Question

A cone with radius 5 and height 12 has its radius doubled. How many times greater is the volume of the larger cone than the smaller​ cone? Use a pencil and paper. Explain how the volume of the cone would change if the radius were halved.

Answer 1

Answer:

$V = \frac{1}{3} \pi (5)^2 (12)= 314.159$

Now if we increase the radius by a factor of 2 the new volume would be:

$V_f = \frac{1}{3} \pi (2*5)^2 (12)= 1256.637$

And we can find the increase factor for the volume like this:

$\frac{V_f}{V}= \frac{1256.637}{314.159}= 4$

Then if we increase the radius by 2 the volume increase by a factor of 4

If we reduce the radius by a factor of 2 then we will have that the volume would be reduced by a factor of 4.

On the figure attached we have an illustration for the cases analyzed we see that when we increase the radius the volume increase and in the other case decrease.

Step-by-step explanation:

For this case we have the following info given:

$r = 5 , h =12$

and we can find the initial volume:

$V = \frac{1}{3} \pi r^2 h$

And replacing we got:

$V = \frac{1}{3} \pi (5)^2 (12)= 314.159$

Now if we increase the radius by a factor of 2 the new volume would be:

$V_f = \frac{1}{3} \pi (2*5)^2 (12)= 1256.637$

And we can find the increase factor for the volume like this:

$\frac{V_f}{V}= \frac{1256.637}{314.159}= 4$

Then if we increase the radius by 2 the volume increase by a factor of 4

If we reduce the radius by a factor of 2 then we will have that the volume would be reduced by a factor of 4.

On the figure attached we have an illustration for the cases analyzed we see that when we increase the radius the volume increase and in the other case decrease.