Volume of a Truncated Right Prism with generic base convex polygonI can find lots of pages online saying the volume of a Truncated Right Triangular Prism is 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 A n ∑ 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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Volume of a Truncated Right Prism with generic base convex polygonI can find lots of pages online saying the volume of a Truncated Right Triangular Prism is       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 &quot;flatten&quot; 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       A    n        ∑          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&#039;s true, but what if the base shape was like a really long kite; wouldn&#039;t the height of the vertex at the bottom tip of the kite skew the average height of the top face?

skarvama · Accepted Answer

Step 1Your formula for the volume cannot be true, in general, for   n  &amp;gt;  3. Here&#039;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  )  ,        V    2    =  (  1  ,  0  ,  a  )  ,        V    3    =  (  0  ,  1  ,  b  )  ,        V    4    =  (  2  ,  2  ,  2  a  +  2  b  )  ,with a and b positive constants. Note that those points all lie in the same plane, because                               V          1                          V          4                    →        =  2  ⋅                              V          1                          V          2                    →        +  2  ⋅                              V          1                          V          3                    →      .Step 2The 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    =  2  a  +  2  b we get:  V  =                    A        1            3        (      h    1    +      h    2    +      h    4    )  +                    A        2            3        (      h    1    +      h    3    +      h    4    )  =            1      3        (  3  a  +  2  b  )  +            1      3        (  2  a  +  3  b  )  =            5      3        (  a  +  b  )  .On the other hand, if your generalised formula were true, we would have:  V  =                              A          1                +                  A          2                    4        (      h    1    +      h    2    +      h    3    +      h    4    )  =            3      2        (  a  +  b  )  .Hence the generalised formula doesn&#039;t work.

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?

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