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Three congruent circles overlap in such a way that each circ [#permalink]

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12 Mar 2013, 00:48

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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?

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

I'd go with approximation with this question. Look at the diagram below:

Notice that triangle ABC is equilateral (all sides are radii of the circles, for example AB and AC are radii of the lower circle and BC is radius of both upper circles). This implies that angle A is 60 degrees, thus the area of sector ABC (yellow region in the lower circle) is 1/6th of the area of the circle, so it's area is \(\frac{8^2\pi}{6}\approx{34}\).

Now, the area of the region we need to find must be more than this but not too much. Only B fits.

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

To get the exact answer, we can just follow Bunuel's solution a little further.

Area of overlap = Area of sector ABC + 2 (Area of sector ABC - Area of triangle ABC) (or you can also look at it as Area of overlap = 3*Area of sector ABC - 2*Area of triangle ABC)

Area of overlap \(= \frac{64\pi}{6} + 2(\frac{64\pi}{6} - \frac{\sqrt{3}*64}{4})\) Area of overlap = \(32(\pi - \sqrt{3})\)
_________________

Re: Three congruent circles overlap in such a way that each circ [#permalink]

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26 Sep 2016, 06:42

Hello Bunuel / Krishna, please critique my logic below:

Let us consider top left circle as A, top right as B, and below circle as C.

As the radius is 8, each circle's area would be 64π , Total area of all 3 circles combined: 64π * 3 = 192π.

As the overlap is between two radiuses of the circles, we can consider that there is 1/2 of overlap between circles A, B and 1/4 overlap between A and C, similar would be the case for all 3 circles. Therefore in total, for each circle, the overlap of the area is 3/4*(64π) = 16π.

So basically, each section above is 16π including the central area.

Re: Three congruent circles overlap in such a way that each circ [#permalink]

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07 Jul 2017, 07:49

VeritasPrepKarishma wrote:

emmak wrote:

Attachment:

2.jpg

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

To get the exact answer, we can just follow Bunuel's solution a little further.

Area of overlap = Area of sector ABC + 2 (Area of sector ABC - Area of triangle ABC) (or you can also look at it as Area of overlap = 3*Area of sector ABC - 2*Area of triangle ABC)

Area of overlap \(= \frac{64\pi}{6} + 2(\frac{64\pi}{6} - \frac{\sqrt{3}*64}{4})\) Area of overlap = \(32(\pi - \sqrt{3})\)

Could you please elaborate on the green highlight. I understand that the area of the over lap in 3 * sector area which is \(32\pi\), however, I was lost with 2* area of triangle. Could you please explain?

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.
_________________

Re: Three congruent circles overlap in such a way that each circ [#permalink]

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15 Sep 2017, 05:35

This is how I approached the problem :

Since radii of all the circles are equal, an equilateral triangle will be formed inside the common region. To find the area of common region, we can simply follow A(common region) = 3* Area of sector inside the common region - 3* Area of equilateral triangle = 3* (60/360) * pi * 8^2 - 3 * (sqrt3 * 8^2/4) Further simplifying, we get option B as answer