Mathematics

In the adjoining figure, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and of the major segment. Take π = 3.14.

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Given,

Radius of the circle = 10 cm

Angle at the centre subtended by a chord AB = 90°.

We know that,

Area of sector OACB = $πr^2 \times \dfrac{90}{360}$

= 3.14 × 10 × 10 × $\dfrac{1}{4}$

= $\dfrac{314}{4}$

= 78.5 cm².

In right angle △OAB,

Area of △OAB = $\dfrac{1}{2}$ × OA × OB

= $\dfrac{1}{2}$ × 10 × 10

= 50 cm².

Area of minor segment = Area of sector OACB – Area of △OAB

= 78.5 - 50

= 28.5 cm².

Area of circle = πr²

= 3.14 × 10 × 10

= 314 cm².

Area of major segment = Area of circle – Area of minor segment

= 314 - 28.5

= 285.5 cm².

Hence, area of sector OACB = 78.5 cm² and area of major segment = 285.5 cm².

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