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 segment of the circle.
Answers
.Area of sector ABCD =πr^2¢/360°
=)22/7×15×15×60°/360°
=)117.85.
Area of ∆ABC.
=1/2r^2sin¢
=)1/2×15×15×√3/2
=)225√3/4
=)97.3125.
minor segment BDC=Area of sector ABDC-Area of triangle ABC.
=117.85-97.3125
=)20.53 ___1)
and again ,Area of circle =πr^2¢
=)22/7×15×15
=)707.14.cm^2.
.
so, major segment BDC
=)Area of circle - minor segment.
=)707.14-20.5
=)686.642 Ans
hope it help you.
@rajukumar1☺
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)2 = √3/4 × 152
= (225√3)/4 cm2 = 97.3 cm2
Angle subtend at the centre by minor segment = 60°
Area of Minor sector making angle 60° = (60°/360°) × π r2 cm2
= (1/6) × 152 π cm2 = 225/6 π cm2
= (225/6) × 3.14 cm2 = 117.75 cm2
Area of the minor segment = Area of Minor sector - Area of equilateral ΔAOB
= 117.75 cm2 - 97.3 cm2 = 20.4 cm2
Angle made by Major sector = 360° - 60° = 300°
Area of the sector making angle 300° = (300°/360°) × π r2 cm2
= (5/6) × 152 π cm2 = 1125/6 π cm2
= (1125/6) × 3.14 cm2 = 588.75 cm2
Area of major segment = Area of Minor sector + Area of equilateral ΔAOB
= 588.75 cm2 + 97.3 cm2 = 686.05 cm2
hope its help you
thanx and be brainly...........................