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

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[insert problem here]
A semicircle has diameter <imath>\overline{AB}</imath> and chord <imath>\overline{CD}</imath> of length <imath>16</imath> parallel to <imath>\overline{AB}</imath>. A smaller semicircle
with diameter on <imath>\overline{AB}</imath> and tangent to <imath>\overline{CD}</imath> 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?
 
<imath>\textbf{(A) } 16\pi \qquad\textbf{(B) } 24\pi \qquad\textbf{(C) } 32\pi \qquad\textbf{(D) } 48\pi \qquad\textbf{(E) } 64\pi</imath>
 


==Solution 1 (Somewhat Cheese)==
==Solution 1 (Somewhat Cheese)==

Revision as of 14:39, 6 November 2025

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

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