Question

Three congruent circles overlap in such a way that each circle intersects the centers of both of the other circles, as shown above. If the radius of each of the circles is 8, what is the area of the central section where all three circles overlap?

Answer 1

Heya Mate

(•) For figure, :p Refer your Pen and a Notebook ( lol )

Now, that you have it...

-> Hold any one centre, and find Area of the Sector enclosed within the bounds !

$sector = \frac{ \frac{\pi}{3} }{2\pi} (\pi {r}^{2} ) = \frac{\pi {r}^{2}}{6}$

✓✓ Careful there buddy ^^'' there's two or in fact, three centers there that gives you the same Area of Sector !

Now, =_= You just found ->

$(area \: of \: region) + 2(area \: of \: trapped \: \triangle)$

So, if you can kindly do some work and get that extras off of you ^^"

-> Area of additional triangle :

$area = \frac{ \sqrt{3} }{4} {r}^{2}$
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Someone's hungry for the Answer ^^"

$(area \: of \: region) = 3( \frac{ \pi{r}^{2} }{6} ) - 2( \frac{ \sqrt{3} }{4} {r}^{2} )$

which is ->

$ar = \frac{ {8}^{2} }{2} (\pi - \sqrt{3} ) = 32(\pi - \sqrt{3} )$
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