2025 AMC 10A Problems/Problem 10

[insert problem here]

Solution 1 (Somewhat Cheese)

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 $r=\frac{16}{2}=8\Rightarrow$ $A=\frac{\pi(8)^2}{2}=\boxed{\textbf{(C) }32\pi}$ ~Bocabulary142857

Solution 2

Let the radius of the larger semicircle be $R$ and that of the smaller one be $r.$ We are looking for $\dfrac{1}{2}\pi(R^2-r^2).$ If we connect the center of the large semicircle to one endpoint of the chord and to the center of the chord, we get a right triangle with legs $8$ and $r$ and hypotenuse $R.$ Hence, $R^2=r^2+8^2\implies R^2-r^2=64\implies\dfrac{1}{2}\pi(R^2-r^2)=\boxed{\text{() }32\pi}.$

~Nioronean

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2025 AMC 10A Problems/Problem 10

Solution 1 (Somewhat Cheese)

Solution 2