Question

一个截锥有水平底面，两个底面半径分别为 $18$ 和 $2$ 。一个球与截锥的上底面、下底面和侧面都相切。这个球的半径是多少？

A truncated cone has horizontal bases with radii $18$ and $2.$ A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?

$6$

$4\sqrt{5}$

$9$

$10$

$6\sqrt{3}$

答案：A

知识点：圆锥球切线

难度评级：1870

解答：

轴截面是梯形 $ABCD$ ，平行边长为 $2$ 和 $18$ ，且有一个内切圆（球的大圆）。从 $B$ 和 $C$ 引出的切线段长度相等，所以斜边 $BC = 18 + 2 = 20$ 。从 $C$ 向下底作垂线，得到一个水平直角边为 $18 - 2 = 16$ 的直角三角形，所以高为 $\sqrt{20^2 - 16^2} = 12$ 。球的半径是高的一半，即 $6$ 。

因此正确答案是 A。

The axial cross-section is a trapezoid $ABCD$ with parallel sides $2$ and $18$ and an inscribed circle (a great circle of the sphere). By equal tangent lengths from $B$ and $C,$ the slant side $BC = 18 + 2 = 20.$ Dropping a perpendicular from $C$ to the bottom base gives a right triangle with horizontal leg $18 - 2 = 16,$ so the height is $\sqrt{20^2 - 16^2} = 12.$ The sphere's radius is half the height, $6.$

Thus, the correct answer is A.

2004 AMC 12B 第 19 题

答案：A

难度评级：1870

解答：

其他年份的第 19 题