What is the surface area of a cube-shaped box that can tightly hold a basketball whose volume is 500/3 cm^3?

Question

What is the surface area of a cube-shaped box that can tightly hold a basketball whose volume is 500/3 cm^3?

Answer 1

The Volume of a Sphere has this formula:

V_sphere = (4/3) π•r³

The edge of the cube (s) is basically the same as the diameter (d) of the basketball. When you get the radius, multiply it by 2 to get the diameter:

(500/3)π = (4/3)πr³

divide both sides by (4/3)π

500/4 = r³

r³ = 125

r = 5 cm.

d = s = 2•5 =10 cm.

The Surface Area of a cube has this formula:

S_cube = 6•s²

plugin the value of s:

=6•10² =600 cm²

Answer 2

For this problem, we will need to know the formula for the volume of a sphere

V = (4/3)πr^3

We know the basketball is a sphere with radius r.

Since the basketball is fit tightly within a cube, that cube must have side length of 2r.

Once we have the side length, we can easily calculate the surface area of the cube.

First, let's find the radius of the basketball. We are given the volume, so plug that into the formula and solve for r.

500/3 = (4/3)πr^3

125 = r^3 [divide both sides by /3]

5 = r [cube root of both sides]

Since the radius is 5, we know that the side length of the cube is 10.

Therefore, the area of a single side of the cube is 10x10 = 100

and there are 6 faces on a cube, so 6x100 = 600

The surface area of the cube is 600 cm^2