AMC 8 2014 Problem 20

AMC8 2014 年第 20 题

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Rectangle ABCD has sides CD=3 and DA=5 . A circle with a radius of 1 is centered at A , a circle with a radius of 2 is centered at B , and a circle with a radius of 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?

长方形 ABCD 的边长 CD＝3，DA＝5．圆 A 的半径为 1，圆 B 的半径为 2，圆 C 的半径为 3．那么在长方形内部但在三个圆外部的区域的面积最接近下面哪个数？

A.3.5
B.4.0
C.4.5
D.5.0
E.5.5

正确答案 · Correct Answer

解析 · Solution

Solution 1

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

\displaystyle \frac{\pi}{4}

\displaystyle \frac{\pi(2)^2}{4}

\displaystyle \frac{\pi(3)^2}{4} = \frac{14\pi}{4}

\displaystyle = \frac{7\pi}{2}

. Therefore the area in the rectangle but outside the circles is

\displaystyle 15- \frac{7\pi}{2}

\pi

is approximately

\dfrac{22}{7},

and substituting that in will give

15-11=\text{(B)4.0}

.5 (People who don't know how to do the area of a sector) Instead of calculating the area of the rectangle minus the area of the 3 quarter-circles that are in it, we calculate the area of the 3 circles, not just the quarter-circles.

3^2\pi

is equal to

9\pi

2^2\pi

squared is equal to

4\pi

, and

1^2\pi

is equal to 1. It is obvious that they are all quarter-circles, as mentioned in solution 1, so you divide each of them by 4. So they become

\pi

\displaystyle \frac{9}{4}\pi

, and

\displaystyle \frac{1}{4}\pi

. Adding them together, it's

\displaystyle \frac{14}{4}\pi

. Now you can calculate

\displaystyle 15- \frac{14}{4}\pi

, and you'll get around 4.00000000000001, which is closest to

\text{(B)4.0}

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