Three congruent circles overlap in such a way that each circ

The area of the overlapping section should be little less than the half of the area of the circle , and only B fits

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.

The only answer closer to 16π is 32(π−3‾√) .

So the answer is B.

Hi VeritasPrepKarishma

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?

Thanks in advance.

Responding to a pm:


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.

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

Hope it helps

Hi Bunuel,

I was wondering could you please explain how you deduced the middle part to be "1/6th of the area of the circle"?

Thank You!

My solution (presented below) follows Bunuel´s idea and one of VeritasKarishma´s comments.

I have decided to post it so that readers may study it explicitly:





\(? = {S_{\Delta {\rm{equil}}}} + \,\,3 \cdot \,{S_{{\rm{blue}}}}\)

\({S_{\Delta {\rm{equil}}}} = {{{r^{\,2}}\sqrt 3 } \over 4}\,\,\,\mathop = \limits^{r\, = \,8} \,\,16\sqrt 3\)

\({S_{{\rm{blue}}}} = {{60} \over {360}}\left( {\pi \cdot {8^2}} \right) - {S_{\Delta {\rm{equil}}}} = {{\pi \cdot {8^2}} \over 6} - \,16\sqrt 3 = 8\left( {{{4\pi } \over 3} - 2\sqrt 3 } \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,3 \cdot \,{S_{{\rm{blue}}}} = 8\left( {4\pi - 6\sqrt 3 } \right)\)


\(? = \,\,16\sqrt 3 + 8\left( {4\pi - 6\sqrt 3 } \right) = 32\left( {\pi - \sqrt 3 } \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{B}} \right)\)


Regards,
Fabio.

OFFICIAL SOLUTION

Answer: (B) This is a great example of when filling in the info you know and adding to the figure given helps point you in the right direction for solving what may initially be a confusing, complex question. You’re told the radius of each circle is 8, and that each circle intersects the centers of both the other circles. Therefore, if you connected the centers, the length of each line segment would be the radius and would create an equilateral triangle with side length 8 within the central segment:



From here you can portion that central area into smaller segments: the equilateral triangle, and three congruent arced segments on each of the three sides (see below for illustration of one of the segments):


Therefore, the area of the central region will be equal to the area of the triangle plus three times the area of that small arced segment. First go ahead and calculate the area of the equilateral triangle using the formula for that area:
s^2root3/4 you know that the area of the triangle is 16root 3.

Next calculate the area of one of the small segments. For the segments, you need to calculate the area of a sector - area of the equilateral triangle. Since each sector is 60° out of the full 360° in a triangle (since they form an equilateral triangle, which must have all angles measuring 60°), the area is 1/6. f the full area of the circle; the radius is 8, so the area of the circle is 64π, so the area of the sector is 32π/3. To find the segment you need to subtract the equilateral triangle from the sector which is: 32π/3−16root3. Since there are three segments, the area of those will be multiplied by 3. Lastly add that to the equilateral triangle.Thus the answer is B

\(32(\pi - \sqrt{3})\)
\(96(\pi - \sqrt{3})\)

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