Given,

Radius = 15 cm

θ = 60°

So,

Area of sector OAPB = \(\dfrac{60°}{360°}×πr^2 \;cm^2 \)

= \(\dfrac{225}{6}
\) πcm^{2}

Now, ΔAOB is equilateral as two sides are

the radii of the circle and hence equal and one angle is 60°

So, Area of ΔAOB = \(\dfrac{\sqrt{3}}{4}×a^2 \)

Or, \( \dfrac{\sqrt{3}}{4}×15^2 \)

Area of ΔAOB = 97.31 cm^{2}

Now, area of minor segment APB = Area of OAPB – Area of ΔAOB

Or, area of minor segment APB = ((\( \dfrac{225}{6}
\))π – 97.31) cm^{2 }= 20.43 cm^{2}

And,

Area of major segment = Area of circle – Area of segment APB

Or, area of major segment = (π×15^{2}) – 20.4 = 686.06 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.