So, M is the midpoint of AB and angle AOM = angle BOM = 1 / 2 x 120^o = 60^o

Let, OM = x cm

So, from Triangle OMA ,OM / OA = cos60^o

or,               x / 21 = 1 / 2         (Therefore , cos60^o = 1 / 2)

or,              x = 21 / 2

So OM = 21 / 2 cm.

Also , AM / OA = sin60^o

AM / 21 = √3 / 2                (Therefore , sin60^o = √3 / 2)

So , AM = 21 √3 / 2 cm.

Therefore ; AB = 2AM = 2 x 21 √3 / 2 cm. = 21 √3 cm.

So , Area of Triangle OAB = 1 / 2 x AB x OM

= 1/2 x 21 √3 x 21 / 2 cm²

= 441 / 4 √3 cm².          .....(2)

Therefore, area of the segment AYB = (462 - 441 / 4 √3) cm²

[Therefore , from (1) , (2)]

= 21 / 4 (88 - √3) cm²

= 271.047 cm²