I can find lots of pages online saying the volume of a Truncated Right Triangular Prism is $\frac{A}{3}(h_1 + h_2 + h_3)$, 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 $\frac{A}{n}\sum_{i=1}^nh_i$, 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?


1 Answer 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: $$ V_1=(0,0,0),\quad V_2=(1,0,a),\quad V_3=(0,1,b), \quad V_4=(2,2,2a+2b), $$ with $a$ and $b$ positive constants. Note that those points all lie in the same plane, because $\vec{V_1V_4}=2\cdot\vec{V_1V_2}+2\cdot\vec{V_1V_3}$

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:

$$V={A_1\over3}(h_1+h_2+h_4)+{A_2\over3}(h_1+h_3+h_4)= {1\over3}(3a+2b)+{1\over3}(2a+3b)={5\over3}(a+b).$$

On the other hand, if your generalised formula were true, we would have: $$ V={A_1+A_2\over4}(h_1+h_2+h_3+h_4)={3\over2}(a+b). $$

Hence the generalised formula doesn't work.


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