MATH 10 MEASURE Self-Test ANSWERS (DETAILED answ ers start on NEXT PAGE.) 1. a. (½) (4u) + [(8u) (u) ] = (4 + 64)u (see diagram ) 1a. Perimeter = 4u + 8u + ½ (u) = (8 + 16)u (½) (u) b. 7.5 u [ 4 6 4 ½ 1 ½ 6 ½ = 7.5 ] (w ork from outside in) (½) (u) Perimeter = ( 5 + 1 0 + 4 5 ) units (Use the Pythagorean theorem.) (½) (4u) (8u) That c. w hole + ½ partial = 10u + (½ )(10u ) = 15u........ d. l w h = 8lwh = 8 (original volume)........ (6+ 1)(8+ 1)(7+ 1) 6 8 7 + 1! It s 6 8 7 + 6 8 + 8 7 + 6 7 + 1 1b.......... a. h (a + b)/ b. r c. (x/60) r........ d. C = r e. V = 4 r f. SA = 4 r........ ))))))) ) 4. a. 4 + 8 u = 4 5 units.. 1/ b. D[( 1,4),(,0)] = 5; D[(,0),(, )] = 5; D[( 1,4),(, )] = 50. Yes: d + d = d. 1 5. Area in square Area in circle = (1 cm) (6 cm) = (144 6 )cm 6. V of prism or cylinder = (Area of base) (Height) = 8000 m 7. a. (a) (0 cm) 0 cm = 6000 cm Volume of pyramid or cone = (1/) (area of base)(height) b. (a) (80 cm ) 0cm = 1600/ cc. ( cc is an alternate abbreviation for cm cubic centimeters) 8. a. V = (a) (5 cm) 5cm = 65 / cm lateral SA = (1/) (10 cm)(5 6 cm) = 5 6 cm b. V = (5 cm) 5cm = 65 cc BTW: Total SA = ( 5 + 5 6 ) cm lateral SA = (10cm) 5cm = 50 cm BTW: Total SA = ( 5 + 50 ) cm radial face part of Circumference 9a. SA = top+ bottom + radial faces + strip of C V = one-sixth of the V of the original w heel SA = (15cm cm) + (1/6) (15cm) + 0 cm /6 V = (1/6) (15cm) cm = 5 /cm 9b. SA = (15cm cm) + (0/60) (15cm) + (1/1) 0 cm V = (0/60) (15cm) cm = 5 /4 cm 0 is 1/1 of the circle, so V is half V of part a 10. SA = (1/) 4 (5cm) + (1/) (10cm) 1cm = 115 cm V = V + V = (1/) (4/) (5cm) + (1/) (5cm) (1cm) = (550 /)cm hemisphere cone o o 11. The relationship betw een C and F is a line going from (0,) to (100,1); o o o o o o that makes a rise of 100 C equal to a rise of 180 F: i.e. 100 C = 180 F, so ( 100 C/180 F ) = 1!. o o o o o o o o o o o 100 C/180 F reduces to 5 C/9 F. And, yes, 5 C/9 F = 1 also. So C = (5 C/9 F) ( F F). o o o (Notice F must be adjusted dow n to 0 before multiplying, so that F w ill end up being 0 C.). 5 o o o o 9 9 o o 5 C = ( / )(F ) a. 7 F = C (18 Reaumur) c. 98.6 F = 7 C F = ( / )C + b. 0 C = 86 F 1. a. cm b. cm c. m d. km e. kg f. g g. g (actually mg, but that' s not in the list) h. t i. ml or cc j. L k. ml or cc l. kl 6 6 1. 1 kl = 1000 L = 1000 1000 ml = 10 ml = 10 cc = 6 10 g = 1000 kg = 1 metric ton. OR: 1 kl = 1 kl 1000L 1000 ml 1 cc 1 g* 1 kg 1 metric ton = 1 metric ton 1 kl 1 L 1mL cc 1000g 1000 kg valid ONLY for w ater at 4 C 14. a. Carpet costs $/yd ; wood costs $4/ft = $4/ft ft/yd ft/yd = $6/yd ; carpet is cheaper. b. 1 ft = (1 in) = 144 in ; 1 yd = ( ft) = 9 ft (Use the method show n in #1.) c. (1 ft) = (1 in) = 178 in ; (1 yd) = ( ft) = 7 ft ; 1 yd = 7 ft = 7 178 in = 46656 in d. 1 m = (100 cm) = 1000000 cm ; (see #1) = 1000000 ml e. (see #1; the conversion at * is valid for w ater only!) f..1 dal g. 00 dam g. 00 cm = 00 cm 1m 1 m =.0m h. 8dl = 8dl 100ml 1cc = 800 cc = 000 m 100cm. 100cm dl ml = 00000 cm 15. ½ Just barely not. The longest dimension in a 4" x7" x7" box is the extreme diagonal, (674) 5.96" 16. New area = 100cm.5.5 = 65 cm
M ATH 1 0 Self-Test M EASURE extended solutions w ith comments 1a. We assume arcs are semi-circular. Perimeters are show n on page )))) 8mm We view this as a half-circular region[i] (with diameter 8mm) 8mm joined to a rectangular region [II] (8mm by 8mm) )))) 6mm from w hich two small semicircular regions [III] have been removed radius of the small circles must be the difference 8mm 6mm= mm. (And must also be ½ of 8mm/.) Area I: radius is ½ of 8mm, or 4mm. Area of half-circular disc is (½ ) (4mm) Area II: area of 8m by 8mm square is 64mm 8mm Area III: area of tw o half-circular cutouts [diameter = 4mm]... is ½ (mm) Thus Total Area of figure is (½ ) (4mm) + 64mm ½ (mm) = (4 + 64)mm 1b......... We start w ith the smallest vertical-horizontal bounded rectangle that........ contains the polygon given (w hich happens to be a triangle, but that s........ not important, this w orks for any polygon w ith vertices on lattice pts)......... Area w ithin polygon = area of rectangle areas of take-aw ay parts:........ (4u)(6u) ½ (4u)(u) ½ (u)(1u) ½ (6u)() = 4 / = 7.5 u 1c......... We cover the region w ith squares......... We count the squares w hich lie entirely w ithin the region outlined......... We count the squares w hich lie partially w ithin the region outlined......... We then make the ESTIMATE:........ Area w ithin outline 10u + (½ )(10)u = 15u (approximate area) 1d. 8m 6m (This is based on the assumption that the squares w hich are partially enclosed average half inside, half outside) The volume of the box is (area of base)(height) = length w idth height 7m = (6m 8m) 7m = 6 m Doubling all the dimensions, w e can compute: NEW V= (1m 16m) 14m = 688 m... that is, 8 6 m In fact, w e could have anticipated this, since V = l w h = 8 l w h = 8 (original volume) If each dimension is increased by 1m, is the volume increased by 1m? Explain. 8m + 1m 7m 6m NO! The increase in volume is much, much greater than that! Just consider, first, the effect of increasing just ONE dimension e.g. increase the length from 8m by 1m to 9m. As illustrated at left, this w ould add a 1m by 6m by 7m slab to one end of the box. That alone, then, adds 4m of volume to the box. Here w e illustrate the addition of a 1m extension to the depth of the box. Just for the old box, this extension results in an additional 1m 8m 7m or 56m, and that does not even take into account the extension of the length from 8m to 9m! To account for that, w e need another 1m by 1m by 7m piece in the corner! Extending the height an additional 1m, from 7m to 8m results in a new slab added to the top of the box, 1m by 6m by 8m to cover the original box, but 1m x 7m x 9m to cover the box w ith its new extended length and depth. Picture it (draw the top extension in) yourself! Here s the entire difference, found algebraically: Original volume, V = l w h, After each dimension is increased by amount a: NEW V = (l+a)( w+a)(h+a) = l w h + l w a + l a h + a w h + l a a + a w a + h a a + a a a Difference: l w a + l a h + a w h + l a a + a w a + h a a + a a a
4. a. Sketch the points ( 4,5) and (4,1) in the plane. Find the distance betw een the points. (-4,5).......... The change in x = 5 1 = 4.......... The change in y = 4 ( 4) = 8.................... 4 + 8 = c.......... (4,1) c = 80.......... c = 8 0) = 1 6 5 = 4 5 (Units) b. Do points ( 1,4), (,0), (, ) lie at the vertices of a right triangle? How do you know? The solution is based on the Pythagorean Theorem: the sum of the squares of the tw o short legs is equal to the square of the hypotenuse if, and ONLY if, the triangle is a right triangle. So find the distances betw een the three pts. They are 5, 5, and sqrt(50). We find the pythagorean relationship holds true (5 + 5 = 5 0 ). So the triangle must be a right triangle. 5. Area inside the square is (1cm). D of the circle = 1cm, so A = (6cm). Now subtract! 6. No matter w hat shape the base of a prism is, the Volume of the prism is just the product: (Area of base) (Height) 7. No matter w hat shape the base of a pyramid is, the Volume of the pyramid is just (one third) the volume of the corresponding prism (See #6 note above!) These ideas apply to cylinders and cones as well. 8a. Volume of cone = ( a) (Area of Base)(Height) [see note above!] Finding the surface area of cone is similar to surface of pyramid. Pyramids have lateral faces that are triangular. The area of each triangle is (½ )(base length)(height)... When these are added up around the pyramid, the total (for lateral surface area) is (½ ) (Perimeter of base) (slant height). [See more details on the answ ers to the Surface Area Quiz).] Similarly, the lateral surface area of a cone is (½)(Perimeter of Base of cone)(slant height of cone) 9. The entire wheel of brie has the shape of a right circular cylinder, with height of only cm. A w edge w hich is one-sixth of the w heel has volume = one-sixth of the w heel. A w edge w ith a central angle of 0 0 1 contains only one-tw elfth of the w heel, since 60 = 1. The Surface Area consists of three parts: The top and bottom of the w edge are surfaces in the shape of a sector of a circle (radius 15cm). Thus the area is (PART) (15cm) In part a, this is (1/6) (15cm). The bottom has the same area. The remaining surface of the w edge might be covered w ith a strip of paper cm high. The length w ould be ((radius) + (1/6) (Circumference) cm Part of Circumference ((15cm) + (1/6) ( (15cm) ) So the total lateral area is (cm) ((15cm) + (1/6) ( (15cm)) 15cm The total surface area of the 1/6-w heel w edge of brie is: (1/6) (15cm) + (cm) ((15cm) + (1/6) ( (15cm)) (For the 0 w edge of brie, replace (1/6) w ith (1/1) in all parts!)
10. Finding the total surface area and volume of an ice cream cone, topped w ith a hemisphere of ice cream, given the diameter of the top of the cone is 10 cm. and the height of the cone is 1 cm. Volume = V + V = (½) ( /) (r) + (a) (r) (h) = (½) ( /) (5cm) + (a) (5cm) (1cm) = (550 /)cm hemisphere 4 cone 4 Surface Area = SA hemisphere + Lateral SA cone = (½) 4 r + (½ ) r (h) = (½) 4 (5cm) + (½ ) (5cm) (1cm) 11. Was/w ill be discussed in class. 1. Don t be intimidated by the number of steps here. See how each multiplication just gives you new units. 1 kl = 1 kl 1000L 1000 ml 1 cc 1g* 1 kg 1 metric ton 1 kl 1 L 1mL cc 1000g 1000 kg = 1 metric ton For instance, if w e pause right here, w e see that w e now have L (liters) and since w e know a conversion from milliliters (ml) to cc to grams*, we continue by converting the liters to milliliters. Then w e need, at THIS POINT, to get from grams to kilograms, then to tons. 1g = 1cc...valid ONLY for w ater at 4 C 14. b. 1 ft = (1 in) = 144 in Using dimensional analysis: 1 ft = 1 ft 1 in 1 in = 144in )))) )))) 1 ft 1 ft * One foot * Using common sense : This 1-foot by 1-foot square is a square foot. One ft 1 1 (ie 144) square inches are required to cover this square foot! ONE square inch 1a. PERIMETER: 1a. We have a semi-circular arc of a circle with D= 8mm, and a full circle with diameter 4mm, 8mm )))) In addition to tw o straight sides of length 8mm each. So the full perimeter is 8mm )))) 6mm P = (½) 8mm + (4mm) + 8mm + 8mm = (8 + 16) mm 1b......... If the sides w ere horizontal and vertical, w e could just count the units........ of length on each side. But none of the sides are horizontal or vertical,........ So w e must compute the length of each side, using the Pythagorean thm................. Each side of the given triangle is the hypotenuse of a right triangle w ith horizontal & vertical sides. For instance, the upper left side of the given triangle is the hypotenuse of the shaded triangle. The shaded triangle is 4 units high & units w ide, so its hypotenuse must be... 5units long. For the upper right side: + 1 = c, so c = 1 0 P = 5 + 1 0 + 4 5 units.
15. Find the longest diagonal inside a 4" by 7" by 7" right rectangular prism. (BOX!) The diagonal on the base of the box: (4") + (7") = c 576 in + 49in 65 in = c 7" D 5 in = c 4" c + (7in) = D (5in) + 49in = c 7" 674in = D 5.9615 in D 16. Explain w hy the answ er to #16 is NOT just.5 times the old area! The scale factor applies to both dimensions that contribute to area: height and w idth. The scale factor is the ratio of new lengths to old, so the ratio of (New height)/(old height)=.5 and the ratio of (New w idth)/(old w idth) is also.5. New Area = (Old Area) (Ratio of New height to Old height) (Ratio of New w idth to Old w idth) = (Old area) (scale factor) (scale factor) If you think in terms of the simplest area (think rectangle), this is obvious. D x by y.5x by.5y It becomes even more obvious when scaled up by a factor which is a whole number, say : x by y x by y: Area is 9 times as great.