2012 AMC 10B Problems/Problem 19

Revision as of 14:45, 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 $180-225/2,$ which turns out to be $135/2,$ which makes the answer (C).