2004 AMC 10A Problem 9

Below is the professionally curated solution for Problem 9 of the 2004 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10A solutions, or check the answer key.

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Concepts:triangle area area decomposition

Difficulty rating: 1330

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

$2$

$4$

$5$

$8$

$9$

Solution:

Let $[ABD]$ be the area shared by both large triangles. Then $[ABE] = [ADE] + [ABD]$ and $[ABC] = [BDC] + [ABD].$

Subtracting, $[ADE] - [BDC] = [ABE] - [ABC].$ Since $\angle EAB$ and $\angle ABC$ are right angles, $[ABE] = \tfrac12(4)(8) = 16, \qquad [ABC] = \tfrac12(4)(6) = 12.$

The difference is $16 - 12 = 4.$

Thus, the correct answer is B.

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