Q8 of 36 Page 249

A chord of a circle, of length 10 cm, subtends a right angle at the centre. Find the areas of the minor segment and the major segment formed by the chord. (π = 3.14)

Length of the chord = 10 cm


Angle of measure = 90o



since triangle ABC is a right-angled triangle


AB, AC is same as the radius of the circle


BC2 = AB2 + AC2


102 = r2 + r2


100 = 2 r2


50 = r2


r = √50


Area of segment =


=


= 39.25 cm2


Area of the triangle = AB× AC


= × (√50) × (√50)


= 25 cm2


Area of minor segment = Area of segment - Area of the triangle


= 39.25 – 25


= 14.25 cm2


Area of major segment = Area of circle - Area of minor segment


= π r2 – 14.25


= 3.14 × (√50)2 – 14.25


= 142.82 cm2


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Figure 13.19


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Figure 13.20