MATH 310 Ë SAM PLE TEST M easure Answ ers Ë 1a. A regular icosahedron has a 60 mL capacity.If a new regular icosahedron is constructed with every edge twice as long as the original, w hat is the capacity of the new icosahedron?If every edge is doubled, then all “lengths” and “heights” and “depths” are doubled, so the volume isdoubled, and doubled again, and doubled again. Thus the new icosahedron has a 480 mL capacity.1b. If the edge of a cube is increased by 2 c m, w hat is t he ef f ect on t he surface area of the cube?This is the “unansw erable one” . Act ually, it is answerable, however:The surf ace area is increased by 6 (2)(2cm)(old edge) + 6(2 cm)(2 cm).This can be demonstrated algebraically . The surface area of t he cube (w ith side “ s” ) is 6s2.With the edge increased by 2cm, the surface area becomes (s + 2cm)2.The difference in the surface area is 6(s + 2cm)2 – 6s 2 = 6(s 2 + 2(2cm)s + (2cm)2) – 6s 2If the edge of a cube is increased by 20 % , w hat is t he ef f ect on t he surface area of the cube?Original SA = 6(s 2) New SA = 6(1.20s)2 = 6s 2 (1.2 )(1.2 (1.2) = 6s 2 (1.72 8) = (old SA) C 1.728The surface area is multiplied by the factor 1.728, and thus increases by 72.8%.(12) 2. Convert each of the following, showing your work.a. 5.2 hm = dmWe need to know only that there are 100 meters in 1 hectometer, and 10 decimeters in a meter.5.2 hm = 5.2 hm 100m 10dm x )))) x )))) = 5200 dm1 hm 1 mb. 53 x106 cm3 = m3We need to know only that there are 100 centimeters in 1 meter.53 x106 cm3 53 x106 cm3 1m 1m 1m 53000000 m3 = x )))) x )))) x )))) = ))))))))))) = 53 m3 100cm 100cm 100cm 1000000c. 5 0 0 mL w at er (at 4°C) = ______ kg.We need to know t hat 1 mL [is 1 cc or cm3] and, if w ater at most dense point, has mass 1 kg.And.. .. w e need to know there are 1000 grams in a kilogram.500 mL w ater@4 °C 500 mL w ater@4 °C 1 cc 1 g 1kg = x )))) x)))))))))) x ))))))) = ,5 kg mL 1cc w ater@4 °C 1000 g3. How many liters of water are needed to fill a tank 1 meter wide, ½ meter high and ½ met er deep? Show t he dimensional analysis that leads to your answer.Volume of tank is 1m x .5m x .5m = .25m3 Now t o convert to lit ers: one must know some equiv alency bet w een lit ers and ordi nary volum e.That might be that 1 cc ( 1 cm3 ) is 1 mL (one cubic centimeter is one milliliter.).25 m3 = .25 m3 (100cm)3 1 mL 1 Liter x ))))))) x )))))))))) x ))))))) = 250 Liters m 3 1cm3 1000 mLOr it might be that you know that 1 cubic met er is 10 00 liters..25 m3 = .25 m3 10 00 Liters x )))))))))) = 250 Liters m 34. Find the volume of the prism shown at right .The BASE of t he prism is a right t riangle wit h height 30dm 50 dmand hypot enuse 50 dm . The third side, w hich is t he basew idth of the t riangle, has length 40 dm.(30dm)2 + w2= (50 dm)2 30dmw2= 160 0 dm 2 60 dm wVolume of prism = (Area of Base) (Height) = ( ½ (40 dm)(30 dm ) ) (60 dm) = (600 dm2) (60 dm) = 36000 dm3 5. The base of a cone is a 160 cm2 region enclosed by a simple closed curve. The base has perimeter 60cm. The height of the cone is 40 cm. Find the volume of the cone. Volume of a cone or pyramid is (1/3) of the volume of the corresponding cylinder...That makes the volume (1/3 ) (Area of Base) (Height)= (1/3) (160 cm2) (40 cm)= 6400 cm3 )))))))3 160 cm2 160 cm2 ±(12 ) 6. Find the surf ace area of the solid illustrated below. Assume all angles betw een connected segments are right angles.We hav e five f aces of t he small box on t op. .. and f iv e full f aces of the large box,plus t he exp osed shoulders o f t he t op of the large box.How ever , w e can c ombine t he t op of the smal l box w it h t he exp osed shoulders o f t he t op of the large box, t o obtain a simple square region, 30 mm by 30 mm, so c ombined area is 900 mm2.Areas of lateral sides of the small box add up to 4 (12mm)(10mm) = 480 mm2 12mmAreas of lateral sides of the large box add up to 12mm 10mm4 (3 0 mm) (40 mm) = 4800 mm2Base of large box (Bot t om! ) has area (30mm)(30mm) = 900 mm230 mm Tot al Surf ace area = 40 mm900 mm2 + 480 mm2 + 4800 mm2 + 900 mm2 =7080 mm 2 30 mm. . . . . . .(– 3, 4) 7. . . . . . . . 8.. . . . . . . . 9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5, – 2) . . . . . . . . . . . . . . . . . . . . . . . . . In all the above, the area shown here . . is one square unit. . .7. Estimat e the area of t he figure in #7.AS SHOWN HERE* : t here are 9 complete square units encl osed w it hin the cu rve,and 14 to 16 addit ional squares needed to cover the part ial sq uares (2 shown) around the border...We assume t he addit ional squares aver age ½ u2 inside t he curve, so t he partial squares should t otal (½ )(14to 16 u2) …