# 1984 AHSME Problems/Problem 8

## Problem

Figure $ABCD$ is a trapezoid with $AB||DC$, $AB=5$, $BC=3\sqrt{2}$, $\angle BCD=45^\circ$, and $\angle CDA=60^\circ$. The length of $DC$ is

$\mathrm{(A) \ }7+\frac{2}{3}\sqrt{3} \qquad \mathrm{(B) \ }8 \qquad \mathrm{(C) \ } 9 \frac{1}{2} \qquad \mathrm{(D) \ }8+\sqrt{3} \qquad \mathrm{(E) \ } 8+3\sqrt{3}$

## Solution

$[asy] unitsize(1.5cm); draw((0,0)--(5,0)--(8,-3)--(-sqrt(3),-3)--cycle); label("A",(0,0),NW); label("B",(5,0),NE); label("C",(8,-3),ENE); label("D",(-sqrt(3),-3),WNW); draw((0,0)--(0,-3)); draw((5,0)--(5,-3)); label("E",(5,-3),S); label("F",(0,-3),S); label("5",(2.5,0),N); label("3\sqrt{2}",(6.5,-1.5),ENE); label("45^\circ",(8,-3),WNW); label("60^\circ",(-sqrt(3),-3),ENE); draw((-.25,-2.75)--(-.25,-3)); draw((-.25,-2.75)--(0,-2.75)); draw((5.25,-3)--(5.25,-2.75)); draw((5.25,-2.75)--(5,-2.75));[/asy]$ Drop perpendiculars from $A$ and $B$ to $CD$, thus forming a 30-60-90 right triangle, a 45-45-90 right triangle, and a rectangle, as shown in the diagram. Since $ABEF$ is a rectangle, we have $EF=AB=5$. Also, from the properties of 45-45-90 right triangles, we have $EB=EC=\frac{BC}{\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{2}}=3$. Again since $ABEF$ is a rectangle, $AF=3$, too. From the properties of 30-60-90 right triangles, we have $DF=\frac{AF}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}$. Therefore, $CD=CE+EF+FD=3+5+\sqrt{3}=8+\sqrt{3}, \boxed{\text{D}}$.

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