Q) If a chord of a circle of radius 10 cm subtends an angle of 60 deg at the centre of the circle, find the area of the corresponding minor segment of the circle. (Use pi = 3.14 and sqrt(3) = 1.73 )

Ans: Let’s draw the diagram for better understanding:

Step 1: Here, we are given that θ  = 600, r = 10 cm

∵ ∠ AOB is 600

Therefore, sum of other two angles:

∠ OAB + ∠ OBA = 1800 – 600

= 1200

Now, OA and OB are radius of the icrcle, hence these are equal.

Therefore, angles opposite to equal sides will also be equal.

∴ ∠ OAB = ∠ OBA

∴ ∠ OAB = ∠ OBA = = 600

Therefore, Δ OAB is an equilateral triangle.

Step 2: Area of minor segment:

From the above diagram, Area of minor segment APB

= Area of sector AOBP – Area of triangle AOB

We know that Area of minor segment APB =

Here, we are given that θ  = 600, r = 10 cm

∴ Area of minor segment APB =

= (3.14) (100) = 52.33 cm2

Next, Area of equilateral Δ AOB =

Here, a = 10 cm, Therefore, area of Δ AOB =

= (100) = 43.25 cm2

Area of minor segment APB = Area of sector AOBP – Area of triangle AOB

= 52.33 – 43.25 = 9.08 cm2

Therefore, the area of minor segment is 9.08 cm2

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