2007 AMC 10A Problems/Problem 24
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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 
. We know 
and 
, so is isosceles, a 
- right , and has 
with length . The area of 
. For obvious reasons, 
, and so the area of 
is also .
(or 
, for that matter) is 
the area of its circle. Thus and both have an area of 
.
Plugging all of these areas back into the original equation yields 
.
See also
| 2007 AMC 10A (Problems • 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 | ||
