AMC10 2020 B

AMC10 2020 B · Q10

AMC10 2020 B · Q10. It mainly tests Fractions, Pythagorean theorem.

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，可以沿着所示的两条半径粘合卷起形成直圆锥的侧面。锥体的体积是多少立方英寸？

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

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

(D) $6\pi\sqrt{5}$ $6\pi\sqrt{5}$

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

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): 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 $$ \frac{1}{3}\pi r^2h=\frac{1}{3}\pi\cdot 3^2\cdot \sqrt{7}=3\pi\sqrt{7}. $$

答案（C）：扇形的弧长为 $\frac{3}{4}\cdot 2\pi \cdot 4 = 6\pi$。这段弧构成圆锥底面的周长，因此圆锥底面半径为 $r=\frac{6\pi}{2\pi}=3$。圆锥的斜高等于扇形的半径。由勾股定理，圆锥的高为 $h=\sqrt{4^2-3^2}=\sqrt{7}$。圆锥的体积为 $$ \frac{1}{3}\pi r^2h=\frac{1}{3}\pi\cdot 3^2\cdot \sqrt{7}=3\pi\sqrt{7}. $$

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