2008 AMC 12B Problems/Problem 25

Problem 25

Let $ABCD$ be a trapezoid with $AB||CD, AB=11, BC=5, CD=19,$ and $DA=7$ . Bisectors of $\angle A$ and $\angle D$ meet at $P$ , and bisectors of $\angle B$ and $\angle C$ meet at $Q$ . What is the area of hexagon $ABQCDP$ ?

$\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \textbf{(E)}\ 36\sqrt{3}$

Solution

Drop perpendiculars to $CD$ from $A$ and $B$ , and call the intersections $X,Y$ respectively. Now, $DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2$ and $DX+CY=19-11=8$ . Thus, $DX-CY=3$ . We conclude $DX=\frac{11}{2}$ and $CY=\frac{5}{2}$ . To simplify things even more, notice that $90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD$ , so $\angle P=\angle Q=90^{\circ}$ .

Also, $\sin(\angle PDA)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}$ So the area of $\triangle APD$ is: $R\cdot c\sin a\sin b =\frac{7\cdot7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}$

Over to the other side: $\triangle BCY$ is $30-60-90$ , and is therefore congruent to $\triangle BCQ$ . So $[BCQ]=\frac{5\cdot5\sqrt{3}}{8}$ .

The area of the hexagon is clearly $[ABCD]-([BCQ]+[APD])$ $=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}=30\sqrt{3},\qquad\boxed{B}$

2008 AMC 12B (Problems • Answer Key • Resources)
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2008 AMC 12B Problems/Problem 25

Problem 25

Solution

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