Sample Test Measure, Answers

MATH 310 Ë Sample TEST Measure Ë Answers 1. Use the draw ing below to illustrate the relationship betw een cubic yards and cubic feet. 2. Show 9 the dimensional analysis for the conversion of 2 cubic yards to cubic feet. 2 yd3 = 54 ft 3 27 cubic feet are required to fill the cubic yard– “ three layers,w ith nine 1-foot-cubes making up each layer” . (The nine one-ft-cubes are visible on the top layer.) Here is the dimensional analysis: 2 yd3 = 2 yd3C )))) 3 ft 3 ft 3 ft 1 yd C )))) 1 yd C )))) 1 yd = 2C3 3 ft 3 = 54 ft 3 3. Convert each of the follow ing, show ing the dimensional analysis. 0.0052 a. 5.2cm = dam 5.2 cm = 5.2cmC )))))) 1 m C 1)))))) dam = 100 cm 10 m b. 0.67 m 3 = 670000 5.2 x10 !3 dam = cm 3 0.67 m 3 = 0.67 m 3 C100cm 100cm 100cm )))))) C ))))) C )))))) = 1 m 1 m 1 m OR 0.67 m 3 = 0.67m 3 C )))))) 100cm 3 = ‰ 1 m  c. 2500 mL w ater = 2500 mL = d. 36 ft 3 = 36 ft 3 = 0.0052 dam 0.67 C100 3 cm 3 = 670000 cm 3 6 .67m 3 C 10 cm 3 = )))))) 1 m3 .67C10 6 cm 3 = 670000cm 3 _____ kL 2500 mLC )))))) 1 L 1 kL )))))) = .0025 kL 1000 mL C 1000 L in3 36 ft 3 12 in 12 in 12 in )))) )))) )))) 1 ft 1 ft 1 ft = 62208 in3 4. Find the mass, in kilograms, of w ater (at 4EC) needed to fill a tank 1 meter w ide, 1 meter high and 1 meter deep, including the dimensional analysis that leads to your answ er. So w e have 1 cubic meter of w ater at its most dense state. We know that 1 cm 3 of such w ater has a mass of 1 gram, so if w e convert this to cubic centimeters, w e can then convert to grams. 1 m 3 = 1 m 3 C 100cm 100cm 100cm 1 g3 1))))) kg = 1000000 kg )))))) C )))))) C )))))) C ))))) C 1000g )))))))))) = 1 m 1 m 1 m cm 1000 1000 kg Notice at this point, w e have 1,000,000 cm 3 ... and at the next point, w e have 1,000,000 grams Of interest: 1000 kg is a M ETRIC TON. 1 cubic meter of w ater (at its most dense) has a mass of one metric ton. How many pounds (approximately) is a metric ton? 5. A REGULAR OCTAHEDRON has 10cm edges. Show your work on each of the following: If w e slice through the regular octahedron on a plane containing four edges (think equator), w e see a square. Find the diagonal length of that square. D 10cm The edges of that square are 10cm each. 10 cm The diagonal is the hypotenuse of a right triangle w ith sides of length 10cm. D2 = (10cm)2 + (10cm)2 D2 = 2 (10cm)2 or D2 = 200 cm 2 D = 10 %& 2 cm D = %& 2& 0& 0 cm or Find the height of the octahedron. There are a number of w ays to do this. The SIMPLEST is to recall that in a regular polyhedron, all edges, angles, vertices are congruent. So the distance from vertex to diametrically opposite vertex is the same for any of the three diametrically opposed pairs of vertices. Therefore, the “ HEIGHT ” of the regular octahedron is the same as the “ DIA M ETER” of the octahedron, 10%& 2 cm or %& 2& 0& 0 cm. 10cm METHOD 2: Using the diagram at right, w e see that the height of the TOP HALF of the octahedron is part of a RIGHT TRIA NGLE w ith HYPOTENUSE 10 cm, and BA SE half the diameter w e found above. h 2 + (5%& 2 cm)2 = (10cm)2 h 2 + 50 cm2 h 2 = h And of course the = 100 cm 2 50 cm2 = 5%& 2 cm HEIGHT OF THE OCTA HEDRON is tw ice this value, or 10%& 2 cm Find the volume of the octahedron. The octahedron’ s volume can be view ed as that of tw o square-based pyramids joined at their bases. Thus volume is one-third of the product of the area of base and the height. V = (a) (Area of Base) (Height) V = (a) (10 cm)2 (10%& 2 cm) V = 1000 %& 2 cm3 ))))))) 3 6. Find the ENTIRE SURFA CE A REA of the SOLID object illustrated below . Assume all curves are circular, and all edges that appear vertical are. 2 0m There are four surfaces to add up: ² 4m (1) The “ roof” : consisting of a 20m circular disc surrounded by a ring extending the disc to 30m diameter. Area is B (15m)2 = 225B m 2 (2) Upper lateral w all: a band* 20Bm long, 4m high. Area = (20Bm )(4m) = 80B m 2 (3) Low er lateral w all: a band* 30Bm long, 22m high. Area = (30Bm)(22m) = 660B m 2 (4) The base (not seen in this view ): same size as the “ roof” . 22m 30m Thus the entire surface area is 225B m 2 + 80B m 2 + 660B m 2 + 225B m 2 = 1190B m 2 SA = * band = long rectangle, such as the label w rapped around a can. 7. Find the volume of the observatory illustrated at right, consisting of a 10-foot high right circular cylinder topped by a hemispherical dome w ith diameter 12 feet. (Ignore t he st eps!) See also http://commons.w ikimedia.org/w iki/Image:Aldershot_observatory_01.JPG The volume of the observatory may be found via Volume of hemispherical top + Volume of cylindrical base The diameter of both is 12 feet. The height of the cylinder is 10 feet. V hemisphere = (½) V sphere = (½) (4 / 3 ) B r3 = (½) (4 / 3 ) B (6ft)3 = 144B ft 3 V cylinder = (Area of Base)(Height) = B r2 (Height) = B ( 6 ft )2 (10 ft ) = 360B ft 3 Volume of the observatory = V hemisphere + V cylinder 8. = 144B ft 3 + = 504B ft 3 360B ft 3 What is the Surface Area (of the exposed surface) of the building above? (Ignore t he st eps! ) As in #7, w e find tw o parts— SA of hemispherical dome & SA of cylinder’ s lateral w all SA dome = (½) of SA sphere = (½) 4 B r2 = (½) 4 B (6 ft)2 = (½) 144 B ft 2 = 72 B ft 2 Lateral SA cylinder = Circumference CHeight = 2 B r h = 2 B (6 ft) (10 ft) = 120 B ft 2 Surface Area = SA dome + LSA cylinder = 72 B ft 2 + 120 B ft 2 = 192 B ft 2 9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 0 .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . In all the above, the area show n here is one square unit. . . . . . . . . . . . . . . . . . . . . . . . . . . 11. . . . . . . . . . . . . . . . . . . . . . . . . (-4 ,-3 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(4 ,– 3 ) . . 9. ESTIM A TE THE A REA of the figure in #9. We can see 3C4 + 2 = 14 square units lie entirely w ithin the curve, and sixteen square units have part w ithin the curve, and part outside it. We estimate that the partial squares w ithin the curve average half the area of each of those sixteen squares, so that half of sixteen square units lie w ithin the curve. Thus our estimate is 14 + (½) 16 square units = 22 square units. 10. FIND THE A REA enclosed by the figure in #10. (Curve turns at points: (7,1) & (6,7) & (1,7) & (2,5) ) We begin w ith the area of the rectangle covering the entire polygon, then subtract the areas of the four peripheral figures that are outside the given curve. A curve = A rect angle – (A Upper right t riangle + A Low er right rect angle + A large t riangle + A Lef t t riangle) = 6C6 u2 = 36 u2 – ( (½) 2C1u2 – ( 1u2 + + 4C1 u2 4 u2 + + (½) 5C4 u2 10 u2 + + (½) 1C6 u2 ) 3 u2 ) = 18 u2 11. FIND THE PERIMETER of the triangle in figure #11 above PERIMETER is distance around triangle = sum of lengths of sides of triangle P = length from (– 4,– 3) to (4,– 3) + length from (4,– 3) to (0,0) + length from (0,0) to (– 4,– 3) To find the latter tw o lengths, w e use the Pythagorean theorem: c2 = (3u)2 + (4u)2 = 25u2 c c 3 c = 5 u ( & c= 5 also ) 4 P= 8 units + 5units + 5 units = 18 units 12. Place the follow ing measurements in order, least to greatest: a. .003 km, 0.5 m, 2 in, 10 cm 2 in < 10 cm < 0.5 m < .003 km .1 m 3 m b. 0.5 qt, 1 gal, 5 dL, 100 mL, 900 cm 3 100 mL < 0.5 qt < 5 dL < 500 mL ½L 900 cm 3 < 1 gal close to 1000 cc 4 qts w hich is 1 Liter close to 4 Liters + 4 13. Find the volume of a 40E w edge of cake cut from a 8" round cake that is 1 5 /B" high. (If the cake averages 25 calories per cubic inch, how many calories is that?) A 40E w edge is 40 / 3 6 0 or 1 / 9 of the cake... The volume of cake is that of a cylinder w ith base radius 4" , height V= 1 / 9 (Area of base) (Height) = = 1 / 9 C B (4 in)2 = 80 15 1 15 /B" . / 9 C B (radius)2 Height /B in / 3 in3 = 26b in3 (A t 2 5 c alories per c ubic inc h, t hat ’ s 6 6 6 .6 6 6 @@@ calories. I see w hy t hey call it .... chocolat e.) 1 4 *. A cylinder w as designed to hold 200 mL. If a new cylinder is designed w ith the diameter doubled, and the height tripled, w hat is the capacity of the new cylinder? V cylinder = B (Radius)2 Height = B (R)2 H Doubling the diameter doubles the radius. The volume of the new cylinder w ith double radius and triple height is V cylinder = B (2 Radius)2 (3Height) = B (2R)2 (3H) = B 4R2 3H = 4C3 B R2 H = tw elve times the original volume. Therefore... The new cylinder holds 12C200mL = 2400 mL 15 . Find the AREA of the figure at right, given it w as constructed by pasting four circles of diameter 4cm atop a square of side 4 cm, so that each circle’ s center is located at a vertex of the square. Adding the area of the square to the area of the four circles counts the area w here the square overlaps the circles tw ice. To correct this, w e subtract the areas of the four quarter-circle overlaps. Area shaded = Area of square + 4C Area of circle – Area of overlap = Area of square + 4C Area of circle – 4C ¼ C Area of circle = Area of square + 3C Area of circle = ( 4 cm ) 2 + 3 C B ( 2 cm )2 = (16 + 12 B ) cm 2 Note: There are several other w orkable methods, including noticing that the shaded region consists of four circular discs, plus the middle section that looks like h.The area contained in that section is (16 !4B )cm 2 (see problem #12, replacing 2m to 4 cm). Total then w ould be 4C B ( 2 cm )2 + (16 ! 4B )cm 2 . Find the perimeter of the figure. The perimeter consists of 4 arcs, each arc 3/4 of the circumference of the circle. P = 4 C 3 / 4 C 2 B ( 2 cm ) = 12 B cm