The correct option is
C 271.95 sq cm Area of the shaded region = Area of semicircle APC - Area of segment AQC
Area of semicircle =
πr22=3.14×15×152=353.25sqcm Area of segment = Area of sector OAQC - Area of triangle OAC
Consider triangle OAC
Since all sides are equal to 30 cm in the triangle, it is an equilateral triangle.
Hence, each angle is equal to 60 degrees.
Hence
∠OAC=∠AOC=∠ACO=60∘.
Now, let the height be h. sin60∘=h30 √32=h30 h=15√3 Now, area of triangle OAC is 12×base×h =12×30×15√3=225√3sq.cm=389.7sq.cm Area of the sector OAQC =
60360×π×r2=16×3.14×30×30=471sq.cm. Hence, area of segment AQC = 471 - 389.7 = 81.3 sq cm
Hence, area of the shaded region = Area of semicircle - area of segment AQC
= 353.25 - 81.3 = 271.95 sq cm.