+0 A right pyramid has a square base. The area of each triangular face is one-third the area of the square face. If the total surface area of the pyramid is $432$ square units, then what is the volume of the pyramid in cubic units?
+1632 Let the square base have side length a. The area of the triangle with slant height b has area ab/2 = (a^2)/3, so 2a/3 = b.
Thus, the area of each surface triangle is (a^2)/3, we have four of them with a base of a^2, so the surface area = (7a^2)/3 = 432
Solving for a, you get a = 36/sqrt(7)
The height of the pyramid, using pythagorean theorem with legs h and a/2 with hypotenuse 2a/3 is:
\(({2a\over{3}})^2-({a\over{2}})^2=h^2\)
Substitution:
\(({24\over{\sqrt{7}}})^2-({18\over{\sqrt{7}}})^2={576\over{7}}-{324\over{7}}=36=h^2\)
Solving for h, we get h = 6.
The volume of a pyramid, being base area * height / 3 would be a^2 * 6 / 3 = 1296*2/7 = 2592/7
