MATH 310 Ë Spring 2005 Ë Test #3 Ë NAME: If possible, show your w ork neatly on t his paper. If necessary, use additional pages wit h careful numbering.(10) 1. A cy lindrical shaf t w it h a 3-met er diam eter w as driven vert ically int o t he ground. We hav e beenunable to measu re the depth of this shaft, but at 11:00 AM, the sunlight just reaches the base ofthe opposit e w all (see illust rat ion). A t the same t ime, a 40 -f t high flagpo le is casting a 2-ft longshadow on lev el gro und. How deep is the shaf t ?sunlightSolution: The light that ends the shadow of the vertical flagpole is arriving at the sameangle as the light ending t he shadow in t he vert ical shaf t. So similar triangles are formed,with the third side of the smaller triangle being the shadow on the horizontal ground, andthe t hird sid e of the larg er t riangle, t he shadow ed horizont al bot t om o f the shaf t . ‘’‘’‘’‘’‘’‘’ ‘’‘’‘’‘’‘’Since cor responding part s of the sim ilar t riangles f orm the same r atios, w e have:2ft h 40ft which reduc es h 20))) = ))) ))) = ))) 3m 2ft to... 3m 1 hSo h = 3m (20 ) = 60 mNote that t he problem reduces to this: Since the flagpole height is 20 t imes the length ofits shadow , the depth of the shaft must be 20 times the length of its shadow . 3 m(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 a500 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 g(8) 3. How many liters of w ater are needed to fill a tank that is 1 meter w ide, half a meter high and half a meter deep?. Show the dimensional analysis that leads to your answer.Volume of tank is 1m x .5m x .5m = .25m3 Now to convert to lit ers: one mu st kno w some equic alenc y 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 3(12) 4. Find the volume of the prism shown at right .The BASE of t he prism is a right t riangle wit h height 30 dm 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 dm2 60dm wVolume of prism = (Area of Base) (Height) = ( ½ (40 dm)(30 dm ) ) (60 dm) = (600 dm2) (60 dm) = 36000 dm3 (5) 5. The base of a cone is a 160 cm2 region enclosed by a simple closed curve. The base has perimet er60 cm. Th e heigh t of t he cone is 4 0 c m. Find t he volume of the co ne. Volume of a cone or pyramid is (1/3) of the volume of the corresponding cylinder...That makes t he volum e (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 fiv e f aces of the small box on t op. .. and f ive f ull f aces of the large box,plus t he exp osed sh oulders of 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 sh oulders o f the t op of t he large box, t o obt ain a simple square region, 30 mm by 30 mm, so combined 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 b ox (underside) 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. . .(5) 7. Estimat e the area of t he figure in #7.There are 9 complete square units enclosed w it hin t he curv e,and 14 to 16 addit ional squares needed …