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AMC 10 Math Olympiad USA Math Olympiad

Area of Region in a Circle | AMC-10A, 2011 | Problem 18

Try this beautiful problem from Geometry: Area of Region in a Circle from AMC-10A, 2011, Problem -18. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based on Area of Region in a Circle.

Area of Region in a Circle – AMC-10A, 2011- Problem 18


Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$. What is the area inside Circle $C$ but outside circle $A$ and circle $B$ ?

  • \(\pi\)
  • \(\frac{3\pi}{2}\)
  • \(2\)
  • \(6\)
  • \(\frac{5\pi}{2}\)

Key Concepts


Geometry

Circle

Rectangle

Check the Answer


Answer: \(2\)

AMC-10A (2011) Problem 18

Pre College Mathematics

Try with Hints


We have to find out the area of the shaded region .Given that three circles with radius \(1\) and Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$.so if we draw a rectangle as shown in given below then we can find out the required region by the area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$

can you finish the problem……..

Now area of the rectangle is \(2\times 1=2\)

Area of the half circle with center (gray shaded region)=\(\frac{\pi (1)^2}{2}\)

The area of the two sectors created by $A$ and $B$(blue region)=\(\frac{2\pi(1)^2}{4}\)

can you finish the problem……..

Therefore, the required region (gray region)=area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$=\(\frac{\pi (1)^2}{2}\)+\(2\times 1=2\)-\(\frac{2\pi(1)^2}{4}\)=\(2\)

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