The figure above shows a polyhedron $ABCDEF$. $ABCD$ is a rectangle with $AB = 30$ and $BC = 50$. $EF$ is parallel to the plane of $ABCD$ and $EF = 36$. Triangles $ABF$ and $CDE$ are congruent with $AF = BF = DE = CE = \sqrt{355} $.
Find the volume of this polyhedron.
My attempt:
Analyzing the figure, if $h$ is the height of points $E$ and $F$ above $ABCD$, then
$ 355 = h^2 + 15^2 + (25 - 18)^2 $
From which $h = 9$
Now taking slices of the polyhedron parallel to $ABCD$, we can express the area of the slice as
$ A(z) = 30 (1 - \dfrac{z}{9}) (50 (1 - \dfrac{z}{9} ) + 36 (\dfrac{z}{9} ) ) $
And this simplifies to
$ A(z) = 30 (1 - \dfrac{z}{9} ) ( 50 - \dfrac{14 z}{9} ) $
And further into,
$ A(z) = 30 (50 - \dfrac{64 z}{9} + \dfrac{14 z^2}{81} ) $
Integrating from $z = 0 $ to $z = 9$ gives the volume as
$ V = \displaystyle \int_0^9 A(z) d z = 30 \left( 450 - \dfrac{64}{9} \left( \dfrac{81}{2} \right) + \dfrac{14}{81} \left( \dfrac{729}{3} \right) \right) = \boxed{6120} $
This shows another way of finding the volume of a composite figure (IE find volume that we can add or subtract to make the calculation easier)
Let $F'$ be the point alone $EF$ that is vertically over (the midpoint of) $AB$.
Likewise, let $E'$ be the point alone $EF$ that is vertically over (the midpoint of) $CD$.
OP's work shows that $ABF'$ is an isosceles triangle with base 30 and height 9, so has area $ 135$.
Volume of $ABF'CDE'$ = Area of $ABF' \times $ distance $BC = 6750$.
Volume of $ABF'F$ = $\frac{1}{3} \times $ Area of $ABF' \times $ distance $FF' = 315$. Likewise for volume of $CDE'$.
Hence, volume of $ABFCDE = 6750 - 2\times 315 = 6120$.