2001 AMC 10 第 21 题

Question

一个直圆柱内接于一个直圆锥，圆柱的直径等于它的高。圆锥直径为

10

，高为

12

，且圆柱与圆锥的轴重合。求圆柱的半径。

A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter

10

and altitude

12,

and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

\dfrac83

\dfrac{30}{11}

3

\dfrac{25}{8}

\dfrac72

在轴截面中，圆柱半径为

r

，高为

2r

。

In an axial cross-section, the cylinder has radius

r

and height

2r

相似三角形给出

\dfrac{12-2r}{r}=\dfrac{12}{5}

。

Similar triangles give

\dfrac{12-2r}{r}=\dfrac{12}{5}

取轴截面。圆锥底面半径为

5

、高为

12

；圆柱截面的宽为

2r

、高为

2r

。

由相似三角形，

\dfrac{12-2r}{r}=\dfrac{12}{5}

，所以

5(12-2r)=12r

，即

60=22r

，从而

r=\dfrac{30}{11}

。

所以正确答案是 B。

Take an axial cross-section. The cone has base radius

5

and height

12;

the cylinder appears as a rectangle of width

2r

and height

2r.

By similar triangles,

\dfrac{12-2r}{r}=\dfrac{12}{5},

so

5(12-2r)=12r,

giving

60=22r

and

r=\dfrac{30}{11}.

Thus, the correct answer is B.

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