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Question
The diagram shows a sector of circle of radius ‘r’ can containing an angle ๐. The area of sector is A cm2 and perimeter of sector is 50 cm. Prove that
(i) ๐ =`360/pi(25/r− 1)`
(ii) A = 25r – r2
Solution
(i) Radius of circle = ‘r’ cm
Angle subtended at centre = ๐
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`
⇒ ๐ =`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)
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