Question

Find the perimeter of a triangular-based pyramid with slant height $18$m and lateral surface area $360m^2$.

Answer 1

The general formula for the lateral surface area of a regular pyramid is $LSA=\frac{1}{2}Pl$ where $P$ represents the perimeter of the base and $l$ is the slant height.

Since the lateral surface area of the pyramid is $P=360$ m${}^2$ and the slant height is $l=18$ m, therefore, the perimeter of the pyramid is:

$LSA=\frac{1}{2}Pl\Rightarrow 360=\frac{1}{2}\times P\times 18\Rightarrow 360=9P\Rightarrow P=\frac{360}{9}=40$

Hence, the perimeter of the pyramid is $40$m${}^2$.