# Class 10 NCERT Solutions Maths Chapter 11 - Areas Related to Circles

## Areas Related to Circles Exercise Ex. 11.1

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

Radius (r) of the circle = 15

Area of sector OPRQ =

=

=

= 117.75 cm^{2}

In ∆OPQ

.... (Since OP = OQ)

\

Area of ∆OPQ =

Area of segment PRQ = Area of sector OPRQ –Area of ∆OPQ

= 117.75 – 97.3125

= 20.4375 cm^{2}

Area of major segment PSQ

= Area of circle – Area of segment PRQ

= 15^{2}p – 20.4375

= 3.14 × 225 – 20.4375

= 686.0625 cm^{2}

### Solution 7

Draw a perpendicular OV on chord ST. It will bisect the chord ST.

SV = VT

In ∆OVS

OV = 6

ST = 2SV =

Area of ∆OST =

=

=

= 36 × 1.73

= 62.28

Area of sector OSUT =

= 150.72

Area of segment SUT = Area of sector OSUT

= 150.72 – 62.28

= 88.44 cm^{2}

### Solution 8

The horse can graze a sector of 90° in a circle of 5 m radius.

i. So area that can be grazed by horse = area of sector OACB

=

=

= 19.63 m^{2}

ii. Area that can be grazed by the horse when the length of rope is 10 m long =

=

= 78.5

Change in grazing area = 78.5 – 19.63 = 58.87 cm^{2}

### Solution 9

### Solution 10

### Solution 11

The figure shows that each blade of the wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.