2016 AMC 8 第 22 题

先试着解答 2016 AMC 8 第 22 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2016 AMC 8 解答,或核对答案

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22.

下图中长方形 DEFADEFA 是一个 3×43 \times 4 长方形,且 DC=CB=BA=1DC=CB=BA=1。“蝙蝠翅膀”(阴影部分)的面积是

Rectangle DEFADEFA below is a 3×43 \times 4 rectangle with DC=CB=BA=1.DC=CB=BA=1. The area of the "bat wings" (shaded area) is

22

2122 \dfrac{1}{2}

33

3123 \dfrac{1}{2}

44

答案:C
知识点:相似三角形面积
难度评级:1640
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文字解答:

IIAD\overline{AD} 的中点,GGEF\overline{EF} 的中点。再设 HHCF\overline{CF}BE\overline{BE} 的交点。

BCE\triangle BCE 的面积是 1214=2\dfrac12\cdot1\cdot4=2。由对称性可知,BCH\triangle BCHEFH\triangle EFH 相似。它们的底边之比为 1:31 : 3,所以高之比也相同。因此 3IH=HG3IH = HG,从而 IH=1IH = 1

所以 [BCH]=1211=12[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12。于是 [ECH]=212=32[\triangle ECH]=2-\dfrac12=\dfrac32。图形关于中线对称,因此蝙蝠双翼的总面积为 232=32\cdot\dfrac32=3

所以正确答案是 C

Define II to be the midpoint of AD\overline{AD} and GG to be the midpoint of EF.\overline{EF}. Also define HH to be the intersection of CF\overline{CF} and BE.\overline{BE}.

The area of BCE\triangle BCE is 1214=2.\dfrac12\cdot1\cdot4=2. By symmetry, we can see that BCH\triangle BCH and EFH\triangle EFH are similar. Since their bases are in a 1:31 : 3 ratio, so are their altitudes. This means that 3IH=HG,3IH = HG, which implies that IH=1.IH = 1.

Therefore, [BCH]=1211=12.[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12. It follows that [ECH]=212=32.[\triangle ECH]=2-\dfrac12=\dfrac32. Since the figure is symmetric, the total area of the bat wings is 232=3.2\cdot\dfrac32=3.

Thus, C is the correct answer.

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