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Question

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+πθ1801)

(ii) Area of the shaded region is r22(tanθπθ180)


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Solution

(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=OBOC=r sec θr

Now, perimeter of shaded region = arc AC + AB + BC

=πrθ180+r tan θ+r sec θr

=r(tan θ+sec θ+πθ1801)

Hence, proved.

(ii) Area of shaded region = Area of triangle OAB - Area of sector OAC

=12×OA×ABπr2×θ360o

=12×r×r tan θπr2×θ360o

=r22(tan θπθ180)

Hence, proved.


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