A right pyramid with a square base has total surface area 432 square units. The area of each triangular face is half the area of the square face. What is the volume of the pyramid in cubic units?
Let the side length of the square base be s. Then the triangular faces have area s2/2, and the total surface area is 4(s2/2)+s2=4s2=432, so s2=108 and s=12. The height of the pyramid is then the length of an altitude of a triangle with base 12 and area 6s2=756. By the Pythagorean Theorem, the height is sqrt(144+756)=sqrt(900)=30.
The volume of the pyramid is (1/3)(122)(30)=1440 cubic units.
Call the area of the base = B
Call the area of each triangular face B / 2
B + 4(B / 2) = 432
B + 2B = 432
3B = 432
B = 144
So ...the side of the base = sqrt (144) = 12
And the area of each triangular face = B / 2 = 72
The slant height of the pyramid =
area of triangular face = (1/2) (slant height) (base side)
72 = (1/2)(slant height)(12)
144 = (slant height) (12)
12 = slant height
Height of pyramid = sqrt [ slant height^2 - (base length / 2)^2 ] = sqrt [ 12^2 - 6^2] =
sqrt (108) = sqrt (36 * 3) = 6sqrt3
Volume of pyramid = (1/3) (base area) (height) = (1/3)(144)(6sqrt 3) = 288sqrt 3 units^3