A regular triangular 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?

since the triangular pyramid has equilateral triangles as its faces, this means that all of its faces are equal, including its triangular base, because the base is a triangle that has for sides the sides of the lateral triangles (which are equilateral triangles, i.e. all sides are equal in size)

So recall that such type of pyramid has a total of 4 faces, three of them are lateral and the fourth one is the base.

The information in the problem tells us that the lateral surface of the pyramid is 72 [tex]cm^2[/tex], that means that the addition of its three lateral triangles renders 72 [tex]cm^2[/tex],

We can then find what is the area of each of the three triangles by simply dividing this number by 3 (recall that all faces in the pyramid are equal)

Each triangular face has an [tex]Area=\frac{72}{3} \,cm^2 =24\,cm^2[/tex]

Then, the total surface area of the triangular pyramid is the addition of the area of its four faces. That is:

Total surface are [tex]= 24\,cm^2+24\,cm^2+24\,cm^2+24\,cm^2=4\,*\,24\,cm^2=96\,cm^2[/tex]