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 $\frac{A}{n}\sum _{i=1}^{n}{h}_{i}$, 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?

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 $\frac{A}{n}\sum _{i=1}^{n}{h}_{i}$, 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?