Question

a cone and a sphere have the same volume. the height of the cone is 96 units. what would be the values for the radius of the cone and the sphere?

Answer 1

The volume $V_c$ of a cone  with base radius $r_c$ and height $h_c$ is

$V_c = \dfrac{1}{3}\pi r_c^2 h_c$

Similarly, the volume $V_s$ of a sphere with radius $r_s$ is

$V_s = \dfrac{4}{3}\pi r_s^3$

We know that $V_c=V_s$ and that $h_c=96$

So, we can set up the following equation:

$\dfrac{96}{3}\pi r_c^2=\dfrac{4}{3}\pi r_s^3$

We can simplify the common denominator 3, and pi appearing on both sides:

$96r_c^2=4r_s^3$

We can divide both sides by 4:

$24r_c^2=r_s^3$

Without further information, this is all we can say: the cubed radius of the sphere is the same as 24 times the squared radius of the cone.

Answer 2

The values for the radius of the cone and the sphere is 24 units.

It is given that a cone and a sphere have the same volume.

It is required to find the radius of the cone and sphere if the height of the cone is 96.

What is volume?

It is defined as a three-dimensional space enclosed by an object or thing.

We know the volume of the cone:

$\rm V_c = \pi R^2\frac{H}{3}$

Where R is the radius of the cone and H is the height of the cone.

And the volume of the sphere:

$\rm V_s = \frac{4}{3} \pi r^3$

Where r is the radius of the sphere.

$\rm V_c = V_s$      (From the question)

$\rm \pi R^2\frac{H}{3}=\rm \frac{4}{3} \pi r^3$

$\rm \pi R^2\frac{96}{3}=\rm \frac{4}{3} \pi r^3$    ( H = 96 units)

$\rm 32R^2=\rm \frac{4}{3} r^3$

$\rm 24R^2= r^3$

We are assuming that R = r

$\rm 24r^2= r^3$

r = 24 units and

R = 24 units.

Thus, the values for the radius of the cone and the sphere is 24 units.

Learn more about the volume here:

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