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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 1

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

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 $h^2= 100-x^2$ 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 $h^2 = 100-4 \implies h = \sqrt{96}$.

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

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

Solution 2

size(7cm);
pair A,B,C,D,CC,DD;
A = (-2,7);
B = (14,7);
C = (10,0);
D = (0,0);
E = (4,7);
draw(A--B--C--D--cycle);
draw(D--E);
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);
label("$D$",D,SW);
label("$21$",(C+D)/2,S);
(Error compiling LaTeX. E = (4,7);
^
1b1ea1b866e85d255ff55009031082b8f710a74e.asy: 9.1: modifying non-public field outside of structure)