Angle made by sector = 40°,

Radius the inner circle = r = 7 cm, and

Radius of the outer circle = R = 14 cm

We know,

Area of the sector = (\(\dfrac{ θ}{360°}
\))×πr^{2}

So, Area of OAC = (\( \dfrac{ 40°}{360°}
\))×πr^{2 }cm^{2}

= 68.44 cm^{2}

Area of the sector OBD = (\( \dfrac{ 40°}{360°}
\))×πr^{2 }cm^{2}

= (\( \dfrac{1}{9}\))×(\( \dfrac{22}{7}\))×7^{2 }= 17.11 cm^{2}

Now, area of the shaded region ABDC = Area of OAC – Area of the OBD

= 68.44 cm^{2} – 17.11 cm^{2 }= 51.33 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.