Here we will first find out the area of semicircle whose diameter is BC and then subtract the area of right angle triangle ABC from it and then we will subtract this result from the area of semicircles whose diameters are AB and AC.

Consider ∆ABC, ∠BAC = 90°

⇒ BC² = 4² + 3²

⇒ BC² = 16 + 9

⇒ BC² = 25

∴ BC = 5 cm

Area of semicircle whose diameter is AC

∴ Radius = 2 cm

∴ Area of semicircle = 2π cm² –eqn2

Area of semicircle whose diameter is AB

∴ Radius = 1.5 cm

∴ Area of semicircle = 1.125π cm²→ eqn3

Area of semicircle whose diameter is BC

∴ Radius = 2.5 cm

∴ Area of semicircle = 3.125π cm²→ eqn4

⇒ Area of triangle PQR = 3×2

∴ Area of triangle PQR = 6 cm²→ eqn5

Now subtract equation 5 from equation 4,

⇒ Area of semicircle excluding ∆ABC = eqn4 – eqn5

⇒ Area of semicircle excluding ∆ABC = 3.125π – 6

∴ Area of semicircle excluding ∆ABC = 3.8214 cm²→ eqn6

Area of shaded region = eqn3 + eqn2 – eqn6

⇒ Area of shaded region = 2π + 1.125π – 3.8214

⇒ Area of shaded region = 3.125π – 3.8214

⇒ Area of shaded region = 9.8214 – 3.8214

∴ Area of shaded region = 6 cm²

Area of shaded region 6 cm².