MATH 3 10 New Improved MEASURE Self-Test ANSW ERS 1. a. (½)CB(4u)2 + [(8u)2 ! B(2u)2] = (4 B+ 64)u2(see diagram `) 1a. ²(½) B(2u)2b. Exactly 7.5 u2 [ 4C6 – 4C3C½ – 1C 3C½ – 6C 3C½ = 7.5 ] ²(½) B(2u)2(work from outside in...) ü (½)CB(4u)2 (8u)2 — That üc. 10u2 + (½ )(10u2) = 15u2. . . . . . . . d. The perim eter is ( 5 + r &1&&0& + r &4&&5& ) units (Use the Pythagorean theorem.) . . . . . . . . 1b. . . . . . . . . 3. a. h@(a + b)/2 b. Br2c. (xo/360o)C Br2. . . . . . . . d. C = 2Br e. V = (4/3)Br3SA = 4Br2. . . . . . . . ))))))) )4. a. r 42 + 82 u = 4r 5 units. 2.b. D[(!1,4),(2,0)] = 5; D[(2,0),(!2,!3)] = 5; D[(!1,4),(!2,!3)] = 501/2 . Yes: d12 + d22 = d32.5. (12 cm)2 ! B(6 cm)2 = (144 ! 36B)cm26. V o f p rism o r cy linder = (Area o f b ase)C(Height) = 8000 m2 7. a. (1/3)C(30 cm)220 cm = 6000 cm3Volume of pyramid or cone = (1/3) (area of base)(height)b. (1/3)C(80 cm2)C20cm = 1600 /3 cc. (“cc” is an alternate abbreviation for cm3 – cubic cen tim eters)8. a. V = (1/3)CB(5 cm)2C25cm = 625 B/3 cm3lateral SA = (1/2)CB(10 cm )(5r &2& &6& cm) = 25r &2&&6& Bcm2b. V = B(5 cm)2C25cm = 625B cc (cc is abbreviation for cm3)lateral SA = B(10cm)C25cm = 250Bcm2 radial faceþ þpart of Circumference9a. SA = top+ bottom + radial faces + strip of C V = one-six th of the V o f t he origin al w heelSA = 2C(15cmC3cm) + 2C(1/6)CB(15cm)2 + 3C30Bcm2/6 V = (1/6)CB(15cm)2C 3cm = 225B/2cm39b. SA = 2C(15cmC3cm) + 2C(30/360)CB(15cm)2 + (1/ 12 )3C30Bcm2V = (30/360)CB(15cm)2C 3cm = 225B/4 cm330 ° is 1/1 2 of the c ircle, so V is half V of part a10. SA = (1/2)C4B(5cm)2 + (1/2)CB(10cm)C13cm = 115Bcm2V = (1/2)C(4/3)B(5cm)3 + (1/3)CB(5cm)2C (12cm) = (550B/3)cm3 11. The relationship between oC and oF is a line going from (0,32) to (100,212); that makes a rise of 100oC equal to a rise of 180oF: i.e. 100oC = 180oF, so ( 100oC/180oF ) = 1!.100oC/180oF reduc es to 5oC/9oF. An d, y es, 5oC/9oF = 1 also. So oC = 5oC/9oFC(oF ! 32oF).(Not ice oF must be adjust ed dow n t o 0 bef ore m ult iplyin g, so th at 3 2oF will end up being 0oC.).C = (5 /9 )(F ! 32) a. 72oF = 22oC (18 °Reaumur) c. 98.6oF = 37oC F = (9/5 )C + 32 b. 30oC = 86oF12. a. cm b. cm c. m d. km e. kg f. g g. g (act ually! mg, but t hat ' s not in t he list )h. t i. mL or cc j. L k. mL or cc l. kL13. 1 kL = 1000 L = 1000C1000 mL = 106mL = 106cc =j 106g = 100 0 kg = 1 metric ton. OR:1 kL = 1 kL C 1000L/kL = 1000 L = 1000L C 100 0 mL/L = 100 000 0 mL = 100 000 0 cc = 100 000 0 g =100 000 0g C 1kg/1000 g = 100 0 kg = 100 0 kg C 1 metric ton/1000 kg = 1 metric tonj valid ON LY f or w at er at 4°C14. a. Carpet cost s $32/yd2; wood costs $4/ft2 = $4/ft2C3ft/ydC3ft/yd = $36/yd2; carpet is cheaper.b. 1 ft2 = (12 in)2 = 144 in2; 1 yd2 = (3 ft)2 = 9 ft2c. (1 ft)3 = (12 in)3 = 1728 in3; (1 yd)3 = (3 ft)3 = 27 ft3; 1 yd3 = 27 ft3 = 27C1728 in3d. 1 m3 = (100 cm)3 = 1000 000 cm3; (see #13 )e. (see #13; t he conversion at * is valid for water only!)f. 300 dam g. 200 cm2 = 200 cm2 C 1m C 1 m. = .02m2h. 3 28 dl = 32 8d l C100mlC1cc = 3000 m 100cm.C100 cm dl ml = 30000 cm = 32800 cc15. Just barely n ot . Th e longest dimen sion in a 2 4" x7 " x7 " box is the ext reme d iagonal, (67 4)½ “ Ñ 25.96" 16. New area = 100cm2 C 2.5 C 2.5 = 625 cm2M AT H 3 1 0 Self -T est M EAS URE ad dit iona l not es reg ardin g solut ions . . . . . . . . . . . . . . 1. a. Find b. appro xim ate a. b. . . . . . . . c. . . . . . . . th e area enclosed by t he 8 4B u2 . . . . . . . . . . . . . . simple closed curve sketched. 6u . . . . . . . . . . . . . . In part a, all curv y part s are 9 A2. . . . . . . . . . . . . . ¼ or ½ circles w ith equal radii. . . . . . . . . . . . . . . One Solution: 78u6 . . . . . . . . . . . . . . For the Area bounded by all those arcs, . . . . . . . . . . . . . . I see tw o full circular regions, plus the region marked A2. . . The radius of each arc is 2 units, so the area of each . . = 1 sq unitcircular region is 4B u2 .The region marked A2 is similar to the shaded region described and illustrated in #5. Each of th e four quarters o f t his region is th e area betw een t he vert ex of a 4-un it sq uare, and t he insc ribedcircle... Area sh aded is area of square ! area of circle... (4 u nit s)2 ! B (2 unit s)2 16 u2 ! 4 B u2So t he t ot al area enclosed by t he heart -shaped f igure c om posed o f c ircular arc s is: 2( 4B u2 ) + ( 16 ! 4B ) u2 This is the ( 4B + 16 ) u2 OLD #1 !b. W ith in the bou ndary , w e see 11 fu lly enc losed squ ares, and 18 part ial squares.We assume t he part ially enc losed squ ares w ould av erage out to being 5 0% contain ed w ith in the boundary , so w e estim ate t he area inside t he bou ndary to be appro xim ately ( 11 + 18 /2 ) u2.c. T o f ind t he area of th e region en closed by t he poly gon joining th e respect ive v ertic es in t he Cart esianpl an e, m ak e s ur e …