Question

The diagram shows a sector of circle of radius ‘r’ can containing an angle 𝜃. The area of sector is A cm2 and perimeter of sector is 50 cm. Prove that

(i) 𝜃 =`360/pi(25/r− 1)`

(ii) A = 25r – r²

 

Answer 1

(i) Radius of circle = ‘r’ cm

Angle subtended at centre = 𝜃

Perimeter = OA + OB + (AB arc)

= r + r +`theta/360^@× 2pir = 2r + 2r [(pitheta)/360^@]`

But perimeter given as 50

`50 = 2r [1 +(pitheta)/360^@]`

⇒`(pitheta)/360^@=50/(2r)− 1`

⇒ 𝜃 =`360^@/pi[25/r− 1]` …..(i)

(ii) Area of sector =`theta/360^@× pir^2`

=`((360^@/pi)(25/r−1))/360^@× pir^2`

=`25/r× r^2 − r^2`

= 25r – r²

⇒ A = 25r – r² …..(ii)