2006 AMC 10B Problems/Problem 19
Contents
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 1
The shaded area is equivalent to the area of sector , minus the area of triangle plus the area of triangle .
Using the Pythagorean Theorem, so .
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
Solution 2
From the pythagorean theorem, we can see that is . Then, . The area of the shaded element is the area of sector minus the areas of triange and triange combined. Because (Using the Base Altitude formula, where and are the bases and and are the altitudes, respectively), we have the area of sector to be
See Also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AMC 10 Problems and Solutions |
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