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Question

# Question 4 Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60∘. (use π=3.14)

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Solution

## Given that, radius of a circle (r) = 12 cm And central angle of sector OBCA (θ)=60∘ ∴ Area of sector OBCA =πr2360×θ [here, OBCA = sector and ABCA = segment] 314×12×12360∘×60∘ =314×2×12 =314×24=75.36cm2 Since, ΔOAB is an isosceles triangle. Let ∠OAB=∠OBA=θ1 and OA = OB = 12cm ∠AOB=θ=60∘ ∴∠OAB+∠OBA+∠AOB=180∘ [∵ Sum of all interior angles of a triangle is 180∘] ⇒θ1+θ1+60∘=180∘ ⇒2θ1=120∘ ⇒θ1=60∘ ∴θ1=θ=60∘ So, the required ΔAOB is an equilateral triangle. Now, area of ΔAOB=√34(side)2 [∵ area of an equilateral triangle=√34(side)2] =√34(12)2 =√34×12×12=36√3cm2 Now, area of the segment of a circle i.e ABCA = Area of sector OBCA - Area of ΔAOB =(75.36−36√3)cm2 Hence, the required of segment of a circle is: (75.36−36√3)cm2.

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