Common area of circles

**geton** · October 31st 2010, 04:03 AM

Two circles have the same radius 1 unit. If each circle passes through the centre of the other circle, what is the area common to the two circles?

Area of two circles is 2pi. But how could I find the common area?

**Unknown008** · October 31st 2010, 04:12 AM

You'll need to make a sketch.

You'll see that the common area can be divided into 6 parts of which there are 2 equilateral triangles and 4 equal segments.

Can you try out to find the area now?

**geton** · October 31st 2010, 05:04 AM

Originally Posted by Unknown008

You'll need to make a sketch.

You'll see that the common area can be divided into 6 parts of which there are 2 equilateral triangles and 4 equal segments.

Can you try out to find the area now?

So area of two equilateral triangles is $\frac {\sqrt 3}{2}.$

Area of 4 segment = $4 \times \frac{1}{2} (\frac {\pi}{3} - \frac{\sqrt 3}{2}) = \frac {2}{3} \pi - \sqrt 3$

Therefore, enclosed area = $\frac {\sqrt 3}{2} + \frac {2}{3} \pi - \sqrt 3 = \frac {2}{3} \pi -\frac {\sqrt 3}{2}$

Is this right?

**earboth** · October 31st 2010, 06:07 AM

Originally Posted by geton

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So area of two equilateral triangles is $\frac {\sqrt 3}{2}.$

Area of 4 segment = $4 \times \frac{1}{2} (\frac {\pi}{3} - \frac{\sqrt 3}{2}) = \frac {2}{3} \pi - \sqrt 3$

Therefore, enclosed area = $\frac {\sqrt 3}{2} + \frac {2}{3} \pi - \sqrt 3 = \frac {2}{3} \pi -\frac {\sqrt 3}{2}$

Is this right? Yes

...

**Unknown008** · October 31st 2010, 06:22 AM

Yup, well done

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