2012 AMC 10B Problems/Problem 19

Problem

In rectangle $ABCD$ , $AB=6$ , $AD=30$ , and $G$ is the midpoint of $\overline{AD}$ . Segment $AB$ is extended 2 units beyond $B$ to point $E$ , and $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$ . What is the area of $BFDG$ ?

$\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}$

Solution

The easiest way to find the area would be to find the area of $ABCD$ and subtract the areas of $ABG$ and $CDF.$ You can easily get the area of $ABG$ because you know $AB=6$ and $AG=15$ , so $ABG$ 's area is $15\cdot 6/2=45$ . However, for triangle $CDF,$ you don't know $CF.$ However, you can note that triangle $BEF$ is similar to triangle $CDF$ through AA. You see that $BE/DC=1/3.$ So, You can do $BF+3BF=30$ for $BF=15/2,$ and $CF=3BF=3(15/2)=45/2.$ Now, you can find the area of $CDF,$ which is $135/2.$ Now, you do $[ABCD]-[ABG]-[CDF]=180-45-135/2=135-135/2=135/2,$ which makes the answer (C).

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2012 AMC 10B Problems/Problem 19

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