# Surface Area of a Pyramid

The surface area of a pyramid consist of the lateral surface which is the area of the number of triangles which form the sides (or faces) of the figure along with area of the base, which may be any polygon.

The lateral area of a right pyramid is the sum of the areas of the triangles forming the faces of the pyramid. In a right pyramid, by definition, these are congruent triangle. Also by definition, the base of a right pyramid is a regular polygon. Therefore, the base of the triangular faces is equal and their altitudes are also equal and are equal to the slant height of the pyramid.

Rules:

1. The lateral area of a right pyramid equals to the perimeter of the base times one-half the slant height.

1. Total surface area = lateral surface area + area of the base
2. Slant height, $l = \sqrt {{r^2} + {h^2}}$

Example:

A pyramid on a square base has four equilateral triangles. For its other faces each edge being $9$cm; find the whole surface.

Solution:

Let $OABCD$ be the pyramid on the square base $ABCD$. As the side face are equilateral triangles of each side $9$cm, therefore, the side of the square base is $9$cm.

Area of the base $= 9 \times 9 = 81$square cm

Area of one side face $= \frac{{{a^2}\sqrt 3 }}{4} = \frac{{9 \times 9 \times 1.732}}{4} = \frac{{140.292}}{4}$

Area of all the four sides faces $= \frac{{140.292}}{4} \times 4 = 140.292$square cm

Area of the whole surface $= 140.292 + 81 = 221.29$square cm.