In the figure (ii) given below, O is the centre of a circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region. (Use π = 3.14)

We know that,

Angle in semi-circle = 90°.

∴ ∠A = 90°.

In right angle △ABC

Using Pythagoras theorem,

⇒ BC² = AC² + AB²

⇒ BC² = 24² + 7²

⇒ BC² = (576 + 49) = 625

⇒ BC = $\sqrt{625}$ = 25 cm.

From figure,

Radius of circle (OB) = $\dfrac{BC}{2} = \dfrac{25}{2}$ = 12.5 cm.

By formula,

Area of △ABC = $\dfrac{1}{2}$ × AB × AC

= $\dfrac{1}{2}$ × 7 × 24

= 84 cm².

Area of circle = πr²

= 3.14 × 12.5 × 12.5

= 490.63 cm².

Area of quadrant COD = $\dfrac{1}{4}πr^2 = \dfrac{1}{4} \times 490.63$ = 122.66 cm².

Area of shaded portion = Area of circle – (Area of △ABC + Area of quadrant COD)

= 490.63 – (84 + 122.66)

= 490.63 – 206.66

= 283.97 cm².

Hence, area of shaded portion = 283.97 cm².