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 16:31, 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