emmak wrote:

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

A. \(16\sqrt{3}\)

B. \(32(\pi-\sqrt{3})\)

C. \(16(\pi+\sqrt{3})\)

D. \(32\pi\)

E. \(32(\pi+\sqrt{3})\)

Responding to a pm:

**Quote:**

A query regarding this question:

Can you explain the region of overlap again?

I thought it was 3 * each small sectors formed by 60deg central angle + area of eq. triangle

Yes, you are right. You mean that the area is the area of triangle + 3* area of the small segment

Here is the problem with that approach - we don't know the area of the small segment. So it involves an additional step of first finding the area of the sector and then subtracting the area of the triangle out of it. Let me explain.

If we consider the area of the sector and try to find the answer in those terms, it will be easier. The sector is made by the sides AB, AC and the arc BC. We know that if we know the central angle, finding the area of such a sector is easy provided we know the radius of the circle.

Area of the sector = (Q/360) * pi*r^2

Of course, if we know the area of the sector, we can subtract the area of the triangle from it and that will give us the area of the segment (between arc BC and side BC). Thereafter, you can use your own approach.

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Karishma

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