Question

一个钝角等腰三角形的两条相等边长度的乘积，等于底边长度与该底边上高的两倍的乘积。这个三角形顶角的度数是多少？

The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?

$105$

$120$

$135$

$150$

$165$

答案：D

知识点：等腰三角形三角形面积三角学

难度评级：1530

解答：

设两条相等边长为 $s$ ，底边为 $b$ ，到底边的高为 $h$ 。题设条件为 $s^2 = 2bh$ 。

三角形面积既等于 $\tfrac12 bh$ ，也等于 $\tfrac12 s^2 \sin\theta$ ，其中 $\theta$ 是顶角。因此 $bh = s^2 \sin\theta$ 。

代入 $s^2 = 2bh$ ，得到 $bh = 2bh\sin\theta$ ，所以 $\sin\theta = \tfrac12$ 。由于三角形为钝角， $\theta = 150^\circ$ 。

所以正确答案是 D。

Let the congruent sides have length $s,$ the base be $b,$ and the height to the base be $h.$ The given condition is $s^2 = 2bh.$

The area equals $\tfrac12 bh$ and also $\tfrac12 s^2 \sin\theta,$ where $\theta$ is the vertex angle. So $bh = s^2 \sin\theta.$

Substituting $s^2 = 2bh$ gives $bh = 2bh\sin\theta,$ so $\sin\theta = \tfrac12.$ Since the triangle is obtuse, $\theta = 150^\circ.$

Thus, the correct answer is D.

2021 AMC 12B Fall 第 8 题

答案：D

难度评级：1530

解答：

其他年份的第 8 题