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

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 $B$ ?

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

Geometry

Circle

Rectangle

But try the problem first...

Answer: \(2\)

Source

Suggested Reading

AMC-10A (2011) Problem 18

Pre College Mathematics

First hint

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........

Second Hint

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........

Final Step

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 $B$=\(\frac{\pi (1)^2}{2}\)+\(2\times 1=2\)-\(\frac{2\pi(1)^2}{4}\)=\(2\)

- https://www.cheenta.com/problem-on-equilateral-triangle-amc-10a-2010-problem-14/
- https://www.youtube.com/watch?v=ruRoSSe1U18

Contents

[hide]

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

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 $B$ ?

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

Geometry

Circle

Rectangle

But try the problem first...

Answer: \(2\)

Source

Suggested Reading

AMC-10A (2011) Problem 18

Pre College Mathematics

First hint

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........

Second Hint

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........

Final Step

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 $B$=\(\frac{\pi (1)^2}{2}\)+\(2\times 1=2\)-\(\frac{2\pi(1)^2}{4}\)=\(2\)

- https://www.cheenta.com/problem-on-equilateral-triangle-amc-10a-2010-problem-14/
- https://www.youtube.com/watch?v=ruRoSSe1U18

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