Difference between revisions of "2012 AMC 10B Problems/Problem 19"
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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>15\cdot 6/2=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=3BF=3(15/2)=45/2.</math> Now, you can find the area of <math>CDF,</math> which is <math>135/2.</math> Now, you do <math>[ABCD]-[ABG]-[CDF]=180-45-135/2=135-135/2=135/2,</math> | 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>15\cdot 6/2=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=3BF=3(15/2)=45/2.</math> Now, you can find the area of <math>CDF,</math> which is <math>135/2.</math> Now, you do <math>[ABCD]-[ABG]-[CDF]=180-45-135/2=135-135/2=135/2,</math> | ||
which makes the answer (C). | which makes the answer (C). | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:57, 10 February 2014
Solution
The easiest way to find the area would be to find the area of and subtract the areas of
and
You can easily get the area of
because you know
and
, so
's area is
. However, for triangle
you don't know
However, you can note that triangle
is similar to triangle
through AA. You see that
So, You can do
for
and
Now, you can find the area of
which is
Now, you do
which makes the answer (C).
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