle Þ 44 The triangle is not a right triangle. 5. 0,, and > 4 8 > 5 > 0 The segment lengths form a triangle. c? a b 4? 0 96? < The triangle is an acute triangle. 6. 0, 5, and 5 Ï Ï ø Ï ø 5 > 5 Ï 8 > 5 > 0 The segment lengths form a triangle. c? a b (5 Ï )? 0 5 5? The triangle is a right triangle. 7. 4, 0, and 6 Ï Ï 4 ø Ï 4 ø > 6 Ï 4 6. > > 4 The segment lengths form a triangle. c? a b (6 Ï 4 )? ? > 476 The triangle is an obtuse triangle. 8. 5, 6, and > 7 > 6 > 5 The segment lengths form a triangle. c? a b 7? ? < 6 The triangle is an acute triangle. 9., 6, and > 0 > 6 6 > The segment lengths form a triangle. c? a b 0? 6 400? The triangle is a right triangle. Geometr Worked-Out Solution Ke 0

8 0. 8, 0, and > 0 > 0 > 8 The segment lengths form a triangle. c? a b? ? < 64 The triangle is an acute triangle.. 5, 0, and < 6 5 > 0 56 > 5 The segment lengths do not form a triangle.. 6, 8, and > 0 6 > 8 8 > 6 The segment lengths form a triangle. c? a b 0? ? The triangle is a right triangle.. 8., 4., and >. 0.4 > > 8. The segment lengths form a triangle. c? a b.? ? > The triangle is an obtuse triangle. 4. ; 0, 4, and Þ C; 4, 7, and 9 c? a b 9? 4 7 8? > 65 The triangle is an obtuse scalene triangle. 6. Multipling all of the lengths of a Pthagorean triple b a constant will produce another Pthagorean triple. 7. C 5 Ï (6 ()) (0 4) 5 4 Ï 5 C 5 Ï (5 6) ( 0) 5 5 Ï 5 C 5 Ï (5 ()) ( 4) 5 Ï (5 Ï 5 )? (4 Ï 5 ) ( Ï ) 5? 80 5 > 9 The triangle is an obtuse triangle. 8. C 5 Ï (5 0) ( ) 5 Ï 6 C 5 Ï ( 5) ( ) 5 Ï 5 C 5 Ï ( 0) ( ) 5 Ï 0 ( Ï 6 )? ( Ï 5 ) ( Ï 0 ) 6? < 0 The triangle is an acute triangle. 9. ()? (5) () 69? The triangle is a right triangle. 0. C? C EF? DF DE (4 Ï 0 )? 4 ( Ï 96 )? ? ? < 88 Triangle C is a right triangle. Triangle DEF is an acute triangle. So, m > m D.. When m 5 908: m m m C m m C m m C When m D < 908: m D m E m F m E m F m D m E m F > So, m m C < m E m F. Copright b McDougal Littell, a division of Houghton Mifflin Compan. 04 Geometr Worked-Out Solution Ke

9 Copright b McDougal Littell, a division of Houghton Mifflin Compan.. 6, 8, and > 8 8 > 6 4 > > > Right triangle: 5 0 cute triangle: < < 0 Obtuse triangle: 0 < < 4. 0 < ( 4) 400 < < < ( 4 9) 0 < ( )( 6) > and > 6 For 0 to be the longest side, 4 < 0, so < 6. The triangle is an acute triangle when < < (6 ) > (4 6) ( ) 6 > > 0 4(4 6 9) > 0 4( )( 9) > 0 > 0 and 9 > 0 > and > 9 The triangle is an obtuse triangle when > 9. Problem Solving 5. c c c 5 64 c ø.8 To be certain that the corners are 908, make sure that the diagonals of the rectangular frame measure.8 inches ft 05 ft 800 ft , ,00 9,05 640,000 Þ 654,06 The triangle is not a right triangle. So, ou do not live directl north of the librar. 7. a. ecause ncd is a right triangle: 5 C C 5 5 C 5 5 C b. C 0 C So, nc is a right triangle. c. Sample answer: Þ 65 4 C The pole is not perpendicular to the ground. To make it perpendicular, tighten the rope to a length of Ï 65, or about 8 feet. Then tie another rope of the same length at the same point on the pole, and stake it in the ground 4 feet from the pole in another direction. 9. a The triangle is a right triangle because 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. b Þ 5 The triangle is not a right triangle because the square of the length of the longest side does not equal the sum of the squares of the lengths of the other two sides. c. No; the second corner does not form a right triangle, so the frame is not sound. 40. Given: In nc, c < a b where c is the length of the longest side. Prove: nc is an acute triangle. c a b C P a R b Q D Geometr Worked-Out Solution Ke 05

10 Statements Reasons. In nc,. Given c < a b where c is the length of the longest side. In npqr, R is a right angle.. a b 5. Pthagorean Theorem. c <. Substitution Propert 4. c < 4. propert of square roots 5. m R Definition of right angle 6. m C < m R 6. Converse of the Hinge Theorem 7. m C < Substitution Propert 8. C is an acute angle. 8. Definition of an acute angle 9. nc is an acute triangle. 4. Given: In nc, c > a b where c is the length of the longest side. Prove: nc is an obtuse triangle. Statements 9. Definition of an acute triangle c a C b Q a Reasons. In nc,. Given c > a b where c is the length of the longest side. In npqr, R is a right angle.. a b 5. Pthagorean Theorem. c >. Substitution Propert 4. c > 4. propert of square roots 5. m R Definition of right angle 6. m C > m R 6. Converse of the Hinge Theorem 7. m C > Substitution Propert 8. C is an obtuse angle. 9. nc is an obtuse triangle. 8. Definition of an obtuse angle 9. Definition of an obtuse triangle a P b R 4. Given: In nlmn, LM is the longest side, and c 5 a b. Prove: nlmn is a right triangle. Statements L c b M a N P Reasons. In nlmn,. Given c 5 a b where c is the longest side. In npqr, R is a right angle.. a b 5. Pthagorean Theorem. c 5. Substitution Propert 4. c 5 4. propert of square roots 5. m R Definition of right angle 6. m N 5 m R 6. Converse of the Hinge Theorem 7. m N Substitution Propert 8. N is an right angle. 8. Definition of a right angle 9. nlmn is a right triangle. b 9. Definition of a right triangle 4. ecause C and DCE are vertical angles, the are congruent. lso C CD 5 C. So, nc,ndec. CE ecause 5 5 9, is a right angle. So, D is also a right angle. 44. a. (7, ) C(4, 4) (0, ) b. Slope of 4 C : m 5 4 (7) 5 Slope of : m 5 0 (7) Slope of 4 C : m 5 0 (4) ecause none of the slopes are negative reciprocals, the triangle is not a right triangle. Q a R Copright b McDougal Littell, a division of Houghton Mifflin Compan. 06 Geometr Worked-Out Solution Ke

11 Copright b McDougal Littell, a division of Houghton Mifflin Compan. Chapter 7, continued 45. c. C 5 Ï (4 (7) (4 ) 5 Ï 5 Ï (0 (7) ( ) 5 5 Ï C 5 Ï (0 (4)) ( 4) 5 5 c 0 a b (5 Ï ) 0 5 ( Ï ) Þ 8 the Converse of the Pthagorean Theorem, the triangle is not a right triangle. d. The answers in (b) and (c) are the same. h G 5 5 D E F C H 4 h ndgf is isosceles, so GDE > GFE and DGE > FGE. GD > CGF b Vertical ngles Congruent Theorem. GC > G b Congruent Supplements Theorem. So, b Similarit Postulate, ngef, ngc. h h ecause 4 5 5, ngcf is a right triangle with m GCF So, HCF > GC and nchf, ngc. h ( h) h h 5 h Using Pthagorean Theorem: h 5 h (h) 5 5h 5 9 h 5 Ï 5 5 Ï Ï Ï Ï 5 5 Mied Review nswers will var. 49. If 0 5 7, then p 5 7 p If 5 5, then 5 5. p 5 5 p If 8 5 9, then p p l 5 8w P 5 l w 5 5 (8w) w 5 5 8w w l 5 8(w) 5 8(7.5) 5 60 The length of the rectangle is 60 feet and the width is 7.5 feet. Quiz (p. 447) Ï Ï ? ? < 85 5 Ï 5 The triangle is an acute triangle. 5. 6? 0 56? > 44 The triangle is an obtuse triangle. 6. (8 Ï 6 )? (6)? > 0 The triangle is an obtuse triangle. Geometr Worked-Out Solution Ke 07

12 7. 9? 0 84? The triangle is a right triangle. 8. ( Ï 7 )? 8 7? The triangle is a right triangle. 9.? ? > 64 The triangle is an acute triangle. Lesson 7. Investigating Geometr ctivit 7. (p. 448). The two smaller triangles have different side lengths than the larger one, but the angle measurements are the same. m 5 m 5 5 m m 5 m 6 5 m 9 m 5 m 4 5 m 7. ecause m 5 m 7, m 5 m 9, and m 5 m 8, the green triangle is similar to the red triangle.. ecause m 8 5 m 5, m 9 5 m 6, and m 7 5 m 4, the red triangle is similar to the blue triangle. 4. a a 5 b b 7. Guided Practice (pp ). E G 5 G 4 F H F H nefg, ngfh, negh GH EG 5 GF EF E G. L 5 J M J K njlm, njmk, nmlk MK LM 5 MJ LJ ø 4.6. In Eample, Theorem 7.7 was used to solve for. The length of the leg is the geometric mean of the length of the hpotenuse of nrpq and the length of the segment of the hpotenuse that is adjacent to. 4. The wall is 9.5 feet high and Mar s ee level from the ground is 5.5 feet, so w feet. 4 d 5 d 5.5 (4)(5.5) 5 d 77 5 d 8.77 ø d Mar would have to stand about 8.77 feet from the wall. 7. Eercises (pp ) Skill Practice. Two triangles are similar if their corresponding angles are congruent and their corresponding side lengths are proportional.. Sample answer: When two similar triangles share a common side length, and ou write a proportion of side lengths that includes the common length, this length is the geometric mean of the other lengths in the proportion.. F H G E G E nfgh, nhge, nfhe 4. M H L K L M M K nlmk, nlnm, nmnk ø 5.7 The length of the altitude is about 5.7 feet. N M M F N 5 H L K Copright b McDougal Littell, a division of Houghton Mifflin Compan. 08 Geometr Worked-Out Solution Ke

13 Copright b McDougal Littell, a division of Houghton Mifflin Compan. Chapter 7, continued () ø. The length of the altitude is about. feet (0) ø 6.7 The length of the altitude is about 6.7 feet. X nxyz, nxzw, nzyw XW ZW 5 ZW YW Q S R S T Y Z X nqrs, nsqt, nrst QR SR 5 SQ TQ Q 0. E F E G G F G nefg, negh, ngfh EF EG 5 EG EH. The length z is the geometric mean of the length of the hpotenuse, w v, and the segment adjacent to the leg of length z, which is v, not w. v z 5 z w v. The length d is the geometric mean of the lengths of the two segments of the hpotenuse, not the two segments whose lengths are e and f. e d 5 d g (9) R T H N Z W S Z H Y W z z z ø (9) 5 5(8) ø (8 ) 8(8 ) ø 9. C; C C D D. 5 C is incorrect; it should be C 5 0. C; C C 5 C DC DC 6DC 5 8 DC So, D a (a 5) 5 8a a 5 54 a 5 b (b ) 5 6 8b b 5 b C D Ï 45 ø 6.7 Geometr Worked-Out Solution Ke 09

14 (6 9) 5 9(5) z 6 5 z z 5 (6 9) 5z 5 (5) 5z 5 00 z , 5 5, z c 0 a b ( Ï 89 ) The triangle is a right triangle. 0 h 89 6 h Ï 89 Ï 89 h 5 0(6) h 5 60 Ï 89 h ø c 0 a b (4 Ï ) The triangle is a right triangle. 4 h 8 h Ï 4 Ï h 5 8() 96 h 5 4 Ï h ø c 0 a b (4 Ï ) Þ 50 The triangle is not a right triangle. 7. C 5 C C C C 5 65 C 5 5 Using Theorem 7.7: C C 5 C DC DC 5(DC) 5 5 DC So, D Using Theorem 7.6: DC D 5 D D 9 D 5 D 6 9 D 5 D 6 9(6) 5 (D) 44 5 (D) 5 D 8. The two smaller triangles are congruent isosceles triangles. ecause the are similar to the larger triangle, and the larger triangle is isosceles, the smaller triangles must also be isosceles. Copright b McDougal Littell, a division of Houghton Mifflin Compan. 0 Geometr Worked-Out Solution Ke

15 Copright b McDougal Littell, a division of Houghton Mifflin Compan. Problem Solving 9. c 5 a b 0.. c c c c ø (.5) 5.5. ø. The height of the roof is about. feet. 7. ft h 5.5 ft h h 5 7. h ø 9.4 Height of monument 5 h 5.5 ø The monument is about 5 feet high. 6 ft Find h. 9.5 ft h 5 6 h Ï Ï h 5 ( Ï 54.5 ) 6h h ø Ï Height of monument 5 h 6 ø The monument is about 5 feet high. This is the same answer as in Eercise 0. The similar triangles on Paul s side of the monument are different than those on our side, but the hpotenuses of the large triangles, which represent the monument s height, are the same.. Given: In nc, C is a right angle. f Prove: c 5 a b D c b C a e Statements. Draw nc. C is a right angle.. Draw a perpendicular from C to.. c a 5 a e and c b 5 b f. Given Reasons. Perpendicular Postulate. Geometric Mean (Leg) Theorem 4. ce 5 a and cf 5 b 4. Cross Product Propert 5. ce b 5 a b 5. ddition Propert of Equalit 6. ce cf 5 a b 6. Substitution Propert of Equalit 7. c(e f ) 5 a b 7. Factoring 8. e f 5 c 8. Segment ddition Postulate 9. c p c 5 a b 9. Substitution Propert of Equalit 0. c 5 a b 0. Simplif.. a. Ever triangle has three altitudes. Triangle EGF has the altitude FH which is perpendicular to EG and crosses through verte F. It also has the altitudes EF and FG which lie on the triangle. oth of these segments cross through a verte and are perpendicular to the side opposite the verte. HG b. FH 5 FH EH 7 FH 5 FH FH Ï 5 5 FH 4. a. Q c. rea 5 (base)(height) S 5 (5 7)( Ï 5 ) ø 5.5 R S R Q S T fter sketching the three triangles so that the corresponding angles and sides have the same orientation, notice which segments are the short legs, long legs, and hpotenuses of the triangles and label them accordingl. b. nqrs, nsrt, nqst c. Segment RS is the geometric mean of RT and RQ, because RT RS 5 RS RQ. T Geometr Worked-Out Solution Ke

16 5. Prove: ncd, nc, ncd, nc, and ncd, ncd Statements. nc is a right triangle. C is a right angle, and CD.. CD is a right angle.. C > CD. Given Reasons. Definition of perpendicular lines. Right ngles Congruence Theorem 4. > 4. Refleive Propert of Congruence 5. ncd, nc 5. Similarit Postulate 6. > 6. Refleive Propert of Congruence 7. ncd, nc 7. Similarit Postulate 8. CD and are complementar. 9. and are complementar. 8. Corollar to the triangle Sum Theorem 9. Corollar to the triangle Sum Theorem 0. CD > 0. Congruent Complements Theorem. CD > CD. Right ngles Congruence Theorem. ncd, ncd. Similarit Postulate 6. Prove: D CD 5 CD D Statements. nc is a right triangle.. Given. CD. Given Reasons. nc, ncd, ncd. Theorem D CD 5 CD D 7. Prove: C 5 C D, C 5 C D Statements. nc is a right triangle.. Given. CD. Given 4. Definition of similar triangles Reasons. nc, ncd. Theorem C 5 C D 4. Definition of similar triangles 5. nc, ncd 5. Theorem C 5 C D 6. Definition of similar triangles 8. a. a 5 0, b 5 5 ab a b 5 (0)(5) (0 5) The harmonic mean of 0 and 5 is. b. a 5 6, b 5 4 ab a b 5 (6)(4) The harmonic mean of 6 and 4 is 8.4. c. Strings whose lengths have the ratio 4 : 6 : will have lengths 4k, 6k, and k, for an constant k. The strings will sound harmonious if 6k 5 (4k)(k) 4k k. (4k)(k) 4k k 5 96k 6k 5 6k So, 6k is the harmonic mean of 4k and k, and the strings will sound harmonious. Mied Review 9. Ï 7 p Ï 5 Ï 54 5 Ï Ï 8 p Ï 0 5 Ï Ï 5 4. Ï p Ï 7 5 Ï 84 5 Ï 4. Ï 8 p Ï 5 Ï Ï Ï 7 p Ï 7 Ï Ï Ï p Ï Ï 5 8 Ï Ï 7 p Ï 7 Ï Ï Ï Ï 46. Ï 4 p Ï 4 Ï 4 5 Ï Ï Ï Line : (, 4), (4, ) m Line : (, 5), (, ) m m p m 5 ()() 5 The lines are perpendicular because the product of their slopes is. 48. Line : (0, ), (, ) m Line : (, ), (, 5) m The lines are parallel because the have the same slope. Copright b McDougal Littell, a division of Houghton Mifflin Compan. Geometr Worked-Out Solution Ke

17 Copright b McDougal Littell, a division of Houghton Mifflin Compan. 49. Line : (, 7), (4, 7) m Line : (5, ), (7, 4) m The lines are neither perpendicular nor parallel, because the product of their slopes is not, and their slopes are not equal. Lesson Guided Practice (pp ). the ase ngles Theorem and the Corollar to the Triangle Sum Theorem, the triangle is a triangle. hpotenuse 5 leg p Ï Ï 5 p Ï 5. the ase ngles Theorem and the Corollar to the Triangle Sum Theorem, the triangle is triangle. hpotenuse 5 leg p Ï 5 Ï p Ï 5 Ï 4 5. the ase ngles Theorem and the Corollar to the Triangle Sum Theorem, the two triangles are triangles. hpotenuse 5 leg p Ï d 5 8 Ï ø. 4. hpotenuse 5 leg p Ï 6 5 p Ï 6 Ï 5 6 Ï p Ï Ï 5 Ï 5 The length of the hpotenuse is Ï. 5. longer leg 5 shorter leg p Ï 5 Ï p Ï 5 6. The equilateral triangle has an altitude that forms the longer leg of two triangles. longer leg 5 shorter leg p Ï h 5 Ï ø h 4 ft 08 hpotenuse 5 p shorter leg 4 5 p h 7 5 h The height of the bod of the dump truck is 7 feet. 8. In a triangle, the shorter side of the triangle is opposite the 08 angle. The longer side of the triangle is opposite the 608 angle. 7.4 Eercises (pp ) Skill Practice. triangle with two congruent sides and a right angle is called an isosceles right triangle.. The Corollar to the Triangle Sum Theorem requires that the acute angles of a right triangle are complementar. ecause the triangle is isosceles, its base angles are congruent. Half of 908 is 458, so each of the acute angles measures hpotenuse 5 leg p Ï 5 7 Ï 4. hpotenuse 5 leg p Ï 5. hpotenuse 5 leg p Ï 5 5 Ï p Ï Ï 5 p Ï 5 5 p C; hpotenuse 5 leg p Ï 7 5 C p Ï 7 Ï 5 C 7 Ï 5 C 7. hpotenuse 5 leg p Ï Ï 5 Ï 5 The corner triangles have leg lengths of inches. The overall side length of the tile is p 5 4 inches. 8. hpotenuse 5 p shorter leg 5 p longer leg 5 shorter leg p Ï 5 9 Ï 9. hpotenuse 5 p shorter leg 5 longer leg 5 shorter leg p Ï Ï 5 Ï 5 5 () 5 6 Geometr Worked-Out Solution Ke

18 0. hpotenuse 5 p shorter leg Ï 5 6 Ï 5 longer leg 5 shorter leg p Ï 5 6 Ï p Ï 5 6() 5 8. a 5 b hpotenuse 5 leg p Ï c 5 a Ï a 7 0 Ï p Ï Ï 5 5 Ï 6 Ï 5 b 7 5 Ï 6 Ï 5 c 7 Ï Ï 0 6 Ï Ï 5 p Ï 5 Ï 0. hpotenuse 5 p shorter leg 5. hpotenuse 5 p shorter leg 4 5 p 5 p longer leg 5 shorter leg p Ï q 5 p p Ï q 5 Ï 6. The altitude s forms the longer leg of two triangles. hpotenuse 5 p shorter leg r 5 8 r 5 8 longer leg 5 shorter leg p Ï s 5 9 Ï t 4 f 5 p d d 5 f longer leg 5 shorter leg p Ï e 5 d p Ï d 5 e Ï p Ï Ï 5 e Ï d Ï p Ï 5 8 e 5 Ï 7 Ï 8 Ï f p p d 8 Ï 5 9 Ï Ï 5 4 Ï e 9 Ï p Ï 5 7 f 8 Ï p 4 Ï 5 8 Ï. hpotenuse 5 p shorter leg longer leg 5 shorter leg p Ï 5 5 p Ï 5 5 Ï 4. hpotenuse 5 leg p Ï Ï 6 5 m Ï Ï 6 Ï 5 m Ï 5 m n 5 m 5 Ï u The triangle formed is a triangle, so each leg has a length of 4. u hpotenuse 5 leg p Ï t 5 4 Ï 8. The upper triangle is a triangle. longer leg 5 shorter leg p Ï 9 Ï 5 e Ï 9 5 e hpotenuse 5 p shorter leg f 5 (9) f 5 8 The lower triangle is a triangle. hpotenuse 5 leg p Ï f 5 g Ï 8 5 g Ï 8 Ï 5 g 8 Ï p Ï Ï 5 g 9 Ï 5 g 9. C; 5, 5 Ï, 0 hpotenuse 0 p shorter leg 0 0 p 5 0 Þ 5 Copright b McDougal Littell, a division of Houghton Mifflin Compan. 4 Geometr Worked-Out Solution Ke

19 Copright b McDougal Littell, a division of Houghton Mifflin Compan. 0. The formula for the longer leg was used instead of the hpotenuse formula. The correct solution is: hpotenuse 5 p shorter leg 5 p The length of the hpotenuse was incorrectl calculated as leg p leg p Ï 5 Ï 5 p Ï 5 p Ï 5 5 Ï. The correct solution is: hpotenuse 5 leg p Ï 5 Ï 5 p Ï 5 Ï bigail s method does work. longer leg 5 shorter leg p Ï Ï 9 Ï 5 Ï p Ï 9 Ï 5 p Ï 5 Her method is algebraicall correct, so the equation simplifies to the same answer found in Eample 5. g f longer leg 5 shorter leg p Ï 0 5 g Ï 0 Ï p Ï Ï 5 g 0 Ï 5 g hpotenuse 5 p shorter leg f 5 0 Ï f 5 0 Ï hpotenuse 5 p shorter leg 4 Ï 5 Ï 5 longer leg 5 shorter leg p Ï 5 p Ï 5 Ï p Ï 5 Ï First triangle: hpotenuse 5 p shorter leg Second triangle: longer leg 5 shorter leg p Ï 5 Ï 4 5 Ï 4 Ï 5 4 Ï 5 6. C 5 Ï ( ()) ( ()) 5 Ï Ï hpotenuse 5 p shorter leg C longer leg 5 shorter leg p Ï C 5 p Ï C 5 Ï Let (, ) represent the coordinates of. 5 Ï ( ) ( ()) 5 Ï ( ) ( ) 9 5 ( ) ( ) 9 ( ) 5 ( ) C 5 Ï ( ()) ( ()) Ï 5 Ï ( ) ( ) 7 5 ( ) ( ) 7 ( ) 5 ( ) 9 ( ) 5 7 ( ) Geometr Worked-Out Solution Ke 5

20 9 5 ( ) Ï 8 4(4)() (4) Ï Ï Point lies in the first quadrant, so its -coordinate is positive., Ï Problem Solving 7. hpotenuse 5 p shorter leg 5 h 5 h h The height of the ramp is 5 feet 6 inches h ft 458 h ft ft ft 458 h ft ft hpotenuse 5 p shorter leg 84 5 p h 4 5 h hpotenuse 5 leg p Ï 84 5 h p Ï 84 Ï 5 h 4 Ï 5 h hpotenuse 5 p shorter leg 84 5 p s 4 5 s longer leg 5 shorter leg p Ï h 5 4 Ï h ø 46 When the angle is 08, the seagull rises 4 feet; when the angle is 458, the seagull rises about 00 feet 0 inches; when the angle is 608, the seagull rises about 46 feet. 9. You could show that all isosceles right triangles are similar to each other b showing that the corresponding angles are congruent. The are all triangles. You could also show that the corresponding side lengths are alwas proportional, because in an isosceles right triangle, the side lengths are alwas,, and Ï. For eample, let,, Ï, and,, Ï, be the side lengths of two isosceles right triangles. The ratios of corresponding side lengths are 5, 5, and Ï Ï ecause ndef is a triangle, DF 5 FE. the Pthagorean Theorem, DF FE 5 DE. This equation can be written as DF 5 DE or FE 5 DE. the Propert of Squares, Ï DF 5 DE and Ï FE 5 DE. So, the hpotenuse is equal to Ï times the length of one leg of the triangle.. 0 in in. 08 h h 0 in in in. The height h divides the equilateral triangle into two triangles. longer leg 5 shorter leg p Ï h 5 0 Ï ø 7. The height of the equilateral triangle is about 7. inches.. K L M Construct njml congruent to njkl. ecause the are congruent, m KJL 5 m LJM 5 08, so m KJM 5 m KJL m LJM ecause njml > njkl, m M 5 m K So, all three angles of njkm measure 608, and the triangle is equilateral. So, all of its side lengths are 5. It is given that the shorter leg of njkl is. So, njkl s hpotenuse is two times the length of its shorter side. the Pthagorean Theorem, JL 5 (), and JL This simplifies to JL 5 Ï, showing that the longer leg is equal to Ï times the length of the shorter leg.. a. The large orange triangles are triangles, because the are isosceles right triangles. The smaller blue triangles are also triangles, because the 458 angles of the orange triangle are complementar with the acute angles of the blue triangle. b. hpotenuse 5 leg p Ï p Ï Ï Ï p Ï Ï 5 Ï 5. ø The square of fabric for the large orange triangles should be about. inches b. inches. Copright b McDougal Littell, a division of Houghton Mifflin Compan. 6 Geometr Worked-Out Solution Ke

21 Copright b McDougal Littell, a division of Houghton Mifflin Compan. c hpotenuse 5 leg p Ï Ï 5 p Ï The square of fabric for the small blue triangles should be.5 inches b.5 inches. 4. a. The first triangle is a triangle, because the legs are both. 5. a. hpotenuse 5 leg p Ï r 5 p Ï r 5 Ï The second triangle is a right triangle with legs and r 5 Ï. Using the Pthagorean Theorem, s 5 r s 5 ( Ï ) s 5 s 5 s 5 Ï. This process can be duplicated to find the unknown lengths of the remaining right triangles. t 5 s u 5 t t 5 ( Ï ) u 5 t 5 u 5 5 t 5 4 u 5 Ï 5 t 5 v 5 u w 5 v v 5 ( Ï 5 ) w 5 ( Ï 6 ) v 5 5 w 5 6 v 5 6 w 5 7 v 5 Ï 6 w 5 Ï 7 b. The onl triangle that is a triangle is the first one, because it is the onl one whose legs are equal in length. c. The third triangle, with sides, Ï,, is the onl triangle. It is the onl one whose side lengths satisf the Triangle Theorem. T 8 8 Z Q S 608 R b. QR > SR, QT > ST, and TR is shared b the two triangles so nrqt > nrst b the Side-Side-Side Congruence Postulate. c. RT is longer than QS. ZR is the longer leg of a triangle, so ZR > ZS. From part (b), nrqt > nrst, so QTR > STR. ecause m QTS 5 908, m QTR 5 m STR ecause nqts is an isosceles right triangle, its base angles measure 458. So, nqzt and ntzs are congruent right isosceles triangles, and QZ 5 TZ 5 ZS. So, ZR TZ > ZS QZ, and RT > QS. Mied Review 6. D is a perpendicular bisector of C, 5 C and D 5 DC C CD 5 D , 4, and 7 5 (5) > 7 > 4 > 4 The side lengths can form a triangle. 40.,, and 9 Ï ò 9 Ï The side lengths cannot form a triangle. 4. 7, 5, and > 8 > 5 6 > 7 The side lengths can form a triangle. 4.,, and 5 Ï 7 (5 Ï 7 ) The side lengths form a right triangle. 4.,, and The side lengths form a right triangle , 0, and Þ 64 The side lengths do not form a right triangle. Geometr Worked-Out Solution Ke 7

22 Quiz (p. 464) Mied Review of Problem Solving (p. 465). 0 C C C 5 C CD D 6 C D 0. W 45 N 4 Treasure (4, 45) (0, 0) (4, 0) Stump E (0, 0) (0, 0) C is the geometric mean of C and CD.. C 5 CD D D D 64 5 D 8 5 D D D 5 D CD D p D 5 8 p 8 D D 5 ø 0.67 D 5 C CD a.. S If the stump is at point (0, 0), the hidden treasure is at point (4, 45). The distance to the treasure from the stump is Ï Ï paces. Jefferson 66 mi Little Rock 546 mi 459 mi tlanta b. 546? ,6? 70,756 0,68 98,6 > 8,47 The pushpins do not form a right triangle. The form an obtuse scalene triangle, because none of the side lengths are equal, and the square of the longest side is greater than the sum of the squares of the other two sides. 6 mi 6 p 5 8 p 0 6 p ø.. hpotenuse 5 leg p Ï 5 8 Ï 4. hpotenuse 5 Ï 0 5 p Ï 0 Ï 5 0 Ï 5 5 Ï 5 5. longer leg 5 shorter leg p Ï Ï 5 a Ï Ï Ï 5 a Ï Ï p Ï Ï 5 a Ï 6 5 a hpotenuse 5 p shorter leg b 5 a b 5 Ï mi W N S E ob s distance 5 (4mi/h)(.5 h) 5 6 mi John s distance 5 (5 mi/h)(.5 h) mi ecause ob ran east and John ran south, their paths are legs of a right triangle. The distance between them at :0.M. was Ï Ï 9.5 ø 9.6 miles > 6 > 8 8 > 6 4 > > > So, the figure is a triangle when < < 4. a Using the Pthagorean Theorem, the triangle is a right triangle when 5 0. Copright b McDougal Littell, a division of Houghton Mifflin Compan. 8 Geometr Worked-Out Solution Ke

23 Copright b McDougal Littell, a division of Houghton Mifflin Compan. b. > 6 8 > 6 64 > 00 > 0 When > 0, can be obtuse, because the square of the length of the longest side is greater than the sum of the squares of the lengths of the other two sides. ut to satisf the Triangle Inequalit Theorem, < < 4. So, is obtuse when 0 < < 4. c. < 6 8 < 6 64 < 00 < 0 When < 0, can be acute, because the square of the length of the longest side of the triangle is less than the sum of the squares of the lengths of the other two sides. ut to satisf the Triangle Inequalit Theorem, < < 4. So, is acute when < < 0. d. The triangle is isosceles when 5 6 or 5 8, because in either case, two of the sides of the triangle are congruent. e. No triangle is possible when or 4, because the values would not satisf the Triangle Inequalit Theorem. 5. a. The purple triangle has ten marble holes. It is an equilateral triangle, because each side of the triangle has four equall-spaced marble holes. b. 8 cm h 8 cm cm The height h forms the longer leg of two triangles. longer leg 5 shorter leg p Ï h 5 8 Ï 5 4 Ï 5 (base)(height) 5 (8)(4 Ï ) 5 6 Ï ø 7.7 The area of the purple triangle is about 7.7 square centimeters. c. There are 6 marble holes in the center heagon. The purple triangle has about the area of the 6 center heagon. area of center heagon # marble holes in heagon d. 5 area of purple triangle # marble holes in triangle (7.7) Using marble holes as an estimate, the area of the heagon is about 69 square centimeters. 6. a Ï ø.6 The length of the plwood is about feet 7 inches. b. ft ft ft ft ft 5 Ï Ï 5 p 6 5 Ï 5 6 Ï ø.66 The length of the support is about foot 8 inches. c. ft a ft ft b Using the Geometric Mean (Leg) Theorem: Ï 5 a Ï a a 5 Ï 5 4 Ï ø. So, the support attaches about. feet up from the bottom of the plwood. Lesson 7.5 ctivit (p. 466) Step Check student s work. Step Sample answer: Triangle djacent Leg Opposite Leg Opposite Leg djacent Leg nc 5 cm.9 cm 0.58 nde 0 cm 5.8 cm 0.58 nfg 5 cm 8.7 cm 0.58 Step The proportions C DE 5 C E and C C 5 DE are true because E nc, nde. The two triangles are similar because their corresponding angles are congruent. Geometr Worked-Out Solution Ke 9

24 Step 4 Sample answer: The ratio of the lengths of the legs in a right triangle is constant for a given angle measure; answers will var. 7.5 Guided Practice (pp ) opp. J. tan J 5 adj. to J opp. K tan K 5 adj. to K ø. opp. J. tan J 5 adj. to J ø 0.5 opp. K tan K 5 adj. to K tan 68 5 opp. adj. tan 68 5 p tan tan 68 ø.8040 ø. 4. tan opp. adj. tan p tan (.486) ø 9. ø longer leg 5 shorter leg p Ï 5 5 Ï tan opp. adj. 5 5 Ï 5 5 Ï 7.5 Eercises (pp ) Skill Practice. The tangent ratio compares the length of the leg opposite the angle to the length of the leg adjacent to the angle.. ll right triangles with an acute angle measuring n8 will have the same ratio of the leg opposite the angle to the leg adjacent to the angle because tangent n8 is a constant. So, the triangles must be similar.. tan ø tan 5 5 ø.967 tan ø 0.97 tan 5 5 ø tan tan 48 5 p tan tan 48 ø ø.8 7. tan p tan 78 ø 5 p ø 7.6 ø 8. tan p tan tan 588 ø.600 ø.7 9. tan t p tan p Using the Theorem: hpotenuse 5 leg p Ï 6 Ï 5 p Ï 6 5 The results are the same using either method. 0. tan Ï 0 Ï p tan Ï p ø 0 ø Using the Theorem: longer leg 5 shorter leg p Ï 0 Ï 5 Ï 0 5 The results are the same using either method.. tan p tan p.7 ø 6.9 ø Using the Theorem: longer leg 5 shorter leg p Ï 5 4 Ï ø 6.9 The results are the same using either method. Copright b McDougal Littell, a division of Houghton Mifflin Compan. tan ø Geometr Worked-Out Solution Ke

25 Copright b McDougal Littell, a division of Houghton Mifflin Compan.. tan 08 5 opp. adj. shorter leg tan 08 5 longer leg shorter leg tan 08 5 shorter leg p Ï 5 Ï 5 Ï ø tan opp. adj. tan leg leg 5. The tangent was incorrectl written as the ratio of the adjacent leg, 8, to the hpotenuse, 8. The tangent of D is the ratio of the length of the opposite side to the length of the adjacent side. It should be tan D The error is that the triangle is not a right triangle. If it were, the other two angles would be complementar, but the are not. So, using the tangent ratio is not possible. 5. In order to use the tangent ratio, ou must know that the triangle is a right triangle, and ou must know either the measurement of the acute angle and one of the legs, or the lengths of the two legs. 6. C; tan p tan tan C; tan 8 5 p tan 8 5 p ø 7.5 ø 8. tan p tan tan ø 7. The other acute angle is tan tan ø.5 9. tan p tan tan 408 ø 0.89 ø 5.5 The other acute angle is tan tan ø.9 0. tan p tan p.445 ø 9. ø The other acute angle is tan 58 5 a a tan tan 88 5 p tan tan 88 ø 0.78 ø 4. rea 5 (base)(height) 5 ()(4.) ø tan p tan tan ø.48 ø. rea 5 (base)(height) ø (.)(6) ø 89.6 Geometr Worked-Out Solution Ke

26 . tan p tan tan 8 7 ø ø The first triangle is a triangle. hpotenuse 5 p shorter leg tan z rea 5 (base)(height) tan z 5 (7.)(7) ø 60.6 z p tan z 5 tan 408 ø ø tan p tan p (0.9657) ø 8 ø ø 8 9 ø 65 ø Perimeter ø tan p tan ø 6 0 ø 6 ø tan 8 6 ø ø 0 Perimeter ø tan p tan tan 688 ø 5 6. ø 6. ø 6. 5 ø.475 ø 6. Perimeter ø p tan 48 5 z p tan 48 5 z 60 p ø z 54.0 ø z tan p tan tan 88 5 z tan z z p tan E 508 D 8 z 5 tan 88 ø ø G C tan p tan ø tan ED ED p tan ED 5 tan 508 ø 0.9 E 5 ED D E ø (0.9) E ø 87.8 E ø 7.0 C 5 C C ø 6 8. C ø C ø.7 F H Copright b McDougal Littell, a division of Houghton Mifflin Compan. Geometr Worked-Out Solution Ke

27 Copright b McDougal Littell, a division of Houghton Mifflin Compan. m GF 5 m G tan FG p tan FG 8.6 ø FG G 5 F FG G ø 8.6 G ø G ø.7 CG 5 C G ø.7.7 ø 9 the S Congruence Postulate, nfg > nfh. So, FH 5 FG ø 8.6. ecause nfg > nfh, H 5 G ø.7. Perimeter 5 E ED DC CG GF FH H ø ø 8 Problem Solving. tan h 8 8 p tan h 555 ø h The height of the Washington Monument is about 555 feet.. tan 58 5 h p tan 58 ø h 55 ø h h 8 ft 788 The height of the roller coaster is about 55 feet.. tan p tan ø The distance from one end of the class to the other is about 6.7 p 5.4 feet. 4. a. tan tan ø The last student is about 6.7 feet from the center of the row. b. tan p tan ø The end of the camera range is about 4. feet from the center of the row. c. The length of the empt space is about feet. d. Number of students 5 Length of empt space feet 4 feet 5.75 Space needed 4 per student ecause each student needs feet of space, no more than students can fit into the etra 7 feet of space. 5. tan 5 a b ; tan 5 b a The tangent of one of the acute angles is the inverse of the tangent of the other acute angle. and are complementar angles. 6. tan 8 5 h 0 0 p tan 8 5 h 0.49 ø h in. (0.49 ft) ft ø 4 in. The height of the E is about 4 inches. 7. a. Maimum height: 0 in. 5.5 ft tan l l p tan l 5 tan 58 ø 9 The maimum horizontal length of one ramp is about 9 feet. b. Total height 5 Number of ramps.5 feet 7.5 ft 5 r p.5 ft r 5 r The least number of ramps needed is, with landings. 7.5 ft 58 9 ft 9 ft 5 ft 5 ft 9 ft c. Total length of base The total length of the base of the sstem of ramps and landings is about 97 feet Geometr Worked-Out Solution Ke

28 8. C 5 πr 80 5 πr 80 π 5 r.7 ø r The radius of the base of the cone is about.7 feet. h s 8.7 h tan 8 ø.7.7 tan 8 ø h 7.9 ø h The height of the cone is about 7.9 feet. s 5 h r s ø s ø.7 s ø 5 The length s of the cone-shaped pile is about 5 feet. Mied Review 9. m m m C m m p m C p The triangle is acute, because all angles are acute. 40. m m m C (5 60) ø 4.8 m 5 8 ø 4.8 m 5 8 ø 4.8 m C 5 (5 60)8 5 5 p ø.48 The triangle is obtuse because m C > m m m C ( 0)8 ( 5)8 ( 0) m 5 ( 0)8 5 (5 0) m 5 ( 5)8 5 ( p 5 5) m C 5 ( 0) The triangle is obtuse because m > m < m The length of the side of the triangle opposite is less than the length of the side opposite and the other sides are congruent. So, b the Converse of the Hinge Theorem, m < m. 4. m 5 m The two triangles are congruent b the SSS Congruence Postulate, so >. 44. m > m Two sides of the two triangles are congruent, and the side of the triangle opposite is greater than the side of the triangle opposite. So, b the Converse of the Hinge Theorem, m > m ( Ï 57 ) ( Ï ) Lesson Guided Practice (pp ) opp. X. sin X 5 5 ZY hp. XY ø opp. Y sin Y 5 5 XZ hp. XY ø opp. X. sin X 5 5 YZ hp. XY opp. Y sin Y 5 5 XZ hp. XY RS 5 RT TS RS 5 9 RS 5 5 RS 5 5 adj. to R cos R 5 hp. adj. to S cos S 5 hp. 5 RT RS TS RS Copright b McDougal Littell, a division of Houghton Mifflin Compan. 4 Geometr Worked-Out Solution Ke

29 Copright b McDougal Littell, a division of Houghton Mifflin Compan. 4. SR 5 ST TR SR SR 5 56 SR 5 4 adj. to R cos R 5 hp. adj. to S cos S 5 hp. 5. cos 58 5 adj. hp. cos 58 ø adj p cos 58 ø adj. 9. p 0.89 ø adj. 5.7 ø adj. 5 TR SR ø ST SR ø The length of the other leg of the triangle formed is about 5.7 feet. 6. sin 88 5 opp. hp. sin p sin sin 88 ø ø You would ski about 556 meters down the mountain. 7. sin 58 5 opp. hp. sin p sin ø The height of the ramp is about 8 feet. cos 58 5 adj. hp. cos p cos ø The length of the base is about.5 feet. 8. Using the Triangle Theorem, the side lengths are, Ï, and sin 08 5 opp. hp cos 08 5 adj hp. 5 Ï ø Eercises (pp ) Skill Practice. The sine ratio compares the length of the opposite side to the length of the hpotenuse.. The side of a right triangle that is adjacent to an acute angle is the leg that is net to the angle. The hpotenuse is the longest side of the right triangle, and is the side opposite the right angle.. sin D sin E sin D sin E 5 7 ø sin D ø 0.58 ø 0.4 sin E ø The student used the adjacent side over the hpotenuse rather than the opposite side over the hpotenuse. The correct statement for the sine of the angle is sin cos X cos Y cos X ø cos Y ø cos X cos Y 5 Ï 6 5 Ï ø sin p sin ø cos p cos ø. cos a a p cos a 5 cos 488 ø 4.9 sin b a a p sin b (4.9) p sin 488 ø b. ø b Geometr Worked-Out Solution Ke 5