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. Unknown error_msg). 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 $\boxed{\textbf{(B) }25}$

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2014 AMC 10B Problems/Problem 21

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