Chapter 12.3, Problem 41HP

To determine

To describe:The method to find surface area of a regular polygonal pyramid with n-gone base, height h units, and an apothem of a units.

Answer to Problem 41HP

The method for finding the surface area of the pyramid is explained here that shows the surface area of pyramid of height “ h ” and apothem “ a ” is given by $S=\frac{1}{2}P\times \sqrt{h^2+a^2}+B$ .

Explanation of Solution

Given data: $y$

The base of regular polygonal pyramid is n sides polygon, height h units, and an apothem of a units.

Concept used:

The area of a triangle is half of the product of its base length and height.

Calculations:

The considered polygon has n sides each of length s (say). Then perimeter of the polygon is $P=ns$ . The area of a triangle AOB formed with one side is $△=\frac{1}{2}AB\times ON$ , $AB=s$ and $ON=a\left(\text{apothem}\right)$ , therefore $△=\frac{1}{2}s\times a$ . Since there are such n triangles in the base so total area of the base of perimeter is

  $B=n\cdot △=n\cdot \frac{1}{2}sa=\frac{1}{2}ns\cdot a=\frac{1}{2}P\cdot a$

The area of a lateral triangle ABZ is

  ${△}_L=\frac{1}{2}\text{AB×MZ}=\frac{1}{2}s\times \sqrt{h^2+a^2}$ , h is the height of the pyramid. There are such n lateral triangles, therefore total lateral area of the pyramid is

  $L=\frac{1}{2}ns\times \sqrt{h^2+a^2}=\frac{1}{2}P\times \sqrt{h^2+a^2}$ .

Hence, surface area of the pyramid is $S=\frac{1}{2}P\times \sqrt{h^2+a^2}+B$

Conclusion:

The method for finding the surface area of the pyramid is explained here that shows the surface area of pyramid of height “ h ” and apothem “ a ” is given by $S=\frac{1}{2}P\times \sqrt{h^2+a^2}+B$ .