Figure shows a sector of a circle, centre O, containing an angle θ ∘. Prove that:
(i) Perimeter of the shaded region is r(tanθ+secθ+πθ180−1)
(ii) Area of the shaded region is r22(tanθ−πθ180)
(i) Length of arc AC =2πr×θ360o=πrθ180
Now, in △ OAB, tan θ=ABOA
⇒tan θ=ABr
⇒AB=r tan θ
Also, sec θ=OBOA=OBr
⇒OB=r sec θ
⇒BC=OB−OC=r sec θ−r
Now, perimeter of shaded region = arc AC + AB + BC
=πrθ180+r tan θ+r sec θ−r
=r(tan θ+sec θ+πθ180−1)
Hence, proved.
(ii) Area of shaded region = Area of triangle OAB - Area of sector OAC
=12×OA×AB−πr2×θ360o
=12×r×r tan θ−πr2×θ360o
=r22(tan θ−πθ180)
Hence, proved.