goyalsau wrote:

See the shaded part only.

ABC is an equilateral triangle. 1 cm each (Radius)

Thus each angle 60 deg.

360 deg of 1 circle = pie*1^2 = pie
60 deg --> pie/6

This is the area of say arc ABC.

Similarly, area of arc ACB is pie/6 and arc BAC is pie/6

Total = arc ABC + arc ACB + arc BAC - 2 * (area of equilateral triangle ABC)

The reason 2 * (area of equilateral triangle ABC) is subtracted as we are considering the area of equilateral triangle ABC each time we are considering the ARC. Actually this triangle should be consider once.

Thus, Total = arc ABC + arc ACB + arc BAC - 2 * (area of equilateral triangle ABC)

= 3*(pie/6) - 2 (sqrt(3)/4)

= pie/2 - sqrt(3)/2

IMO
A

The centers of three circles form a equilateral triangle. Area of shaded region will be area of equilateral triangle + 3 times area of curved surface (Area of sector - Area of Triangle formed by)
Area of sector = 1/6(pie)
Area of equilateral triangle = sqrt(3)/4 (side)^2 = sqrt(3)/4
Thus area of 1 curved surface (this is the small portion) = Area of sector - Area of equilateral triangle
= (pie)/6 - sqrt(3)/4 ................ (1)

Area of shaded region = Area of triangle + 3 (1)
= sqrt(3)/4 + 3[(pie)/6 - sqrt(3)/4]
= [(pie) - sqrt(3)]/2
Option A.

Hi

This question took me some time to solve.The answer according to me is A.

Here is why.

Join AB,AC, and BC. Now all of them are equal since all the circle have the same radius=1. This means this is an equilateral triangle.

Hence angle ABC=ACB=BAC=60.

Now the measure of any sector in a circle is given by the formula : x/360 * area of sector. Here each sector has an equal measure of 60 and each circle has area of pi*1=pi.

Hence area of each sector= 1/6 * pi=pi/6.

Now remember triangle ABC is an equilateral triangle. To measure the area of an equilateral triangle we need height.Since an perpendicular dropped from a vertex say A in an equilateral triangle splits the base into equal halves and creates two right triangles with angle measure 30-60-90( the ratio of sides is 1:2:root3). We can infer that height is 1/2 root 3 the base is 1.Hence the area of equilateral triangle is 1/2 *b*h=1/2*1*(root 3/2) = root 3/4.

Consider this question a set theory question.The area of shaded portion is equal to sum of each sector 3 times - sum of equilateral triangle 2 times ( we minus 2 times to take into account the triple counting of this area in adding each sector 3 times).

Remember we need to add each sector 3 times to take into account the area not include by the equilateral triangle.

This gives 3 (pi/6) - 2 (root 3/4)= pi/2- root 3/2 = A

What is the OA?

Btw what is the source of this question.It was definitely not an easy one.

Thanks Guys, Thanks a lot........

You all are right, The solution that is given in the book is This, ( i am not able to understand Why use Sin theta )

Please Guys It will be real Help IF you can explain from there this sin theta comes

I found a formula on segment of a circle ,

You can plug the value of r in it and multiply it by 3 because we have 3 segments from 3 circles add the area of equilateral triangle you will directly get the answer,

r^2 { ( pie * theta ) / 360 - ( sin theta )/2 }

r = 1 , theta = 60 degree ,

pie / 6 - sqrt ( 3 ) / 4 { Area of one shaded region - equilateral triangle )

multiply by 3

Pie / 2 - 3 * Sqrt ( 3 ) / 4 + sqrt ( 3 ) / 4

Pie / 2 - 2 Sqrt ( 3 ) / 4

Pie / 2 - Sqrt ( 3 ) / 2


I Got these questions from one of my friend

Gmat doesn't test trigonometry.Don't worry about sines and cosines...

beat_gmat_09 wrote:The centers of three circles form a equilateral triangle. Area of shaded region will be area of equilateral triangle + 3 times area of curved surface (Area of sector - Area of Triangle formed by)
Area of sector = 1/6(pie)
Area of equilateral triangle = sqrt(3)/4 (side)^2 = sqrt(3)/4
Thus area of 1 curved surface (this is the small portion) = Area of sector - Area of equilateral triangle
= (pie)/6 - sqrt(3)/4 ................ (1)

Area of shaded region = Area of triangle + 3 (1)
= sqrt(3)/4 + 3[(pie)/6 - sqrt(3)/4]
= [(pie) - sqrt(3)]/2
Option A.

I makes it really easy area of the euqai triangle + area of the arch (whihc is area of the secotor - arear of the equi triangle)