2018 AMC 8 第 15 题

先试着解答 2018 AMC 8 第 15 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 2018 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。

15.

如下图,在两个较小圆中,每个较小圆的一条直径都是大圆的一条半径。如果两个较小圆的总面积为 11 平方单位,那么阴影区域的面积是多少平方单位?
In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of 11 square unit, then what is the area of the shaded region, in square units?
14\dfrac{1}{4}
13\dfrac{1}{3}
12\dfrac{1}{2}
11
π2\dfrac{\pi}{2}
答案:D
知识点:圆面积长度、面积与体积的缩放关系
小提示:
比较小圆半径与大圆半径
Compare the radius of a smaller circle to the radius of the larger circle.
大提示:
面积按长度比例的平方缩放
Area scales by the square of the scale factor.
视频讲解:
解答视频缩略图
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文字解答:
设大圆面积为 AA。
每个小圆的直径等于大圆半径,所以每个小圆的半径是大圆半径的一半。
用符号表示,若大圆半径为 rlr_l,每个小圆半径为 rsr_s,则 rs=12rlr_s = \dfrac12 r_l 因为大圆面积为 A=πrl2A=\pi r_l^2,所以每个小圆面积为 πrs2=π(12rl)2=14(πrl2)=14A。\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A\end{align*}\text{。}两个小圆总面积为 11 平方单位,于是 2⋅14A=12\cdot \dfrac14 A=1 平方单位,所以 A=2A=2 平方单位。
阴影面积等于大圆面积 (A)(A) 减去两个小圆的总面积 (1)(1),所以阴影面积为 A−1=2−1=1A - 1=2-1=1 平方单位。
所以正确答案是 D。
Let AA be the area of the large circle.
Since the diameter of each of the two smaller circles is itself the radius of the larger circle, the radius of each smaller circle is half that of the larger circle.
Symbolically, if we allow rlr_l to be the radius of the large circle and rsr_s to be the radius of each of the smaller circles: rs=12rlr_s = \dfrac12 r_l As the area of the larger circle is equal to A=πrl2,A=\pi r_l^2, the area of the smaller circles are equal to πrs2=π(12rl)2=14(πrl2)=14A.\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A.\end{align*} As the area of two of these smaller circles combined is equal to 11 square unit, then it follows that 2⋅14A=12\cdot \dfrac14 A=1 square unit, implying that A=2A=2 square units.
As the area of the shaded region is equal to the area of the larger circle (A)(A) minus the combined area of the two smaller circles (1),(1), the area of the shaded region is A−1=2−1=1 A - 1=2-1=1 square unit.
Thus, the correct answer is D
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