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
OR
has the same height as which is
We already know that
Therefore the area of is
Since
Therefore the sum of the areas is
Then the area of the shaded area becomes
~mathboy282
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 triangle and triangle combined. Below is an image to help.
Using the Base Altitude formula, where and are the bases and and are the altitudes, respectively, . The area of sector is of circle . The area of circle is , and therefore we have the area of sector to be .
Solution 3 (Using Answer Choices)
Like the first solutions, you find that the area of sector is . We also know that the triangles will not be in terms of . Looking at the answers, choices and both contain . However, based on the diagram, we observe that the answer must be less than . Only consists of a value less than .
Solution 4
We calculate the shaded area by subtracting the unshaded area in the quarter circle from a quarter circle, which consists of two sectors and two 30-60-90 triangles minus a square. The area of the shaded region is
~thatmathsguy
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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