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Difference between revisions of "2009 AMC 10A Problems/Problem 6"
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Thus the answer is <math>\frac{1}{2}\Rightarrow \fbox{A}</math>
 
Thus the answer is <math>\frac{1}{2}\Rightarrow \fbox{A}</math>
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 +
==Solution==
 +
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
 +
 +
<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 21:41, 20 July 2018

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 radiuses. 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 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 $2$ 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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