2012 AMC 8 第 24 题

先试着解答 2012 AMC 8 第 24 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 2012 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

24.

一个半径为 22 的圆被分成四段全等圆弧。将这四段圆弧连接成图中的星形。星形面积与原圆面积之比是多少?
A circle of radius 22 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?
4−ππ\dfrac{4-\pi}{\pi}
1π\dfrac{1}\pi
2π\dfrac{\sqrt2}{\pi}
π−1π\dfrac{\pi-1}{\pi}
3π\dfrac{3}\pi
答案:A
知识点:圆面积面积分割
小提示:
将四段圆弧重新放回原圆中思考。
Rearrange the four circular arcs back inside the original circle.
大提示:
将星形与连接四个切分点形成的正方形比较。
Compare the star with the square formed by joining the four cut points.
视频讲解:
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文字解答:
原圆面积为 π⋅22=4π\pi\cdot2^2=4\pi。
连接四个四分之一圆弧端点形成一个正方形。这个正方形的两条对角线分别是 44 和 44,所以面积为 12⋅4⋅4=8\frac12\cdot4\cdot4=8。
圆内但正方形外的部分面积为 4π−84\pi-8。这四块与正方形内但星形外的四块全等。
因此星形面积为 8−(4π−8)=16−4π8-(4\pi-8)=16-4\pi。所求比值为 16−4π4π=4−ππ\dfrac{16-4\pi}{4\pi}=\dfrac{4-\pi}{\pi}。所以正确答案是 A。
The area of the original circle is π⋅22=4π\pi\cdot2^2=4\pi.
Join the four quarter-circle endpoints to form a square. The square has diagonals 44 and 44, so its area is 12⋅4⋅4=8\frac12\cdot4\cdot4=8.
The part inside the circle but outside this square has area 4π−84\pi-8. Those four pieces are congruent to the pieces inside the square but outside the star.
Thus the star area is 8−(4π−8)=16−4π8-(4\pi-8)=16-4\pi. The desired ratio is 16−4π4π=4−ππ\dfrac{16-4\pi}{4\pi}=\dfrac{4-\pi}{\pi}. Thus, the answer is A .
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