Difference between revisions of "2002 AMC 10A Problems/Problem 25"
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Revision as of 13:44, 11 February 2008
Problem
In trapezoid 
with bases 
and 
, we have 
, 
, 
, and 
. The area of is
Solution
It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend 
and 
to meet at point 
:
![[asy] size(250); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13); draw(A--B--C--D--cycle); draw(D--F--C,dashed); label("\(A\)",A,S); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,W); label("\(E\)",F,N); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,E); label("12",(B+C)/2,WSW); [/asy]](http://latex.artofproblemsolving.com/0/8/f/08fd9a3001f04eb9624f966ea35af9f5984fd7a8.png)
Since 
we have 
, with the ratio of proportionality being 
. Thus
So the sides of 
are 
, which we recognize to be a 
right triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),
![\[[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{210} \Rightarrow \mathrm{(C)}.\]](http://latex.artofproblemsolving.com/c/d/d/cdd5bcb19508a5509f631e8bb65ff675d29c72fe.png)
See also
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last problem | |
| 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 | ||
