SOLUTION: A regular triangular pyramid with a slant height of 9 m has a volume equal to 50 cu. m. Find the lateral surface area of the pyramid.

Question 1175766: A regular triangular pyramid with a slant height of 9 m has a volume equal to 50 cu. m. Find the lateral surface area of the pyramid.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Let s be the side length of the equilateral triangle which is the base of the pyramid.

The given slant height of 9 is the hypotenuse of a right triangle in which the legs are the height h of the pyramid and a segment from the center of the triangular base to the middle of one edge. The segment from the center of the base to the middle of one edge is 1/3 of the altitude of the triangular base, which is sqrt(3)/2 times the side length s. That makes the length of that segment .

The Pythagorean Theorem with that right triangle then gives us the equation

[1]

The volume of the pyramid is given as 50.

where B is the area of the base. The base is an equilateral triangle with side length s; its area is . So

So now we have the equation

[2]

s is a parameter we are using in our analysis; our objective is to find the height, h. To do that, we need to eliminate s (or, actually, s^2) between the two equations we have.

Solve [2] for s^2...

[3]

... and substitute in [1]

[4]

That's the ugly part of the analysis. I leave the rest to you.

(1) Use a graphing calculator or some other utility to solve equation [4] for the height h.
(2) Use that value of h in [3] to find the side length s.
(3) The lateral surface area is the area of three congruent isosceles triangles each with height 9 and base length s.

Note you can verify your solution by seeing that the values you find for h and s satisfy equation [1].

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