# Difference between revisions of "2014 AMC 10B Problems/Problem 21"

## Problem

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 $A$ 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

$[asy] draw((0,0)--(33,0)--(23, -9.79795897)--(2, -9.79795897)--(0,0)); label("A", (0,0), NW); label("B", (33,0), NE); label("C", (23,-9.79795897), SE); label("D", (2, -9.79795897), SW); draw((2, 0)--(2, -9.79795897)); draw((23, 0)--(23,-9.79795897)); label("E", (2,0), N); label("F", (23,0), N); [/asy]$

In the diagram, $\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}$. Denote $\overline{AE} = x$ and $\overline{DE} = h$. In right triangle $AED$, we have from the Pythagorean theorem: $x^2+h^2=100$. Note that since $EF = DC$, we have $BF = 33-DC-x = 12-x$. Using the Pythagorean theorem in right triangle $BFC$, we have $(12-x)^2 + h^2 = 196$.

We isolate the $h^2$ term in both equations, getting \begin{align*}h^2 &= 100-x^2 \\h^2 &= 196-(12-x)^2\end{align} (Error compiling LaTeX. ).

Setting these equal, we have $100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2$. Now, we can determine that $h^2 = 100-4 \implies h = \sqrt{96}$. The two diagonals are $\overline{AC}$ and $\overline{BD}$. Using the Pythagorean theorem again on $\bigtriangleup AFC$ and $\bigtriangleup BED$, we can find these lengths to be $\sqrt{96+529} = 25$ and $\sqrt{96+961} = \sqrt{1057}$. Obviously, $25$ is the shorter length, and thus the answer is $\boxed{\textbf{(B) }25}$

solution by crastybow