2IIM

Here is a question from mensuration

1. Two horses are tied to opposite vertices of a rectangular field of length and breadth 6 and 6√2 meters. If they are tied with rope of length 6 meters, what is the common area that both horses can graze?

a. 6 π - 9√3
b. 12 π - 18√3
c. π - 2√3
d. 16 π -8√3
e. Cannot be calculated

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The area that one horse can graze is (90/360)*π*6*6. i.e. 9π.
Therefore the both horses can together graze 18π sq.units.
The common area of grazing will be formed by three sides each of which will be equal to 6. Therefore the area of this portion should be (1/2)*6*(√3/2)*6 = 9√3.

Angle wise the area of the portion outside the common area of grazing should be (30/360)*π*6*6. i.e. 3π. (Since the angle will be 30deg)

Therefore the common of area of grazing = 18π -(2*3π)-(2*9√3) =12 π - 18√3.
The answer i have given is based on few assumptions.
Please correct me if the answer is wrong. It will be better if someone comes up with a more clear & convincing solution. Thank You

please explain it again i didnt get it