Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline{CD}$ of length $21$. The other two sides are of lengths $10$ and $14$. The angles at $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$? $\textbf{(A) } 10\sqrt{6} \qquad$ $\textbf{(B) } 25 \qquad$ $\textbf{(C) } 8\sqrt{10} \qquad$ $\textbf{(D) } 18\sqrt{2} \qquad$ $\textbf{(E) } 26$
Solution
Place $A=(0,0)$ and $B=(33,0)$. Since $CD=21$ and both bottom angles are acute, write $D=(x,h)$ and $C=(x+21,h)$ for some $0<x<12$.
The horizontal projections of the legs are $x$ and $12-x$. Since their lengths are $10$ and $14$, subtracting their squared-length equations gives $$\left|(12-x)^2-x^2\right|=14^2-10^2=96,$$ so $x=2$ or $x=10$. In either case the leg with horizontal projection $2$ has length $10$, and hence $h^2=10^2-2^2=96$.
The diagonals have horizontal projections $21+x$ and $33-x$. Thus the shorter one has horizontal projection $21+\min(x,12-x)=23$.
Its length is $\sqrt{23^2+h^2}=\sqrt{529+96}=25$, answer (B).
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 $ABCD$?
