Question

a. A 5in radius ball is in a
box such that all six sides of
the box are touching the ball.
What percent of the space in
the box is being taken up by
the ball?

Answer 1

Answer:

52.33%

Step-by-step explanation:

Since, all six sides of the box are touching the ball.

So, all the three dimensions of the box are equal.

Hence, it is a cubical box.

Side length of the box (l)

= diameter of the ball

= 2*radius of the ball

= 2*5

= 10 In.

$V_{ball} = \frac{4}{3} \pi r^3$

$V_{ball} = \frac{4}{3} \times 3.14(5)^3$

$V_{ball} = \frac{4}{3} \times 3.14\times 125$

$V_{ball} = \frac{1,570}{3}\: In^3$

$V_{box} =l^3$

$V_{box} =(10)^3$

$V_{box} =1000\: In^3$

Percentage of the space in the box occupied by the ball

$= \frac{V_{ball}}{V_{box} } \times 100$

$= \frac{\frac{1,570}{3}}{1000} \times 100$

$= \frac{1570}{3000} \times 100$

$= \frac{157}{3}$

$= 52.33\%$