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| − | ==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 ADE</math> and <math>\triangle BDC</math>?
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| − | <center>[[Image:AMC10_2004A_9.gif]]</center>
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| − | <math> \mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 4 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ } 9 </math>
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| − | ==Solution==
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| − | Let the area of <math>[ADE]=a</math>, <math>[ADB]=b</math>, <math>[DBC]=c</math>.
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| − | <math>[ADE]-[DBC]=a-c=a+b-c-b=[ABE]-[ABC]=\dfrac{8\cdot 4}{2}-\dfrac{6\cdot 4}{2}=4\Rightarrow \boxed{\mathrm{(B)}}</math>
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| − | == See also ==
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| − | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131320 AoPS topic]
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| − | {{AMC10 box|year=2004|ab=A|num-b=8|num-a=10}}
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