**Frustum of a pyramid and frustum of a cone**

The formula for *frustum of a pyramid* or *frustum of a cone* is given by

$V = \dfrac{h}{3} \left[ \, A_1 + A_2 + \sqrt{A_1A_2} \, \right]$

Where:

*h* = perpendicular distance between *A*_{1} and *A*_{2} (*h* is called the altitude of the frustum)

*A*_{1} = area of the lower base

*A*_{2} = area of the upper base

Note that *A*_{1} and *A*_{2} are parallel to each other.

**Derivation:**

$V_1 = \frac{1}{3}A_1 y$

$V_2 = \frac{1}{3}A_2 (y - h)$

$V = V_1 - V_2 = \frac{1}{3}A_1 y - \frac{1}{3}A_2 (y - h)$

$V = \frac{1}{3}A_1 y - \frac{1}{3}A_2 y + \frac{1}{3}A_2 h$

$V = \frac{1}{3} \, \left[ \, (A_1 - A_2) y + A_2 h \right]$ ← Equation (1)

By similar solids:

$\dfrac{A_2}{A_1} = \left( \dfrac{y - h}{y} \right)^2$

$\sqrt{\dfrac{A_2}{A_1}} = 1 - \dfrac{h}{y}$

$\dfrac{h}{y} = 1 - \sqrt{\dfrac{A_2}{A_1}} = 1 - \dfrac{\sqrt{A_2}}{\sqrt{A_1}}$

$\dfrac{h}{y} = \dfrac{\sqrt{A_1} - \sqrt{A_2}}{\sqrt{A_1}}$

$\dfrac{y}{h} = \dfrac{\sqrt{A_1}}{\sqrt{A_1} - \sqrt{A_2}}$

$y = \dfrac{\sqrt{A_1}}{\sqrt{A_1} - \sqrt{A_2}}\,h = \left( \dfrac{\sqrt{A_1}}{\sqrt{A_1} - \sqrt{A_2}} \, \times \, \dfrac{\sqrt{A_1} + \sqrt{A_2}}{\sqrt{A_1} + \sqrt{A_2}}\right)\,h$

$y = \dfrac{A_1 + \sqrt{A_1A_2}}{A_1 - A_2}\,h$

Substitute y to Equation (1),

$V = \frac{1}{3} \, \left[ \, (A_1 - A_2) \left( \dfrac{A_1 + \sqrt{A_1A_2}}{A_1 - A_2}\,h \right) + A_2 h \right]$

$V = \frac{1}{3} \, \left[ \, (A_1 + \sqrt{A_1A_2})h + A_2 h \, \right]$

$V = \frac{1}{3} \, \left[ \, A_1 + \sqrt{A_1A_2} + A_2 \, \right] \, h$

$V = \dfrac{h}{3} \left[ \, A_1 + A_2 + \sqrt{A_1A_2} \, \right]$