During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

# 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 $h^2 = 100-4 \implies h = 4\sqrt{6}$.

$[asy] size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("D",D,SW); label("E",DD,SE); label("F",CC,SW); draw(C--CC); draw(D--DD); label("21",(CC+DD)/2,N); label("2",(A+DD)/2,N); label("10",(CC+B)/2,N); label("4\sqrt{6}",(C+CC)/2,W); label("4\sqrt{6}",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); [/asy]$

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

• Or, alternatively, one can notice that the two triangles have the same height but $\bigtriangleleup AFC$ (Error compiling LaTeX. ! Undefined control sequence.) has a shorter base than $\bigtriangleup BED$.

## Solution 2

$[asy] size(7cm); pair A,B,C,D,E; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(12,1),E); label("14",(2,1),E); label("12",(A+E)/2,N); label("21",(E+B)/2,N); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("D",D,SW); label("E",E,N); [/asy]$

The area of $\Delta AED$ is by Heron's, $4\sqrt{9(4)(3)(2)}=24\sqrt{6}$. This makes the length of the altitude from $D$ onto $\overline{AE}$ equal to $4\sqrt{6}$. One may now proceed as in Solution $1$ to obtain an answer of $\boxed{\textbf{(B) }25}$.

## See Also

 2014 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Invalid username
Login to AoPS