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). | ||
| + | == See Also == | ||
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| + | {{AMC10 box|year=2012|ab=B|num-b=18|num-a=20}} | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:49, 15 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 ![$[ABCD]-[ABG]-[CDF]=180-45-135/2=135-135/2=135/2,$](http://latex.artofproblemsolving.com/d/e/6/de622a8dff1693f7739536ccd6a0c49ad37b77af.png)
which makes the answer (C).
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
