Question

下图显示了一辆大型公交车驾驶员一侧的雨刷。

雨刷臂 $\overline{AB}$ 绕点 $A$ 来回转动，扫过 $60^\circ$ 的弧，并且关于过 $A$ 的竖直线对称。雨刷片 $\overline{CD}$ 在其中点处连到 $B$ ，并且在雨刷臂运动时始终保持竖直。雨刷臂长 $3$ 英尺，雨刷片高 $3.5$ 英尺。雨刷清洁的挡风玻璃面积是多少平方英尺？答案四舍五入到百分位。（假设挡风玻璃是平的竖直平面。）

The windshield wiper on the driver's side of a large bus is depicted below.

Arm $\overline{AB}$ pivots back and forth around point $A,$ sweeping out an arc of $60^\circ,$ symmetric about the vertical line through $A.$ The wiper blade $\overline{CD}$ is attached to $B$ at its midpoint and stays vertical as the arm moves. The arm is $3$ feet long, and the wiper blade is $3.5$ feet tall. What is the area of the windshield cleaned by the wiper, in square feet, to the nearest hundredth? (Assume that the windshield is a flat vertical surface.)

$9.68$

$10.14$

$10.50$

$11.32$

$12.00$

答案：C

知识点：面积弧

难度评级：1690

解答：

把 $A$ 放在原点。则 $B = (3\sin\theta, 3\cos\theta)$ ，其中 $\theta \in [-30^\circ, 30^\circ]$ ，所以 $B$ 的水平坐标范围是 $[-1.5, 1.5]$ ，宽度为 $3$ 。每个水平位置都有一条高度为 $3.5$ 的竖直雨刷片经过，所以由卡瓦列里原理，清洁面积为 $3.5 \times 3 = 10.5$ 平方英尺。

所以正确答案是 C。

Put $A$ at the origin. Then $B = (3\sin\theta, 3\cos\theta)$ for $\theta \in [-30^\circ, 30^\circ],$ so the horizontal coordinate of $B$ ranges over $[-1.5, 1.5],$ a width of $3.$ At each horizontal position exactly one vertical blade of height $3.5$ passes through, so by Cavalieri's principle the cleaned area is $3.5 \times 3 = 10.5$ square feet.

Thus, the correct answer is C.

2025 AMC 12B 第 12 题

答案：C

难度评级：1690

解答：

其他年份的第 12 题