TEST Measure, Answers

MATH 310 Ç TEST Measure SAMPLE ONLY Á 1a. È Answers É Illustrate the relationship betw een square inches and square feet. 1 in 2 ü In the sketch w e can see that 1 square foot is 144 square inches. Dimensional analysis says: 1 ft 2 1 ft 2 12 in 12 in = ))) x )))) x )))) = 144 sq in ft ft 1b. Which tile price is less expensive: $5 ' ft 2 , or $0.10 ' inch2 ? .10 .10 144 in2 ))2 = ))) x )))) = 2 in in ft 2 .10C144 ))))) = $14.40 per square foot NOT cheap !! ft 2 1 foot = 12 inches But 1 ft 2 = 144 in2 Tile at $5 per sq ft is much less expensive than tile at $14.40 per sq ft. 1c. Show the dimensional analysis for the conversion of 10 cubic yards to cubic feet. 10 yd3 2. 10 yd3 ))))) = 3 ft 3 ft x ))) x )))) x yd yd 3ft yd )))) 270 ft 3 = Convert each of the follow ing, show ing the dimensional analysis. a. .22km = 22 dam 0.22 km = .22 km x 1000m ))))) )))))) x 1dam ))))) = km 10m 0.22 km = .22 km x 100dam ))))) )))))) = km 5000 cm3 = b. 5000 cm3 c. 22 dam 22 dam ...If you happen to know 100 dam = I km. m3 0.005 3 1m 1m 1m = = 5000 cm x )))))) x ))))) x ))))) 100cm 100cm 100cm 356 g w ater = 356 g = 356 g x ...Since w e know 1000m = 1km and 10m = 1 dam 5000= 0.005 m 3 ))))) 6 10 0.356 L 1 mL x )))))) 1L ))))) 1g 1000 mL = 0.356 L Water at its most dense: 1 g = 1 cc = 1 mL .... or 1000cc = 1 dm 3 = 1 L = 1 kg 3. Find the mass of w ater (at 4EC) needed to fill a box-tank 2m w ide, 2m high and 2m deep, including the dimensional analysis that leads to your answ er. Volume is (2m)(2m)(2m) = 8 m3 8 m3 = 8 m3 ))))) x 10dm m )))))) 10dm 10dm x ))))) x ))))) = m m 8000dm3 = 8000L = 8000 kg OR 8000 dm 3 = 8000 dm 3 If w e know : 1 cc = 1 g = 1 mL* (* for WATER at 4EC) x 10 cm ‰ 1 dm  )))))) 3 = 8000000cm3 = 8000000 g Or 8000 kg Or 8 metric tons If w e know 1 dm3 = 1 L and 1L* = 1000 g or 1 kg. 4. A toy pyramid has a square base that is 6in on a side. The pyramid is 4in high. Find the volume of this pyramid, show ing every detail of the w ork, w ith proper units. V pyramid = (a) (Area of base)(Height) 4" = (a) (6 in @ 6 in) (4 in) = (a) (36 in2 ) (4 in) 3" 6" 48 in3 = Find the surface area of the entire pyramid. Show your w ork. 6" 4" l We w ill need the height (l) of the triangular lateral faces of the pyramid. 3" The height of the pyramid is measured perpendicular to the base, rising from the center, assuming this is a right pyramid. So the segment joining the height and the slant height l completes a right triangle w ith dimensions show n. Since this is a right triangle, by the Pythagorean theorem, l2 = (4" ) 2 l2 = 25 in2 (3" ) 2 + l 4" ² one of the four lateral faces of the pyramid 5" l = 5 in 3 “ Surface Area of Pyramid = = 6. 6 in Area of square base + Area of the four lateral triangular faces (6 in) (6 in) = 36 in2 = 96 in2 + 4 @ ( (½ ) (6 in) (5 in) + 60 in2 A rectangular building in the shape of a box, 50m long by 30m deep, by 20m high, is topped by a hemispherical copper dome w ith diameter 20m. Find the entire surface area of the building w hich comes into contact w ith the air. Surface Area = Front+ Back 20 m + Tw o “ Sides” = 2@(50m)(20m) + 2@(30m)(20m) + (50m)(30m) – = 2@1000 m2 + 2@ 600 m2 = ( 4700 + 100 ð ) m 30m 50 m + Flat roof less covered part + (½) Dome 2 + 1500 m2 ð (10m)2 – 100 ðm 2 + (½) 4 ð (10m)2 + 200 ð m2 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . In all the above, the area show n here is one square unit. (3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9. . (!5 ,2 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(5 ,2 ) . . . . .(5 ,!2 ) . 7. Estimate the area inside the figure in #7. There are 22 square units clearly contained entirely inside the figure, and 16 squares each of w hich is partially inside, partially outside the figure. Assuming that these “ partial squares” average half inside (& half outside) the figure, w e estimate the area inside the figure to be about 22 u2 + (½) 16 u2 30 u2 = ( In f ac t t he area inside is ex ac t ly 3 0 u 2 , as t he f igure is 5 unit s high at ev ery point across it s 6 -unit w idt h ! ) (5) 8. Find the area enclosed by the figure in #8. (Curve turns at points: (1,1) & (6,4) & (1,7) & (2,5) ) We w ork from the outside in. The area of the rectangle w hich just encloses the figure is 5u @ 6u = 30 u2 From that w e subtract the four triangular regions show n. Area inside polygon = 30 u2 — ( (½) 1@2 + (½) 1@4 + (½) 5@3 + (½) 5@3 ) u2 ) = 30 u2 — 18 u2 (5) = 12 u2 9. Find the PERIMETER of the triangle in figure #9 above P = 10 u + 4 u + = 10 u + 4 u + Distance from (!5,2) to (5,2) . 10u %& 1& 1& 6 u 4u = (14 + %& 1& 1& 6 )u c2 = 10 2 + 4 2 c2 = 116 c = %& 1& 1& 6 (unit s) c < — — — 17 mm (6) 10 . — — — > Find the AREA of the figure at right, given all arcs are semicircular. This area comprises tw o semicircular regions. The larger has diameter 17 mm, the smaller has diameter 7 mm (17 ! 10 mm). The radii are 8.5 mm and 3.5 mm, respectively. 5m m Area enclosed = Area of karger half-circular region + Area of smaller half-circle = ( ½ ) ð R2 + ( ½) ð r2 = ( ½ ) ð(8.5 mm) 2 + ( ½ ) ð (3.5 mm) 2 = ( ½ ) ð 72.25 mm 2 + ( ½ ) ð 12.25 mm) 2 = 42.25 ð mm 2 5mm (4) 11. Find the area w ithin a 30E sector of a circle w ith radius 6m. The area w ithin the ent ire circle of radius 6m is A 30E sector contains 30 1 /3 6 0 = ð (6m) 2 or 36ð m 2 / 1 2 of the circle. 30E 6m So the area w ithin a 30E sector must be 1 / 1 2 of 36ð m 2 = (2) 3ð m 2 ð r2 12. Write an expression for the Area w ithin a circle of radius r: Write an expression for the Perimeter of a circle of radius r: 2ðr 4 ) ð r3 3 Write an expression for the Surface Area of a sphere of radius r: 4 ð r 2 Write an expression for the Volume of a sphere of radius r: (6) 13. Find the area of the shaded region (w ithin the circle, outside the square). (given the circular curve is a circle, passing through vertices of the square.) Area shaded = Area w ithin circle, but outside of square = Area inside circle = ð r2 ! = ð 8 m2 ! = (4) ! 4m Area of square ( 4m ) (4 m) ... but w hat is r? 16 m2 ! 16) m2 (8ð 4m Well, of course r is the radius of the circle. But w here is r In this picture? Hint: don’ t look at the red radius, look at the dotted diagonal. How long is the diagonal? %& 3& 2 = 2%& 8 14. Given the 10m high prism at left has an octagonal base w ith perimeter 44m and area 49 m2 , a. find the volume of the prism. V prism = (Area of base) (Height) = ( 49 m2 ) ( 10 m) = 490 m 3 b. find the surface area of the prism. SA prism = Area of base + Area of top + Area of Lateral sides = 49 m 2 + 49 m 2 = 49 m 2 + 49 m 2 = (4) Picture a + Perimeter @ Height— piece of paper w rapped all the + 44m @ 10m w ay around the side... 538 m2 15. The right circular cone show n has capacity 42 mL. A new cone is being designed. If the diameter and the height are both tripled, w hat w ould be the capacity? Diameter tripled ÷ radius is tripled ÷ area of base is multiplied by 9 Since volume of cone is (area of base) (height)/3 New volume is (9@old area of base)(3 old height) /3 = 27@ original volume If the diameter is doubled, and the height is cut in half, w hat w ould be the capacity? Diameter doubled ÷ radius is doubled ÷ area of base is quadrupled (@4) New volume is (4@old area of base) ( (½) old height) /3 = double original volume.