Question

图中， $\angle EAB$ 和 $\angle ABC$ 都是直角， $AB = 4$ 、 $BC = 6$ 、 $AE = 8$ ，且 $\overline{AC}$ 与 $\overline{BE}$ 交于 $D$ 。 $\triangle ADE$ 和 $\triangle BDC$ 的面积之差是多少？

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 $\triangle BDC?$

$2$

$4$

$5$

$8$

$9$

答案：B

知识点：三角形面积面积分割

难度评级：1330

解答：

设 $[ABD]$ 是两个大三角形共有的面积。则 $[ABE] = [ADE] + [ABD]$ ，且 $[ABC] = [BDC] + [ABD]$ 。

利用 $\angle EAB$ 和 $\angle ABC$ 都是直角，相减得 $\begin{aligned} &[ADE] - [BDC] \\ &= [ABE] - [ABC]. \end{aligned}$ $\begin{aligned} [ABE] &= \tfrac12(4)(8) = 16, \\ [ABC] &= \tfrac12(4)(6) = 12. \end{aligned}$

因此差为 $16 - 12 = 4$ 。

所以正确答案是 B。

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 第 9 题

答案：B

难度评级：1330

解答：

其他年份的第 9 题