pyramid with a trapezoid as a base. Prove that there exists a pyramid SABCD with a given trapezoid ABCD as a base (BC||AD; the trapezoid's lateral sides AB and CD are not parallel) such that the pyramid's lateral faces SAB and SCD are both orthogonal to the base plane .
I figured out the case where the base of the pyramid is a triangle. In that case, two sides can be orthogonal to the base plane. I dont think it is possible for all three sides of the pyramid to be orthogonal though. As for the case where the base is a trapezoid, I have a hard time seeing it.
 A: Let $A = (2,0,0)$, $B = (1,0,0)$, $C = (0,1,0)$, $D = (0,2,0)$, and $S = (0,0,2)$. 
Then, $ABCD$ is a trapezoid (with $BC \parallel AD$ and $AB \not\parallel CD$) which lies in the $xy$-plane. 
Also, $SAB$ lies in the $xz$-plane and $SCD$ lies in the $yz$-plane, which are each orthogonal to the $xy$-plane. 
A: If you have the solution where the base of the pyramid is a triangle with two sides orthogonal in the base plane you are there.  Take the two lateral sides as the two sides that need to be perpendicular in the base plane.  Extend the other side of the triangle and the zero length side  at the opposite vertex of the triangle by the same amount and you have your solution with a trapezoid.
A: Here are three ways to do it.  The proof that these work is left as an exercise to the reader.


*

*Imagine a trapezoid in a plane.  Now imagine the planes containing the lateral sides and orthogonal to the trapezoid plane.  These planes intersect at some line orthogonal to the trapezoid plane.  Choose any point on that line as the apex of the pyramid.

*Let $E$ be the point of intersection of the lines through $AB$ and $CD$.  Then let $S$ be any point on the line through $E$ orthogonal to the trapezoid plane.

*Let $E$ be as above, and suppose without loss of generality that $BC$ is shorter than $AD$. Consider a pyramid with base $ADE$ and apex $S$, such that faces $AES$ and $DES$ are both orthogonal to the base.  Now cut through the pyramid with the plane through $BCS$.  One side of the cut will be pyramid $SABCD$, which has the desired property.
