Answers to Exercises. c 2 2ab b 2 2ab a 2 c 2 a 2 b 2
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1 Answers to Eercises CHAPTER 9 CHAPTER LESSON 9. CHAPTER 9 CHAPTER. c 9. cm. cm. b 5. cm. d 0 cm 5. s cm. c 8.5 cm 7. b cm 8.. cm 9. 0 cm 0. s.5 cm. r cm. 7 ft. 5 m.. cm 5.,, 5. 8 m 7. The re of the lrge squre is re of tringle re of smll squre. c b (b ) c b b b c b b b c 9 8. Smple nswer: Yes, ABC XYZ b SSS. Both tringles re right tringles, so ou cn use the Pthgoren Theorem to find tht CB ZY cm cm Mrk the unnmed ngles s shown in the figure below. B the Liner Pir Conjecture, p 0 80, so p 0. B AIA, m q. B the Tringle Sum Conjecture, q p n 80. Substitute m q nd p 0 to get m 0 n 80. m n 0. m 0 p q n Or use the Eterior Angle Conjecture to get q n 0. B AIA, q m. Substituting, m n 0.., b 7, c 0, d, e 90, f 7, g 7, h 7, n 7, r, s 7, t 7, u, v 7. Possible eplntion: B the Tngent Conjecture, the qudrilterl contining g hs two right ngles formed b rdii intersecting tngent lines. Using c 0 nd the Qudrilterl Sum Conjecture, g , so g 7. The ngle mesures e nd f nd the mesure of the inscribed ngle tht intercepts the rc with mesure u sum to 80. Using e 90 nd f 7, the inscribed ngle mesures. Using the Inscribed Angle Conjecture, u.. T T Answers to Eercises ANSWERS TO EXERCISES 05
2 Answers to Eercises LESSON 9.. es. es. no. no 5. no. no 7. no 8. No, the given lengths re not Pthgoren triple cm 0. units. 7. m., 8, 0. 0 cm.. ft cm. 0 m 7. cm 7b. 7.8 cm 8. Smple nswer: The numbers given stisf the Pthgoren Theorem, so the tringle is right tringle; but the right ngle should be inscribed in n rc of 80. Thus the tringle is not right tringle. 9. Smple nswer: (BD) 7; (BC) (BD) 9 08; then (AB) (BC) (AC) ( 08 ), so ABC is right tringle b the Converse of the Pthgoren Theorem. 0. centroid. 0. Becuse mdcf 90, mdce (90 ). Becuse DCE is isosceles, mdec (90 ). md 80 (90 ).Becuse D is centrl ngle,.therefore,.. The pth from C to M to T lies on stright line nd therefore must be shorter thn the pth from C to A to T. P C. 790 squre units A U M T B 0 ANSWERS TO EXERCISES
3 USING YOUR ALGEBRA SKILLS Answers will vr. Possible nswer: The length of the hpotenuse of n isosceles right tringle with legs of length units is 8 units. This is the sme s,or. 7. The hpotenuse represents 5 units. In the second right tringle, the legs hve lengths nd, so the hpotenuse hs length 0.The length of the hpotenuse is twice the length of the hpotenuse of the smller tringle, so Possible nswer: A right tringle with legs of lengths nd units hs hpotenuse of length 0 units. A right tringle with legs of lengths nd units hs hpotenuse of length Possible nswer: A right tringle with legs of lengths nd units hs hpotenuse of length units. 0 0.,. Becuse is twice s long s,, so Answers to Eercises ANSWERS TO EXERCISES 07
4 Answers to Eercises. 7 cm. cm. 0 cm, 5 cm. 0 cm, 0 cm 5. cm, 7 cm. 7 cm 7. cm cm, 00 cm 9. cm 0., LESSON 9.. A tringle must hve sides whose lengths re multiples of,, nd. The tringle shown does not reflect this rule.. Possible nswer: Use nd possible nswer: 8. Possible nswer: Use 5 nd 5.. c Strt with the Pthgoren Theorem. c Combine like terms. c Tke the squre root of both sides m 9.7 m m Construct n isosceles right tringle with legs of length, construct tringle with legs of lengths nd,nd construct right tringle with legs of lengths nd Ares:.5 cm,8 cm,.5 cm ; tht is, the sum of the res of the semicircles on the two legs is equl to the re of the semicircle on the hpotenuse.. 7. chih. Etend the rs tht form the right ngle. m m5 80 b the Liner Pir Conjecture, nd it s given tht m5 90. m 90. m m m m m m m 90. m m b AIA. m m CDA, AEC, AEB, BFA, BFC 5b. MDB, MEB, MEC, MFC, MFA ANSWERS TO EXERCISES
5 LESSON 9.. No. The spce digonl of the bo is bout.5 in.. 0 m. 50 km/h. 8 ft 5. re: 0 m ; cost: $700. surfce re of prism 7 80 cm.8 cm ; surfce re of clinder 78 cm 5.0 cm 7. cm 0.8 cm ft;. lb 9. 0 ft-lb 9b. 0 lb 9c. 0 lb 0. bout. ft... units. 8 cm 5. Drw rdii CD nd CB. ABC ADC 90. For qudrilterl ABCD, 5 90 mc 90 0, so mc. BD but 0.., 7. SAA 8. orthocenter PO Answers to Eercises ANSWERS TO EXERCISES 09
6 LESSON units. 5 units. units. 5 m units. isosceles 7. rectngle prllelogrm H D A C B 8 0 E F b. Circle A: ( ) ( ) ; Circle B: ( ) c. ( h) ( k) (r). ( ) 5. Center is (0, ), r 9.. ( ) ( ) 8 5. units 5b. 7 units 5c. ( ) ( ) z (z ). 8.5 units 7. units 8. n (n ) (n ) (n ) k, m.,, Answers to Eercises 9. kite K 8 J 8 0 L 0. squre I G. 9 cm. The ngle of rottion is pproimtel 77. Connect two pirs of corresponding points. Construct the perpendiculr bisector of ech segment. The point where the perpendiculr bisectors meet is the center of rottion.. An long digonl of regulr hegon divides it into two congruent qudrilterls. Ech ngle of regulr hegon is ( ) 80 0, nd the digonl divides two of the 0 ngles into 0 ngles. Look t the digonl s trnsversl. The lternte interior ngles re congruent, thus the opposite sides of regulr hegon re prllel. 8 P M O 0 0 N (0, ) (, ) (, ) Circle A (, ) Circle B 0 ANSWERS TO EXERCISES
7 LESSON cm 5.5 cm. (8 ) m 9. m. 5 cm cm. 0 cm 77.0 cm 5. m 5. m. (5 8) cm 0.5 cm 7. cm 9. cm 8. (0 8) m 8.5 m 9. 0 cm 0. Possible proof: Given: Circle C with tngents AM nd AN Show: AM AN M. cm. cm cm 8.8 cm. 7 cm 7. Inscribed circle: cm. Circumscribed circle: cm. The re of the circumscribed circle is four times s gret s the re of the inscribed circle ( ) ( ) 0. The dimeter is the trnsversl, nd the chords re prllel b the Converse of the Prllel Lines Conjecture. The chords re congruent becuse the cn be shown to be the sme distnce from the center. (Drw perpendiculr from ech chord to the center nd use AAS nd CPCTC.) Smple construction: C N Add MC,NC nd AC to the digrm s shown. B the Tngent Conjecture, M nd N re right ngles, so AMC nd ANC re right tringles. Using the Pthgoren Theorem, (AM) (MC) (AC) nd (AN) (NC) (AC).Solvingfor AM nd AN, AM (AC) (MC) nd AN (AC).BecuseMC (NC) nd NC re rdii of the sme circle, the hve the sme lengths. So, substitute MC for NC in the second eqution: AN (AC) (MC) AM.Therefore, AM AN.. 8 m. cm 08 cm. 9 cm A. Becuse crpenter s squre hs right ngle nd both rdii re perpendiculr to the tngents, squre is formed. The rdius is 0 in., therefore the dimeter is 0 in. 0 in. d 0 in. 0 in. 0 in. 0 in. b. Possible nswer: Mesure the circumference with string nd divide b.. 5 Answers to Eercises ANSWERS TO EXERCISES
8 Answers to Eercises CHAPTER 9 REVIEW. 0 cm. 0 cm. obtuse. cm 5.,., cm. cm 8. d cm 7.0 cm 9. cm 0. 7 in. in. cm 75. cm. ( ) cm.8 cm..8 cm. isosceles right 5. No. The closest she cn come to cmp is 0 km.. No. The 5 cm digonl is the longer digonl. 7.. km; 8 min 8. es 9. 9 ft 0. 5 ft. 50 mi. 707 m. nd 8. m 5.. No. If ou reflect one of the right tringles into the center piece, ou ll see tht the re of the kite is lmost hlf gin s lrge s the re of ech of the other tringles. Etr The qurter-circle gives the mimum re. Tringle: s 5 A s Squre: _ s A s s s Qurter-circle: s s r r s A s s s s s s s s 5 _ s s m 0. 0 Or students might compre res b ssuming the short leg of the tringle is. The re of ech tringle is then 0.87 nd the re of the kite is.7.. ; 0; 0. The rule is n if n is even, but 0 if n is odd. ANSWERS TO EXERCISES
9 . in./s. in./s. true. true 5. Flse. The hpotenuse is of length.. true 7. flse; AB ( ) ( ) 8. Flse. A glide reflection is combintion of trnsltion nd reflection. 9. Flse. Equilterl tringles, squres, nd regulr hegons cn be used to crete monohedrl tesselltions. 0. true. D. B. A. C 5. C. D 7. See flowchrt below. 8. W B 8-bll 9. cm ; 7.7 cm cm.9 cm 5. bout 55.9 m 5. bout.5 cm 5. 8 cm 5. ft N S A Cue bll E Answers to Eercises 7. (Chpter 9 Review) ABCD is rectngle Given D B Definition of rectngle ABCD is prllelogrm Definition of rectngle DA CB Definition of prllelogrm AC AC 5 DAC BCA AIA Conjecture 7 ABC CDA SAA Congruence Conjecture Sme segment ANSWERS TO EXERCISES
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