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, $\sin(\angle PAD)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}$ 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 $[ABCD]-([BCQ]+[APD])$ $=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}=30\sqrt{3}\implies\boxed{B}$

Alternate Solution

Let $AP$ and $BQ$ meet $CD$ at $X$ and $Y$ , respectively.

Since $\angle APD=90^{\circ}$ , $\angle ADP=\angle XDP$ , and they share $DP$ , triangles $APD$ and $XPD$ are congruent.

By the same reasoning, we also have that triangles $BQC$ and $YQC$ are congruent.

Hence, we have $[ABQCDP]=[ABYX]+\frac{[ABCD]-[ABYX]}{2}=\frac{[ABCD]+[ABYX]}{2}$ .

If we let the height of the trapezoid be $x$ , we have $[ABQCDP]=\frac{\frac{11+19}{2}\cdot x+\frac{11+7}{2}\cdot x}{2}=12x$ .

Thusly, if we find the height of the trapezoid and multiply it by 12, we will be done.

Let the projections of $A$ and $B$ to $CD$ be $A'$ and $B'$ , respectively.

We have $DA'+CB'=19-11=8$ , $DA'=\sqrt{DA^2-AA'^2}=\sqrt{49-x^2}$ , and $CB'=\sqrt{CB^2-BB'^2}=\sqrt{25-x^2}$ .

Therefore, $\sqrt{49-x^2}+\sqrt{25-x^2}=8$ . Solving this, we easily get that $x=\frac{5\sqrt{3}}{2}$ .

Multiplying this by 12, we find that the area of hexagon $ABQCDP$ is $30\sqrt{3}$ , which corresponds to answer choice $\boxed{B}$ .

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2008 AMC 12B Problems/Problem 25

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Problem 25

Solution

Alternate Solution

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