# The area of a truncated pyramid with irregular top and bottom surface, given the height $h$

This question is inspired by the question The volume for truncated pyramid with irregular base.

Given that we have the top and bottom surface area ($$A_1$$, $$A_2$$) of a pyramid, ad the height of the truncated pyramid is given by $$H$$, how can we find the surface area of the pyramid, given the height from the bottom $$h$$? Without a loss of generality we can assume that $$A_2$$ is always bigger than $$A_1$$.

I can prove that in two limiting cases where the surface is circle and rectangle, the area $$A_h$$ at height $$h$$ is somehow proportional to the square of $$h$$, ie: $$A_h \varpropto h^2$$. But is there a general formula connecting $$A_h$$, $$h$$, $$H$$, $$A_1$$ and $$A_2$$?

• The combination of the bottom surface and a horizontally shifted copy of the top surface describes a new truncated pyramid with a generally different area. Commented Dec 24, 2018 at 11:57
• @random, what does your comment has to do with my question? Commented Dec 25, 2018 at 9:13
• You are asking for the area of a truncated pyramid, which is to be derived from only its height and the areas of the upper and lower surface. Such a derivation is impossible if there are truncated pyramids that fit that description, but have different total areas. For the class of truncated pyramids described in the comment, the area of their vertical projection on the base plane can be arbitrarily large, with the real area being at least twice as large. Commented Dec 25, 2018 at 10:34
• If your real question is described in its last paragraph instead of in the header then the answer would be that the computation additionally requires the height of the not yet truncated pyramid or the distance between the top and bottom surface. Commented Dec 25, 2018 at 10:43
• @random you are right. I have modified the question , we do know the $H$ after all Commented Dec 25, 2018 at 11:21

$$A_h$$ is proportional with the square of the distance from its plane to the virtual top of the pyramid at height $$H_{top}$$, so from $$(\frac{H_{top}-H}{H_{top}})^2=\frac{A_1}{A_2}$$ it follows that $$H_{top}=\frac H{1-\sqrt{\frac{A_1}{A_2}}}$$ and $$A_h=(1-\frac h{H_{top}})^2A_2$$.
• $A_h$ is proportional with the square of the distance... but how am I going to know for sure that this is true for any shape $A$? Any proof for that? Commented Dec 25, 2018 at 23:46