#### Question

In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.

#### Solution

BC is the diameter of the circle passing through the centre O.

Now,

∆ABC is a right angles triangle, right angled at A. (Angle subtended by the diameter on the circumference of the circle is 90^{°})

In right ∆ABC,

BC^{2} = AB^{2} + AC^{2}

= (7)^{2} + (24)^{2 }

= 49 + 576

= 625

∴ BC = 25 cm

Also,

∠COD + ∠BOD = 180^{°} (Linear pair angles)

⇒∠COD = 180^{°} − 90^{°} = 90^{°}

Now,

Area of the shaded region = Area of sector OCABDO − Area of ∆ABC

`= 270^@/360^@ xx pi((BC)/2)^2 - 1/2 xx AB xx AC` (∵ 360^{°} - 90^{°} = 270^{°} )

`= 3/4 xx 3.14 xx (25/2)^2 - 1/2 xx 7 xx 24`

= 367.97 - 84

= 283.97 cm^{2} (Approx)

Hence the area of shaded region is approximately 283.97 cm^{2}