McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
9. Perfect Squares
Continue to next subchapter

Exercise 49 Page 69

Practice makes perfect
a To begin we need to convert the depth, or height, into feet so that all measurements use the same units.
42÷12= 3.5Next, we can substitute the given values for depth and volume into the volume of a prism. Depth=& 3.5 Volume=& 1750 V=& B h ⇕& 1750=&B ( 3.5) Finally, we divide the equation by the depth. The result will be the surface area.
1750=B3.5
500=B
Therefore, the surface area of the pool is 500 ft^2.
b We can find the dimensions of the pool using the formula for the area of a rectangle and the answer from Part A.
Area=& 500 Length=& w+5 Width=& w A= l w ⇒& 500= (w+5) wNext, we solve for w.
500=(x+5)(w)
500=w^2+5w
0=w^2+5w-500
0=(w+25)(w-20)
Finally, we set each factor equal to 0 and solve for w. rcl (w+25)=0 &⇒ & w=-25 (w-20)=0 &⇒ & w=20 The width cannot be negative so we can disregard -25. Therefore, the width is 20ft. The length is 5ft greater than the width, or 25ft. We were given the depth, which is 3.5ft
c Since the volume is doubled, we can double any single dimension.

50 * 20 * 3.5 = 3500 25 * 40 * 3.5 = 3500 25 * 20 * 7 = 3500 There are other possible dimensions as well these are just some examples.

d We need to start by calculating the volume of Model C. We will multiply the length and width both by 2.

25(2) * 20(2) * 3.5 ⇕ 50 * 40 * 3.5 = 7000 Now, we can create a ratio using the volume of Model A and Model C. 1750/7000 ⇒ 1 : 4