SOLUTION: A regular triangular pyramid with a slant height of 9 m has a volume equal to 50 m^3. Find the lateral area of the pyramid. Answer:84.65 sq.m. How to solve? Please explain.

Question 1117233: A regular triangular pyramid with a slant height of 9 m has a volume equal to 50 m^3. Find the lateral area of the pyramid.
Answer:84.65 sq.m.
How to solve? Please explain.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Let the altitude of the pyramid be h. Then consider the right triangle with that altitude as one leg and the slant height of 9 as the hypotenuse. Then the other leg of that right triangle has length .

That other leg of the right triangle is one-third the length of the altitude of the triangular base; the length of that altitude is times the side length s of the triangular base. So

We can find the height of the pyramid by knowing that its volume is 50:

volume = one-third base times height

A graphing calculator shows two solutions to that equation; to several decimal places the solutions are h=0.35695 and h=8.81621.

So there will be two different pyramids that give the correct volume. The one with the very small height is probably not the intended one, so from here I will assume the height is 8.81621.

Substituting that value in the expression for the side length of the triangular base,

Now the lateral surface area consists of three congruent triangles, each with base 6.26822 and height 8.81621. The total surface area is

This is very close to the given answer. Since I kept 5 or more decimal places in all my calculations, I suspect that the given answer 84.65 was obtained keeping fewer decimal places in the calculations.
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