Difference between revisions of "2002 AIME I Problems/Problem 10"
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| − | In the diagram below, angle <math>ABC</math> is a right angle. Point <math>D</math> is on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects angle <math>CAB</math>. Points <math>E</math> and <math>F</math> are on <math>\overline{AB} and < | + | In the diagram below, angle <math>ABC</math> is a right angle. Point <math>D</math> is on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects angle <math>CAB</math>. Points <math>E</math> and <math>F</math> are on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>AE=3</math> and <math>AF=10</math>. Given that <math>EB=9</math> and <math>FC=27</math>, find the integer closest to the area of quadrilateral <math>DCFG</math>. |
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Revision as of 17:08, 25 September 2007
Problem
In the diagram below, angle 
is a right angle. Point 
is on 
, and 
bisects angle 
. Points 
and 
are on 
and 
, respectively, so that 
and 
. Given that 
and 
, find the integer closest to the area of quadrilateral 
.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Solution
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