McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
8. Congruent and Similar Solids
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Exercise 5 Page 867

Begin by identifying the radius of the larger ball.

≈ 220 893.2 cm^3

Practice makes perfect

We have two different sizes of exercise balls. Let d_1 be the diameter of the larger ball and let d_2 be the diameter of the smaller ball.

We have been told that the ratio of the diameters is 15:11. With this ratio, we can write the following proportion. d_1/d_2=15/11 To find the volume of the larger ball, we need to identify its radius. Therefore, we should first find its diameter. Given that the diameter of the smaller ball is 55 centimeters, we can substitute d_2=55 into the proportion.
d_1/d_2=15/11
d_1/55=15/11
Solve for d_1
55d_1/55=825/11
d_1=825/11
d_1=75
Now that we have found the diameter as d_1=75 centimeters, we can immediately identify the radius of the larger ball as 37.5 centimeters. Since the ball is a sphere, we will use the following formula to find the volume of a sphere. \begin{gathered} V = \dfrac{4 \pi r^3} 3 \end{gathered} Let's substitute 37.5 for r into the formula and simplify it!
V = \dfrac{4 \pi r^3} 3
V = \dfrac{4 \pi ({\color{#0000FF}{37.5}})^3} 3
V = 220 893.23345...
V≈ 220 893.2
The volume of the larger ball is approximately 220 893.2 cm^3.