a bowling ball measures 264 cm in circumference. what's the volume of the smallest cube that will hold this ba?

C = πd = 264 cm

r = 84.03380995... cm

Let e = 84.034cm

V = e^3 = 593419.9772... cm^3

593420 cm^3 would be the best estimate.

Volume Of A Bowling Ball

The smallest cube that can hold the bowling ball has the same side length as the diameter of the bowling ball.

Since the balling ball measures 264 cm in circumference, we see that:

πd = 264 ==> d = 264/π.

Hence, the smallest cube must have sides 264/π cm in length.

Therefore, the volume of the smallest cube is:

V = s^3 = (264/π)^3 = 18399744/π^3 ≈ 593,420 cm^3.

I hope this helps!

C/d = pie (C=circumference, d=diameter)

solving for d=C/pie

after finding the diameter the cube is defined as

therefore knowing length of one side and knowing that a cube have same sides calculations would be

d^3

C= (pi)d

d= C/(pi)

d=264/(pi)

The cube must have 264/(pi) as its dimension

V= [264/(pi)]^3

V= 18399744 / 31

V = 593540 cc