Radius, r = 12 cm

Now, draw a perpendicular OD on chord AB and it will bisect chord AB.

So, AD = DB

Now, the area of the minor sector = (\( \dfrac{θ}{360°}\))×πr^{2}

= (\( \dfrac{120}{360}\))×(\( \dfrac{22}{7}\))×12^{2}

= 150.72 cm^{2}

Consider the ΔAOB,

∠ OAB = 180°-(90°+60°) = 30°

Now, cos 30° = \( \dfrac{AD}{OA}\)

\( \dfrac{\sqrt{3}}{2}\) = \( \dfrac{AD}{12}\)

Or, AD =\(6 \sqrt{3}\) cm

We know OD bisects AB. So,

AB = 2×AD = \(12 \sqrt{3}\) cm

Now, sin 30° = \( \dfrac{OD}{OA}\)

Or, \( \dfrac{1}{2}=\) \( \dfrac{AD}{12}\)

OD = 6 cm

So, the area of ΔAOB = \( \dfrac{1}{2}\) × base × height

Here, base = AB = \( 12 \sqrt{3}\) and

Height = OD = 6

So, area of ΔAOB = \( \dfrac{1}{2}\)×\( 12 \sqrt{3}\)×6 = \( 36 \sqrt{3}\) cm = 62.28 cm^{2}

Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB

= 150.72 cm^{2}– 62.28 cm^{2 }= 88.44 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.