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

size(7cm);
pair A,B,C,D,CC,DD;
A = (-2,7);
B = (14,7);
C = (10,0);
D = (0,0);
E= (0, 7);
F= (10, 7)
CC = (10,7);
DD = (0,7);
draw(A--B--C--D--cycle);
//label("33",(A+B)/2,N);
label("21",(C+D)/2,S);
label("10",(A+D)/2,W);
label("14",(B+C)/2,E);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
draw(C--CC); draw(D--DD);
 (Error making remote request. Unknown error_msg)

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 $h^2= 100-x^2$ and $h^2 = 196-(12-x)^2$ .

Setting these equal, we have $100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2$ . Now, we can determine that $h^2 = 100-4 \implies h = \sqrt{96}$ .

$[asy] size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); draw(C--CC); draw(D--DD); label("21",(CC+DD)/2,N); label("$2$",(A+DD)/2,N); label("$10$",(CC+B)/2,N); label("$\sqrt{96}$",(C+CC)/2,W); label("$\sqrt{96}$",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$

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 $\sqrt{96+961} = \sqrt{1057}$ . Since $\sqrt{96+529}<\sqrt{96+961}$ , $25$ is the shorter length, so 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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