Radius of the quadrant = 3.5 cm = \( \dfrac{7}{2}\) cm

**(i)** Area of quadrant OACB = \((\dfrac{πR^2}{4})
\) cm^{2}

= \( \dfrac{22}{7}\times \dfrac{7}{2}\times\dfrac{7}{2}\times \dfrac{1}{4}
\) cm^{2}

= \(\dfrac{77}{8}\) cm^{2}

**(ii)** Area of triangle BOD = \( \dfrac{1}{2}\times \dfrac{7}{2}\times 2\) cm^{2}

= \( \dfrac{7}{2}\) cm^{2}

Area of shaded region = Area of quadrant – Area of triangle BOD

= (\( \dfrac{77}{8}\))-(\( \dfrac{7}{2}\)) cm^{2 }= \( \dfrac{49}{8}\)cm^{2}

= 6.125 cm^{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.