The volume of a cone is 25/3 pi cm³. What is the volume of a sphere if its radius is the same as the cones and the height of the cone is equal to the spheres diameter?

Answer 1

First, let's write the formula for the volume of a cone:

\[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \]

where \( V_{\text{cone}} \) is the volume of the cone, \( r \) is the radius of the base of the cone, and \( h \) is the height of the cone.

Given that the volume of the cone is \( 25/3 \) \( \pi \) cm³, we can equate:

\[ \frac{25}{3} \pi = \frac{1}{3} \pi r^2 h \]

Now, we are told that the height of the cone, \( h \), is equal to the diameter of the sphere. Since the diameter is twice the radius (\( d = 2r \)), we have \( h = 2r \). Let's substitute \( h \) with \( 2r \) into the equation:

\[ \frac{25}{3} \pi = \frac{1}{3} \pi r^2 (2r) \]
\[ \frac{25}{3} \pi = \frac{2}{3} \pi r^3 \]

Now we can solve for \( r^3 \):

\[ r^3 = \frac{25}{3} \pi \times \frac{3}{2 \pi} \]
\[ r^3 = \frac{25}{2} \]

Once we have \( r^3 \), we can find the volume of the sphere using the formula for the volume of a sphere:

\[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \]

Now, let's plug in \( r^3 = \frac{25}{2} \):

\[ V_{\text{sphere}} = \frac{4}{3} \pi \times \frac{25}{2} \]
\[ V_{\text{sphere}} = \frac{4}{3} \times \frac{25}{2} \pi \]
\[ V_{\text{sphere}} = \frac{4 \times 25 \pi}{3 \times 2} \]
\[ V_{\text{sphere}} = \frac{100 \pi}{6} \]
\[ V_{\text{sphere}} = \frac{50 \pi}{3} \]

Therefore, the volume of the sphere with the same radius as the cone is \( \frac{50}{3} \pi \) cm³.