Question

A square pyramid has sides of 8 m. Each face is an equilateral triangle. What is the lateral area of the pyramid?

Answer 1

Answer:

The lateral area is equal to

$LA=64\sqrt{3}\ m^{2}$

Step-by-step explanation:

In this problem the lateral area is equal to the area of one equilateral triangle multiplied by $4$

To find the area of one equilateral triangle calculate the height

The area of the triangle is equal to

$A=\frac{1}{2}bh$

we have

$b=8\ m$

Applying the Pythagoras theorem

$h^{2}=8^{2} -4^{2} \\h^{2}=64-16\\ h=\sqrt{48} \\ h=4\sqrt{3}\ m$

The area of one triangle is equal to

$A=\frac{1}{2}(8)(4\sqrt{3})\ m^{2}$

so

The lateral area is equal to

$LA=4\frac{1}{2}(8)(4\sqrt{3})=64\sqrt{3}\ m^{2}$