# Prove that the volume of a truncated isosceles triangular pyramid is `V = (1/3)*(A1 + A2 + sqrt(A1*A2))*h` where A1 and A2 denote the area of the end triangles.

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### 1 Answer

The volume of a pyramid is `(1/3)*B*h` where B is the area of the base and h is the height of the pyramid.

For the truncated pyramid, let A1 and A2 denote the area of the lower and upper end triangles respectively and h be the height of the pyramid. Let the height of the complete pyramid be h+x and the height of the portion cut at the top be x.

The area of the truncated pyramid is `(1/3)*(A1*(h+x) - A2*x)`

= `(1/3)*(A1*h + x*(A1 - A2))`

Now the area A1 and A2 can be expressed as a relation of h and x as follows: `(A1)/(A2) = (h +x)^2/x^2`

=> `(h+x)/x = sqrt(A1)/sqrt (A2)`

=> `h/x = sqrt(A1)/sqrt(A2) - 1`

=> `h/x = (sqrt(A1) - sqrt(A2))/sqrt(A2)`

=> `x = h*sqrt (A2)/(sqrt A1 - sqrt A2)`

Substituting this for x in the relation for the volume of the truncated pyramid derived earlier gives

V = (1/3)*(A1*h + x*(A1 - A2))

=> `(1/3)*(A1*h + h*sqrt (A2)/(sqrt A1 - sqrt A2)*(A1 - A2))`

`A1 - A2 = (sqrt(A1) - sqrt(A2))(sqrt(A1) + sqrt(A2))`

=> `(1/3)*(A1*h + h*sqrt (A2)/(sqrt A1 - sqrt A2)*(sqrt(A1) - sqrt(A2))(sqrt(A1) + sqrt(A2)))`

=> `(1/3)*(A1*h + h*sqrt (A2)*(sqrt(A1) + sqrt(A2)))`

=> `(h/3)*(A1 + A2 + sqrt (A1*A2))`

**This proves that the volume of a truncated pyramid is** `(h/3)*(A1 + A2 + sqrt (A1*A2))`

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