Question

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 $\triangle BDC?$

$2$

$4$

$5$

$8$

$9$

Answer: B

Concepts:triangle area area decomposition

Difficulty rating: 1330

Solution:

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

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

The difference is $16 - 12 = 4.$

Thus, the correct answer is B.

2004 AMC 10A Problem 9

Answer: B

Difficulty rating: 1330

Solution:

Problem 9 in Other Years