Q) In the given figure, chord AB subtends an angle of 120 ⁰ at the centre of the circle with radius 7 cm.
Find: (i) Perimeter of major sector OACB
(ii) Area of the shaded segment, if area of A OAB = 21.2 cm ²

(Q 30 – 30/3/3 – CBSE 2026 Question Paper)

Ans:

(i) Perimeter of major sector OACB

Method 1:

We are given that ∠ OAB = 120 ⁰

∴ external ∠ OAB = 360 ⁰ – 120 ⁰ = 240 ⁰

Now, Arc length of major sector OACB = 2 π r = 2 π r

Given that r = 7 cm

∴ Arc length of major sector OACB = x 2 x () (7)

= x 2 x 22

= cm = 29.33 cm

∴ Perimeter of Major Sector OACB

= Arc length of major sector OACB + 2 x Radius length

= 29.33 + 2 x 7 = 43.33 cm

Therefore, Perimeter of Major Sector OACB is 43.33 cm.

Method 2:

We are given that ∠ OAB = 120 ⁰

∴ Arc length of minor sector OAB = 2 π r

= 2 π r =

Now, Arc length of major sector OACB

= Perimeter of circle – Arc length of minor sector OAB

= 2 π r – =

Given that r = 7 cm

∴ Arc length of major sector OACB

= =  

=  = cm = 29.33 cm

∴ Perimeter of Major Sector OACB

= Arc length of major sector OACB + 2 x Radius length

= 29.33 + 2 x 7 = 43.33 cm

Therefore, Perimeter of Major Sector OACB is 43.33 cm.

(ii) Area of the shaded segment:

Area of minor sector OAB = π r ²  =

= = = 51.33 cm ²

Given, Area of Δ OAB = 21.2 cm ²

Now Area of the shaded segment = Area of sector OAB – Area of Δ OAB

= 51.33 – 21.2 = 30.13 cm ²

Therefore, the area of the shaded segment is 30.13 cm ².

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