Question

The volume of a sphere whose diameter is 18 centimeters is _ cubic centimeters. If it’s diameter we’re reduced by half, it’s volume would be _ of its original volume

Answer 1

Answer:

First Part

Given that

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

We have that

$Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (\frac{Diameter}{2})^{3} = \frac{4}{3} \pi 9^{3} = 972\pi cm^{3} \approx 3053.63 cm^{3}$

Second Part

Given that

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

If the Diameter were reduced by half we have that

$Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (\frac{r}{2}) ^{3} = \frac{\frac{4}{3} \pi r^{3}}{8}$

This shows that the volume would be $\frac{1}{8}$ of its original volume

Step-by-step explanation:

First Part

Gather Information

$Diameter = 18cm$

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

Calculate Radius from Diameter

$Radius = \frac{Diameter}{2} = \frac{18}{2} = 9$

Use the Radius on the Volume formula

$Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi 9^{3}$

Before starting any calculation, we try to simplify everything we can by expanding the exponent and then factoring one of the 9s

$Volume = \frac{4}{3} \pi 9^{3} = \frac{4}{3} \pi 9 * 9 * 9 = \frac{4}{3} \pi 9 * 9 * 3 * 3$

We can see now that one of the 3s can be already divided by the 3 in the denominator

$Volume = \frac{4}{3} \pi 9 * 9 * 3 * 3 = 4 \pi 9 * 9 * 3$

Finally, since we can't simplify anymore we just calculate it's volume

$Volume = 4 \pi 9 * 9 * 3 = 12 \pi * 9 * 9 = 12 * 81 \pi = 972 \pi cm^{3}$

$Volume \approx 3053.63 cm^{3}$

Second Part

Understanding how the Diameter reduced by half would change the Radius

$Radius =\frac{Diameter}{2}\\\\If \\\\Diameter = \frac{Diameter}{2}\\\\Then\\\\Radius = \frac{\frac{Diameter}{2} }{2} = \frac{\frac{Diameter}{2}}{\frac{2}{1}} = \frac{Diameter}{2} * \frac{1}{2} = \frac{Diameter}{4}$

Understanding how the Radius now changes the Volume

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

With the original Diameter, we have that

$Volume = \frac{4}{3}\pi (\frac{Diameter}{2}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{2^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{2 * 2 * 2} = \frac{4}{3}\pi \frac{Diameter^{3}}{8}$

If the Diameter were reduced by half, we have that

$Volume = \frac{4}{3}\pi (\frac{Diameter}{4}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{4^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 4 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 2 * 2 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{8 * 8} = \frac{\frac{4}{3}\pi\frac{Diameter^{3}}{8}}{8}$

But we can see that the numerator is exactly the original Volume!

This shows us that the Volume would be  $\frac{1}{8}$ of the original Volume if the Diameter were reduced by half.