AMC 8 2000 Problem 25

AMC8 2000 年第 25 题

四边形与多边形★★★★☆

The area of rectangle

ABCD

is 72 units squared. If point

A

and the midpoints of

\overline{BC}

and

\overline{CD}

are joined to form a triangle, the area of that triangle is

矩形

ABCD

的面积是 72 平方单位。若连接点

A

与

\overline{BC}

、

\overline{CD}

的中点构成一个三角形，那么这个三角形的面积是多少？

正确答案 · Correct Answer

解析 · Solution

解法1

常规解法 · 取方便边长面积与形状无关，取 AB=CD=12、BC=AD=6（面积 72）。设

BC

中点

M

、

CD

中点

N

：

[ADN]=\tfrac12\cdot6\cdot6=18

，

[MNC]=\tfrac12\cdot3\cdot6=9

，

[ABM]=\tfrac12\cdot12\cdot3=18

。故 [AMN]=72-18-9-18=27。

Solution 1

Elementary · pick a convenient side length Area is independent of shape, so take AB=CD=12 and BC=AD=6 (area 72). With

M

the midpoint of

BC

and

N

the midpoint of

CD

[ADN]=\tfrac12\cdot6\cdot6=18

[MNC]=\tfrac12\cdot3\cdot6=9

[ABM]=\tfrac12\cdot12\cdot3=18

. Then [AMN]=72-18-9-18=27.

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