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Difference between revisions of "2008 AMC 12B Problems/Problem 25"
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==Solution==
 
==Solution==
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Drop perpendiculars to <math>CD</math> from <math>A</math> and <math>B</math>, and call the intersections <math>X,Y</math> respectively. Now, <math>DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2</math> and <math>DX+CY=19-11=8</math>. Thus, <math>DX-CY=3</math>.
 +
We conclude <math>DX=\frac{11}{2}</math> and <math>CY=\frac{5}{2}</math>.
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To simplify things even more, notice that <math>90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD</math>, so <math>\angle P=\angle Q=90^{\circ}</math>.
  
 +
Also, <cmath>\sin(\angle PDA)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}</cmath>
 +
So the area of <math>\triangle APD</math> is: <cmath>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}</cmath>
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 +
Over to the other side: <math>\triangle BCY</math> is <math>30-60-90</math>, and is therefore congruent to <math>\triangle BCQ</math>. So <math>[BCQ]=\frac{5\cdot5\sqrt{3}}{8}</math>.
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 +
The area of the hexagon is clearly <math>[ABCD]-([BCQ]+[APD])</math><cmath>=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}=30\sqrt{3},\qquad\boxed{B}</cmath>
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}}
 
{{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}}

Revision as of 16:25, 2 March 2008

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 $[ABCD]-([BCQ]+[APD])$\[=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}=30\sqrt{3},\qquad\boxed{B}\]

See Also

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
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All AMC 12 Problems and Solutions
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