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 - Mathematics

Sum

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

Solution

(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)

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 13 Areas Related to Circles
Exercise 13.2 | Q 27 | Page 26