2004 AMC 10A 第 9 题

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

小提示：

向两个小三角形都加上

\triangle ABD

，会得到两个较大的三角形。

Adding

\triangle ABD

to each of the two triangles produces two larger triangles

大提示：

减去共有面积后，

[ADE] - [BDC]

等于

[ABE] - [ABC]

。

Subtracting the shared area makes

[ADE] - [BDC]

equal to

[ABE] - [ABC]

解答：

设

[ABD]

是两个大三角形共有的面积。则

[ABE] = [ADE] + [ABD]

，且

[ABC] = [BDC] + [ABD]

。

相减得

\begin{aligned} &[ADE] - [BDC] \\ &= [ABE] - [ABC] \end{aligned}\text{。}

由于

\angle EAB

和

\angle ABC

都是直角，

\begin{aligned} [ABE] &= \tfrac12(4)(8) = 16, \\ [ABC] &= \tfrac12(4)(6) = 12 \end{aligned}\text{。}

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

小提示：

大提示：

解答：

其他年份的第 9 题