**Q) AB is a chord of a circle centred at O such that ∠AOB=60˚. If OA = 14 cm then find the area of the minor segment. (take √3 =1.73)**

**Ans: **

Let the chord AB cut the circle in 2 parts. Sector APB is minor segment and AQB is major one.

Now in △AOB, radius OA=OB=14 cm

Therefore, ∠OAB = ∠OBA [identity: angles opposite to equal side]

Given that ∠AOB=60°

Therefore, ∠AOB + ∠OAB + ∠OBA = 180°

or 60° + 2 ∠OAB = 180°

or 2 ∠OAB = 120°

∠OAB = 60° and ∠OBA = 60°

Since All angles are 60°, therefore △OAB is an equilateral triangle

Now, area of minor segment APB = Area of sector OAPB – Area of △OAB

= π OA^{2 } OA^{2 }^{ }

= OA^{2 }

= ( 14 x 14) ^{ }

= 196 x (0.5238 – 0.4325)

= 196 x 0.0913 = **17.897 cm ^{2}**

**Hence, the area of minor segment is 17.897 cm ^{2 }**

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