Side of square = OA = AB = 20 cm

Radius of the quadrant = OB

OAB is right angled triangle

By Pythagoras theorem in ΔOAB,

OB^{2 }= AB^{2}+OA^{2}

⇒ OB^{2 }= 20^{2 }+20^{2}

⇒ OB^{2 }= 400+400

⇒ OB^{2 }= 800

⇒ OB= \(20 \sqrt{2}\) cm

Area of the quadrant = \(\dfrac{πR^2}{4}
\) cm^{2 }

=\(\dfrac{3.14}{4}×(20\sqrt{2})^2
\) ^{ }cm^{2 }= 628cm^{2}

Area of the square = 20×20 = 400 cm^{2}

Area of the shaded region = Area of the quadrant – Area of the square

= 628-400 cm^{2 }= 228cm^{2}

Prove that the area of a circular path of uniform width hsurrounding a circular region of radius r is πh(2r + h).

A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.

The outer circumference of a circular race-track is 528 m. The track is everywhere 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre

A circular pond is of diameter 17.5 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of ₹25 per square metre (Use π = 3.14)

A path of width 3.5 m runs around a semicircular grassy plot whose perimeter is 72 m. Find the area of the path.