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A chord of a circle subtends an angle ЁЭЬГ 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`
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Let radius of circle = r
Area of circle = ЁЭЬЛЁЭСЯ2
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`× ЁЭР┤ЁЭР╡ × ЁЭСВЁЭСА
`= 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`(ЁЭСОЁЭСЯЁЭСТЁЭСО ЁЭСЬЁЭСУ ЁЭСРЁЭСЦЁЭСЯЁЭСРЁЭСЩЁЭСТ)
`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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