In Fig. 7, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and radius OP while arc PBQ is a semi-circle drawn on PQ ad diameter with centre M. If OP = PQ = 10 cm show that area of shaded region is `25(sqrt3-pi/6)cm^2`.
Given: OP = OQ = 10 cm
It is known that tangents drawn from an external point to a circle are equal in length.
OP = OQ = 10 cm
Therefore, ∆ABC is an equilateral triangle.
Area of part II = Area of the sector − Area of the equilateral triangle POQ
Area of the semicircle on diameter PQ = Area of part II + Area of part III
∴ Area of the shaded region (part III)