The volume of the frustum of a cone is equivalent to the sum of the volume of three cones, having for their common altitude and for their several bases, the bases of the frustum and a mean proportional between the bases. The volume of the frustum of the cone is $\frac{1}{4} $$\pi$ times the frustum of the pyramid.

**Formula:**

**Volume of the frustum of a cone =** $\frac{1}{3}$$\pi$h($r^2 + rR + R^2$)

Where r and R are the radii of ends and h is height of frustum of a cone.**Example:**Find the volume of a frustum of a cone with height 5 cm, and radii 3 cm, 7 cm.

Step 1:Height of frustum of a cone (h) = 5 cm

Radius of top part (r) = 3 cm

Radius of base (R) = 7 cm

Step 2:Volume of the frustum of a cone = $\frac{1}{3}$$\pi$h($r^2 + rR + R^2$)=

$\frac{1}{3}$$\pi$ * 5($3^2 + 3 * 7 + 7^2$)

=

$\frac{1}{3}$$\pi$ * 5(9 + 21 + 49)

=

$\frac{1}{3}$$\pi$ * 5(79)

=

$\frac{1}{3}$$\pi$ * 395

= 131.7 $\pi$.

Hence the volume of the frustum of a cone is 131.7$\pi$ cm

^{3}.