AMC12 2020 B

AMC12 2020 B · Q9

AMC12 2020 B · Q9. It mainly tests Circle theorems, 3D geometry (volume).

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?

一个半径为 4 英寸的圆的三刻度扇形连同其内部，可以沿所示的两条半径粘合，形成一个右圆锥的侧面。何为该圆锥的体积（立方英寸）？

(A) $3\pi\sqrt{5}$ $3\pi\sqrt{5}$

(B) $4\pi\sqrt{3}$ $4\pi\sqrt{3}$

(D) $6\pi\sqrt{3}$ $6\pi\sqrt{3}$

(E) $6\pi\sqrt{7}$ $6\pi\sqrt{7}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

The length of the circular portion of the sector is $\frac{3}{4} \cdot 2\pi \cdot 4 = 6\pi$. That arc forms the circumference of the base of the cone, so the radius of the base of the cone is $r = \frac{6\pi}{2\pi} = 3$. The slant height of the cone is the radius of the sector. By the Pythagorean Theorem the cone's height is $h = \sqrt{4^2 - 3^2} = \sqrt{7}$. The volume of the cone is $$\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \cdot 3^2 \cdot \sqrt{7} = 3\pi\sqrt{7}.$$

扇形圆弧长为 $\frac{3}{4} \cdot 2\pi \cdot 4 = 6\pi$。此弧形成圆锥底面周长，故底面半径 $r = \frac{6\pi}{2\pi} = 3$。圆锥的斜高为扇形半径 4。由勾股定理，高 $h = \sqrt{4^2 - 3^2} = \sqrt{7}$。体积为 $$\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \cdot 3^2 \cdot \sqrt{7} = 3\pi\sqrt{7}.$$

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