Question

The angle opposite to the base edge of a square pyramid on its lateral face is 60${}^{\circ }$. If the slant height of the pyramid is 40$cm$, what is the lateral surface area of the pyramid?

• A. $\frac{3200}{\sqrt{3}}cm$​​​
• B. $\frac{800}{\sqrt{3}}cm$​​​
• C. $\frac{400}{\sqrt{3}}cm$​​​
• D. $\frac{6400}{\sqrt{3}}cm$​​​

Answer 1

The correct option is D

$\frac{6400}{\sqrt{3}}cm$​​​

Since the triangular face of a pyramid is isosceles, we have AC bisecting BD.

Thus, using trigonometric ratios in $△ACD$ where $\angle CAD=\frac{\angle BAD}{2}={30}^{\circ }$ we have

$tan{30}^{\circ }=\frac{CD}{AC}=\frac{CD}{40}$

$⟹\frac{1}{\sqrt{3}}=\frac{CD}{40}$

$⟹CD=\frac{40}{\sqrt{3}}cm⟹Baseedge=2\times \frac{40}{\sqrt{3}}cm=\frac{80}{\sqrt{3}}cm$

Thus, Area of $△ACD=\frac{1}{2}\times \frac{80}{\sqrt{3}}cm\times 40=\frac{1600}{\sqrt{3}}cm$

Now, the lateral surface area of the pyramid = $4\times Areaofthe△ACD$ =$4\times \frac{1600}{\sqrt{3}}cm=\frac{6400}{\sqrt{3}}cm$​​​