Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $21$ and $\overline{CD}$ of length $7$. The other two sides are of lengths $13$ and $15$. The angles at $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$?

1. Set $A=(0,0)$, $B=(21,0)$, $D=(x,h)$, $C=(x+7,h)$. Suppose $AD=13$ and $BC=15$: $x^2+h^2=169$ and $(14-x)^2+h^2=225$, so $196-28x=56$, $x=5$, $h=12$. (With the legs swapped the trapezoid is the mirror image, with the same diagonals.) So $D=(5,12)$, $C=(12,12)$, and both lower angles are acute.
2. The diagonals have lengths $AC=\sqrt{12^2+12^2}=12\sqrt2$ and $BD=\sqrt{16^2+12^2}=20$.
3. Since $12\sqrt2<20$, the shorter diagonal is $12\sqrt2$.