AMC8 1997

AMC8 1997 · Q24

AMC8 1997 · Q24. It mainly tests Circle theorems, Ratios in geometry.

Diameter ACE is divided at C in the ratio 2 : 3. The two semicircles, ABC and CDE, divide the circular region into an upper (shaded) region and a lower region. The ratio of the area of the upper region to that of the lower region is

直径 ACE 在 C 点按 2 : 3 的比例分割。两个半圆 ABC 和 CDE 将圆形区域分成上部（阴影）区域和下部区域。上部区域面积与下部区域面积的比值为

(A) 2:3 2:3

(B) 1:1 1:1

(D) 9:4 9:4

(E) 5:2 5:2

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let the diameter of the large circle equal 10. Then the ratio is: $\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ R+S\end{array}\right)-\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ S\end{array}\right)+\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ T\end{array}\right)$ $\dfrac{\dfrac{1}{2}\pi 5^{2}-\dfrac{1}{2}\pi 2^{2}+\dfrac{1}{2}\pi 3^{2}}{\dfrac{1}{2}\pi 5^{2}-\dfrac{1}{2}\pi 3^{2}+\dfrac{1}{2}\pi 2^{2}}=\dfrac{15\pi}{10\pi}=\dfrac{3}{2}\ \text{or}\ 3:2$ $\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ T+U\end{array}\right)-\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ T\end{array}\right)+\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ S\end{array}\right)$

答案（C）：设大圆的直径为 10，则比值为： $\left(\begin{array}{c}\text{半圆}\\ R+S\text{ 的面积}\end{array}\right)-\left(\begin{array}{c}\text{半圆}\\ S\text{ 的面积}\end{array}\right)+\left(\begin{array}{c}\text{半圆}\\ T\text{ 的面积}\end{array}\right)$ $\dfrac{\dfrac{1}{2}\pi 5^{2}-\dfrac{1}{2}\pi 2^{2}+\dfrac{1}{2}\pi 3^{2}}{\dfrac{1}{2}\pi 5^{2}-\dfrac{1}{2}\pi 3^{2}+\dfrac{1}{2}\pi 2^{2}}=\dfrac{15\pi}{10\pi}=\dfrac{3}{2}\ \text{或}\ 3:2$ $\left(\begin{array}{c}\text{半圆}\\ T+U\text{ 的面积}\end{array}\right)-\left(\begin{array}{c}\text{半圆}\\ T\text{ 的面积}\end{array}\right)+\left(\begin{array}{c}\text{半圆}\\ S\text{ 的面积}\end{array}\right)$

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