Difference between revisions of "2006 AMC 10B Problems/Problem 19"
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== Solution == | == Solution == | ||
− | The shaded area is | + | The shaded area is equivalent to the area of sector <math>DOE</math>, minus the area of triangle <math>DOE</math> plus the area of triangle <math>DBE</math>. |
Using the Pythagorean Theorem: | Using the Pythagorean Theorem: |
Revision as of 17:31, 11 May 2008
Problem
A circle of radius is centered at
. Square
has side length
. Sides
and
are extended past
to meet the circle at
and
, respectively. What is the area of the shaded region in the figure, which is bounded by
,
, and the minor arc connecting
and
?
Solution
The shaded area is equivalent to the area of sector , minus the area of triangle
plus the area of triangle
.
Using the Pythagorean Theorem:
Clearly, and
are
triangles with
.
Since is a square,
.
can be found by doing some subtraction of angles.
So, the area of sector is
.
The area of triangle is
.
Since ,
.
So, the area of triangle is
.
Therefore, the shaded area is