Question

please answer class 10th ch 12 exercise 12.2 question no 6 of ncert of maths cbse
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Answer 1

Answer:

6. 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)

Solution:

Ncert solution class 10 chapter 12-4

Given,

Radius = 15 cm

θ = 60°

So,

Area of sector OAPB = (60°/360°)×πr2 cm2

= 225/6 πcm2

Now, ΔAOB is equilateral as two sides are the radii of the circle and hence equal and one angle is 60°

So, Area of ΔAOB = (√3/4) ×a2

Or, (√3/4) ×152

∴ Area of ΔAOB = 97.31 cm2

Now, area of minor segment APB = Area of OAPB – Area of ΔAOB

Or, area of minor segment APB = ((225/6)π – 97.31) cm2 = 20.43 cm2

And,

Area of major segment = Area of circle – Area of segment APB

Or, area of major segment = (π×152) – 20.4 = 686.06 cm2