Question

The angle opposite to the base edge of a square pyramid on its lateral face is 30${}^{\circ }$. If the slant height of the pyramid is $20cm$, what is the lateral surface area of the pyramid? (Note that $tan{15}^{\circ }=0.268$)

Answer 1

Since the triangular face of a pyramid is isosceles, we have AC bisecting BD. Thus, using trigonometric ratios in $△ACD$ where $\angle CAD=\frac{\angle BAD}{2}={15}^{\circ }$, we have
$tan{15}^{\circ }=\frac{CD}{AC}=\frac{CD}{20}$
$⟹0.268=\frac{CD}{20}$
$⟹CD=0.268\times 20=5.36cm⟹Baseedge=2\times 5.36cm=10.72cm$$\left[1mark\right]$
Thus, Area of $△ABD=\frac{1}{2}\times 10.72\times 20=107.2c{m}^2$
Now, the lateral surface area of the pyramid = $4\times Areaofthe△ABD$
= $4\times 107.2=428.8c{m}^2$​​​$\left[1mark\right]$