"The volume of a pyramid with height $h$ and square base of a side length $a$ is $V=\frac{1}{3}\cdot h\cdot a^2$.
Prove that the volume of a truncated pyramid is $V=\frac{1}{3}\cdot h(a^2+ab+b^2)$ where $h$ is the height and $b$ and $a$ are the length of the sides of the square top and bottom.
I need help with the question in the picture above. I know that the answer will involve creating a proportion and eliminating a variable but I'm not sure how. Please help!!
Best Answer
Conventionally ......
$V_t = \frac{1}{3}\cdot H \cdot a^2 - \frac{1}{3}\cdot H(\frac{b}{a})\cdot b^2$
But the H is not the height h of a truncated pyramid. That would be ...
$h = H\frac{a-b}{a}$.
So, $H = \frac{ha}{a-b}$
Substituting this for H in the conventional equation
$V_t = \frac{1}{3}\cdot (\frac{ha^3}{a-b}) - \frac{1}{3}\cdot (\frac{ha}{a-b})(\frac{b}{a})\cdot b^2$
$V_t=\frac{1}{3}\cdot h(\frac{a^3}{a-b} - \frac{b^3}{a-b})$
$V_t=\frac{1}{3}\cdot h(\frac{a^3-b^3}{a-b})$
$V_t=\frac{1}{3}\cdot h(\frac{(a-b)(a^2+ab+b^2)}{a-b})$
$V_t=\frac{1}{3}\cdot h(a^2+ab+b^2)$