Difference between revisions of "2007 AMC 10A Problems/Problem 24"
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Revision as of 20:25, 1 January 2025
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 . 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.