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Difference between revisions of "2012 AMC 10B Problems/Problem 19"

(Created page with "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 A...")
 
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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 5. 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=15/2. 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).
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The easiest way to find the area would be to find the area of <math>ABCD</math> and subtract the areas of <math>ABG</math> and <math>CDF.</math> You can easily get the area of <math>ABG</math> because you know <math>AB=6</math> and <math>AG=15</math>, so <math>ABG</math>'s area is <math>45</math>. However, for triangle <math>CDF,</math> you don't know <math>CF.</math> However, you can note that triangle <math>BEF</math> is similar to triangle <math>CDF</math> through AA. You see that <math>BE/DC=1/3.</math> So, You can do <math>BF+3BF=30</math> for <math>BF=15/2,</math> and <math>CF=15/2.</math> Now, you can find the area of <math>CDF,</math> which is <math>135/2.</math> Now, you do <math>180-225/2,</math> which turns out to be <math>135/2,</math> which makes the answer (C).

Revision as of 14:40, 15 January 2013

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 $45$. 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 $CF=15/2.$ 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).

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