AMC10 2004 A

AMC10 2004 A · Q9

AMC10 2004 A · Q9. It mainly tests Angle chasing, Area & perimeter.

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 $\triangle BDC$?

在图中，$\angle EAB$ 和 $\angle ABC$ 是直角，$AB = 4$，$BC = 6$，$AE = 8$，且 $AC$ 和 $BE$ 相交于 $D$。$\triangle ADE$ 和 $\triangle BDC$ 的面积差是多少？

(A) 2 2

(B) 4 4

(D) 8 8

(E) 9 9

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $x$, $y$, and $z$ be the areas of $\triangle ADE$, $\triangle BDC$, and $\triangle ABD$, respectively. The area of $\triangle ABE$ is $\frac{1}{2}(4)(8) = 16 = x + z$, and the area of $\triangle BAC$ is $\frac{1}{2}(4)(6) = 12 = y + z$. The requested difference is $x - y = (x + z) - (y + z) = 16 - 12 = 4$.

设 $x$、$y$ 和 $z$ 分别为 $\triangle ADE$、$\triangle BDC$ 和 $\triangle ABD$ 的面积。$\triangle ABE$ 的面积是 $\frac{1}{2}(4)(8) = 16 = x + z$，$\triangle BAC$ 的面积是 $\frac{1}{2}(4)(6) = 12 = y + z$。所求差值为 $x - y = (x + z) - (y + z) = 16 - 12 = 4$。

Topics