2004 AMC 10A Problems/Problem 9

Problem

In the figure, $\angle EAB$ and $\angle ABC$ are right angles. $AB=4, BC=6, AE=8$ , and $AC$ and $BE$ intersect at $D$ . What is the difference between the areas of $\triangle ADE$ and $\triangle BDC$ ?

$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 4 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ } 9$

Solution

Let the area of $[ADE]=a$ , $[ADB]=b$ , $[DBC]=c$ .

$[ADE]-[DBC]=a-c=a+b-c-b=[ABE]-[ABC]=\dfrac{8\cdot 4}{2}-\dfrac{6\cdot 4}{2}=4\Rightarrow \boxed{\mathrm{(B)}}$

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2004 AMC 10A Problems/Problem 9

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