Surface Area of a Pyramid
The surface area of a pyramid consists 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 triangles. 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 they are equal to the slant height of the pyramid.
Rules:
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The lateral area of a right pyramid is equal to the perimeter of the base times one-half the slant height.
\[\therefore {\text{lateral surface area }} = {\text{ }}\frac{{{\text{perimeter }} \times {\text{ slant height}}}}{2}\]
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Total surface area = lateral surface area + area of the base
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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 is $$9$$ cm. Find the whole surface.
Solution:
Let $$OABCD$$ be the pyramid on the square base $$ABCD$$. As the side faces are equilateral triangles with each side being $$9$$ cm, therefore the side of the square base is $$9$$ cm.
The area of the base $$ = 9 \times 9 = 81$$ square cm
The area of one side face $$ = \frac{{{a^2}\sqrt 3 }}{4} = \frac{{9 \times 9 \times 1.732}}{4} = \frac{{140.292}}{4}$$
The area of all four side faces $$ = \frac{{140.292}}{4} \times 4 = 140.292$$ square cm
The area of the whole surface $$ = 140.292 + 81 = 221.29$$ square cm.