AMC 8 1987 Problem 22

AMC8 1987 年第 22 题

圆★★★☆☆
ABCDABCD is a rectangle, DD is the center of the circle, and BB is on the circle. If AD=4 and CD=3 , then the area of the shaded region is between
ABCDABCD 是一个矩形，DD 是圆的圆心，BB 在圆上。如果 AD=4 且 CD=3，则着色区域的面积在以下哪两个数之间？
AMC 8 1987 Problem 22 figure
  1. A.4 and 54\text{ and }5
  2. B.5 and 65\text{ and }6
  3. C.6 and 76\text{ and }7
  4. D.7 and 87\text{ and }8
  5. E.8 and 98\text{ and }9

正确答案 · Correct Answer

D

解析 · Solution

Solution 1

Since ABCDABCD is a rectangle, BC = AD = 4. So by the Pythagorean theorem, BD=BC2+CD2=32+42=5BD = \sqrt{BC^2 + CD^2} = \sqrt{3^2+4^2} = 5. So, the circle has radius 5 and total area 25π25\pi. Consequently, the quarter circle bounded by ∠D\angle D has area 25π4\displaystyle \frac{25\pi}{4}.
Since ABCDABCD has area 3⋅4=123\cdot 4=12, the shaded area is 25π4−12≈25⋅3.144−12=7.625\displaystyle \frac{25\pi}{4} - 12 \approx \frac{25 \cdot 3.14}{4} - 12 = 7.625, so the area is between 7 and 8\text{7 and 8}.

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