A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find area of both the segments.
Answers
Answer:
Step-by-step explanation:
Given :
Radius of circle,(OA,OB),r = 5√2 cm
Chord of a circle, AB = 10 cm
AL = AB/2
AL = 10/2
AL = 5 cm
Let ∠AOB = 2θ, Then, ∠AOL = ∠AOL = θ
In ∆OLA,
sin θ = Perpendicular /Hypotenuse = AL/OA
sin θ = 5/5√2 = 1/√2
sin θ = 1/√2
sin θ = sin 45°
[sin 45° = 1/√2]
θ = 45°
∠AOB = 2θ
∠AOB = 2 × 45° = 90°
∠AOB = 90°
Angle at the centre of a circle, θ = 90°
Area of the minor segment ,A = {πθ/360 - sin θ /2 cos θ/2 }r²
A = {90°π/360° - sin 90°/2 cos 90°/2 }× (5√2)²
A = {π/4 - sin 45°cos 45°} × 50
A = {π/4 - 1/√2 × 1/√2} × 50
A = {π/4 - 1/2 } × 50
A = {50π/4 - 50/2}
A = {(50 ×3.14)/4 - 25}
A = 39.25 - 25
A = 14.25 cm²
Area of the minor segment = 14.25 cm²
Area of circle = πr²
= 3.14 × (5√2)²
= 3.14 × 50
= 157.15
Area of circle = 157.15 cm²
Area of major segment = Area of circle - Area minor segment
Area of major segment = 157.15 - 14.25
Area of major segment = 142.75 cm²
Hence, the area of the minor segment is 14.25 cm² and Area of major segment is 142.75 cm².
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