MATH 310 TEST Measure April Answers

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1 MATH 310 TEST Measure April Answers 1. Use t he draw ing below to illust rat e the relat ionship betw een cubic yards and cubic f eet. 2. Show the dimensional analysis for the conversion of 2 cubic yards to cubic feet. 2 yd 3 = 54 ft 3 27 cubic feet are required to f ill 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 yd 3 = 2 yd 3 3 ft 3 ft 3 ft )))) )))) )))) 1 yd 1 yd 1 yd = ft 3 = 54 ft 3 3. Convert each of the follow ing, show ing t he dimensional analysis. a. 5.2cm = dam 5.2 cm = 5.2cm 1 m 1 dam )))))) )))))) = 5.2 x10 3 dam = dam 100 cm 10 m b m 3 = cm m 3 = 0.67m 3 100cm 100cm 100cm )))))) ))))) )))))) = cm 3 = cm 3 1 m 1 m 1 m OR 0.67 m 3 = 0.67m cm.67m cm 3 )))))) = )))))) = cm 3 = cm 3 1 m 1 m 3 c ml w at er = kl 2500 ml = 2500 ml 1 L 1 kl )))))) )))))) =.0025 kl 1000 ml 1000 L 4. Find the mass, in kilograms, of w ater (at 4 C) needed t o f ill a t ank 1 met er w ide, 1 met er high and 1 meter deep, including the dimensional analysis that leads to your answ er. So we 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 cent imeters, w e can then convert t o grams. 1 m 3 = 1 m 3 100cm 100cm 100cm 1 g 1 kg kg )))))) )))))) )))))) ))))) ))))) = )))))))))) 1 m 1 m 1 m cm g 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: kg is a METRIC TON. How many pounds (approximately) is a metric ton?

2 5. A REGULAR OCTAHEDRON has 10cm edges. Show your w ork on each of the follow ing: If w e slice t hrough the regular octahedron on a plane cont aining f our edges (think equat or), w e see a square. Find the diagonal lengt h of that square. D 10cm The edges of that square are 10cm each. 10 cm The diagonal is the hypotenuse of a right triangle with sides of length 10cm. D 2 = (10cm) 2 + (10cm) 2 D 2 = 2 (10cm) 2 or D 2 = 200 cm 2 D = 10 2 cm or D = cm 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 t he dist ance from vertex t o diamet rically opposite vertex is t he same for any of the three diametrically opposed pairs of vert ices. Therefore, the HEIGHT of the regular octahedron is t he same as the DIAMETER of t he octahedron, 10 2 cm or cm. 10cm METHOD 2: Using t he 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 BASE half the diamet er w e found above. h 2 + (5 2 cm) 2 = (10cm) 2 h cm 2 = 100 cm 2 h 2 = 50 cm 2 h = 5 2 cm And of course the HEIGHT OF THE OCTAHEDRON is tw ice this value, or 10 2 cm Find the volume of the octahedron. The oct ahedron s volume can be v iew ed as t hat of tw o square-based pyramids joined at their bases. Thus volume is one-third of the product of t he area of base and the height. V = (a) (Area of Base) (Height) V = (a) (10 cm) 2 (10 2 cm) V = cm 3 ))))))) 3

3 6. Find the ENT IRE S URFA CE A REA of the SOLID object illustrated below. Assume all curv es are circular, and all edges that appear vertical are. 20m There are four surfaces to add up: 4m (1) The roof : consist ing of a 20m circular disc surrounded by a ring extending the disc to 30m diameter. Area is (15m) 2 = 225 m 2 22m (2) Upper lateral wall: a band* 20 m long, 4m high. Area = (20 m )(4m) = 80 m 2 (3) Lower lateral wall: a band* 30 m long, 22m high. Area = (30 m)(22m) = 66 0 m 2 30m (4) The base (not seen in this view ): same size as the roof. Thus the entire surface area is SA = 225 m m m m 2 = 1190 m 2 * 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 it h diamet er 12 feet. (Ignore the steps!) See also The volume of the observatory may be found via Volume of hemispheric al top + Volume of cylindric al base The diameter of bot h is 12 feet. The height of the cylinder is 10 feet. V hemisphere = (½) V sphere = (½) ( 4 / 3 ) r 3 = (½) ( 4 / 3 ) (6ft) 3 = 144 ft 3 V cyli nder = (Area of Base)(Height) = r 2 (Height) = ( 6 ft ) 2 (10 ft ) = 360 ft 3 Volume of the observatory = V hemisphere + V cyli nder = 144 ft ft 3 = 504 ft 3 8. What is the Surf ace A rea (of the ex posed surface) of the building above? (Ignore the steps!) As in #7, w e find tw o parts SA of hemispherical dome & SA of cylinder s lateral wall SA dome = (½) of SA sphere = (½) 4 r 2 = (½) 4 (6 ft) 2 = (½) 144 ft 2 = 72 ft 2 Lateral SA cyli nder = Circumf erence Height = 2 r h = 2 (6 ft) (10 ft) = 120 ft 2 Surface A rea = SA dome + LSA cyli nder = 72 ft ft 2 = 192 ft 2

4 (4, 3) (-4, -3) In all the above, the area shown here.. is one square unit ESTIMATE THE AREA of t he figure in #9. We can see = 14 square units lie entirely w ithin t he curve, and six teen square units have part w ithin t he curve, and part outside it. We estimate that the partial squares w ithin t he curve average half t he area of each of those six teen squares, so t hat half of sixteen square units lie w it hin the curv e. Thus our estimate is (½) 16 square units = 22 square units. 10. FIN D T HE A REA enclosed by the figure in #10. (Curve turns at points: (7,1) & (6,7) & (1,7) & (2,5) ) We begin w it h t he area of the rectangle covering the entire polygon, t hen subtract the areas of t he four peripheral f igures t hat are out side t he giv en curve. A curve = A rectangle (A Upper right triangle + A Low er right rect angle + A large triangle + A Left t riangle ) = 6 6 u 2 ( (½) 2 1u u 2 + (½) 5 4 u 2 + (½) 1 6 u 2 ) = 36 u 2 ( 1u u u u 2 ) = 18 u FIND THE PERIMETER of t he 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) t o (0,0) + length from (0,0) to ( 4, 3) To find the latter two lengths, we use the Pythagorean theorem: c 2 = (3u) 2 + (4u) 2 = 25u 2 c c 3 c = 5 ( & c= 5 also ) P = 8 units + 5units + 5 units = 18 units 12. Find the area of the shaded region (w ithin t he square, out side the circle). 2m Area shaded = Area inside square Area inside inscribed circle = (side) 2 (radius) 2 2m = 2 m 2m (1m) 2 = 4 m 2 m 2 = ( 4 ) m 2

5 13. Find the volume of a 40 w edge of cake cut from a 8" round cake that is 15 / " high. (If the cake averages 2 5 calories per cubic inch, how many calories is that?) A 4 0 w edge is 40 / 360 or 1 / 9 of the cake... The volume of cake is that of a cylinder w ith base radius 4", height 15 / ". V = 1 / 9 (Area of base) (Height) = 1 / 9 (radius) 2 Height = 1 / 9 (4 in) 2 15 / in = 80 / 3 in 3 = 26b in 3 (At 25 calories per cubic inch, that s calories. I see w hy they call it... chocolate.) 14*. A cylinder w as designed t o hold 2 00 ml. If a new cylinder is designed w it h t he diamet er doubled, and the height tripled, w hat is the capacity of the new cylinder? V cyli nder = (Radius) 2 Height = (R) 2 H Doubling t he diameter doubles t he radius. The volume of the new cylinder with double radius and triple height is V cyli nder = (2 Radius) 2 (3Height) = (2R) 2 (3H) = 4R 2 3H = 4 3 R 2 H = tw elve times t he original v olume. Therefore... The new cylinder holds mL = 2400 ml 15. Find the AREA of the figure at right, given it w as constructed by pasting four c ircles of diameter 4cm atop a square of side 4 cm, so that each circle s cent er is locat ed at a vert ex of the square. Adding t he area of t he square to the area of t he four circles counts the area w here t he square overlaps the circles t w ice. To correct this, w e subtract t he areas of t he four quarter-circle overlaps. Area shaded = Area of square + 4 Area of circle Area of overlap = Area of square + 4 Area of circle 4 ¼ Area of circle = Area of square + 3 Area of circle = ( 4 cm ) ( 2 cm ) 2 = ( ) cm 2 Note: There are several other w orkable methods, including noticing that the shaded region consist s of four circular discs, plus t he middle section that looks like.the area cont ained in that sect ion is (1 6 4 )cm 2 (see problem #12, replacing 2m t o 4 cm). Tot al then would be 4 ( 2 cm ) 2 + (16 4 )cm 2. Find the perimeter of the f igure. The perimet er consists of 4 arcs, each arc 3/4 of the circumference of the circle. P = 4 3 / 4 2 ( 2 cm ) = 12 cm