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Difference between revisions of "2009 AMC 10A Problems/Problem 6"
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==Solution==
 
==Solution==
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The solution above is the standard method of solving it, but we can also realize something about their radiuses. If we see, we know that the smaller circle has a radius of <math>2</math> while the larger semicircle has a radius of <math>4</math>. We know for a fact that if we have <math>2</math> normal full circles and one has a radius that is double the other, the smaller circle will always be <math>4</math> times smaller than the larger one (if you didn't know, now you know). Now, we have a semi-circle, the same thing applies but the larger circle is in half, so, therefore, the semi-circle (which is bigger than the smaller full circle), will be double the size of the small circle that is inscribed in it. Double means that <math>2</math> of the small full circles will be able to fit the larger semi-circle. So, therefore, the shaded parts and the small full circle are the same, making <math>2</math> parts, so the shaded parts are <math>\frac{1}{2}</math> of the whole semi-circle
+
The solution above is the standard method of solving it, but we can also realize something about their radii. If we see, we know that the smaller circle has a radius of <math>2</math> while the larger semicircle has a radius of <math>4</math>. We know for a fact that if we have <math>2</math> normal full circles and one has a radius that is double the other, the smaller circle will always be <math>4</math> times smaller than the larger one (if you didn't know, now you know). Now, we have a semicircle, so the same thing applies but the larger circle is cut in half, so therefore, the semicircle (which has the bigger radius), will be double the size of the small circle that is inscribed in it. Double means that <math>2</math> of the small full circles will be able to fit the larger semi-circle. So, therefore, the shaded parts and the small full circle are the same, making <math>2</math> parts, so the shaded parts are <math>\frac{1}{2}</math> of the whole semi-circle
  
−
<math>\frac{1}{2}\Longrightarrow \fbox{A}</math>  
+
<math>\frac{1}{2}\Longrightarrow \fbox{A}</math>
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2009|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2009|ab=A|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:18, 29 November 2020

Problem

A circle of radius $2$ is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded?

[asy] unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; filldraw(Arc((0,0),4,0,180)--cycle,gray,black); filldraw(Circle((0,2),2),white,black); dot((0,2)); draw((0,2)--((0,2)+2*dir(60))); label("$2$",midpoint((0,2)--((0,2)+2*dir(60))),SE); [/asy]

$\mathrm{(A)}\ \frac{1}{2} \qquad \mathrm{(B)}\ \frac{\pi}{6} \qquad \mathrm{(C)}\ \frac{2}{\pi} \qquad \mathrm{(D)}\ \frac{2}{3} \qquad \mathrm{(E)}\ \frac{3}{\pi}$

Solution

Area of the circle inscribed inside the semicircle $= \pi r^2 \Rightarrow \pi(2^2) = 4 \pi .$ Area of the larger circle (semicircle's area x 2)$= \pi r^2 \Rightarrow \pi(4^2)= 16 \pi$ (4, or the diameter of the inscribed circle is the same thing as the radius of the semicircle). Thus, the area of the semicircle is $\frac{1}{2}(16 \pi) \Rightarrow 8 \pi .$ Part of the semicircle that is unshaded is $\frac{4 \pi}{8 \pi} = \frac{1}{2}$ Therefore, the shaded part is $1 - \frac{1}{2} = \frac{1}{2}$

Thus the answer is $\frac{1}{2}\Rightarrow \fbox{A}$

Solution

The solution above is the standard method of solving it, but we can also realize something about their radii. If we see, we know that the smaller circle has a radius of $2$ while the larger semicircle has a radius of $4$. We know for a fact that if we have $2$ normal full circles and one has a radius that is double the other, the smaller circle will always be $4$ times smaller than the larger one (if you didn't know, now you know). Now, we have a semicircle, so the same thing applies but the larger circle is cut in half, so therefore, the semicircle (which has the bigger radius), will be double the size of the small circle that is inscribed in it. Double means that $2$ of the small full circles will be able to fit the larger semi-circle. So, therefore, the shaded parts and the small full circle are the same, making $2$ parts, so the shaded parts are $\frac{1}{2}$ of the whole semi-circle

$\frac{1}{2}\Longrightarrow \fbox{A}$

See also

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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