AMC8 2014

AMC8 2014 · Q20

AMC8 2014 · Q20. It mainly tests Circle theorems, Area & perimeter.

Rectangle $ABCD$ has sides $CD=3$ and $DA=5$. A circle of radius $1$ is centered at $A$, a circle of radius $2$ is centered at $B$, and a circle of radius $3$ is centered at $C$. Which of the following is closest to the area of the region inside the rectangle but outside all three circles?

矩形 $ABCD$ 的边长为 $CD=3$ 和 $DA=5$。以点 $A$ 为圆心，半径为 $1$ 画圆；以点 $B$ 为圆心，半径为 $2$ 画圆；以点 $C$ 为圆心，半径为 $3$ 画圆。以下哪个数值最接近矩形内部但不在这三个圆内部的区域的面积？

(A) 3.5 3.5

(B) 4.0 4.0

(D) 5.0 5.0

(E) 5.5 5.5

Answer

Correct choice: (B)

正确答案：（B）

Solution

The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle. The area of the rectangle is $3\cdot5 =15$. The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$. Therefore the area in the rectangle but outside the circles is $15-\frac{7\pi}{2}$. $\pi$ is approximately $\dfrac{22}{7},$ and substituting that in will give $15-11=\boxed{\text{(B) }4.0}$

矩形内部但不在圆内的区域面积等于矩形面积减去矩形内部所有三个四分之一圆的面积。矩形的面积是 $3 \times 5 = 15$。三个四分之一圆的总面积是 $\frac{\pi}{4} + \frac{\pi (2)^2}{4} + \frac{\pi (3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$。因此，矩形内部但不在圆内的区域面积是 $15 - \frac{7\pi}{2}$。取 $\pi \approx \frac{22}{7}$，代入得到 $15 - 11 = \boxed{\text{(B) }4.0}$。

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