Question

Jesse 沿两条半径剪开一个半径为 $12$ 的圆形纸片，形成两个扇形，其中较小扇形的圆心角为 $120$ 度。他用每个扇形作为一个圆锥的侧面，做成两个圆锥。较小圆锥的体积与较大圆锥的体积之比是多少？

Jesse cuts a circular paper disk of radius $12$ along two radii to form two sectors, the smaller having a central angle of $120$ degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?

$\dfrac{1}{8}$

$\dfrac{1}{4}$

$\dfrac{\sqrt{10}}{10}$

$\dfrac{\sqrt{5}}{6}$

$\dfrac{\sqrt{10}}{5}$

答案：C

知识点：圆锥体积勾股定理

难度评级：1800

小提示：

每个扇形的弧长会成为对应圆锥底面的周长，从而求出底面半径

The arc length of each sector becomes the circumference of that cone’s base, giving its base radius

大提示：

每个圆锥的母线长都是 $12$ ，比较 $\tfrac13\pi r^2 h$ 前，先用勾股定理求出各自的高

Each cone has slant height $12,$ so find each height with the Pythagorean theorem before comparing $\tfrac13\pi r^2 h$

解答：

每个扇形形成的圆锥母线长为 $12$ 。较小扇形的弧长为 $\tfrac{120}{360}\cdot2\pi\cdot12=8\pi$ ，因此底面半径为 $4$ ，高为 $\sqrt{12^2-4^2}=8\sqrt2$ 。

较大扇形（圆心角 $240^\circ$ ）的弧长为 $16\pi$ ，底面半径为 $8$ ，高为 $\sqrt{12^2-8^2}=4\sqrt5$ 。

体积之比为 $\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}\text{。}$

因此正确答案是 C。

Each sector forms a cone with slant height $12.$ The smaller sector’s arc length is $\tfrac{120}{360}\cdot2\pi\cdot12=8\pi,$ so its base radius is $4$ and its height is $\sqrt{12^2-4^2}=8\sqrt2.$

The larger sector (central angle $240^\circ$ ) has arc length $16\pi,$ base radius $8,$ and height $\sqrt{12^2-8^2}=4\sqrt5.$

The ratio of volumes is $\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}.$

Thus, the correct answer is C.

2012 AMC 12B 第 15 题

答案：C

难度评级：1800

小提示：

大提示：

解答：

其他年份的第 15 题