# 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== | ||

[[1986 AJHSME Problems]] | [[1986 AJHSME Problems]] |

## Revision as of 19:47, 25 January 2009

## 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 .