The perimeter of a sheet of paper in the shape of a quadrant

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The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Its area would be (π = 22/7)

If the radius of circle be r cm, then
Perimeter of quadrant OAB = OA + AB + OB

= r +	2πr	+ r	πr	+ 2r
	4		2

∴	πr	+ 2r = 75
	2

⇒ r		π	+ 2		= 75
		2

⇒ r		22	+ 2		= 75
		7 × 2

⇒ r ×	50	= 75
	14

⇒ r =	75 × 14	= 21 cm
	50

∴ Required area =	πr²
	4

=		22	× 21 × 21		sq. cm.
		7 × 4

= 346.5 sq.cm.