How To Find The Area Of A Segment Of A Circle

On Aug 26, 2018

How To Find The Area Of A Segment Of A Circle

Area of the sector OPRQ = Area of the segment PRQ + Area of ∆OPQ

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⇒ Area of segment PRQ = (left{ frac{pi }{360}times theta -sin frac{theta }{2}cos frac{theta }{2} right} )

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Area Of A Segment Of A Circle With Examples

Example 1: Find the area of the segment of a circle,given that the angle of the sector is 120º and the radius of the circle is 21 cm. (Take π = 22/7)

Sol. Here, r = 21 cm and π = 120

area-of-a-segment-of-a-circle-2

∴ Area of the segment

= (left{ frac{pi }{360}times theta -sin frac{theta }{2}cos frac{theta }{2} right} )

( =left{ frac{22}{7}times frac{120}{360}-sin 60{}^text{o}cos 60{}^text{o} right}c )

( =left{ frac{22}{21}-frac{1}{2}times frac{sqrt{3}}{2} right}c )

( =left{ frac{22}{21}times -times frac{sqrt{3}}{4} right}c )

( =left( 462-frac{441}{4}sqrt{3} right)=frac{21}{4}(88-21sqrt{3})text{ c}^{text{2}}} )

Example 2: A chord AB of a circle of radius 10 cm makes a right angle at the centre of the circle. Find the area of major and minor segments (Take π = 3.14)

Sol. We know that the area of a minor segment of angle θº in a circle of radius r is given by

(A=left{ frac{pi }{360}times theta -sin frac{theta }{2}cos frac{theta }{2} right} )

segment-of-a-circle-3

Here, r = 10 and θ = 90º

( therefore A=left{ frac{3.14times 90}{4}-sin 45{}^text{o},,cos 45{}^text{o} right} )

( Rightarrow A=left{ frac{3.14}{2}-frac{1}{sqrt{2}}times frac{1}{sqrt{2}} right} )

A = {3.14 × 25 – 50} cm² = (78.5 – 50) cm²

= 28.5 cm²

Area of the major segment = Area of the circle – Area of the minor segment

= {3.14 × 10² – 28.5} cm²

= (314 – 28.5) cm² = 285.5 cm²

Example 3: The diagram shows two arcs, A and B. Arc A is part of the circle with centre O and radius of PQ. Arc B is part of the circle with centre M and radius PM, where M is the mid-point of PQ. Show that the area enclosed by the two arcs is equal to ( 25left{ sqrt{3}-frac{pi }{6} right}text{ c}^{text{2}}} )

Sol. We have,

Area enclosed by arc B and chord PQ = Area of semi-circle of radius 5 cm

area-of-a-segment-of-a-circle-4

( =frac{1}{2}times pi times =frac{25pi }{2}c )

Let ∠MOQ = ∠MOP = θ

In ∆OMP, we have

( sin theta =frac{PM}{OP}=frac{5}{10}=frac{1}{2} )

⇒ θ = 30º ⇒ ∠POQ = 2θ = 60º

∴ Area enclosed by arc A and chord PQ.

= Area of segment of circle of radius 10 cm and sector containing angle 60º

( =left{ frac{pi times 60}{360}-sin 30{}^text{o}times cos 30{}^text{o} right}times text{ } )

( left[ because A=left{ frac{pi theta }{360}-sin frac{theta }{2}cos frac{theta }{2} right} right] )

( =left{ frac{50pi }{3}-25sqrt{3} right}c )

Hence, Required area

( =left{ frac{25pi }{2}-left( frac{50pi }{3}-25sqrt{3} right) right} )

( =left{ 25sqrt{3}-frac{25pi }{6} right} )

( =25left{ sqrt{3}-frac{pi }{6} right}text{ c}^{text{2}}} )