ABCP is a quadrant of a circle of radius 14 cm. with AC as diameter, a semicircle is drawn. Find the area of the shaded portion

Explanation

In the right angled-triangle ABC
We have `AC^(2) = 14^(2) + 14^(2)`
`rArr AC^(2) = 14^(2) + 14^(2)`
`rArr AC = 14 sqrt2`
`rArr (AC)/(2) = 7 sqrt2`
Area of shaded region = Area of ABCQA - Area of quadrant ABCPA
and area of ABCQA = Area of `Delta ABC +` Area of semicircle ACQA
Area of quadrant ABCPA `= (1)/(2) xx (14)^(2) + (1)/(2) xx pi xx (7 sqrt2)^(2) = 98 + 154 = 252 cm^(2)`
`= (1)/(4) xx (22)/(7) xx (14)^(2) = 154 cm^(2)`
Area of shaded portion `= 252 - 154 = 98 cm^(2)`