Question

3. A regular square pyramid has a height of 8m. If the slant height makes an angle of 45º with the base, find the lateral area of the pyramid.

Answer 1

Consider a regular square pyramid

Let $a=$ the base edge, $h=$ the altitude, and $s=$ the slant height.

The lateral area $A_L$ of a regular square pyramid is

$$A_L=\dfrac{1}{2}(4a)s$$

The slant height makes an angle $\alpha=45\degree$ with the base.

From the right triangle

$$\sin\alpha=\dfrac{h}{s}, \tan \alpha=\dfrac{h}{a/2}$$

$$s=\dfrac{h}{\sin\alpha}=\dfrac{8}{\sin45\degree}=8\sqrt{2}(m)$$

$$a=\dfrac{2h}{\tan\alpha}=\dfrac{2(8)}{\tan45\degree }=16(m)$$

$$A_L=2as=2(16)(8\sqrt{2})=256\sqrt{2}(m^2)$$

$A_L=256\sqrt{2}m^2$