2016 AMC 8 第 22 题

Question

下图中长方形

DEFA

是一个

3 \times 4

长方形，且

DC=CB=BA=1

。“蝙蝠翅膀”（阴影部分）的面积是

Rectangle

DEFA

below is a

3 \times 4

rectangle with

DC=CB=BA=1.

The area of the “bat wings” (shaded area) is

2

2 \dfrac{1}{2}

3

3 \dfrac{1}{2}

4

答案：C

知识点：相似三角形面积

难度评级：1640

小提示：

使用形成蝙蝠翅膀的两条对角线的交点

Use the intersection point of the two diagonals that form the bat wings.

大提示：

相似三角形给出顶部被切掉的小三角形的高度

Similar triangles give the height of the small triangle cut off at the top.

视频讲解：

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文字解答：

设

I

为

\overline{AD}

的中点，

G

为

\overline{EF}

的中点。再设

H

为

\overline{CF}

与

\overline{BE}

的交点。

\triangle BCE

的面积是

\dfrac12\cdot1\cdot4=2

。由对称性可知，

\triangle BCH

与

\triangle EFH

相似。它们的底边之比为

1 : 3

，所以高之比也相同。因此

3IH = HG

，从而

IH = 1

。

所以

[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12

。于是

[\triangle ECH]=2-\dfrac12=\dfrac32

。图形关于中线对称，因此蝙蝠双翼的总面积为

2\cdot\dfrac32=3

。

所以正确答案是 C。

Define

I

to be the midpoint of

\overline{AD}

and

G

to be the midpoint of

\overline{EF}.

Also define

H

to be the intersection of

\overline{CF}

and

\overline{BE}.

The area of

\triangle BCE

is

\dfrac12\cdot1\cdot4=2.

By symmetry, we can see that

\triangle BCH

and

\triangle EFH

are similar. Since their bases are in a

1 : 3

ratio, so are their altitudes. This means that

3IH = HG,

which implies that

IH = 1.

Therefore,

[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12.

It follows that

[\triangle ECH]=2-\dfrac12=\dfrac32.

Since the figure is symmetric, the total area of the bat wings is

2\cdot\dfrac32=3.

Thus, C is the correct answer.