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2025 AMC 10A Problems/Problem 10: Difference between revisions

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~Nioronean, <imath>\LaTeX</imath> by Tacos_are_yummy_1
~Nioronean, <imath>\LaTeX</imath> by Tacos_are_yummy_1
==See Also==
{{AMC10 box|year=2025|ab=A|before=[[2024 AMC 10B Problems]]|after=[[2025 AMC 10B Problems]]}}
* [[AMC 10]]
* [[AMC 10 Problems and Solutions]]
* [[Mathematics competitions]]
* [[Mathematics competition resources]]
{{MAA Notice}}

Revision as of 14:42, 6 November 2025

Problem

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.

[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(-3,0), B=(3,0); fill(Arc((0,0),3,0,180)--cycle,palered); fill(Arc((-1.125,0),0.75,0,180)--cycle,white); draw(Arc((0,0),3,0,180),black); draw(Arc((-1.125,0),0.75,0,180),black); draw((-3,0) -- (-1.875,0),black); draw((-0.375,0) -- (3,0),black); draw((-2.895, 0.75) -- (2.895,0.75), black); dot((-3,0)); dot((3,0)); dot((-2.925, 0.75)); dot((2.925, 0.75)); label("$16$",midpoint((-2.925, 0.75)--(2.925, 0.75)),N); label("$A$",A,S); label("$B$",B,S); label("$C$",(-2.925, 0.75),W); label("$B$",(2.925, 0.75),E); [/asy]


What is the area of the resulting figure, shown shaded?

$\textbf{(A) } 16\pi \qquad\textbf{(B) } 24\pi \qquad\textbf{(C) } 32\pi \qquad\textbf{(D) } 48\pi \qquad\textbf{(E) } 64\pi$


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, $R^2=r^2+8^2\implies R^2-r^2=64\implies\dfrac{1}{2}\pi(R^2-r^2)=\boxed{\text{(C) }32\pi}.$

~Nioronean, $\LaTeX$ by Tacos_are_yummy_1

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
2024 AMC 10B Problems 		Followed by
2025 AMC 10B Problems
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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