2003 AMC 10B Problems/Problem 19
- The following problem is from both the 2003 AMC 12B #16 and 2003 AMC 10B #19, so both problems redirect to this page.
Contents
Problem
Three semicircles of radius are constructed on diameter of a semicircle of radius . The centers of the small semicircles divide into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?
Solution 1
By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into of a circle with radius and two equilateral triangles with side length . This gives the area of the white region as . The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to .
Thus the answer is .
Note
The reason why it is of a circle and why the triangles are equilateral are because, first, the radii are the same and they make up the equilateral triangles.
Secondly, the reason it is of a circle is because the middle sector has a degree of and thus of a circle.
The other two have areas of of a triangle each.
Therefore, the total fraction of the circle(since they have the same radii) is
~mathboy282
Solution 2(Answer Choices)
Answer choices A and B are impossible since the area is obviously not a circle of radius 1 minus a triangle.
Answer choices C and D are also impossible since they are adding but we are subtracting.
That leaves us with only answer choice
~mathboy282
Video Solution by WhyMath
~savannahsolver
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 12 Problems and Solutions |
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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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