Additional Practice

12.3.AP-10

A cone has a radius of 39 centimeters and a height of 52 centimeters.

a. What is the volume of the cone in terms of [tex]x[/tex]?

b. If the radius is now half the size and the height is the same, how has the volume of the cone changed?

a. The volume of the cone is
(Simplify your answer. Type an exact answer in terms of [tex]x[/tex].)



Answer :

Sure, let's work through this step-by-step:

### Part (a): Finding the Volume of the Cone

The formula for the volume [tex]\( V \)[/tex] of a cone is given by:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where [tex]\( r \)[/tex] is the radius and [tex]\( h \)[/tex] is the height of the cone.

Given:
- Radius [tex]\( r = 39 \)[/tex] cm
- Height [tex]\( h = 52 \)[/tex] cm

Substituting these values into the formula, we get:
[tex]\[ V = \frac{1}{3} \pi (39)^2 (52) \][/tex]

Now, let's simplify the expression:

[tex]\[ 39^2 = 1521 \][/tex]
[tex]\[ 39^2 \cdot 52 = 1521 \cdot 52 = 79132 \][/tex]

So,
[tex]\[ V = \frac{1}{3} \pi \cdot 79132 = \frac{79132}{3} \pi \][/tex]

This simplifies to:
[tex]\[ V = 26377.333 \pi \][/tex]

Therefore, the volume of the cone in terms of [tex]\( \pi \)[/tex] is:
[tex]\[ V = 26377.333 \pi \][/tex]

### Part (b): Volume of the Cone with Half the Radius

When the radius is halved, the new radius [tex]\( r' \)[/tex] is:
[tex]\[ r' = \frac{39}{2} = 19.5 \][/tex] cm

The height remains the same, [tex]\( h = 52 \)[/tex] cm.

Now, using the same volume formula for cones:
[tex]\[ V' = \frac{1}{3} \pi (r')^2 h \][/tex]

Substituting the new radius:
[tex]\[ V' = \frac{1}{3} \pi (19.5)^2 (52) \][/tex]

Let's calculate [tex]\( (19.5)^2 \)[/tex]:
[tex]\[ 19.5^2 = 380.25 \][/tex]

Next, we find:
[tex]\[ 380.25 \cdot 52 = 19773 \][/tex]

So,
[tex]\[ V' = \frac{1}{3} \pi \cdot 19773 = \frac{19773}{3} \pi \][/tex]

This simplifies to:
[tex]\[ V' = 6591 \pi \][/tex]

Now, let's compare the two volumes:
- Original volume [tex]\( V = 26377.333 \pi \)[/tex]
- New volume [tex]\( V' = 6591 \pi \)[/tex]

To see how the volume has changed, we take the ratio of the new volume to the original volume:
[tex]\[ \text{Ratio} = \frac{V'}{V} = \frac{6591 \pi}{26377.333 \pi} = \frac{6591}{26377.333} \approx \frac{1}{4} \][/tex]

Therefore, when the radius is halved and the height remains the same, the volume of the cone is reduced to approximately one-fourth of its original volume.