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Question
In Fig. 9, is shown a sector OAP of a circle with centre O, containing ∠θ. AB is perpendicular to the radius OQ and meets OP produced at B. Prove that the perimeter of shaded region is
`r[tantheta+sectheta+(pitheta)/180-1]`
Solution
Perimeter of shaded region = AB + PB + arc length AP...(1)
Arc length AP = `theta/360xx2pir=(pithetar)/180" ....(2)"`
In right angled ΔOAB,
`tan theta=(AB)/r=>AB=rtan theta" ....(3)"`
`sec theta=(OB)/r =>OB=rsec theta`
OB = OP + PB
∴ r secθ=r+PB
∴ PB = r secθ - r.....(4)
Substitute (2), (3) and (4) in (1), we get
Perimeter of shaded region = AB+PB+ arc length AP
`=rtantheta+rsectheta-r+(pithetar)/180`
`=r[tantheta+sec theta+(pitheta)/180-1]`
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