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NCERT Solutions for Class 10 Maths Exercise 12.2 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

**NCERT solutions for Maths ****Area Related to Circles**** ****Download as PDF**

## NCERT Solutions for Class 10 Maths Area Related to Circles

**Unless stated otherwise, take **

**1. Find the area of a sector of a circle with radius 6 cm, if angle of the sector is **

**Ans. **Here, = 6 cm and

Area of sector =

= = cm^{2}

**2. Find the area of a quadrant of a circle whose circumference is 22 cm.**

**Ans. **Given, = 22

cm

We know that for quadrant of circle,

Area of quadrant =

= = cm^{2}

NCERT Solutions for Class 10 Maths Exercise 12.2

**3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.**

**Ans. **Here, = 14 cm and

Area swept =

= =

NCERT Solutions for Class 10 Maths Exercise 12.2

**4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment, (ii) major segment. **

**Ans. (i)**Here, = 10 cm and

Area of minor sector =

= =

Area of OAB =

= =

Area of minor segment = Area of minor sector – Area of OAB

= 78.5 – 50 =

**(ii)** For major sector, radius = 10 cm and

Area of major sector =

= =

NCERT Solutions for Class 10 Maths Exercise 12.2

**5. In a circle of radius 21 cm, an arc subtends an angle of **** at the centre. Find:**

**(i) the length of the arc.**

**(ii) area of the sector formed by the arc.**

**(iii) area of the segment formed by the corresponding chord.**

**Ans. **Given, = 21 cm and

**(i) **Length of arc =

= = 22 cm

**(ii) **Area of the sector =

= =

**(iii) **Area of segment formed by corresponding chord

= Area of OAB

Area of segment = 231 – Area of OAB……….(i)

In right angled triangle OMA and OMB,

OM = OB[Radii of the same circle]

OM = OM [Common]

OMA OMB [ RHS congruency]

AM = BM[By CPCT]

M is the mid-point of AB and AOM = BOM

AOM = BOM

= AOB =

Therefore, in right angled triangle OMA,

OM = cm

Also,

AM = cm

AB = 2 AM = = 21 cm

Area of OAB =

= =

Using eq. (i),

Area of segment formed by corresponding chord =

NCERT Solutions for Class 10 Maths Exercise 12.2

**6. A chord of a circle of radius 15 cm subtends an angle of **** at the centre. Find the area of the corresponding segment of the circle. **

**Ans. **Here, = 15 cm and

Area of minor sector=

=

=

For, Area of AOB,

Draw OMAB.

In right triangles OMA and OMB,

OA = OB[Radii of same circle]

OM = OM[Common]

OMA OMB [RHS congruency]

AM = BM[By CPCT]

AM = BM = AB and

AOM = BOM = AOB =

In right angled triangle OMA,

OM = cm

Also,

AM = cm

2 AM = = 15 cm

AB = 15 cm

Area of AOB =

= =

= =

Area of minor segment = Area of minor sector – Area of AOB

= 117.75 – 97.3125 = c

And, Area of major segment = Area of minor segment

= 706.5 – 20.4375 =

NCERT Solutions for Class 10 Maths Exercise 12.2

**7. A chord of a circle of radius 12 cm subtends an angle of **** at the centre. Find the area of the corresponding segment of the circle.**

**Ans. **Here, = 15 cm and

Area of corresponding sector=

=

=

For, Area of AOB,

Draw OMAB.

In right triangles OMA and OMB,

OA = OB[Radii of same circle]

OM = OM[Common]

OMA OMB [RHS congruency]

AM = BM[By CPCT]

AM = BM = AB and

AOM = BOM = AOB =

In right angled triangle OMA,

OM = 6 cm

Also,

AM = cm

2 AM = = cm

AB = cm

Area of AOB =

= =

= =

Area of corresponding segment = Area of corresponding sector – Area of AOB

= 150.72 – 62.28 =

NCERT Solutions for Class 10 Maths Exercise 12.2

**8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find: **

**(i) the area of that part of the field in which the horse can graze.**

**(ii) the increase in the grazing area if the rope were 10 m long instead of 5 cm. **

**Ans. (i)**Area of quadrant with 5 m rope

=

= =

**(ii)** Area of quadrant with 10 m rope

=

= =

The increase in grazing area

= 78.5 – 19.625

=

NCERT Solutions for Class 10 Maths Exercise 12.2

**9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find: **

**(i) the total length of the silver wire required.**

**(ii) the area of each sector of the brooch.**

**Ans. (i)**Diameter = 35 mm

Radius = mm

Circumference = =

= 110 mm……….(i)

Length of 5 diameters = = 175 mm……….(ii)

Total length of the silver wire required

= 110 + 175 = 285 mm

**(ii)** mm and

The area of each sector of the brooch =

= =

NCERT Solutions for Class 10 Maths Exercise 12.2

**10. An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.**

**Ans. **Here, = 45 cm and

Area between two consecutive ribs of the umbrella

=

=

=

NCERT Solutions for Class 10 Maths Exercise 12.2

**11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of **** Find the total area cleaned at each sweep of the blades.**

**Ans. **Here, = 25 cm and

The total area cleaned at each sweep of the blades

=

=

=

NCERT Solutions for Class 10 Maths Exercise 12.2

**12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle **** to a distance of 16.5 km. Find the area of the sea over which the ships are warned. **

**Ans. **Here, = 16.5 km and

The area of sea over which the ships are warned =

= =

NCERT Solutions for Class 10 Maths Exercise 12.2

**13. A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Ts. 0.35 per cm**^{2}.

^{2}.

**Ans. ** = 28 cm and

Area of minor sector =

=

= =

For, Area of AOB,

Draw OMAB.

In right triangles OMA and OMB,

OA = OB[Radii of same circle]

OM = OM[Common]

OMA OMB [RHS congruency]

AM = BM[By CPCT]

AM = BM = AB and AOM = BOM = AOB =

In right angled triangle OMA,

OM = cm

Also,

AM = 14 cm

2 AM = = 28 cm

AB = 28 cm

Area of AOB =

= =

= =

Area of minor segment = Area of minor sector – Area of AOB

= 410.67 – 333.2 =

Area of one design =

Area of six designs = =

Cost of making designs = = Rs. 162.68

NCERT Solutions for Class 10 Maths Exercise 12.2

**14. Tick the correct answer in the following:**

**Area of a sector of angle (in degrees) of a circle with radius R is:**

**(A) **

**(B) **

**(C) **

**(D) **

**Ans. (D)** Given, = R and

Area of sector =

=

=

=

## NCERT Solutions for Class 10 Maths Exercise 12.2

NCERT Solutions Class 10 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 10 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 10 Maths have total 15 chapters. 10 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 10 solutions PDF and Maths ncert class 10 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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