Difference between revisions of "1986 AJHSME Problems/Problem 23"
5849206328x (talk | contribs) (New page: ==Problem== The large circle has diameter <math>\text{AC}</math>. The two small circles have their centers on <math>\text{AC}</math> and just touch at <math>\text{O}</math>, the center o...) |
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==Solution== | ==Solution== | ||
− | {{ | + | The small circle has radius <math>1</math>, thus its area is <math>\pi</math>. |
+ | |||
+ | The large circle has radius <math>2</math>, thus its area is <math>4\pi</math>. | ||
+ | |||
+ | The area of the semicircle above <math>AC</math> is then <math>2\pi</math>. | ||
+ | |||
+ | The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is <math>\pi</math>. This means that the area of the shaded part is <math>2\pi-\pi=\pi</math>. | ||
+ | This is equal to the area of a small circle, hence the correct answer is <math>\boxed{\text{(B)}\ 1}</math>. | ||
==See Also== | ==See Also== | ||
− | [[ | + | {{AJHSME box|year=1986|num-b=22|num-a=24}} |
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:32, 3 July 2013
Problem
The large circle has diameter . The two small circles have their centers on and just touch at , the center of the large circle. If each small circle has radius , what is the value of the ratio of the area of the shaded region to the area of one of the small circles?
Solution
The small circle has radius , thus its area is .
The large circle has radius , thus its area is .
The area of the semicircle above is then .
The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is . This means that the area of the shaded part is . This is equal to the area of a small circle, hence the correct answer is .
See Also
1986 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.