AMC10 2025 A

AMC10 2025 A · Q10

AMC10 2025 A · Q10. It mainly tests Circle theorems, Area & perimeter.

A semicircle has diameter $\overline{AB}$ and chord $\overline{CD}$ of length $16$ parallel to $\overline{AB}$. A smaller semicircle with diameter on $\overline{AB}$ and tangent to $\overline{CD}$ is cut from the larger semicircle, as shown below. What is the area of the resulting figure, shown shaded?

一个半圆直径为$\overline{AB}$，弦$\overline{CD}$长为$16$且平行于$\overline{AB}$。一个较小的半圆直径在$\overline{AB}$上且与$\overline{CD}$相切，从较大的半圆中切掉，如下图所示。阴影所示图形的面积是多少？

(A) 16\pi 16\pi

(B) 24\pi 24\pi

(D) 48\pi 48\pi

(E) 64\pi 64\pi

Answer

Correct choice: (C)

正确答案：（C）

Solution

Notice that the size of the smaller semicircle is not specified, and there is no additional information that hints at any specific size for it. Hence, we can shrink the small semicircle until its area is arbitrarily small and negligible, leaving us with a semicircle with a diameter of $16$. The area of the semicircle is given by $\frac{\pi r^2}{2}$, so we have $r=\frac{16}{2}=8\Rightarrow$$A=\frac{\pi(8)^2}{2}=\boxed{\text{(C) }32\pi}$

注意较小半圆的大小没有指定，也没有额外信息暗示其具体大小。因此，我们可以将小半圆收缩到面积任意小且可忽略，从而剩下直径为$16$的半圆。半圆面积为$\frac{\pi r^2}{2}$，所以$r=\frac{16}{2}=8\Rightarrow A=\frac{\pi(8)^2}{2}=\boxed{\text{(C) }32\pi}$

Topics