2020 AMC 12B 第 9 题

Question

一个半径为

4

英寸的圆的四分之三扇形及其内部，可以沿图中所示的两条半径粘合卷成一个直圆锥的侧面。这个圆锥的体积是多少立方英寸？

A three-quarter sector of a circle of radius

4

inches together with its interior can be rolled up to form the lateral surface of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?

3\pi \sqrt{5}

4\pi \sqrt{3}

3\pi \sqrt{7}

6\pi \sqrt{3}

6\pi \sqrt{7}

答案：C

知识点：圆锥体积勾股定理

难度评级：1470

小提示：

扇形的弧长会成为圆锥底面的周长。

The arc length of the sector becomes the circumference of the cone’s base

大提示：

扇形半径

4

会成为圆锥的母线长；先求底面半径，再求圆锥高。

The sector radius

4

becomes the slant height; find the base radius, then the cone’s height

解答：

扇形弧长为

\tfrac34 \cdot 2\pi \cdot 4 = 6\pi

，它等于底面周长。因此

2\pi r = 6\pi

，所以

r = 3

。

母线长是

4

，因此由勾股定理可得高为

h = \sqrt{4^2 - 3^2} = \sqrt{7}

。圆锥体积为

\frac13 \pi r^2 h = \frac13 \pi \cdot 9 \cdot \sqrt{7} = 3\pi\sqrt{7}\text{。}

所以正确答案是 C。

The sector’s arc length is

\tfrac34 \cdot 2\pi \cdot 4 = 6\pi,

which becomes the base circumference:

2\pi r = 6\pi,

so

r = 3.

The slant height is the sector radius

4,

so the height is

h = \sqrt{4^2 - 3^2} = \sqrt{7}.

The volume is

\frac13 \pi r^2 h = \frac13 \pi \cdot 9 \cdot \sqrt{7} = 3\pi\sqrt{7}.

Thus, the correct answer is C.

2020 AMC 12B 第 9 题

答案：C

难度评级：1470

小提示：

大提示：

解答：

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