Prove the truncated pyramid volume formula
"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!!
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$\begingroup$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
$V_t = \frac{1}{3}\cdot (\frac{ha^3}{a-b}) - \frac{1}{3}\cdot (\frac{ha}{a-b})(\frac{b}{a})\cdot b^2$
$V_t=\frac{1}{3}\cdot h(\frac{a^3}{a-b} - \frac{b^3}{a-b})$
$V_t=\frac{1}{3}\cdot h(\frac{a^3-b^3}{a-b})$
$V_t=\frac{1}{3}\cdot h(\frac{(a-b)(a^2+ab+b^2)}{a-b})$
$V_t=\frac{1}{3}\cdot h(a^2+ab+b^2)$
$\endgroup$ $\begingroup$We call it a truncated pyramid because you can construct the same shape by taking an ordinary pyramid (with just one square face), laying the square face of the pyramid on a horizontal plane, using another horizontal plane to cut through the pyramid, and then discarding the part of the pyramid above the cut.
The piece that gets cut off and discarded is an ordinary pyramid, similar in shape to the original pyramid but smaller.
One way to figure out the volume of a truncated pyramid is to deduce what ordinary pyramid you would have had to start with in order to make a truncated pyramid with edge length $a$ on its bottom square face and edge length $b$ on its top square face. Assuming $b < a,$ the original pyramid would have had to have a square face of edge length $a,$ from which you cut away a pyramid with square face of edge length $b.$The ratio of heights of the two pyramids is $a:b,$ and the difference in heights is the height of the truncated pyramid that's left after you cut away the smaller pyramid, namely $h.$
If you can deduce the heights of the two ordinary pyramids (the one with square edge $a$ and the other with square edge $b$), you can use the given formula to get the volume of each pyramid, from that deduce the volume of the truncated pyramid (after you remove the smaller pyramid from the larger one), and simplify to get the desired formula.
Of course this derivation is only good when $b < a,$ but you should then quickly be able to show that the same formula also works when $b > a$ and when $b = a.$
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