A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and $\sqrt{3}$ = 1.73)

We know that,  formula for the area of the sector of a circle

Area of the sector = $\frac{\theta }{{360}^0}$ $\times$ ${\mathrm{\pi r}}^2$

 Area of the segment  = Area of the sector - Area of the corresponding  triangle

Here, r = 12 cm, θ = 120⁰

Area of the sector OAYB = $\frac{\theta }{{360}^0}$ $\times$ ${\mathrm{\pi r}}^2$

Area of the triangle AOB = $\frac{1}{2}$ $\times$ Base $\times$Height

So, For finding the area of ΔAOB, draw OM ⊥ AB then find base AB and height OM using the figure as shown above.

Area of sector OAYB =120⁰/360⁰ $\times$ ${\mathrm{\pi r}}^2$

= 1/3 $\times$ 3.14 $\times$ (12)²

= 150.72 cm²

Draw a perpendicular OM from O to chord AB

In ΔAOM and ΔBOM

AO = BO = r (radius of circle)

OM = OM (common side)

∠OMA = ∠OMB = 90⁰ (perpendicular OM drawn)

Hence,  ΔAOM ≅ ΔBOM 

∠AOM = ∠BOM (By CPCT)

Therefore, ∠AOM = ∠BOM = $\frac{1}{2}$ ∠AOB = 60⁰

In ΔAOM,

AM/OA = sin 60⁰          and OM/OA = cos 60⁰

AM/12 cm = $\sqrt{3}$/2         and OM/12 cm = $\frac{1}{2}$

AM = $\sqrt{3}$/2 $\times$ 12 cm     and OM = $\frac{1}{2}$ $\times$ 12 cm

AM = 6$\sqrt{3}$ cm and OM = 6 cm

Thus AB = 2 AM

= 2 $\times$ 6$\sqrt{3}$ 

= 12 $\sqrt{3}$ cm

Area of ΔAOB = $\frac{1}{2}$$\times$ AB $\times$ OM

= $\frac{1}{2}$ $\times$ 12$\sqrt{3}$  $\times$ 6 

= 36 $\times$ 1.73 

= 62.28 cm²

Area of segment AYB = Area of sector OAYB - Area of ΔAOB

= 150.72  - 62.28 

= 88.44 cm²