# 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 $$\frac{A}{3}(h_1 + h_2 + h_3)$$, 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?

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: $$V={A_1+A_2\over4}(h_1+h_2+h_3+h_4)={3\over2}(a+b).$$