|
|
| (2 intermediate revisions by 2 users not shown) |
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #redirect [[2004 AMC 12A Problems/Problem 8]] |
| − | In the figure, <math>\angle EAB</math> and <math>\angle ABC</math> are right angles. <math>AB=4, BC=6, AE=8</math>, and <math>AC</math> and <math>BE</math> intersect at <math>D</math>. What is the difference between the areas of <math>\triangle ABC</math> and <math>\triangle BDC</math>?
| |
| − | | |
| − | <center>[[Image:AMC10_2004A_9.gif]]</center>
| |
| − | | |
| − | <math> \mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 4 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ } 9 </math>
| |
| − | | |
| − | ==Solution==
| |
| − | {{solution}}
| |
| − | == See also ==
| |
| − | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131320 AoPS topic]
| |
| − | {{AMC10 box|year=2004|ab=A|num-b=8|num-a=10}}
| |
