A chord of a circle subtends an angle 60 deg at the centre. If the length of the chord is 100 cm, find the area of major segment

Dear Student,

Please find below the solution to the asked query :

Consider a circle with center O and a chord AB whch substends an angle of 60⁰ at the center.

To find : Area of major segment

Construction : from O draw OM perpendicular to AB.

Area of major segment = area of major arc + area if triangle ABO

Area of major arc = theta / 360⁰ * pi * r²

= 300⁰ / 360^{0 *} 3.14 * 100 * 100

= 5 / 6 * 314 * 100 

= 157000 / 6

For area of triangle we know that a perpendicular drawn from a center to the chord bisects the chord and also bisects the angle subtended by the chord at the center.

Therefore  AM = BM and angle AOM = 30⁰

In triangle AMO

Sin30⁰ = AM / AO

AM = 1 / 2 * 100

AM = 50cm

Then AB = 2AM = 100cm

cos30⁰ = OM / AM

OM = root3 * 50

OM = 50root3

Area of triangle ABO = 1 / 2 * b * h

= 1 / 2 * 100 * 50root3

= 2500root3

Area of major segment = area of major arc + area if triangle ABO

 = 157000 / 6 + 2500root3

 = 157000 + 15000root3 / 6

= 157000 + 15000(1.732) / 6

= 157000 + 25980 / 6

= 30496.6 cm²

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