2005 AMC 10A 第 12 题

先试着解答 2005 AMC 10A 第 12 题,然后核对你的答案与精心整理的解答,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 2005 AMC 10A 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。

12.

图中形状称为三叶形,由围绕全等等边三角形的边画圆形扇形构成。若三叶形的水平底边长为 22,它的面积是多少?
The figure shown is called a trefoil and is constructed by drawing circular sectors about sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length 2?2?
13π+32\dfrac{1}{3}\pi + \dfrac{\sqrt{3}}{2}
23π\dfrac{2}{3}\pi
23π+34\dfrac{2}{3}\pi + \dfrac{\sqrt{3}}{4}
23π+33\dfrac{2}{3}\pi + \dfrac{\sqrt{3}}{3}
23π+32\dfrac{2}{3}\pi + \dfrac{\sqrt{3}}{2}
答案:B
知识点:扇形面积分割等边三角形
小提示:
底边长 22 跨过两个半径,所以每个圆形扇形半径为 11。
The base of length 22 spans two radii, so each circular sector has radius 11
大提示:
将四个等边三角形和四个圆弓形重新组合成四个半径为 11、圆心角为 60∘60^\circ 的扇形。
Rearrange the four equilateral triangles and four circular segments into four 60∘60^\circ sectors of radius 11
解答:
底边长 22 等于两个半径,所以半径为 11。这个三叶形由四个等边三角形和四个圆弓形组成,它们可以重新拼成四个半径为 11、圆心角为 60∘60^\circ 的扇形,总面积为 4⋅60360π(1)2=23π4 \cdot \dfrac{60}{360}\pi (1)^2 = \dfrac{2}{3}\pi。
所以正确答案是 B。
Since the base 22 equals two radii, the radius is 1.1. The trefoil is made of four equilateral triangles and four circular segments, which reassemble into four 60∘60^\circ sectors of a circle of radius 1.1. Their total area is 4⋅60360π(1)2=23π.4 \cdot \dfrac{60}{360}\pi (1)^2 = \dfrac{2}{3}\pi.
Thus, the correct answer is B.
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