A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find the areas of both the segments. [Take π = 3.14.]

Consider the figure shown above.

In this, the triangle AOB is an isosceles triangle. So here we will construct a perpendicular bisector from O on AB and as this triangle is isosceles therefore this perpendicular will also act as median and angle bisector.

Therefore,

Draw a perpendicular bisector from O which meets AB at D and bisects AB, as ABO is an isosceles triangle therefore OD acts as a median.

So, consider right angle triangle AOD right angled at D

Let ∠AOD = θ ⇒ Perpendicular = AD and Hypotenuse = AO = R

Given Radius of circle = R = 5√2 cm

Length of chord AB = 10 cm, AD = 5 cm

⇒ θ = 45°

∴ ∠AOD = 45°

Area of minor segment = Area of sector – Area of right angle triangle

→ eqn1

Where R = radius of the circle and θ = central angle of the sector

∴ Area of sector = 39.25 cm^{2}

Area of right angle triangle = 1/2 × base × height

As this is isosceles right-angle triangle

∴ height = base = 5√2 cm

Area of right angle triangle = 1/2 ×5√2×5√2 = = 25 cm^{2}

Put the value of area of sector and area of right angle triangle in equation 1,

⇒ Area of minor segment = 39.25 -25

= 14.25 cm^{2}

Area of major segment = πR^{2} – area of minor segment

⇒ Area of major segment = 157 – 14.25 = 142.75 cm^{2}

Area of major segment is 142.75 cm^{2} and of minor segment is 14.25 cm ^{2}.

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