Question

a regular rectangular pyramid has a base and lateral faces that are congruent equilateral triangles. it has a lateral surface area of 72 square centimeters. what is the surface area of the regular triangular pyramid?

Answer 1

A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)

Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.

The lateral faces are equilateral triangles of side length x.

The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18  cm^2.

now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.

by the Pythagorean theorem: 

$h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x$

thus, 

$Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2$

thus:

$x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\ \sqrt{3} }{3}=24 \sqrt{3}$       (cm^2)

but $x^{2}$ is exactly the base area, since the base is a square of sidelength = x cm.


So, the total surface area = base area + lateral area =  $24 \sqrt{3}+72$   cm^2


Answer: $24 \sqrt{3}+72$   cm^2