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 $$\sqrt{3} = 1.73$$)

Asked by Abhisek | 1 year ago |  88

##### Solution :-

Given,

θ = 60°

So,

Area of sector OAPB = $$\dfrac{60°}{360°}×πr^2 \;cm^2$$

$$\dfrac{225}{6}$$ 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 = $$\dfrac{\sqrt{3}}{4}×a^2$$

Or, $$\dfrac{\sqrt{3}}{4}×15^2$$

Area of ΔAOB = 97.31 cm2

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

Or, area of minor segment APB = (($$\dfrac{225}{6}$$)π – 97.31) cm= 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

Answered by Pragya Singh | 1 year ago

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