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प्रश्न
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use π = 3.14.]
A chord of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding minor segment. [Use π = 3.14]
उत्तर १
Let AB be the chord of a circle subtending an angle of 90° at the centre O of the circle.
Area of sector = `theta/(360^circ) x pi^2`
`= (90^circ)/(360^circ) xx 314 /100 xx 10xx10` cm2
= `1/4 xx 314` cm2
`= 157/2` cm2
= 78.5 cm2
Corresponding minor segment = ΔAOB
= 78.5 cm2 − `[1/2 xx10 xx10]`
= 78.5 cm2 − 50 cm2
= 28.5 cm2
उत्तर २
Area of minor segment = Area of sector AOBC − Area of right triangle AOB
`=theta/360^circpi("OA")^2 - 1/2xx` OA× OB `
`= 90^circ/360^circxx3.14(10)^2 - 1/2xx10xx10`
= 78.5 − 50
= 28.5 cm2
Hence, the area of minor segment is 28.5 cm2
Notes
Students should refer to the answer according to their question and preferred marks.
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