Question

You have just opened a small business selling sports equipment . You are packaging basketballs inside boxes that are cubes . The volume of the box is 729 cubic feet . The basketball fits perfectly in the box without additional room at the point of the circumference of the ball . What is the volume of the basketball ? What is the volume of the space not occupied by the ball ?

Answer 1

Answer:

Part A) The volume of the basketball is $V=121.5\pi\ ft^3$

Part B) The volume of the space not occupied by the ball is $V=(729-121.5\pi)\ ft^3$

Step-by-step explanation:

step 1

Find the length side of the cube (box)

we know that

The volume of a cube is equal to

$V=b^3$

where

b is the length side of the cube

we have

$V=729\ ft^3$

so

$b^3=729$

$b=\sqrt[3]{729}\ ft$

$b=9\ ft$

step 2

Find the volume of the basketball

we know that

The volume of a sphere (basketball) is equal to

$V=\frac{4}{3}\pi r^{3}$

we have that

The diameter of the basketball is equal to the length side of the cube

so

$r=9/2=4.5\ ft$

substitute

$V=\frac{4}{3}\pi (4.5)^{3}$

$V=121.5\pi\ ft^3$

step 3

Find the volume of the space not occupied by the ball

we know that

The volume of the space not occupied by the ball is equal to the volume of the box minus the volume of the ball

so

$V=(729-121.5\pi)\ ft^3$