Question

$\huge\red{h}\pink{e}\green{l}blue{l} o$ ...

A CHORD OF A CIRCLE OF RADIUS IS 15CM SUBTENDS AN ANGLE OF 60° AT CENTRE . FIND THE AREAS OF THE CORRESPONDING MINOR AND MAJOR SEGMENTS OF THE CIRCLE .
( USE π = 3.14 AND √3 = 1.73 )

(DRAW THE PIC OF THE QUESTION)
....

@ MAYA KASHYAP ✌️✌️✌️

Answer 1

Here is your solution

Given:-

Radius of the circle = 15 cm

in Δ AOB 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 cm^2 

= 97.3 cm^2

Angle subtend at the centre by minor segment = 60°

Area of Minor sector making angle 60° = (60°/360°) × πr^2 cm^2

=> (1/6) × 152 π  cm^2

=>  225/6 π  cm^2

=> (225/6) × 3.14 cm^2 

=> 117.75  cm^2

Area of the minor segment = Area of Minor sector - Area of equilateral Δ AOB

Area of minor segment =Area of Minor sector-Area of equilateral Δ AOB

=> 117.75cm^2-97.3 cm^2 

Area of minor segment = 20.4 cm^2

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

Area of the sector making angle 300° = (300°/360°) × π r^2 cm^2

=> (5/6) × 152 π  cm^2

=> 1125/6 π  cm^2

=> (1125/6) × 3.14 cm^2 

=> 588.75  cm^2

Area of major segment = Area of Minor sector + Area of equilateral Δ AOB

=>588.75cm^2+97.3 cm^2 

Area of major segment=>686.05 cm^2

Hope it helps you