# Volume of a Pyramid

The cube as shown in the figure illustrates the calculation of the volume of the pyramid. The opposite vertices of the cube were connected, the lines of connection, meeting at the center $O$. This divided the cube into six congruent pyramids.

The volume of any one the six pyramids. Such as $OABCD$ equals one-sixth of the volume of the cube and, therefore, the volume of one pyramid equals the area of the base $ABCD$ times one-third of $PQ$.

Thus, in this special case that of a right square pyramids whose altitude equals one-half the length of a side of the base, we see that the volume of the pyramid equals the area of the base times one-third the altitude; or in other words the volume of the pyramid equals one-third the volume of a prism of the same base and altitude.

Rule:

The volume of the pyramid equals the area of the base times one-third the altitude. i.e. $V = \frac{1}{3}Ah$, $A$ being area of base.

Example:

Find the volume of a pyramid whose base is an equilateral triangle of side $1$m and whose height is $4$m.

Solution:

Since, Base of pyramid is an equilateral triangle of side $1$m

Area of the base $= \frac{{{a^2}\sqrt 3 }}{4} = \frac{{1 \times 1.732}}{4} = 0.43$square m

Volume of pyramid $= \frac{1}{3} \times {\text{ area of base }} \times {\text{ height}} = \frac{1}{3} \times 0.43 \times 4 = 0.58$ cubic m.