Integral of a triangular pyramid

[doctordiddy]

Homework Statement

Find the volume of a pyramid with height 24 and with base an equilateral triangle with side 11.

Homework Equations

The Attempt at a Solution

So I know the relationship

h/x=b/l

where h is my height 24
x is simply x or height above base
b is my base 11
l is the length of the base at height x

i also know the integral is

∫₀²⁴(A(x))dx

I am having trouble finding A(x)

I believe that it you would use the area of the equilateral triangle

so i know area of the triangle is base*height/2

I also know that base is l, and height is unknown, but i do know that

l^2=(l/2)^2 + z^2 where i take z as the height of the equilateral triangle

then z= √(l^2 -(l/2)^2 )

so area of the triangle would finall be

[(l^2)√(l^2 - (l/2)^2)]/2

I would finally plug in l=bx/h from the identity h/x=b/l

However, when i do this, i end up with an integral i am unable to solve. I feel like there is another way to do this, can anyone help?

edit: I mean an easier way using integrals, not volume of a pyramid please

thanks

 

[SteamKing]

What kind of a triangle is an equilateral triangle?

 

[haruspex]

[(l^2)√(l^2 - (l/2)^2)]/2

You've one too many l factors in there - it's an area, not a volume. You should be able to simplify the term inside the square root (bringing the l outside). Once you've done that the integral should be straightforward.