2012 AMC 12B 第 15 题

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：

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