2007 AMC 10A Problems/Problem 24
Problem
Circles centered at and each have radius , as shown. Point is the midpoint of , and . Segments and are tangent to the circles centered at and , respectively, and is a common tangent. What is the area of the shaded region ?
Solution
The area we are trying to find is simply . Obviously, . Thus, is a rectangle, and so its area is .
Since is tangent to circle , is a right triangle. We know and , so is an isosceles right triangle, and has with length . The area of . By symmetry, , and so the area of is also .
(or , for that matter) is the area of its circle since is 45 degrees and forms a right triangle. Thus and both have an area of .
Plugging all of these areas back into the original equation yields .
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.