Difference between revisions of "2009 AMC 10A Problems/Problem 6"
VelaDabant (talk | contribs) (New page: == Problem == A circle of radius <math>2</math> is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's ...) |
(Edited to show how solution was dervied.) |
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==Solution== | ==Solution== | ||
− | <math>\ | + | Area of the circle inscribed inside the semicircle <math>= \pi r^2 \Rightarrow \pi(2^2) = 4 \pi .</math> |
+ | Area of the larger circle (semicircle's area x 2)<math>= \pi r^2 \Rightarrow \pi(4^2)= 16 \pi</math> (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 <math>\frac{1}{2}(16 \pi) \Rightarrow 8 \pi .</math> | ||
+ | Part of the semicircle that is unshaded is <math>\frac{4 \pi}{8 \pi} = \frac{1}{2}</math> Therefore, the shaded part is <math>1 - \frac{1}{2} = \frac{1}{2}</math> | ||
+ | |||
+ | Thus the answer is <math>\frac{1}{2}\Rightarrow \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}} |
Revision as of 09:20, 21 March 2009
Problem
A circle of radius 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?
Solution
Area of the circle inscribed inside the semicircle Area of the larger circle (semicircle's area x 2) (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 Part of the semicircle that is unshaded is Therefore, the shaded part is
Thus the answer is
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 |