Difference between revisions of "2012 AMC 10B Problems/Problem 19"
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<math>\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}</math> | <math>\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}</math> | ||
| − | + | [[Category: Introductory Geometry Problems]] | |
==Solution== | ==Solution== | ||
Revision as of 10:47, 13 August 2014
Problem
In rectangle 
, 
, 
, and 
is the midpoint of 
. Segment 
is extended 2 units beyond 
to point 
, and 
is the intersection of 
and 
. What is the area of 
?
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. 
