2 square units. The trapezoid. 15. There are many possible solutions. Sample answer: 16 cm. 16 cm

Transcription

1 LESSON 8.1 EXERCISES CHAPTER m. Use te Rectngle Are Conjecture. A b (19)(1) 8 m cm. A b (9.3)(4.5) cm yd. A b, so 96 yd b , so b 8 yd cm. A b, so , so 1 cm ft. First use te given perimeter nd eigt to find te lengt of te bse. P b, so 40 b (7) b 14. Terefore, b 6, nd b 13 ft. Now use te lengt of te bse nd te eigt to find te re of te rectngle: A b (13)(7) 91 ft m. Te sded re is te difference between te re of te lrge rectngle, wic s bse lengt 1 m nd eigt 1 m, nd te re of te smll (unsded) rectngle, wic s bse lengt m nd eigt m. Ten te sded re is (1)(1) (10)(7) m in.. Use te Prllelogrm Are Conjecture. A b (1)(8) 96 in cm. First use te given re nd eigt to find te lengt of te bse, nd ten use te lengt of te bse nd te eigt to find te perimeter. A b 508 b 44 b 57 P b P (57) (48) 10 cm 9. A 4 ft. Te sded region is one of two congruent tringles tt mke up prllelogrm. Are of sded region 1 (re of prllelogrm) 1 b 1 (1)(7) 4 ft. 10. Fctor 48 in two different wys. For emple, nd (Oter fctoriztions re possible.) Smple nswer: 4 cm 1 cm 48 cm 8 cm 6 cm 48 cm squre units. Te sded tringle is lf of prllelogrm wit bse lengt 3 units nd eigt units, so its re is 1 (3)() 3 squre units squre units. Te sded tringle cn be divided into two rigt tringles, ec of wic is lf rectngle. Are of tringle on left 1 (3)(4) 6 squre units Are of tringle on rigt 1 ()(4) 4 squre units Sded re squre units squre units. Te trpezoid cn be subdivided into two rigt tringles nd rectngle. Ec of te tringles is lf of rectngle. Are of tringle on left 1 ()(3) 3 squre units Are of rectngle (1)(3) 3 squre units Are of tringle on rigt 1 (1)(3) squre units Are of trpezoid squre units 14. Look for wys to fctor 64. Use one fctor for te lengt of te bse nd te oter for te eigt. Smple nswer: 64 cm 15. Tere re mny possible solutions. Smple nswer: 16 cm 16 cm 16 cm 4 cm 4 cm 16 cm 64 cm 16 cm 8 cm 8 cm m. Te re of ec pnel is (1)(0.7) 0.7 m, nd tere re (11)(3) 33 pnels, so te totl re of te rc is 33(0.7) 3.1 m m. Two of te wlls will be rectngles wit dimensions 4 m by 3 m, nd te oter two wlls will be rectngles wit dimensions 5.5 m by 3 m. Terefore, te totl re of te four wlls is (4)(3) (5.5)(3) 57 m m. For constnt perimeter, te rectngle of mimum re is squre. Tis cn be found by eperimenttion, s in te Project Mimizing Are. Terefore, te lengt of ec side of te pen will be m, nd te re will be (5)(5) 65 m. 150 CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

2 First find te re of te footbll field: A (53)(100) 5300 yd. Te re of ec squre is 1 yd, so te probbility tt te mt club wins is 10 yd yd tiles. First convert te dimensions of te wll from feet to inces: 4 ft 4(1 in.) 48 in., 7 ft 7(1 in.) 84 in. Te re of te wll is (48)(84) 403 in, nd te re of ec tile is (6)(6) 36 in, so te number of tiles required is 40 3 in 3 6 in squre units. Drw qudrilterl ABCD on grp pper. A (0, 0) From te grp, observe tt ABCD is prllelogrm wit bse 6 units nd eigt 16 units. Are of ABCD b (6)(16) 96 squre units.. 3 squre units. Drw qudrilterl EFGH on grp pper. E (0, 0) y y D (8, 16) H (6, 4) B (6, 0) G (8, 0) F (6, 4) C (14, 16) From te grp, observe tt EFGH is kite. Drw digonls HF nd EG to divide EFGH into four rigt tringles, ec of wic is lf of rectngle. Te re of EFGH is te sum of te res of tese four tringles. Are of EFGH 1 (6)(4) 1 (6)(4) 1 ()(4) 1 ()(4) squre units cm. Te dsed segments sow tt te trpezoid cn be subdivided into two rigt tringles nd rectngle, were ec of te rigt tringles is lf of rectngle. Are of trpezoid 1 (5)(0) (1)(0) 1 (1)(0) cm 4.. Smllest: cm, lrgest: cm. To find te smllest possible re, use te smller mesurement for bot bse nd eigt: A b (15.6 cm)(1.3 cm) cm. To find te lrgest possible re, use te lrger mesurement for bot bse nd eigt: A b (15.7 cm)(1.4 cm) cm. b. Answers will vry. Smple nswer: bout 193 cm. Tis is bout lfwy between te smllest nd lrgest possible res. c. Answers will vry. Te smllest nd lrgest re vlues differ t te ones plce, so te digits fter te deciml point re insignificnt compred to te effect of te limit of precision in te mesurements. 5.. In one Oio Str block, te sum of te res of te red ptces is 36 in, te sum of te res of te blue ptces is 7 in, nd te re of te yellow ptc is 36 in. b. Te complete quilt requires 4 blocks. Te re of te quilt is (7)(84) 6048 in, nd te re of ec squre is (1)(1) 144 in, so te number of squres is in 1 44 in 4. Notice tt ec dimension of te quilt is multiple of 1 in, so 4 squres will cover te re ectly ; ; c. About 1814 in of red fbric, bout 369 in of blue fbric, nd bout 1814 in of yellow fbric. Te border requires 5580 in (if it does not need te etr 0%). Red fbric: Tere re 4 quilt squres, ec wit 36 in of red fbric; lso, multiply by 1. (10%) to llow for te etr 0%: 4(36)(1.) 1814 in. Blue fbric: Tere re 4 quilt squres, ec wit 7 in of blue fbric; lso, multiply by 1. (10%) to llow for te etr 0%: 4(7)(1.) 369 in. Yellow fbric: Te sme mount of yellow fbric is needed s red, so te mount of yellow fbric is lso bout 1814 in. Border: From te digrm, you cn see tt te re of te border cn be clculted by dding te res of two rectngles wit dimensions 114 in. by 15 in. nd two rectngles wit dimensions 7 in. by 15 in. Terefore, te re of te border is (15)(114) (7)(15) in. 15 in. 15 in. 84 in. 7 in cm ; (36 64) cm. Te re of te lrge squre equls te sum of te res of te smll squres. Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press

3 7. 76, b 5, c 104, d 5, e 76, f 47, g 90, 43, k 104, nd m 86. For reference, let P be te center of te circle. First, mdbc 90 (Angles Inscribed in Semicircle Conjecture), nd lso mpba 90 (Tngent Conjecture). Terefore, bot DBP nd ABC re complements of PBC, so b 5. Net, AB AC (Tngent Segments Conjecture), so ABC is isosceles wit BAC s te verte ngle. Ten b (5 ) 180 (Tringle Sum Conjecture), so 76. Net, DPB is isosceles wit verte ngle DPB becuse two of its sides re rdii, so d 5 (Isosceles Tringle Conjecture). Ten e (5 ) 180 (Tringle Sum Conjecture), so e 76, nd c d (Tringle Eterior Angle Conjecture); or c e c (Liner Pir Conjecture), so c 104. Also, k 104 becuse tis rc is intercepted by te centrl ngle of mesure c (definition of te mesure of n rc). Net, f 1 (94 ) 47 (Inscribed Angle Conjecture), nd g 90 (Angles Inscribed in Semicircle Conjecture). Ten f g 180 (Tringle Sum Conjecture), so 43, nd m (43 ) 86 (Inscribed Angle Conjecture). (Also, m becuse DEC is semicircle.) 8. Copy AM. Construct line perpendiculr to AM t M. Construct 60 ngle (n ngle of ny equilterl tringle), nd bisect it to form 30 ngle tt s AM s one of its sides. Copy tis ngle on te oter side of AM. Etend te oter sides of tese ngles to intersect te perpendiculr troug M. Te two intersection points will be te oter two vertices of te required equilterl tringle. Smple construction: c. PROJECT Project sould stisfy te following criteri: Student provides cler descriptions, constrints, eplntions, nd predictions. Student includes grps similr to tese: A M 9.. b. 15 CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

4 A 1 b 39 1 (13) 78 (13) cm 6. 9 ft. Use te Tringle Are Conjecture. A 1 b Etr credit Oter grps nd reltionsips re presented. EXTENSIONS A. Results will vry. B. Tere re two nswers: 4 by 4 nd 3 by 6. Possible eplntion: For te perimeter to be equl to te re, tere must be te sme number of unit squres in te rectngle s tere re unit lengts on te sides. If you count one squre of re for ec unit of lengt, you will count ec squre on te corner twice, for totl of four etr squres. Te perimeter will be equl to te re only if te re of tese four etr squres equls te re of te internl squres te squres not counted in te perimeter. Tis ppens only in rectngles tt re 4 by 4 or 3 by 6. C. Approimte res: Tennessee: 4,000 mi ; Ut: 85,000 mi ; Wyoming: 98,000 mi. LESSON 8. EXERCISES 1. 0 cm. Use te Tringle Are Conjecture. A 1 b 1 (8)(5) 0 cm m. A 1 b 1 (11)(9) 49.5 m squre units. Use te Kite Are Conjecture wit d nd d A 1 d 1 d 1 (5)(4) 300 squre units. You cn lso find te re of te kite by dding te res of te four rigt tringles tt re formed by te digonls. If you use tis metod, te re of te kite is 1 (1)(9) 1 (1)(16) squre units cm. Use te Trpezoid Are Conjecture wit b 1 14 cm nd b 6 cm. A 1 (b 1 b ) 1 (6)(14 6) 60 cm cm. Use te Tringle Are Conjecture (b)(7) 63 b(7) b 9 ft ft. Use te Kite Are Conjecture wit d 1 BU ft nd d LE. A 1 d 1 d 40 1 (8)d 40 (14)d d LE 30 ft 8. 5 cm. Use te Trpezoid Are Conjecture wit b 1 13 cm nd b 7 cm. A 1 (b 1 b ) 50 1 (13 7) cm m. Use te Trpezoid Are Conjecture wit b 1 4 m nd b b. A 1 (b 1 b ) (9)(4 b) 360 (9)(4 b) 40 4 b b 16 m cm. Let b represent te unmrked side of te tringle. Te eigt to tis side is 4 cm. Find b, nd ten use tis side lengt to find te perimeter of te tringle. A 1 b 94 1 b(4) 94 b(1) b 77 cm P cm Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press

5 11. 1 cm. Let b 1 10 cm nd let b represent te lengt of te unmrked bse of te trpezoid. First use te perimeter nd te known side lengts to find b. P b 6 38 b b 4 cm Now use te Trpezoid Are Conjecture to find. A 1 (b 1 b ) 04 1 (10 4) 04 (17) 1 cm ft, y 10.8 ft. Te re of te tringle cn be clculted in tree different wys (using ec bse nd its corresponding ltitude), but ll of tem must give te sme re. Notice tt two of te ltitudes lie outside te tringle. A 1 (15)() 1 (5)(y) 1 (6)(9) 7 1 (15)() 7 1 (5)(y) 7 (15)() 54 (5)(y) y ft 10.8 ft 13. Find two positive integers wose product is times 54, or 108. Two suc integers re 9 nd 1. In te pir of tringles sown below, one uses 9 cm s te lengt of te bse nd 1 cm s te eigt, wile te oter uses 1 cm s te lengt of te bse nd 9 cm s te eigt. Smple nswer: 1 cm 14. Coose fctor of 56 for te eigt. Divide 56 by tt number to get te verge of te lengts of te two bses. For emple, if you coose 8 cm for te eigt, te verge of te lengts of te bses will be 7 cm, nd one possibility is to mke te bses ve lengts 10 cm nd 4 cm. Smple nswer: 4 cm 8 cm 9 cm 9 cm 1 cm 7 cm 7 cm 10 cm 9 cm 15. Becuse lf te product of te two digonls is 109, coose two numbers wose product is twice 109, or 184, to be te two digonls. One possible coice is 4 nd 91. You cn use 4 cm s te lengt of te digonl tt is bisected by te oter digonl. You cn split te lengt of 91 cm in mny wys, corresponding to different positions were te digonls intersect. Te figures below sow two wys to do tis. Smple nswer: 46 cm 35 cm 1 cm 45 cm 56 cm 16. Te lengt of te bse of te tringle equls te sum of te lengts of bot bses of te trpezoid. cm 1 cm 3 cm 3 cm 7 cm 9 cm Drw ltitude PQ of TPR. Ten te re of TPR 1 (TR)(PQ) 1 (TR)(AR) 1 (re of rectngle ARTY). Notice tt tis result does not depend on te position of P, wic cn be ny point on AY. 18. More tn lf, becuse te top crd completely covers one corner of te bottom crd. Notice tt tis is like te figure in Eercise 17, but te covered re of te crd ere is lrger frction of te rectngle (or crd) tn in Eercise 17, so te top crd covers more tn lf te bottom crd in. of bls wood nd 960 in of Mylr. Find te sum of te lengts of te digonls to find te mount of bls wood, nd te totl re of te kite nd flps to find te mount of Mylr. Te sum of te lengts of te digonls is (15) in. To find te re of te kite, substitute 30 for d 1 nd for d in te kite re formul, A 1 d 1 d. A 1 (30)(56) 840, so te re of te kite is 840 in. Ec of te four flps is trpezoid wit eigt 1 in.; two ve bses of lengts 1 in. nd 5 in., wile te oter two ve bses of lengts 35 in. nd 39 in. Find te totl re of ll te flps: 1 1(1 5) 1 1(35 39) , so te re is 10 in. Te totl mount of Mylr needed is in. 154 CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

6 b. 56 in. (or less, if e tilts te kite). Becuse te lengt of te sorter digonl is 30 in., te esiest wy to cut te kite out of 36-inc-wide piece of Mylr is to plce te sorter digonl long te widt of te unrolled Mylr. Ten te lengt of Mylr needed would simply be te lengt of te longer digonl of te kite, wic is 56 in singles (to cover n re of 900 ft ). To find te totl re to be covered wit singles, dd te res of te two congruent trpezoids nd te two congruent tringles. Are of two trpezoids: 1 15(0 30) 750, so te sum of te res of te two trpezoids is 750 ft , so te Are of two tringles: 1 sum of te res of te two tringles is 150 ft. Terefore, te totl re of te roof is ft. Becuse ec single covers 0.5 ft, Crystl 900 ft sould buy 0. 5 ft 3600 singles. 1. Te isosceles tringle is rigt tringle becuse te ngles on eiter side of te rigt ngle re complementry. If you use te trpezoid re formul, te re of te trpezoid is 1 ( b)( b). If you dd te res of te tree tringles, te re of te trpezoid is 1 b 1 c 1 b 1 c b.. Given: trpezoid ABCD wit eigt. Are of ABD 1 b 1. Are of BCD 1 b. Are of D trpezoid sum of res of two tringles 1 b 1 1 b 1 (b 1 b ) squre units. Count te complete nd frctionl squres inside te figure. Tere re 7 wole squres, 7 lf-squres, nd 1 tree-qurter squre. Add tese to get te re of te figure: A 7(1) , 4 so te re of te figure is squre units squre units. Enclose te tringle in squre. Find te re of te originl tringle by subtrcting te res of te tree surrounding rigt tringles from te re of te squre. Are of squre 4 16 Are of tringle I 1 (4)() 4 Are of tringle II 1 ()(3) 3 Are of tringle III 1 (4)(1) A b 1 Terefore, te re of te originl tringle is 16 (4 3 ) squre units m. First find te lengt of te bse of te rectngle. Let b represent te lengt of te bse. I b II III B C A b, so 64 b(4), nd b 11 m. Now find te perimeter of te rectngle: P b (11) (4) 70 m cm. Te prllelogrm s two sides of lengt 10 cm. Let b represent te lengt of ec of te oter two sides. Using te given perimeter, b (10) 5 cm. Ten b 16 cm, nd A b (16)(9) 144 cm. 7. A 88 ft, P 144 ft. Divide te figure into two rectngles nd use subtrction to find te unmrked side lengts. 18 ft Are of lrger rectngle b (30)(18) 540 ft Are of smller rectngle b (4)(1) 88 ft Are of sded figure 540 ft 88 ft 88 ft In clculting te perimeter, note tt te dsed segment is not prt of te originl figure, nd its lengt sould not be included. Perimeter of sded figure ft 8. A 1440 cm, P 0 cm. Divide te figure into two prllelogrms nd use subtrction to find te unmrked side lengts. 30 cm 30 ft 30 ft 54 ft 8 cm 40 cm 40 cm 6 ft 4 ft 4 ft 11 cm 40 cm 9 cm 8 cm 10 cm 1 ft 40 cm 9 cm Are of lrger prllelogrm b (40)(8) 110 ft Are of smller prllelogrm b (40)(8) 30 ft Are of sded figure 110 ft 30 ft 1440 ft In clculting te perimeter, note tt te dsed segment seprting te two prllelogrms is not prt of te originl figure, nd its lengt sould not be included. Perimeter of sded figure cm 9.. Incenter. Te construction mrks sow tt tis is te point of concurrency of te ngle bisectors. b. Ortocenter. Te construction mrks sow tt tis is te point of concurrency of te ltitudes. c. Centroid. Te construction mrks sow tt tis is te point of concurrency of te medins. Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press

7 30. 34, b 68, c 68, d 56, e 56, f 90, g 34, 56, m 56, n 90, nd p 34. For reference, let O be te center of te circle. First, mbc (56 ) 11 (Inscribed Angle Conjecture). Becuse DBC is semicircle, b Net, 1 b (Inscribed Angle Conjecture), so 34. By te definition of te mesure of n rc, c b, so c 68. Now look t BAO nd CAO. Notice tt OB OC (ll rdii of circle re congruent), AB AC (Tngent Segments Conjecture), nd AO AO (sme segment), so BOA COA by SSS. Terefore, d e by CPCTC. Observe tt c d e 180, so 68 d 180. Ten d 11, so d 56, nd lso e 56. By te Tngent Conjecture, OC AC, so f 90. Net, g e f 180 (Tringle Sum Conjecture), so g , nd g 34. Net, 1 (11 ) 56 (Inscribed Angle Conjecture) nd m 1 (11 ) 56 (Inscribed Angle Conjecture). Now, by te Intersecting Cords Conjecture (see Lesson 6.5, Eercise 16), n 1 (b 11 ) 1 (68 11 ) 1 (180 ) 90. Finlly, p n (Tringle Sum Conjecture), so p Tere re two types of vertices. One type of verte is surrounded by two equilterl tringles, ten two regulr egons. Te oter type of verte is surrounded by n equilterl tringle, ten regulr egon, ten noter equilterl tringle, nd ten noter regulr egon. PROJECT You migt gter some dt points nd use qudrtic regression rter tn finding nd grping function epression. Projects sould stisfy te following criteri: Te epression nd eqution re equivlent to A (10 ). Te grp sows te mimum re t.5. Etr credit Te project includes n eplntion of wy te mimum re is 1.5 m. One reson, not using te grp, is tt if tere were twice s muc fencing vilble nd no brn, ten te region of mimum re would be squre (s discovered in Eercise 18 in Lesson 8.1); te brn wll cn be considered line of symmetry, cutting te mount of fencing in lf but mking te spe lf squre, wit widt 1 4 te totl mount of fencing. EXTENSIONS A. In tringle, ec midsegment is one-lf te lengt of side. If you consider one side s te bse nd lbel its lengt s b, ten lengt of midsegment 1 b. Ten re of tringle 1 b (midsegment)(eigt). In trpezoid, tere is one midsegment, nd its lengt is lf te sum of te lengts of te bses, tt is, lengt of midsegment 1 (b 1 b ). Ten re of trpezoid 1 (b 1 b ) 1 (b 1 b ) (midsegment)(eigt). B. Possible nswers (ll use trpezoid ABCD wit long bse AB, sort bse CD, b 1 AB, b CD, nd s te perpendiculr eigt between te bses): Are of trpezoid re(abc) re(cda) A 1 b 1 1 b 1 (b 1 b ) Construct point E on AB so tt ED is prllel to BC. Terefore, re of trpezoid re(aed) re(prllelogrm EBCD) 1 (b 1 b ) b 1 (b 1 b ). A Construct DE AB, wit points E nd F on AB. Let c AE. Ten FB b 1 b c. Terefore, re of trpezoid re(aed) re(rectngle EFCD) re(fbc) 1 c b 1 (b 1 b c) 1 (b 1 b ). c A E b 1 F Strt s you did in te previous metod, ten duplicte AED s DEA nd FBC s FCB, forming rectngle ABFE. Terefore, re of trpezoid re(rectngle ABFE) re(dea) re(cfb) b 1 1 c 1 (b 1 b c) 1 (b 1 b ). E' A D D D D E b b 1 b b 1 b b b 1 C C C C B B B F' B C. Approimte res: Cliforni: 164,000 mi ; Tes: 69,000 mi ; Nevd: 111,000 mi. 156 CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

8 LESSON 8.3 EXERCISES ,95 ft. Ec room s two wlls wit dimensions 14 ft by 10 ft nd two wlls wit dimensions 16 ft by 10 ft, wile te ceiling is 14 ft by 16 ft. Find te sum of te res of tese rectngles: (14)(10) (16)(10) (14)(16) 84, so te totl re to be pinted for one room is 84 ft, nd te totl for 148 rooms is 148(84) 11,95 ft. b. 44 gllons of bse pint nd 488 gllons of finising pint. Find te number of gllons of ec kind of pint seprtely. Bse pint: 11,95 ft 1 gl 5 00 ft 43.9 gl; round up to 44 gl. Finising pint: 11,95 ft 1 gl 50 ft gl; round up to 488 gl.. He sould buy t lest four rolls of wllpper. Two wlls re rectngles wit dimensions 11 ft by 10 ft. Te oter two wlls re rectngles wit dimensions 13 ft by 10 ft. Find te sum of te res of tese rectngles: ft, so te totl surfce re to be ppered is 480 ft. Now find te re of ec roll: (.5)(50) 15, so te re of ec roll is 15 ft. Finlly, find te number of rolls of wllpper tt is needed: 480 ft 1 roll 1 5 ft 3.84 rolls; round up to 4 rolls. If pper cut off t te corners is wsted, e ll need 5 rolls. Ec 11-ft wll requires ft strips, rounded up to 5 strips. Ec 13-ft wll requires ft strips, rounded up to 6 strips. Te totl needed is (5 10) (6 10) 0 ft, wic requires ft rolls, rounded up to 5 rolls ft ; 776 ft more surfce re. First find te re of 65,000 rectngulr cells: 65,000(1.5)(.75) 3,437.5, so te totl re of te cells is 3,437.5 in. Convert tis re to squre feet. Tere re 1 in. in foot, so tere re 1(1) 144 in in squre foot. 3,437.5 in 1 ft 14 4 in 155 ft. If te cells re only 1% efficient, 50% more re will be needed tn if tey re 18% efficient, so te dditionl surfce re tt would need to be covered would be 0.5(155) 776 ft First find te re of te kite. Te lengts of te digonls re 40 ft nd 70 ft, so te re of te kite is 1 (40)(70) 1400 ft. Find te mount of selnt needed for one ppliction: 1400 ft 1 gl 4 00 ft 3.5 gl. Hrold will mke 6 pplictions of selnt (twice yer for tree yers), so e will need 6(3.5) 1 gl of selnt. Terefore, e sould buy 1 one-gllon continers ft ; $1,780. First find te re of one flowerbed. Ec flowerbed is trpezoid wit eigt 7 ft nd bses of lengts 1 ft nd 0 ft, so its re is 1 (7)(1 0) 11 ft. Ten te totl re of te tree flowerbeds is 3(11) 336 ft, nd te cost will be $ ($5) $1, $760. Te esiest wy to find te totl re to be crpeted is to find te re of te complete rectngle t te top of te figure nd ten subtrct te res of te btrooms. Tis wy, you don t ve to figure out te dimensions of te llwy. Lengt of rectngle ft Widt of rectngle ft Totl re to be crpeted totl re of tree bedrooms nd llwy (7 17) ( ) 336 ft Convert to squre yrds. Tere re 3 ft in yrd, so tere re 3(3) 9 squre feet in squre yrd. 336 ft 1 yd 9 ft yd ; round up to 38 yd becuse crpeting, pdding, nd instlltion re priced per squre yrd. Te cost for crpeting, pdding, nd instlltion is $14 $3 $3 $0 per squre yrd, so te cost for 38 yd will be 38($0) $ terr cott tiles, 1107 blue tiles; $1, First find te totl re of te entrywy nd kitcen. Look t te rectngle tt contins te entrywy, kitcen, living room, nd dining room, nd t te two smller rectngles contining te entrywy nd kitcen (on te left) nd living room nd dining room (on te rigt). From te dimensions of tese rooms, you cn see te totl re of te entrywy nd kitcen is (10)() 0 ft. Becuse 1-footsqure tiles re to be used for te entrywy nd kitcen, 0 tiles will be needed. Te dimensions of te btrooms re 6 ft by 10 ft nd 7 ft by 9 ft, so te totl re of te two btrooms is ft. Te tiles for te btroom floors re 4-inc-squre tiles, so tere re 9 of tese tiles per squre foot: 1 4 3; Find te number of blue tiles needed: 13 ft 9 tiles 1 ft 1107 tiles. Finlly, find te cost of te tiles. Terr cott tiles: 0($5) $1,100 Blue tiles: 1107($0.45) $ Totl cost: $1,100 $ $1, cm. Move te smll sded wedge from te upper rigt squre to fit wit te sded qurter circle in te lower rigt squre, completing second sded squre. Te new sded region is rectngle wit bse 1 cm nd eigt 6 cm, so A b cm. Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press

9 9. AB 16.5 cm, BD 15.3 cm. First clculte te re of ABC using BC s te bse nd AF s te corresponding ltitude. Are of ABC 1 b 1 (BC)(AF) 1 (16.0)(10.5) 84.0 cm Te re of te tringle is te sme for ny coice of bse, so we cn use tis result to find AB nd BD. First find AB. Notice tt CE is te ltitude to AB. A 1 b A 1 (AB)(CE) Substitute AB for b nd CE for (AB)(10.) Substitute 84 for A nd 10. for CE (AB)(10.) Multiply by. AB 16.5 cm Divide by 10. nd round to te nerest tent. Now find BD. Notice tt BD is te ltitude to AC. A 1 b B. Moving P doesn t cnge te eigt of PDC or te lengt of DC, so te re of PDC won t cnge. C. Moving P doesn t cnge te lengt of DC, so te re of te trpezoid doesn t cnge. D. mpcd mb (CA Conjecture), so ma mpcd mcpd 180, regrdless of te position of P. IMPROVING YOUR VISUAL THINKING SKILLS Tis problem cn be done in undreds of wys. One unusul wy, if disconnected pieces re llowed, is to rerrnge pieces of te tringle to mke prllelogrm or rectngle of te sme re, divide te resulting rectngle into four pieces, nd ten rerrnge tose pieces bck into tringle. A 1 (AC)(BD) Substitute AC for b nd BD for (11.0)(BD) Substitute 84 for A nd 11.0 for AC (11.0)(BD) Multiply by. BD 15.3 cm Divide by 11.0 nd round to te nerest tent cm by eiter metod. Te first step in using Hero s formul is to find te semiperimeter. s b c Ten substitute into Hero s formul: A s(s )(s )(s b) c 0(0 0 8)()( ) , so te re of te tringle is 60 cm. Using te stndrd tringle re formul, A 1 b 1 (15)(8) 60, so by tis formul, te re is 60 cm. 11. AO BO becuse ll rdii of circle re congruent, so AOB is isosceles. Terefore, ma 0 nd maob 140. mab 140 nd mcd 8 (definition of rc mesure). mac mbd becuse prllel lines intercept congruent rcs on circle E. None of te vlues cnge. A. Moving P doesn t cnge te eigt of ABP, so te re of ABP won t cnge. EXTENSIONS A. Results will vry. Divide te figure into more nd more pieces nd dd te resulting res. Te res will pproc some number (te ctul re) s te number of pieces increses. B. Results will vry. USING YOUR ALGEBRA SKILLS Combine te res of te pieces to find te re of te lrge rectngle, wic gives te product. ( 5)( 1) CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

10 ( 3)( 4) Here is wy to rrnge te pieces to mke rectngle Combine te res of te pieces to find te re of te lrge rectngle, wic gives te product. ()( 7) Te bse nd eigt of te rectngle give te fctors, so 11 1 ( 3)( 4) Add te res of te four smll rectngles to find te re of te lrge rectngle, wic gives te product. ( 15)( 11) Combine te res of te pieces to find te re of te lrge rectngle, wic gives te product (3 )( 5) (3)( 1) Here is wy to rrnge te pieces to mke rectngle. 1 Te bse nd eigt of te rectngle give te fctors, so 6 3 (3)( 1). 5. ( 5)( 3) Here is wy to rrnge te pieces to mke rectngle Add te res of te four smll rectngles to find te re of te lrge rectngle, wic gives te product. (3 7)(4 5) Add te res of te four smll rectngles to find te re of te lrge rectngle, wic gives te product. ( 4) Te bse nd eigt of te rectngle give te fctors, so 8 15 ( 5)( 3). Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press

11 10. ( 5)( 5) of te lower rigt region. Tey must lso dd to 10 in order to get sum of 10 from te oter two regions. One of tese numbers must be negtive nd te oter positive becuse teir product is negtive. Te two numbers tt work re 1 nd. Add te res of te four smll rectngles to find te re of te lrge rectngle, wic gives te product. 1 1 ( 5)( 5) ( b) b b b b b b b ( b) b b b b b 1. b b Te fctors re te bse nd eigt of te rectngle, so 10 4 ( 1)( ). 15. ( 5)( 4). Drw rectngle digrm wit in te upper left region nd 0 in te lower rigt region. Te remining two vlues must multiply to 0 becuse tey re te dimensions of te lower rigt region. Tey must lso dd to 1 in order to get sum of (or 1) from te oter two regions. One of tese numbers must be negtive nd te oter positive becuse teir product is negtive. Te two numbers tt work re 5 nd 4. 5 b 5 b b b ( b)( b) b b b b 13. ( 15)( 4). Drw rectngle digrm wit in te upper left region nd 60 in te lower rigt region. Te remining two vlues must multiply to 60 becuse tey re te dimensions of te lower rigt region. Tey must lso dd to 19 in order to get sum of 19 from te oter two regions. Bot of tese numbers must be positive becuse teir product is positive nd teir sum is positive. Te two numbers tt work re 15 nd Te fctors re te bse nd eigt of te rectngle, so 0 ( 5)( 4). 16. ( 3). Here we need to find two numbers wose product is 9 nd wose sum is 6. Te numbers tt work re 3 nd ( 3)( 3) or ( 3) 17. ( 6)( 6). Here we need to find two numbers wose product is 36 nd wose sum is 0. Te numbers tt work re 6 nd Te fctors re te bse nd eigt of te rectngle, so ( 15)( 4) ( 1)( ). Drw rectngle digrm wit in te upper left region nd 4 in te lower rigt region. Te remining two vlues must multiply to 4 becuse tey re te dimensions ( 6)( 6) 160 CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

12 18. ( 7)( 7). Te upper left region will ve n re of 4 nd te lower rigt region n re of 49, wile te sum of te res of te oter two regions must be ( 7)( 7) Originl eqution. ( 4)( 1) 0 Fctor te left-nd side. Ceck: 4 0 or 1 0 Use te zero product property. 4 or 1 Solve for. (4) 5(4) 4? ? (1) 5(1) 4? ? Originl eqution Subtrct 30 to get 0 on te rigt-nd side. ( 10)( 3) 0 Fctor te left-nd side. Ceck: or 3 0 Use te zero product property. 10 or 3 Solve for. (10) 7(10)? ? (3) 7(3)? ? ( 6) 5 4 Originl eqution Multiply on te left-nd side. Add (5 4) to bot sides. ( 8)( 3) 0 Fctor te left-nd side. Ceck: 8 0 or 3 0 Use te zero product property. 8 or 3 Solve for. (8)[(8) 6]? 5(8) 4 8? (3)[(3) 6]? 5(3) 4 8(3)? ( )( ) 9 8 Originl eqution Multiply on te left-nd side. Add ( 9 8) to bot sides. ( 1)( 4) 0 Fctor te left-nd side or or or Use te zero product property. Subtrct 1 from bot sides of te first eqution. 4 Solve for. Ceck: 1 1? ? [(4) ][(4) ]? (4) 9(4) 8 (6)? Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press

13 3.. b. b 1 ; b 4 c. 1 [ ( 4)] 48 d. 1 [ ( 4)] 48 1 ( 4) 48 1 ( 4) ( 8)( 6) or or 6 Te eigt cnnot be 8 feet, so te only vlid solution is 6 nd Te eigt is 6 feet, one bse is 6 feet, nd te oter bse is 10 feet. Ceck: 1 (6 10) IMPROVING YOUR REASONING SKILLS 1. O sould ply in b1, to block X from getting te squre b1-4-d5-e.. Te lst move forces X to ply in c4. Tis cretes two squre options for X, c4-d-b1-3 nd c4-c- -4. O cn t block bot of tese in one move, so X will win. 3. b; c1. O 5 in b will force X 6 in b3. Ten O 7 in c1 will crete two squre options for O, c1-b-c3-d nd c1-b1-b-c, so O will win. 4. X 7, X 9, O 10, X 11, X 13, nd O 14 were forced; X will win. X 7 ws forced by O in c3-c4-d4; X 9 ws forced by O in d-c3-d4; O 10 ws forced by X in e3-c-b4; X 11 ws forced by O in d5-d4-c4; X 13 ws forced by O in b-c3-d; O 14 ws forced by X in c5-e3-c1. Now wit O in 3-c4-d, X is forced to ply X 15 in b1. Tt will crete two squre options for X, -b3-c-b1 nd d1-b1-b3-d3, so X will win cm. Te figure is pentgon, so n 5. A 1 sn 19, (107.5)(5) 39,775 (537.5) 3 9, cm cm. Use te formul A 1 P (38.6)P P P cm cm. A 1 sn 1 (3)(4.4)(5) 33 cm cm. Use te formul A 1 P (9.6)P P P 63 cm cm. A 1 P 1 (1)(81.6) 490 cm m. Use te formul A 1 P (9)P P Te perimeter is bout 57.6 m ft. Use te formul A 1 sn to find te vlue of s. 0,000 1 (80)s(0) 0, s s 5 ft. Te lengt of ec side is bout 5 ft cm. Construct circle wit rdius 4 cm. Mrk si 4-centimeter cords round te circle. s LESSON 8.4 EXERCISES cm. Te figure is eptgon, so n 7. A 1 sn 1 (4.9)(4)(7) 09 cm Use te formul A 1 sn to pproimte te re of te egon. By construction, s 4 cm. Mesure te potem wit ruler: 3.5 cm. Terefore, A 1 sn 1 (3.5)(4)(6) 4 cm. 16 CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

14 cm. Drw circle wit rdius 4 cm, nd use your protrctor to form five congruent centrl ngles, ec of mesure Use te sides of tese ngles to form five rdii of te circle. Use protrctor nd te Tngent Conjecture to drw tngents to te circle t ec of te points were te five rdii touc te circle. Te tngent segments form regulr pentgon circumscribed bout te circle. Becuse ec rdius is perpendiculr to tngent, te rdii form potems of te pentgon, so 4 cm, s required. Mesure one of te tngent segments to find te lengt of side of te pentgon: s 5.8 cm. Ten A 1 sn 1 (4)(5.8)(5) 58 cm. 11. Wen squre wit side s is divided into four isosceles tringles wit vertices t te center of te squre, te bse of ec tringle is s nd te potem is s nd n 4, so A 1 sn 1 1 s s(4) s. 1. It is impossible to increse its re becuse regulr pentgon mimizes te re. Any drgging of te vertices decreses te re. (Subsequent drgging to spce tem out more evenly cn increse te re gin, but never beyond tt of te regulr pentgon.) cm. First find te re of te complete octgon. A 1 sn 1 (0)(16.6)(8) 138 cm Te sded re is 6 8, or 3, 4 of te octgon, so its re is 3 4 of te re of te octgon: 3 4 (138) 996 cm. Becuse te given mesurements for te octgon re pproimte, te sded re is pproimtely 996 cm cm. Te re of te egonl donut is te difference between te re of te lrge egon nd te re of te smller egon. In regulr egon, te distnce from te center to ec verte is equl to te lengt of ec side, so in te smll egon, s r 8 cm, nd in te lrge egon, s r 16 cm. Te given mesurements re pproimte, so te res clculted from tese mesurements will lso be pproimte. Are of lrge egon 1 sn 1 ( 6.9)(16)(6) 66.4 cm Are of smll egon 1 sn 1 (6.9)(8)(6) cm Are of donut (re of lrge egon) (re of smll egon) 497 cm 15. Totl surfce re 13,680 in 95 ft ; cost $8,075. To find te totl surfce re, divide te lower portion of te kitcen floor pln into tree rectngles nd te upper portion into two rectngles nd one regulr octgon. 60 in. 4 in. 48 in. 60 in. In te octgon, s 4 in., 1 (60) 30 in., nd n 8. Lower portion: (4)(60) (4)(138) (4)(7) 6480 in Upper portion: (4)(60) (10)(4) 1 (30)(4)(8) 700 in Te totl surfce re of te countertops is 6,480 7,00 13,680 in. Convert to squre feet: 13,680 in 1 ft 14 4 in 95 ft. Te cost of te countertop will be 95($85) $8, squre units. Divide te qudrilterl into two tringles, A nd B, s sown in tis figure. Find te res of ec of tese tringles, nd dd to find te re of te qudrilterl. y 4 in. (, 6) A 36 in. 36 in. B 144 in. 10 in. 4 in. 138 in. 186 in. y 1_ 5 y 10 4 in. 48 in. 60 in. 4 in. 7 in. To find te re of tringle A, use te side long te y-is s te bse. Becuse te y-intercept of line Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press

15 y 1 5 is 5, te lengt of tis bse is 5 units. Te two lines intersect t (, 6), so te lengt of te ltitude to tis bse is units. (Tis ltitude will fll outside te tringle.) Terefore, re(tringle A) 1 b 1 (5)() 5 squre units. To find te re of tringle B, use te side long te -is s te bse. Becuse te -intercept of te line y 10 is 5, te lengt of tis bse is 5 units. Te two lines intersect t (, 6), so te lengt of te ltitude to tis bse is 6 units. Terefore, re(tringle B) 1 b 1 (5)(6) 15 squre units. Tus, te totl re of te qudrilterl is squre units squre units. As in Eercise 16, divide te qudrilterl into two tringles, A nd B, by connecting te origin to te point were te two lines intersect, wic is (6, 4) in tis cse. y y 4_ 1 3 A (6, 4) y 1_ 6 3 B Let A be te upper tringle. Te y-intercept of te line y is 6, so te vertices of tis tringle re (0, 0), (0, 6), nd (6, 4). Use te side long te y-is s te bse. Te lengt of te ltitude to tis bse is 6, so re(tringle A) 1 (6)(6) 18 squre units. Let B be te lower tringle. Te -intercept of te line y is 9, so te vertices of tis tringle re (0, 0), (9, 0), nd (6, 4). Te lengt of te ltitude to tis bse is 4, so re(tringle B) 1 (9)(4) 18 squre units. Tus, te totl re of te qudrilterl is squre units. 18. Conjecture: Te tree medins of tringle divide te tringle into si tringles of equl re. Argument: Tringles 1 nd ve equl re becuse tey ve equl bses nd te sme eigt. Becuse te centroid divides ec medin into tirds, you cn sow tt te eigt of tringles 1 nd is 1 3 te eigt of te wole tringle. Ec s n re 1 6 te re of te wole tringle. By te sme rgument, te oter smll tringles lso ve res of 1 6 te re of te wole tringle. 19. nw ny. Write te epressions for te perimeter of te first few figures nd look for pttern. Ec figure s two sides of lengt, nd te number of sides of lengts w nd y increses by 1 ec time, so te perimeter of te nt figure is nw ny cm. Te re of pentgon GHIJK is te sum of te res of rectngle KGHI nd tringle HIJ: Are of rectngle KGHI b (30)(15) 450 cm Are of HIJ 1 b 1 (9)(1) 54 cm Are of pentgon GHIJK 450 cm 54 cm 504 cm cm. Te re of te sded region is te difference between te re of prllelogrm FELA nd prllelogrm CDLB. By subtrction, BL 44 cm 3 cm 1 cm. BL is bse of te prllelogrm. Are of prllelogrm FELA b (44)(4) 1056 cm Are of prllelogrm CDLB b (1)(18) 16 cm Are of sded region 1056 cm 16 cm 840 cm IMPROVING YOUR VISUAL THINKING SKILLS A: 9 841; B: ; C: ; D: ; E: 5 65; F: 16 56; G: 18 34; H: 4 576; I: 9 81; J: 7 49; K: 15 5; L: 4; M: 17 89; N: 6 36; O: 11 11; P: ; Q: 35 15; R: 8 64; S: EXTENSIONS A. Possible nswer: Consider regulr polygon wit point t te center nd divided into tringles wit ec side of te polygon s bse, were is te potem, s is te lengt of ec side, nd n is te number of sides in te regulr polygon. If te polygon s n odd number of sides, te tringles cn be rrnged into trpezoid wit long bse n 1 s, sort bse n 1 s, eigt, nd re n 1 s n 1 s 1 ns n 1 s 1 s s n 1 s Odd number of sides Regulr pentgon (n 5) 164 CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

16 If te polygon s n even number of sides, te tringles cn be rrnged into prllelogrm wit bse n s, eigt, nd re 1 ns. s B. Results will vry. LESSON 8.5 EXERCISES 1. 9 in. A r (3) 9 in. s s n_ Even number of sides Squre (n 4). 49 cm. A r (7) 49 cm m. A r (0.5) 0.5 m 0.79 m cm. A r 9 cm, so r 9, nd r 3 cm. (Note tt r must be positive becuse it represents te lengt of segment; te negtive squre root of 9, wic is 3, is not relevnt in tis sitution.) 5. 3 in. A r 3 in, so r 3, nd r 3 in m. A r m, so r in. C r 1 in, so r 6 in. Ten, A r (6) 36 in. 0.5 m m. C r 314 m, so r 3 14 m. (Keep tis number stored in your clcultor.) Ten A r m. 9. (5 48) 30.5 squre units. Te re of te sded region is te difference between te re of te circle nd te re of te rectngle. Circle: r 5, so A 5. Rectngle: Te dimensions re (3) 6 by (4) 8, so A Te re of te sded region is (5 48) squre units. 10. (100 18) 186 squre units. Te re of te sded region is te difference between te re of te circle nd te re of te tringle. Circle: r 10, so A 100. Tringle: Use te orizontl segment t te top of te tringle s te bse. Its lengt is (8) 16, nd te corresponding eigt is Terefore, A 1 (16)(16) 18. Te re of te sded region is (100 18) squre units. s n_ 11. A r 34 cm, so r cm. r 18 cm m. A r (16) m ,310 km. A r (60) 3,600 11,310 km m. A r (7) m times. Find te res of te two circulr regions nd ten compre te res. Muscle wit rdius of 3 cm: A 9 cm Muscle wit rdius of 6 cm: A 36 cm strengt of second muscle Rtio: 3 6 cm strengt of first muscle 9 cm 4 Terefore, te second (lrger) muscle is 4 times s strong s te first (smller) muscle. 16. A r becuse te 100-gon lmost completely fills te circle. Observe ow, s te number of sides of regulr polygon increses, more nd more of te circle is filled. r Hegon cm. Te figure is rombus, so use te Prllelogrm Are Conjecture: A b (4)(19) 456 cm. r ft. Use te Tringle Are Conjecture. Te segment of lengt 9 ft is te ltitude to te side of te tringle wit lengt 8 ft. A 1 b 1 (8)(9) 36 ft. Dodecgon 19. Te tringles ve equl re wen te point is t te intersection of te two digonls. Tere is no oter loction t wic ll four tringles ve equl re. 0. mde 4 48 (Inscribed Angle Conjecture). mde 48 (Prllel Lines Intercepted Arcs Conjecture) Te mesure of te unmrked bse ngle of te isosceles tringle is 8 (Isosceles Tringle Conjecture). Add te ngle mesures of te lrge rigt tringle: 90 (38 8 ) 8 184, but tis is impossible becuse te sum of te ngles of ny tringle must be 180 (Tringle Sum Conjecture). Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press

17 . 6 cm IMPROVING YOUR VISUAL THINKING SKILLS 3. 7 Tere re 3 isosceles tringles wit verte ngle t ec corner, 4 wit verte ngle t ec side midpoint, nd 8 wit verte ngle in te center: Tere re 84 possible wys to select tree points from te grid: out of 84 gives probbility of LESSON 8.6 EXERCISES 18 cm 1 cm 4 cm 1. 6 cm. Te sded region is sector of te circle wit 60. A sector 36 0 r (6) Te re of te sded region is 6 cm cm. Te sded region is sector of te circle wit A sector 36 0 r (8) Te re of te sded region is 64 3 cm cm. Te sded region is sector of te circle wit A sector 36 0 r Te re of te sded region is 19 cm. (16) ( ) cm. Te sded region is segment of te circle wit 90. Te re of te segment is te re of te sector minus te re of te tringle. A segment 36 0 r 1 b () Te re of te sded region is ( ) cm. 5. (48 3) cm. Te re of te sded region is te sum of te res of te sector, wic is 3 4 of te circle, nd te tringle, so A Te re of te sded region is (48 3) cm cm. Te sded region is n nnulus of te circle. A nnulus R r (7) (4) Te re of te sded region is 33 cm cm. Te sded region is n nnulus of te circle. Te rdius of te lrger circle is 1 (10) 5 cm. A nnulus R r (5) () Te re of te sded region is 1 cm cm. Te sded region is of n nnulus of te circle. A 5 6 (1) (9) 5 6 (144 81) 5 6 (63) Te re of te sded region is 10 5 cm. 9. r 6 cm. Te sded region is sector of te circle wit 10. A sector 36 0 r r 36 r r 6 Te rdius is 6 cm. 10. r 7 cm. Te sded region is n nnulus of te circle. Te rdius of te lrger circle is 1 (18) 9 cm. A nnulus R r 3 81 r r 49 r 7 cm Te sded region is sector of te circle wit nd r 4 cm. A sector 36 0 r (4) Te sded region is 36 0 of n nnulus of te circle. A 36 0 (10 8 ) CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

18 13. 4 cns. Te pizz slice is sector of te circle. 36 Are of slice 3 60 (0) Te re of te pizz slice is 40 ft. Now find te number of cns of tomto suce needed to cover tis slice: 40 ft 1 cn 3 ft cns. Round up to 4 cns. 14. $448. Te pt round te circulr fountin cn be considered n nnulus of lrger circle. Te rdius of te smller circle is m, nd te rdius of te lrger circle is m. Are of pt (5.5) (4) Te re of te pt is 14.5 m. Now find te cost to pve tis pt: 14.5 m $10 1 m $14.5 $ b. c. d. b. (144 36) cm ; 78.54%. Te rdius of ec circle is cm, so te re of te four circles is 4 (3) cm. Terefore, s in 17, te re of te sded region is (144 36) cm, nd te re of te circles is bout 78.54% of te re of te squre. c. (144 36) cm ; 78.54%. Te rdius of ec circle is cm, so te re of te nine circles is 9() cm. Terefore, s in 17, te re of te sded region is (144 36) cm, nd te re of te circles is bout 78.54% of te re of te squre. d. (144 36) cm ; 78.54%. Te rdius of ec circle is cm, so te re of te siteen circles is 16(1.5) cm. Terefore, s in 17, te re of te sded region is (144 36) cm, nd te re of te circles is bout 78.54% of te re of te squre m. Combine two conjectures bout trpezoids to find n lternte formul for te re of trpezoid. By te Trpezoid Are Conjecture, A 1 (b 1 b ). By te Trpezoid Midsegment Conjecture, te lengt of te midsegment is 1 (b 1 b ). Terefore, by substitution, A (midsegment)(eigt). Tus, te re of trpezoid wit eigt 15 m nd midsegment wit lengt 3 m is (3)(15) 480 m. 19. AB 17.0 cm, AG 6.6 cm. First clculte te re of ABC using AC s te bse nd BH s te corresponding ltitude. Are of ABC 1 b 1 (7.1)(1.0) 4.6 cm Te re of te tringle is te sme for ny coice of bse, so we cn use tis result to find AB nd AG. First find AB. Notice tt CE is te ltitude to AB. 16. Smple nswer: A 1 b A 1 (AB)(CE) Substitute AB for b nd CE for. 17. Te re of ec squre is cm. Subtrct te re of te circle or circles from te re of te squre to find te re of te sded region.. (144 36) cm ; 78.54%. Te rdius of te circle is 6 cm, so its re is 36 cm, nd te re of te sded region is (144 36) cm. Now compre te re of te circle to te re of te squre. re of circle re of squre 3 6 cm 144 cm 78.54% AB 5.0 Substitute 4.6 for A nd 5.0 for. 85. AB 5.0 Multiply by. AB 17.0 cm Divide by 5.0. Now find AG. Notice tt AG is te ltitude to BC. A 1 b A 1 (BC)(AG) Substitute BC for b nd AG for (13.0)AG Substitute 4.6 for A nd 13.0 for BC. Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press

19 AG Multiply by. AG 6.6 cm Divide by 13.0 nd round to te nerest tent True. If r, ten r 48 cm. 1. True. If 36 0 n 4, ten n 15.. Flse. It could be rombus. 3. True. Tringle Inequlity Conjecture. IMPROVING YOUR REASONING SKILLS Degrees in n Acute Angle of n Isosceles Rigt Tringle. 7 Sides of Heptgon Degrees in ec Angle of Rectngle 4. 5 Digonls in Pentgon EXTENSIONS A. Reserc results will vry. B. Tis metod works becuse R r (R r) (R r)(r r) (R r ) R r. C. R r is te verge of te two circumferences. EXPLORATION GEOMETRIC PROBABILITY II PROJECT Project sould stisfy te following criteri: Predictions mention possible outcomes nd most likely outcomes. Irregulrities in weigt nd spe of dice re minimized. Te istogrm s rnge reflects te minimum nd mimum possible outcomes nd s te pproimte correct spe. Simultion results rougly mtc te teoreticl probbility. Report mkes connection between te number of trils nd ow closely te outcomes mtc te epected vlues. Etr credit R R r Unusul dice re eplored. r Report compres teoreticl outcomes for severl kinds of dice. LESSON 8.7 EXERCISES cm. Te solid is cube. Ec of te si fces is squre wit re (5)(5) 5 cm, so te surfce re of te cube is 6(5) 150 cm cm. Te solid is rectngulr prism, so it s si fces: two squres wit sides of lengt 37 cm, nd four rectngles, ec wit dimensions 37 cm by 9 cm. To find te surfce re of te prism, dd te res of te si fces: (37 37) 4(37 9) , so te surfce re of te rectngulr prism is 4070 cm cm. Te solid is tringulr prism, so it s five fces: one rectngle wit bse of lengt 6 cm nd eigt 7 cm, one rectngle wit bse of lengt 8 cm nd eigt 7 cm, one rectngle wit bse of lengt 10 cm nd eigt 7 cm, nd two tringles, ec wit bse of lengt 8 cm nd eigt 6 cm. To find te surfce re of te prism, dd te res of te five fces: (6)(7) (8)(7) (10)(7) , so te surfce re of te tringulr prism is 16 cm cm. Te solid is squre pyrmid. Use one of te formuls from Investigtion 1 on pge 464 of your book. Here 5 (te potem connects te center of te squre to te midpoint of its side), l 1, nd P nb 4(10) 40. Substitute tese vlues in te formul SA 1 P(l ) to obtin SA 1 40(1 5) 340. Tus, te surfce re of te squre pyrmid is 340 cm cm. Te solid is cone. To find its surfce re, use te formul from Investigtion on pge 465 of your book. Substitute 3 for r nd 8 for l in te formul SA rl r to obtin SA (3)(8) (3) Tus, te surfce re of te cone is pproimtely cm cm. Te solid is cylinder. To find its surfce re, use te formul from Emple B on pges of your book. Substitute 7 for r nd 0 for in te formul SA r (r) Tus, te surfce re of te cylinder is pproimtely cm cm. Te solid is egonl prism, so it s eigt fces: two egonl fces (te bses) nd si congruent rectngulr lterl fces wit sides of lengt 14 cm nd eigt 7 cm. To find te sum of te res of te two egonl bses, substitute 1.1 for, 14 for s, nd 6 for n in te formul A 1 sn, nd multiply by to obtin 1 (1.1)(14)(6) Becuse 1.1 is n pproimte vlue, te totl re of te egons is pproimtely 168 CHAPTER 8 Discovering Geometry Solutions Mnul 008 Key Curriculum Press

20 cm. Now find te sum of te res of te si rectngulr lterl fces: 6(14)(7) 588, so te totl re of te lterl fces is 588 cm. (Anoter wy to tink of tis is to imgine unwrpping te si rectngles into one rectngle. Te lterl re of tis unwrpped rectngle is te eigt times te perimeter, wic gives te sme result.) Terefore, te surfce re of te egonl prism is bout cm cm. Te solid is pyrmid wose bse is regulr pentgon, so it s si fces: te pentgonl bse nd te five tringulr lterl fces. To find te re of te bse, substitute 11 for, 16 for s, nd 5 for n in te formul for te re of regulr polygon: A 1 sn 1 (11)(16)(5) 440. Terefore, te re of te bse is pproimtely 440 cm. Find te re of ec tringulr lterl fce: 1 sl 1 (16)(15) 10, so te re of ec lterl fce is 10 cm, nd te totl re of te five lterl fces is cm. Tus, te surfce re of te pyrmid is pproimtely cm cm. Drw nd lbel ll te fces of tis solid. 8 4 Top nd bottom 9 C 8 Outer surfce C 4 To find te re of te bottom surfce, find te re of n nnulus wit outer rdius 4 nd inner rdius : (4) () Te re of te top surfce is te sme, so te sum of te res of te top nd bottom surfces is 4. Te re of te outer lterl surfce is 8 9 7, nd te re of te inner lterl surfce is Terefore, te surfce re of te solid is cm cm. Te solid is rectngulr prism wit ole troug its center. First find te res of te rectngulr fces. Te top nd bottom fces ec ve re lw, so te sum of te res of tese two fces is (8)(4) 64 cm. Te side fces ec ve re w, so te sum of te res of tese two fces is (10)(4) 80 cm. Net find te res of te front nd bck fces by subtrcting te re of circle wit rdius cm from tt of rectngle wit re l. Te sum of te res of tese two fces is (l r ) [(8)(10) () ] Now find te surfce re of te ole. Notice tt te ole is cylinder wit rdius cm nd eigt 4 cm. Te lterl surfce re of tis cylinder is r ()(4) 16. Add ll te res tt you ve found to obtin te totl surfce re of te solid: (160 8) 16 9 Inner surfce Te surfce re of te solid is pproimtely 39.1 cm. 11. Are of squre 4 re of trpezoid 4 re of tringle 1. $1,570. Find te cost of te pint nd ten te cost of te singles. Pint: Use te mesurements given in te end view to find te re of ec end wll. Te re of ec end wll is te sum of te res of rectngle, trpezoid, nd tringle: (30)(4) 1 (1)(1 30) 1 (.5)(1) ft. Tere re two end wlls, so te sum of te res is (987) 1974 ft. Ec of te side wlls is rectngle wit b 40 ft nd 4 ft, so ec side wll s n re of (40)(4) 960 ft, nd te sum of te res of tese two wlls is ft. Terefore, te totl surfce re of ll te wlls is ft. Find te number of gllons of pint needed to pint tese wlls: 3894 ft 1 gl 50 ft 15.6 gl. Becuse te pint must be purcsed in gllons, round up to 16 gl. Te cost of 16 gl of pint t $5 per gllon is 16 $5 $400. Singles: Find te totl surfce re of te roof, wic is mde up of two rectngles wit dimensions 40 ft by 15 ft nd two rectngles wit dimensions 40 ft by 6.5 ft. (40)(15) (40)(6.5) , so te totl surfce re of te roof is 170 ft. Find te number of bundles of singles needed for te roof: 170 ft 1 bundle 1 00 ft 17. bundles. Becuse te singles must be purcsed in complete bundles, round up to 18 bundles. Te cost of 18 bundles of singles t $65 per bundle is 18($65) $1,170. Te totl cost for pinting te wlls nd putting new singles on te roof is $400 $1,170 $1, Smple nswer: 14. Smple tiling: / /4 4 Discovering Geometry Solutions Mnul CHAPTER Key Curriculum Press