Here AB be the chord which is subtending an angle 90° at the center O.

It is given that the radius (r) of the circle = 10 cm

**(i)** Area of minor sector = \(\dfrac{90}{360°}×πr^2
\)

= \( (\dfrac{1}{4})×(\dfrac{22}{7})×10^2\)

Or, Area of minor sector = 78.5 cm^{2}

Also, area of ΔAOB = \( \dfrac{1}{2}\)×OB×OA

Here, OB and OA are the radii of the circle i.e. = 10 cm

So, area of ΔAOB = \( \dfrac{1}{2}×10×10\)

= 50 cm^{2}

Now, area of minor segment = area of minor sector – area of ΔAOB

= 78.5 – 50

= 28.5 cm^{2}

**(ii)** Area of major sector = Area of circle – Area of minor sector

= (3.14×10^{2})-78.5

= 235.5 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.