AMC12 2025 B

AMC12 2025 B · Q12

AMC12 2025 B · Q12. It mainly tests Area & perimeter, Geometry misc.

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.)

大型巴士驾驶侧的雨刮器如图所示。臂$\overline{AB}$围绕点$A$来回摆动，扫过一个以$A$为中心垂直线的$60^{\circ}$对称弧。雨刮刀片$\overline{CD}$附着在$B$的中点，并随着臂的移动保持垂直。臂长3英尺，雨刮刀片高3.5英尺。雨刮器清洁的挡风玻璃面积有多少平方英尺，保留到小数点后两位？（假设挡风玻璃是一个平坦的垂直表面。）

(A) 9.68 9.68

(B) 10.14 10.14

(D) 11.32 11.32

(E) 12.00 12.00

Answer

Correct choice: (C)

正确答案：（C）

Solution

The area cleaned by the wiper follows a thickened curve with vertical ends, where the curve is a $60^\circ$ arc with radius $3$. Since the sides are vertical, by Cavalieri's Principle, the area is equivalent to a rectangle with side lengths $3.5$ and the distance between the two vertical ends. Let $B'$ be the result of point $B$ at its leftmost point after it has swept $60^\circ$. Then $ABB'$ is equilateral so $BB' = AB = 3$. So the area is $3.5 \cdot 3 = \boxed{\textbf{(C) } 10.50}$. Note: if you aren't aware of Cavalieri's Principle, you can still recognize this by slicing the top curved part of the shape and moving it to the bottom, creating a rectangle with the same area as the original shape.

雨刮器清洁的面积是一个带有垂直端的加粗曲线，其中曲线是半径为3的$60^\circ$弧。由于两侧是垂直的，根据卡瓦列里原理，该面积等价于一个长宽分别为3.5和两个垂直端之间距离的矩形。设$B'$是点$B$扫过$60^\circ$后到达的最左点。那么$ABB'$是等边三角形，故$BB'=AB=3$。因此面积是$3.5\cdot3=\boxed{\textbf{(C) }10.50}$。注：如果你不了解卡瓦列里原理，你仍然可以通过将形状顶部曲线部分切下并移动到底部来识别出，这会形成一个与原形状面积相同的矩形。

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