The correct option is
B 240.76 cm2In the given figure,
Drop a perpendicular from O to side QR.
It is given that PR=28 cm
∴OR=PR2=282=14 cm
So, the radius of circle is 14 cm
Also,
PS=QR=14√3
Now, after construction, in△RTO
∠OTR=900,TR=14√32=7√3,OR=14 cm
Using pythagorus,
OT=√142−7√32=√196−147=7 cm
Area of triangle ROQ
=12×OT×QR=12×7×14√3 cm2
=49√3 cm2
Area enclosed by of segment QR and minor arc QR=
Area of minor sector ROQ− area of ΔROQ
=120360×π(14)2−49√3
=205.25−84.87=120.38 cm2
Area of shaded region =2×(Area of minor sector ROQ)
=2×120.38=240.76 cm2