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

1 Answer

Solution :-

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  pi = 3.14 and √(3) = 1.73 )

Given,

Radius = 15 cm

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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