Question

如图，一个边长为 $3$ 、 $4$ 、 $5$ 的直角三角形，其三个顶点是三个两两外切圆的圆心。这些圆的面积之和是多少？

The vertices of a $3$ – $4$ – $5$ right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of these circles?

$12\pi$

$\dfrac{25\pi}{2}$

$13\pi$

$\dfrac{27\pi}{2}$

$14\pi$

答案：E

知识点：相切圆方程组圆面积

难度评级：1330

解答：

若 $r, s, t$ 是三个顶点处圆的半径，则 $r + s = 3,\ r + t = 4,\ s + t = 5$ 。三式相加得 $r + s + t = 6$ ，所以 $r = 1,\ s = 2,\ t = 3$ 。

面积之和为 $\pi(1^2 + 2^2 + 3^2) = 14\pi$ 。

因此，正确答案是 E。

If $r, s, t$ are the radii at the vertices, then $r + s = 3,\ r + t = 4,\ s + t = 5.$ Adding all three gives $r + s + t = 6,$ so $r = 1,\ s = 2,\ t = 3.$

The sum of the areas is $\pi(1^2 + 2^2 + 3^2) = 14\pi.$

Thus, the correct answer is E.

2006 AMC 12A 第 13 题

答案：E

难度评级：1330

解答：

其他年份的第 13 题