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NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Areas Related to Circles

NCERT Solutions for Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.2 are designed to help students understand and apply the formulas for calculating areas related to circles, including sector and segment areas, in a variety of geometrical and real-world problems. This exercise delves deeper into finding the areas of segments, and solving problems with a combination of circles and other geometric shapes.

Exercise 12.2 focuses on:

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Areas Related to Circles

Area Related to Circles

Medium

Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Area Related to Circles

Medium

\begin{array}{l} Find the area of the shaded region in the following \\ figure, if radii of the two concentric circles with \\ centre O are 7 cm and 14 cm respectively and \\ ∠ AOC = 40°. \end{array}

Area Related to Circles

Medium

In the following figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Area Related to Circles

Medium

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the following figure). If ∠AOB = 30°, find the area of the shaded region

Area Related to Circles

Medium

In the following figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Area Related to Circles

Medium

On a square handkerchief, nine circular designs each of radius 7 cm are made (see the following figure). Find the area of the remaining portion of the handkerchief.

Area Related to Circles

Medium

\begin{array}{l} {The area of an equilateral triangle ABC is 17320.5 cm}^{2} . \\ With each vertex of the triangle as centre, a circle is \\ drawn with radius equal to half the length of the side \\ of the triangle (see the following figure) . Find the \\ area of the shaded region. (Use π = 3.14 and \sqrt{3} = 1.73205) \end{array}

Area Related to Circles

Medium

In the following figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Area Related to Circles

Medium

The following figure depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

the distance around the track along its inner edge
the area of the track.

Area Related to Circles

Medium

In the following figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.

Area Related to Circles

Medium

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the following figure. Find the area of the design (shaded region).

Area Related to Circles

Medium

From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the following figure. Find the area of the remaining portion of the square.

Area Related to Circles

Medium

Find the area of the shaded region in the following figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Area Related to Circles

Medium

Find the area of the shaded region in the following figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Area Related to Circles

Medium

Find the area of the shaded region in the following figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Area Related to Circles

Medium

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

Area Related to Circles

Medium

\begin{array}{l} Tick the correct answer in the following: \\ Area of a sector of angle p (in degrees) of a circle \\ with radius R is \\ (A) \frac{P}{180} \times 2 π R                               (B) \frac{P}{180} \times π R^{2} \\ (C) \frac{P}{360} \times 2 π R                              (D) \frac{P}{720} \times 2 π R^{2} \end{array}

Area Related to Circles

Medium

\begin{array}{l} A round table cover has six equal designs as shown \\ in the following figure. If the radius of the cover is \\ 28 cm, find the cost of making the designs at the \\ {rate of ₹ 0.35 per cm}^{2} . (Use \sqrt{3} = 1.7) \end{array}

Area Related to Circles

Medium

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Area Related to Circles

Medium

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Area Related to Circles

Medium

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

Area Related to Circles

Medium

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 the following figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Area Related to Circles

Medium

\begin{array}{l} 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 the following 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 m. (Use π = 3.14) \end{array}

Area Related to Circles

Medium

\begin{array}{l} A chord of a circle of radius 12 cm subtends an \\ angle of 120° at the centre. Find the area of the \\ corresponding segment of the circle. \\ (Use π = 3.14 and \sqrt{3} = 1.73) \end{array}

Area Related to Circles

Medium

\begin{array}{l} A c h o r d o f a c i r c l e o f r a d i u s 15 c m s u b t e n d s a n a n g l e o f \\ 60 ° a t t h e c e n t r e . F i n d t h e a r e a s o f t h e c o r r e s p o n d i n g \\ m i n o r a n d m a j o r s e g m e n t s o f t h e c i r c l e . \\ (U s e π = 3.14 a n d \sqrt{3} = 1.73) \end{array}

Area Related to Circles

Medium

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

the length of the arc
area of the sector formed by the arc
area of the segment formed by the corresponding chord

Area Related to Circles

Medium

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 sector. (Use π = 3.14)

Area Related to Circles

Difficult

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

Area Related to Circles

Medium

Calculate the area of the designed region in the following figure common between the two quadrants of circles of radius 8 cm each.

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Areas Related to Circles

Finding the area of a sector and a segment of a circle.
Solving problems related to sector area and segment area in complex scenarios.
Applying the concepts of arc length and central angles to real-life applications.

The solutions are presented in a step-by-step manner, helping students apply their knowledge efficiently and prepare for Class 10 exams with confidence.

Q1. Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.

Difficulty Level: Easy
What is the known/given?

Radius of the circle = 6 cm, Angle of the sector = 60°

What is unknown?

Area of the sector of the circle.

Reasoning:
The formula for the area of the sector of an angle

$\theta$

$θ$ is:

$\text{Area of the sector} = \frac{\theta}{360} \times \pi r^2$

$Area of the sector = 360 θ \times π r^{2}$

Solution:
Given:

$\theta = 60^\circ$

$θ = 6 0^{\circ}$ ,

$r = 6 \, \text{cm}$

$r = 6 cm$

$\text{Area of the sector} = \frac{60}{360} \times \pi \times 6^2 = \frac{1}{6} \times \pi \times 36 = 6\pi \, \text{cm}^2$

$Area of the sector = 36060 \times π \times 6^{2} = 61 \times π \times 36 = 6 π cm^{2}$

$\text{Area of the sector} = 6 \times 3.14 = 18.84 \, \text{cm}^2$

$Area of the sector = 6 \times 3.14 = 18.84 cm^{2}$

Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Difficulty Level: Medium
What is the known/given?

The circumference of the circle = 22 cm.

What is unknown?

Area of the quadrant of the circle.

Reasoning:
First, find the radius of the circle from the circumference, then use the formula for the area of a quadrant.

Solution:
Circumference of the circle:

$C = 2\pi r$

$C = 2 π r$

Given

$C = 22 \, \text{cm}$

$C = 22 cm$ ,

$22 = 2\pi r$

$22 = 2 π r$

$r = \frac{22}{2\pi} = \frac{22}{6.28} \approx 3.5 \, \text{cm}$

$r = 2 π 22 = 6.2822 \approx 3.5 cm$

Area of the circle:

$A = \pi r^2 = 3.14 \times (3.5)^2 = 3.14 \times 12.25 = 38.465 \, \text{cm}^2$

$A = π r^{2} = 3.14 \times (3.5)^{2} = 3.14 \times 12.25 = 38.465 cm^{2}$

Since a quadrant is one-fourth of the circle:

$\text{Area of the quadrant} = \frac{1}{4} \times 38.465 = 9.616 \, \text{cm}^2$

$Area of the quadrant = 41 \times 38.465 = 9.616 cm^{2}$

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Difficulty Level: Medium
What is the known/given?

Length of the minute hand = 14 cm

What is unknown?

Area swept by the minute hand in 5 minutes.

Reasoning:
Since the minute hand completes one rotation in 60 minutes, find the area swept by the minute hand in 60 minutes and then use unitary method to find the area in 5 minutes.

Solution:
Length of the minute hand (radius) = 14 cm

Area swept by the minute hand in 60 minutes (full rotation) = Area of the circle with radius 14 cm:

$A = \pi r^2 = 3.14 \times (14)^2 = 3.14 \times 196 = 615.44 \, \text{cm}^2$

$A = π r^{2} = 3.14 \times (14)^{2} = 3.14 \times 196 = 615.44 cm^{2}$

Area swept by the minute hand in 1 minute:

$\text{Area in 1 minute} = \frac{615.44}{60} = 10.257 \, \text{cm}^2$

$Area in 1 minute = 60615.44 = 10.257 cm^{2}$

Area swept in 5 minutes:

$\text{Area in 5 minutes} = 5 \times 10.257 = 51.285 \, \text{cm}^2$

$Area in 5 minutes = 5 \times 10.257 = 51.285 cm^{2}$

Q4. A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding:

(i) Minor Segment
(ii) Major Sector

Difficulty Level: Medium
What is the known/given?

Radius of the circle = 10 cm, Angle subtended by the chord = 90°

What is unknown?

Area of minor segment and major sector.

Reasoning:
(i) Area of the sector =

$\frac{\theta}{360} \times \pi r^2$

$360 θ \times π r^{2}$
(ii) Area of the segment = Area of the sector – Area of the corresponding triangle.

Solution:
(i) Area of the minor segment (subtended by 90°):
Area of the sector

$\theta = 90^\circ$

$θ = 9 0^{\circ}$ :

$\text{Area of the sector} = \frac{90}{360} \times \pi \times (10)^2 = \frac{1}{4} \times 3.14 \times 100 = 78.5 \, \text{cm}^2$

$Area of the sector = 36090 \times π \times (10)^{2} = 41 \times 3.14 \times 100 = 78.5 cm^{2}$

Area of the triangle (right-angled triangle) with base and height = 10 cm:

$\text{Area of triangle} = \frac{1}{2} \times 10 \times 10 = 50 \, \text{cm}^2$

$Area of triangle = 21 \times 10 \times 10 = 50 cm^{2}$

Therefore, area of the minor segment:

$\text{Area of minor segment} = 78.5 - 50 = 28.5 \, \text{cm}^2$

$Area of minor segment = 78.5 - 50 = 28.5 cm^{2}$

(ii) Area of the major sector:
The total area of the circle is:

$\text{Area of circle} = \pi r^2 = 3.14 \times (10)^2 = 314 \, \text{cm}^2$

$Area of circle = π r^{2} = 3.14 \times (10)^{2} = 314 cm^{2}$

Area of the major sector:

$\text{Area of major sector} = \text{Area of circle} - \text{Area of minor sector} = 314 - 78.5 = 235.5 \, \text{cm}^2$

$Area of major sector = Area of circle - Area of minor sector = 314 - 78.5 = 235.5 cm^{2}$

Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. 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

Difficulty Level: Hard
What is the known/given?

Radius of the circle = 21 cm, Angle subtended by the arc = 60°

What is unknown?

Length of the arc, Area of the sector, Area of the segment.

Reasoning:
(i) Length of the arc:

$\text{Length of the arc} = \frac{\theta}{360} \times 2\pi r$

$Length of the arc = 360 θ \times 2 π r$

(ii) Area of the sector:

$\text{Area of the sector} = \frac{\theta}{360} \times \pi r^2$

$Area of the sector = 360 θ \times π r^{2}$

(iii) Area of the segment = Area of the sector – Area of the corresponding triangle.

Solution:
(i) Length of the arc:

$\text{Length of the arc} = \frac{60}{360} \times 2 \times 3.14 \times 21 = \frac{1}{6} \times 132 = 22 \, \text{cm}$

$Length of the arc = 36060 \times 2 \times 3.14 \times 21 = 61 \times 132 = 22 cm$

(ii) Area of the sector:

$\text{Area of the sector} = \frac{60}{360} \times 3.14 \times 21^2 = \frac{1}{6} \times 3.14 \times 441 = 231 \, \text{cm}^2$

$Area of the sector = 36060 \times 3.14 \times 2 1^{2} = 61 \times 3.14 \times 441 = 231 cm^{2}$

(iii) Area of the segment:
The area of the triangle can be found using the formula for the area of a triangle (with sides

$r$

$r$ ,

$r$

$r$ , and the angle

$60^\circ$

$6 0^{\circ}$ ):

$\text{Area of the triangle} = \frac{1}{2} r^2 \sin(\theta) = \frac{1}{2} \times 21^2 \times \sin(60^\circ)$

$Area of the triangle = 21 r^{2} sin (θ) = 21 \times 2 1^{2} \times sin (6 0^{\circ})$

$\text{Area of the triangle} = \frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2} = 183.5 \, \text{cm}^2$

$Area of the triangle = 21 \times 441 \times 2 3 = 183.5 cm^{2}$

Area of the segment:

$\text{Area of the segment} = 231 - 183.5 = 47.5 \, \text{cm}^2$

$Area of the segment = 231 - 183.5 = 47.5 cm^{2}$

Q6. A chord of a circle of radius 12 cm subtends an angle of 120° at the center. Find the area of the corresponding segment of the circle.

Difficulty Level: Hard
What is the known/given?

A chord of a circle with radius

$r = 12 \, \text{cm}$ $r = 12 cm$ subtends an angle

$\theta = 120^\circ$ $θ = 12 0^{\circ}$ at the center.

What is unknown?

Area of the segment of the circle.

Reasoning:
In a circle with radius

$r$

$r$ and angle at the center with degree measure

$\theta$

$θ$ :
(i) Area of the sector is given by

$\frac{\theta}{360} \times \pi r^2$

$360 θ \times π r^{2}$
(ii) Area of the segment = Area of the sector – Area of the corresponding triangle.

Solution:
Given:

$r = 12 \, \text{cm}, \theta = 120^\circ$

$r = 12 cm, θ = 12 0^{\circ}$

(i) Area of the sector:

$\text{Area of the sector} = \frac{120}{360} \times \pi \times (12)^2 = \frac{1}{3} \times 3.14 \times 144 = 150.72 \, \text{cm}^2$

$Area of the sector = 360120 \times π \times (12)^{2} = 31 \times 3.14 \times 144 = 150.72 cm^{2}$

(ii) Area of the triangle:
To find the area of the triangle, we use the formula for the area of a triangle with two known sides and the included angle:

$\text{Area of triangle} = \frac{1}{2} \times r^2 \times \sin(\theta)$

$Area of triangle = 21 \times r^{2} \times sin (θ)$

$\text{Area of triangle} = \frac{1}{2} \times 12^2 \times \sin(120^\circ) = \frac{1}{2} \times 144 \times \frac{\sqrt{3}}{2} = 62.28 \, \text{cm}^2$

$Area of triangle = 21 \times 1 2^{2} \times sin (12 0^{\circ}) = 21 \times 144 \times 2 3 = 62.28 cm^{2}$

(iii) Area of the segment:

$\text{Area of the segment} = \text{Area of the sector} - \text{Area of the triangle}$

$Area of the segment = Area of the sector - Area of the triangle$

$\text{Area of the segment} = 150.72 - 62.28 = 88.44 \, \text{cm}^2$

$Area of the segment = 150.72 - 62.28 = 88.44 cm^{2}$

Q7. 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. 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 m.

Difficulty Level: Medium
What is the known/given?

Length of the side of the square grass field = 15 m
Rope length = 5 m (for part 1) and 10 m (for part 2).

What is unknown?

(i) Area of the field the horse can graze with a 5 m rope.
(ii) Increase in grazing area if the rope were 10 m long.

Reasoning:
(i) The horse can graze the area of a sector of a circle with radius 5 m and angle 90° (since the peg is at the corner of the square).
(ii) For the 10 m rope, the grazing area increases as it corresponds to a sector of a circle with radius 10 m.

Solution:
(i) Area grazed with a 5 m rope:
The angle subtended by the rope at the center is

$90^\circ$

$9 0^{\circ}$ , so the area of the sector is:

$\text{Area} = \frac{\theta}{360} \times \pi r^2 = \frac{90}{360} \times 3.14 \times (5)^2 = \frac{1}{4} \times 3.14 \times 25 = 19.625 \, \text{m}^2$

$Area = 360 θ \times π r^{2} = 36090 \times 3.14 \times (5)^{2} = 41 \times 3.14 \times 25 = 19.625 m^{2}$

(ii) Area grazed with a 10 m rope:
The angle remains

$90^\circ$

$9 0^{\circ}$ , but the radius is now 10 m:

$\text{Area} = \frac{90}{360} \times 3.14 \times (10)^2 = \frac{1}{4} \times 3.14 \times 100 = 78.5 \, \text{m}^2$

$Area = 36090 \times 3.14 \times (10)^{2} = 41 \times 3.14 \times 100 = 78.5 m^{2}$

Increase in grazing area:

$\text{Increase} = 78.5 - 19.625 = 58.875 \, \text{m}^2$

$Increase = 78.5 - 19.625 = 58.875 m^{2}$

Q8. A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters, which divide the circle into 10 equal sectors. Find:

(i) The total length of the silver wire required.
(ii) The area of each sector of the brooch.

Difficulty Level: Medium
What is the known/given?

Diameter of the circle = 35 mm
The wire is used to make 5 diameters, dividing the circle into 10 equal sectors.

What is unknown?

(i) Total length of the silver wire required.
(ii) Area of each sector of the brooch.

Reasoning:
(i) The total length of the silver wire required is the sum of the circumference of the circle and the lengths of the 5 diameters.
(ii) The area of each sector is

$\frac{360^\circ}{10} = 36^\circ$

$10 360 ^{\circ} = 3 6^{\circ}$ .

Solution:
(i) Total length of the silver wire required:
Circumference of the circle =

$\pi \times d = 3.14 \times 35 = 109.9 \, \text{mm}$

$π \times d = 3.14 \times 35 = 109.9 mm$
Length of 5 diameters =

$5 \times 35 = 175 \, \text{mm}$

$5 \times 35 = 175 mm$
Total length of the silver wire = Circumference + 5 diameters =

$109.9 + 175 = 284.9 \, \text{mm}$

$109.9 + 175 = 284.9 mm$

(ii) Area of each sector:
Since the circle is divided into 10 equal sectors, the angle of each sector is

$36^\circ$

$3 6^{\circ}$ .
Area of each sector = ( \frac{36}{360} \times \pi \times r^2 = \frac{1}{10} \times 3.14 \times \left(\frac{35}{2}\right)^2 = \frac{1}{10} \times 3.14 \times 17.5^2 = 96.2 , \text{mm}^2

Q9. An umbrella has 8 ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius 45 cm, find the area between two consecutive ribs of the umbrella.

Difficulty Level: Medium
What is the known/given?

An umbrella has 8 ribs, and the umbrella is assumed to be a flat circle with a radius of 45 cm.

What is unknown?

The area between two consecutive ribs of the umbrella.

Reasoning:
Since there are 8 equal spaced ribs, the angle between two consecutive ribs at the center of the circle is

$\frac{360^\circ}{8} = 45^\circ$

$8 360 ^{\circ} = 4 5^{\circ}$ .
The area between two ribs is the area of the sector formed by a central angle of 45°.

Solution:
Area between two consecutive ribs = Area of a sector with central angle 45°:

$\text{Area of the sector} = \frac{45}{360} \times \pi r^2 = \frac{1}{8} \times 3.14 \times (45)^2 = 1585.5 \, \text{cm}^2$

$Area of the sector = 36045 \times π r^{2} = 81 \times 3.14 \times (45)^{2} = 1585.5 cm^{2}$

Q10. An umbrella has 2 wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Difficulty Level: Medium
What is the known/given?

Length of each wiper blade = 25 cm
Each wiper sweeps through an angle of 115°.

What is unknown?

Total area cleaned at each sweep of the blades.

Reasoning:
The area cleaned by each wiper is the area of a sector formed by an angle of 115° with a radius of 25 cm. Since there are two wipers, the total area cleaned is double the area cleaned by one wiper.

Solution:
Area cleaned by each wiper = Area of a sector with central angle 115°:

$\text{Area of the sector} = \frac{115}{360} \times \pi r^2 = \frac{115}{360} \times 3.14 \times (25)^2 = 3.14 \times 625 \times \frac{115}{360} = 231.49 \, \text{cm}^2$

$Area of the sector = 360115 \times π r^{2} = 360115 \times 3.14 \times (25)^{2} = 3.14 \times 625 \times 360115 = 231.49 cm^{2}$

Total area cleaned by both wipers =

$2 \times 231.49 = 462.98 \, \text{cm}^2$

$2 \times 231.49 = 462.98 cm^{2}$

FAQs: Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.2

Q1. What is the focus of Exercise 12.2?
Answer:
Exercise 12.2 focuses on calculating the area of a sector and a segment of a circle and solving word problems that involve these areas.

Q2. How do I calculate the area of a sector?
Answer:
The area of a sector of a circle is calculated using the formula:

$A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2$

$A_{sector} = 360 ^{\circ} θ \times π r^{2}$

where

$\theta$

$θ$ is the central angle of the sector in degrees and

$r$

$r$ is the radius of the circle.

Q3. How is the area of a segment of a circle calculated?
Answer:
To calculate the area of a segment:

Find the area of the sector using the formula for the area of a sector.
Subtract the area of the triangle formed by the chord and the radius from the area of the sector.

Q4. How do I solve problems involving sectors and segments of circles?
Answer:
To solve problems:

Identify the given values such as radius, central angle, or chord length.
Use the sector area formula and segment area formula to find the required areas.
If needed, calculate the area of the triangle formed by the chord and subtract it from the sector's area.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide step-by-step methods to solve complex geometry problems involving sectors and segments of circles. By practicing these solutions, students can develop a clear understanding of the concepts and solve problems confidently in the Class 10 exams.

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Areas Related to Circles

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Areas Related to Circles

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Areas Related to Circles

FAQs: Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.2

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NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Areas Related to Circles

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Areas Related to Circles

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Areas Related to Circles

FAQs: Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.2

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