In the given figure, ABMC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Given: ABMC is a quadrant of radius 14 cm.

In ∆ABC,

∠A = 90°, AB = AC = 14 cm

By Pythagoras theorem,

BC² = AB² + AC²

⇒ BC² = 2× 14×14

⇒ BC = 14√2 cm

Now,
Area of ∆ABC

⇒ Area of ∆ABC

⇒ Area of ∆ABC = 98cm²

We know that the area of the minor sector

⇒ Area of ABMC

⇒ Area of ABMC = 154cm²

So, area of ABMC = area of ∆ABC + area of BCM

⇒ Area of BCM = area of ABMC - area of ∆ABC

⇒ Area of BCM = 154 – 98 = 56cm²

Also, we have a semicircle made at BC in the figure.

Diameter of the semicircle = BC = 14√2 cm

⇒ Radius of the semicircle = 7√2 cm

Area of semicircle

⇒ Area of semicircular plot

⇒ Area of semicircular plot = 154 cm²

Area of the shaded region = Area of the semicircle - Area of BCM

⇒ Area of the shaded region = 154 – 56 = 98 cm²