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2012 AMC 10B Problems/Problem 19

Revision as of 14:50, 15 January 2013 by N1000 (talk | contribs)

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).

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