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.

$\dfrac83$

$\dfrac{30}{11}$

$3$

$\dfrac{25}{8}$

$\dfrac72$

答案：B

知识点：相似圆锥圆柱

难度评级：1680

解答：

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.

2001 AMC 10 第 21 题

答案：B

难度评级：1680

解答：

其他年份的第 21 题