A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)

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asked Mar 8, 2016 in Class X Maths by bhai Basic (90 points)

1 Answer

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answered Mar 9, 2016 by Pranav lunawat Mature (287 points)
 
Best answer

Radius of the circle = 15 cm

ΔAOB is isosceles as two sides are equal.

∴ ∠A = ∠B

Sum of all angles of triangle = 180°

∠A + ∠B + ∠C = 180°

⇒ 2 ∠A = 180° - 60°

⇒ ∠A = 120°/2

⇒ ∠A = 60°

Triangle is equilateral as ∠A = ∠B = ∠C = 60°

∴ OA = OB = AB = 15 cm

Area of equilateral ΔAOB = √3/4 × (OA)= √3/4 × 15

                                          = (225√3)/4 cm= 97.3 cm2


Angle subtend at the centre by minor segment = 60°

Area of Minor sector making angle 60° = (60°/360°) × π rcm2

                                                                                     = (1/6) × 15π  cm2 =  225/6 π  cm2

                                                  =  (225/6) × 3.14 cm= 117.75  cm2

Area of the minor segment = Area of Minor sector - Area of equilateral ΔAOB

                                            = 117.75  cm2 - 97.3 cm= 20.4 cm2


Angle made by Major sector = 360° - 60° = 300°

Area of the sector making angle 300° = (300°/360°) × π rcm2

                                                   = (5/6) × 15π  cm2 =  1125/6 π  cm2

                                                  =  (1125/6) × 3.14 cm= 588.75  cm2

Area of major segment = Area of Minor sector + Area of equilateral ΔAOB

                                            = 588.75  cm2 + 97.3 cm= 686.05 cm2

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