2004 AMC 12A Problem 8

Below is the professionally curated solution for Problem 8 of the 2004 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12A solutions, or check the answer key.

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Concepts:triangle area area decomposition

Difficulty rating: 1370

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 $x,$ $y,$ and $z$ be the areas of $\triangle ADE,$ $\triangle BDC,$ and $\triangle ABD,$ respectively.

Then $\triangle ABE$ has area $\tfrac12 \cdot 4 \cdot 8 = 16 = x + z,$ and $\triangle ABC$ has area $\tfrac12 \cdot 4 \cdot 6 = 12 = y + z.$

The requested difference is $x - y = (x + z) - (y + z) = 16 - 12 = 4.$

Thus, the correct answer is B.

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