posted by mhy .
a regular triangular pyramid with a slant height of 9 m has a volume equal to 50 m³. Find the lateral areas of the pyramid.
Answer is 81 square root of 3 sq.m
If the triangular base has area A and the pyramid has height h, the volume of the pyramid is 50 = 1/3 Ah so you have 150 = Ah and h=150/A. The area of the base is A=b^2*sqrt (3) /4 where b is the length of each side of the base.
So you have a h = 150 / (b^2*sqrt (3) /4) = 200sqrt (3) / b^2
The edge length e, slant heights, and base length b satisfy e^2 = h^2 + b^2 /3 and s^2 = h^2 +b^2 /12. e^2 -s^2 = b^2 /4
Substituting into the equation s^2 = h^2 +b^2 /12 for h^2 = 120000b^4:
9^2 =120000 / b^4 +b^2 /12. This gives b = 6.26836.
Each of the three lateral triangles has an area BS /2 so the total lateral area of the pyramid is 3bs/2 = 3 (6.26836) (9) /2 = 84.62286.
You could find h using 9^2 = h^2 +6.26836^2 /12, then he = 8.81621.