Question

A chord of a circle of radius 14 cm subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle. `[User pi22/7 and sqrt3=1.73]`

Answer 1

Let O be the centre of the circle and AB be the chord that subtends an angle of 120° at the centre. 

Here, OM ⊥ AB. 

Radius of the circle, r = 14 cm 

Area of the minor segment = Area of sector OAB − area of ΔAOB 

`text{Area of sector OAB}=O/ /360^@ xx pi r^2`

`=120^@/360^@xx22/7xx(14 cm)^2`

`=1/3xx22/7xx196 cm^2`

`=616/3 cm^2`

Now ,OM ⊥ AB.

`∴AM =MB =1/2AB` [Perpendicular from the centre to the bisects the chord]

`rArr AB =2AM `

⇒ AB = 2AM 

In ΔOAM and ΔOBM: 

OA = OB [Radii of the same circle] 

OM = OM [Common] 

∠OMA = ∠OMB [Each 90°] 

∴ ΔOAM ≅ ΔOBM [RHS congruence criterion] 

⇒ ∠AOM = ∠BOM [C.P.C.T] 

`∴∠AOM =∠BOM= 120^@/2=60^@`

In ΔAOD: 

`rArr sqrt3/2=(AM)/(OA), cos 60^@=(OM)/(OA)`

`rArr sqrt3/2=(AM)/(14 cm),1/2 = (OM)/(14cm)`

`∴ AB=2AM=14sqrt3cm`

`\text{Area} (Δ AOB)=1/2xxABxxOM=1/2xx14sqrt3xx7cm^2=49sqrt3cm^2`

`\text{Area of minor segment}=616/3 cm^2 -49 sqrt3 cm^2`

`=205.33cm^2-49xx1.73cm^2`

`=205.33cm^2-84.77 cm^2`

`=120.56 cm ^2`

Thus, the area of the minor segment is 120.56 cm².