Ok - Here is how I did this question. I have put the alphabets in the figure above:

So lets say o --> Centre of the circle

P--> Left hand side where two circle meets

A--> Left hand corner of the Triangle

M--> Where the bottom two circle meets

Q--> Right Hand Side where two circle meets

To answer this question one has to be look at the diagram closely and should be familiar with the area of sector formula, area of circle formula and area of triangle formula.

So if you see my naming conventions there are three sectors in the figure which are:

OPQ, APM and QBM.

From the figure its clear that Area of Shaded area = Area of triangle - Area of 3 sectors which are named above.

Now we have to figure out what kind of triangle is this.

So, OA=AP+OP.

and AP=OP {because the radius of circles are always equal). let say r be the radius. So AP=OP=r and OA = 2r

Same applies to other two sides of the Triangle. Therefore, we can conclude that the Triangle is an equivalent triangle and the sum of all the three angles is 180 degrees and each angle is of 60 degree in that case. Therefore angle O is of 60 degrees.

Area of sector = Pi.r^2(angle)/360 = pi (r)^2 * 60/360 {Substituuing the values of r and angle which are 2r and 60} ==> pi r^2 /6

Area of one sector => pi r^2 /6

Area of three sectors ==>pi r^2/2

Area of Equilateral triangle {Triangle is equilateral because every angle is 60 degrees and sides corresponding to equal angles are also equal)

Therefore area of equilateral triangle = Side^2 * Square root3/4

Area of shaded region = Side^2 * Square root3/4 minus pi r^2/2.

Plug in the all values of above to give you area 100\sqrt{3} - 50pi. when r=10 you will get this value

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Best Regards,

E.

MGMAT 1 --> 530

MGMAT 2--> 640

MGMAT 3 ---> 610