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Question
A chord of a circle subtends an angle ๐ at the centre of circle. The area of the minor segment cut off by the chord is one eighth of the area of circle. Prove that 8 sin`theta/2 "cos"theta/2+pi =(pitheta)/45`
Solution
Let radius of circle = r
Area of circle = ๐๐2
AB is a chord, OA, OB are joined drop OM ⊥ AB. This OM bisects AB as well as ∠AOB.
∠AOM = ∠MOB =`1/2(0) =theta/2` AB = 2AM
In ΔAOM, ∠AMO = 90°
`"Sin"theta/2=(AM)/(AD)⇒ AM = R."sin"theta/2` AB = 2R sin`theta/2`
`"Cos"theta/2=(OM)/(AD)⇒ OM = R"cos"theta/2`
Area of segment cut off by AB = (area of sector) – (area of triangles)
=`theta/360× pir^2 −1/2`× ๐ด๐ต × ๐๐
`= r^2 [(pitheta)/360^@−1/2. 2"rsin"theta/2. R" cos"theta/2]`
`= R^2 [(pitheta)/360^@− "sin"theta/2. "cos"theta/2]`
Area of segment =`1/2`(๐๐๐๐ ๐๐ ๐๐๐๐๐๐)
`r^2 [(pitheta)/360− "sin"theta/2." cos"theta/2] =1/8pir^2`
`(8pitheta)/360^@− 8 "sin"theta/2. "cos"theta/2= pi`
`8 "sin"theta/2. "cos"theta/2+ pi =(pitheta)/45`
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