# For a generic base convex polygon, does the same argument apply? Is the volume equal to A/n sum_{i=1}^{n} h_i, where n is the number of vertices of the base/top?

Volume of a Truncated Right Prism with generic base convex polygon
I can find lots of pages online saying the volume of a Truncated Right Triangular Prism is $\frac{A}{3}\left({h}_{1}+{h}_{2}+{h}_{3}\right)$, where A is the area of the base and ${h}_{i}$ is the height of vertex i of the top face. Or, in other words, to find the volume you "flatten" the top face by finding the average height of its 3 vertices, and then the volume is just that of a regular prism.
My question is: for a generic base convex polygon, does the same argument apply? Is the volume equal to $\frac{A}{n}\sum _{i=1}^{n}{h}_{i}$, where n is the number of vertices of the base/top? Is there a proof for this anywhere?
My intuition says it's true, but what if the base shape was like a really long kite; wouldn't the height of the vertex at the bottom tip of the kite skew the average height of the top face?
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skarvama
Step 1
Your formula for the volume cannot be true, in general, for $n>3$. Here's a counter-example with $n=4$.
Take a truncated right quadrangular prism with its base on the $z=0$ plane and its top vertices given by:
${V}_{1}=\left(0,0,0\right),\phantom{\rule{1em}{0ex}}{V}_{2}=\left(1,0,a\right),\phantom{\rule{1em}{0ex}}{V}_{3}=\left(0,1,b\right),\phantom{\rule{1em}{0ex}}{V}_{4}=\left(2,2,2a+2b\right),$
with a and b positive constants. Note that those points all lie in the same plane, because $\stackrel{\to }{{V}_{1}{V}_{4}}=2\cdot \stackrel{\to }{{V}_{1}{V}_{2}}+2\cdot \stackrel{\to }{{V}_{1}{V}_{3}}$.
Step 2
The volume of this solid can be computed dividing it into two truncated triangular prisms with a plane passing through y-axis and ${V}_{4}$. Both their bases have unit area, hence applying the formula for the triangular case with ${h}_{1}=0$, ${h}_{2}=a$, ${h}_{3}=b$, ${h}_{4}=2a+2b$ we get:
$V=\frac{{A}_{1}}{3}\left({h}_{1}+{h}_{2}+{h}_{4}\right)+\frac{{A}_{2}}{3}\left({h}_{1}+{h}_{3}+{h}_{4}\right)=\frac{1}{3}\left(3a+2b\right)+\frac{1}{3}\left(2a+3b\right)=\frac{5}{3}\left(a+b\right).$
On the other hand, if your generalised formula were true, we would have:
$V=\frac{{A}_{1}+{A}_{2}}{4}\left({h}_{1}+{h}_{2}+{h}_{3}+{h}_{4}\right)=\frac{3}{2}\left(a+b\right).$
Hence the generalised formula doesn't work.