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

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~MK
~MK
==Video Solution by SpreadTheMathLove==
https://www.youtube.com/watch?v=dAeyV60Hu5c


==Video Solution by Daily Dose of Math==
==Video Solution by Daily Dose of Math==

Revision as of 15:17, 7 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); pair 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("$D$",(2.925, 0.75),E); [/asy]

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

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

Video Solution

https://youtu.be/l1RY_C20Q2M

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{\text{(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, $\LaTeX$ and writing by Tacos_are_yummy_1

Solution 3

The problem doesn't restrict where the smaller semicircle is along the larger semicircle's diameter. Therefore, we can assume that the two semicircles are concentric. Let the center of both semicircles be $O$, and let $CD$ be tangent to the smaller semicircle at $T$. Let the radius of the smaller semicircle be $x$, and let the radius of the larger semicircle be $r$. If we mirror the diagram over $AB$, we can see that we have two concentric circles. We are trying to find $\pi(\frac{r^2-x^2}{2})$. By Power of a Point on $T$, we can see that \[64 = (r + x)(r - x) = r^2 - x^2.\] Thus, $\pi(\frac{r^2 - x^2}{2}) = 32\pi \implies \boxed{\text{(C) }32\pi}.$ ~vinceS

Solution 4

The problem doesn't define where the chord is within the circle. So, let's place the chord such that when lines are drawn from its ends to the center of the larger semicircle, an equilateral triangle with side length $16$ is formed.

[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; fill(Arc((0,0),3,0,180)--cycle,palered); fill(Arc((0,0),2.598,0,180)--cycle,white); draw(Arc((0,0),3,0,180),black); draw(Arc((0,0),2.598,0,180),black); draw((-2.598,0) -- (2.598,0), white); draw((-1.5,2.598) -- (1.5,2.598),black); draw((0,0) -- (-1.5, 2.598), black); draw((0,0) -- (1.5, 2.598), black); draw((0,0) -- (0, 2.598), black); draw((-3,0) -- (-2.598, 0), black); draw((3,0) -- (2.598, 0), black); dot((0,0)); dot((0,2.598)); dot((-1.5,2.598)); dot((1.5,2.598)); dot((3,0)); dot((-3,0)); label("$16$",midpoint((0, 2.598)--(0, 2.598)),N); label("$8\sqrt{3}$",midpoint((0, 1.4)--(0, 1.4)),E); label("$16$",midpoint((-0.9, 1.3)--(-0.9, 1.3)),W); label("$16$",midpoint((0.9, 1.3)--(0.9, 1.3)),E); label("$A$",midpoint((-3, 0)--(-3, 0)),W); label("$B$",midpoint((3, 0)--(3, 0)),E); label("$C$",midpoint((-1.5,2.7)--(-1.5,2.7)),N); label("$D$",midpoint((1.5,2.7)--(1.5,2.7)),N); [/asy]

In this definition, the radius of the larger semicircle is $16,$ giving it an area of $\frac{\pi\cdot16^2}{2} = 128\pi.$

The height of the equilateral triangle is $8\sqrt{3},$ which is equal to the radius of the smaller semicircle. This gives the smaller semicircle an area of $\frac{\pi\cdot(8\sqrt{3})^2}{2} = 96\pi.$

The total shaded area is the difference of these semicircles, or $128\pi - 96\pi = \boxed{\text{(C) } 32\pi}.$

~chisps

Solution 5

Let the radius of the larger semicircle be $R$ and the smaller one $r.$ We are asked to compute $\frac{R^2-r^2}{2} \cdot \pi$. Since a diameter is always perpendicular and bisects a chord, by drawing a diameter and applying Power of a Point Theorem, this yields $(R+r)(R-r)=8^2$ $\implies$ $R^2-r^2=64$. Therefore, the answer is $\frac{64}{2} \pi = \boxed{\text{(C) } 32\pi}.$ ~hxve

Solution 6

Since the problem does not restrict where the chord is, we can simply let the chord be the base of the semicircle. Therefore, the area is simply $\frac{\pi\cdot8^2}{2} = 32\pi.$

~metrixgo

Solution 7

We can move this small semicircle to the middle of the big semicircle. Let $r$ be the radius of the small semicircle. By Pythagorean Theorem, the line draw from the midpoint of the diameter of the big semicircle to the big semicircle at the height of the small semicircle (see the diagram) has length $\sqrt{r^2 + 64}$. This is a radius of the big semicircle. So the shaded area would be \[\frac{\pi \cdot (\sqrt{r^2+64})^2 - \pi r^2}{2} = \frac{\pi(r^2 + 64 - r^2)}{2} = \frac{64\pi}{2} = \boxed{\text{(C) } 32\pi}.\] Diagram:

[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real R = 3; pair A = (-R,0), B = (R,0); fill(Arc((0,0),R,0,180)--cycle,palered); draw(Arc((0,0),R,0,180),black); real h = 0.75; pair C = (-sqrt(R^2 - h^2), h); pair D = ( sqrt(R^2 - h^2), h); draw(C--D,black); label("$C$",C,W); label("$D$",D,E); label("$16$", midpoint(C--D), N); real r = h; pair O = (0,0); pair Q = (0,r); fill(Arc(O,r,0,180)--cycle,white); draw(Arc(O,r,0,180),black); draw(A--B,black); draw(O--Q); draw(O--D); draw(Q--D, dashed); label("$A$",A,S); label("$B$",B,S); label("$r$", midpoint(O--Q), W); label("$\sqrt{r^{2}+64}$", midpoint(O--D), S); label("$8$", midpoint(Q--D), N); [/asy]

~JerryZYang

Chinese Video Solution

https://www.bilibili.com/video/BV1bj2uBxEkU/

~metrixgo

Video Solution (In 1 Min)

https://youtu.be/8VWMNAx55g0?si=IqLseEtKLl2joUNU ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution by Daily Dose of Math

https://youtu.be/gPh9w3X3QSw

~Thesmartgreekmathdude

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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