Crucial Practice Problems for CBSE Class 10 Maths Chapter 11: Areas Related to Circles
Class 10 Maths Chapter 11 Important Questions Areas Related to Circles are given here based on the latest pattern of CBSE for 2024-2025. Students who are preparing for the board exams can practice Areas Related to Circles Important Questions to score full marks for the questions from this chapter. Along with the Important Questions for Class 10 Maths Chapter 11 related to Circles, we have also provided Extra Questions of Chapter 11 Class 10 Maths. Students can refer to the solutions whenever they get stuck while solving a problem. Also, in the end, some practice questions are provided for students to boost their preparation for the exam.
Areas Related to Circles chapter contains many formulas and concepts thus, it is important from the examination perspective, since most of the questions and objective-based questions will come in the exam from this chapter. Students can also refer to Chapter 11 Maths Class 10 Important Questions while preparing for their exam. NCERT Solution is also available at Vedantu to help the students in scoring full marks in the board exam.
Area related to circles includes the area of a circle, segment, sector, angle and length of a circle are provided in this pdf. In this chapter, we will discuss the concepts of the perimeter (also known as circumference) and area of a circle and apply this knowledge in finding the areas of two special parts of a circular region (or a circle) known as sector and segment. You can download Maths NCERT Solutions Class 10 and NCERT Solution for Class 10 Science to help you to revise the complete Syllabus and score more marks in your examinations.






Study Important Questions for Class 10 Maths Chapter 11- Area Related to Circles
Very Short Answer Questions (1 Mark)
Unless stated otherwise, take
1. The radii of two circles are
Ans: Let us suppose R, radius of the circle whose circumference is equal to the circumference of the two circles with radius
Thus,
2. The circumference of a circular field is
a.
b.
c.
d.
Ans: b.
Circumference of a circle is given by,
3. The circumference of a circle exceeds its diameter by
a.
b.
c.
d.
Ans: d.
Let the diameter of the circle be
Circumference of a circle is given by,
4. Area of the sector of angle
a.
b.
c.
d. None of these
Ans: b.
Area of the sector of a circle
5. Area of a sector of angle
a.
b.
c.
d.
Ans: d.
Area of the sector of a circle
6. If the sum of the circumferences of two circles with radii
a.
b.
c.
d. None of these
Ans: a.
Circumference of a circle is given by,
Thus, according to the given condition,
7. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
a.
b.
c.
d.
Ans. b.
The perimeter of a circle is given by,
The perimeter of a square is given by,
Given,
Ratio of area of circle to area of square is
Hence, the ratio of their areas is,
8. The circumference of a circular field is
a.
b.
c.
d.
Ans. Circumference of a circle is given by,
9. The area of a circle is
a.
b.
c.
d.
Ans. c.
The area of a circle is given by,
10. If the perimeter and area of circle are numerically equal, then the radius of the circle is
a.
b.
c.
d.
Ans. a.
Perimeter / Circumference of a circle is given by,
Area of a circle is given by,
Given,
11. The radius of a circle is
a.
b.
c.
d.
Ans. d.
Area of a circle is given by,
12. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii
a.
b.
c.
d.
Ans. d.
Let
13. The circumference of a circle is
a.
b.
c.
d.
Ans. a.
Circumference of a circle is given by,
Thus, area
Very Short Answer Questions (2 Marks)
1. The radii of two circles are
Ans: Area of a circle is given by,
Let
Thus, we have:
2. Figure depicts an archery target marked with its five scoring area from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is
Ans: Given, diameter of the region representing Gold score as
Thus, radius of the Gold scoring region,
Area of a circle is given by,
Area of the Gold scoring region,
Area of scoring region = Area of previous circle – Area of next circle.
Area of Blue scoring region,
Area of Black scoring region,
Area of White scoring region,
3. The wheels of a car are of diameter
Ans: Given, diameter of car wheel
Distance covered by the wheel in one revolution is equal to the circumference of the wheel.
Thus, distance covered by the wheel in one revolution is equal to
Given, the distance covered in
Hence, distance covered by the wheel in
Number of revolutions
Number of revolutions
Hence, number of revolutions
4. Find the area of a sector of a circle with radius
Ans: Given, Angle of the sector is
Radius,
Area of the sector of a circle
5. Find the area of a quadrant of a circle whose circumference is
Ans: Given, the circumference of the circle as
We know, the circumference of a circle is given by,
Therefore,
For the quadrant of a circle,
Area of quadrant is given by,
6. The length of the minute hand of a clock is
Ans: Given, length of the minute hand (radius) ,
Area swept by the minutes hand in
7. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (Use
i. minor segment
Ans: Given, Radius,
Area of minor sector,
Area of
Area of minor segment can be obtained as,
Area of minor segment
ii. Major Segment
Ans: For major sector, radius =
8. A horse is tied to a peg at one corner of a square shape grass field of side
i. the area of that part of the field in which the horse can graze.
Ans: Area of a sector of a circle is given by,
Area of the part of the field where the horse can graze is in the form of a quadrant with radius,
For the quadrant of a circle,
Thus, the area is given as,
ii. the increase in the grazing area if the rope were
Ans: Area of quadrant with
Thus, the grazing area for
Hence, Increase in area
9. A brooch is made with silver wire in the form of a circle with diameter
i. The total length of the silver wire required.
Ans: Given, diameter of silver wire circle as
Thus, radius is
Circumference of a circle is given by,
Therefore, the circumference of the circle made by silver wire is,
As the wire is used in making
Total length of the silver wire required would be the sum of circumference of the circle and the length of
Hence, adding equation (1) and (2), we get
ii. The area of each sector of the brooch.
Ans: Given, diameter of silver wire circle as
Thus, radius is
Angle of each sector,
10. An umbrella has
Ans: Let umbrella be a flat circle with radius,
Given, the umbrella has
Area between two consecutive ribs of the umbrella can be obtained using the formula,
11. A car has two wipers which do not overlap. Each wiper has a blade of length
Ans: Given, a car has two wipers with a blade length of
Let us suppose radius,
Angle,
The two wipers do not overlap.
Hence, area cleaned at each sweep is
12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle
Ans: Given, a lighthouse spreads a red light over a sector of angle
Let us suppose radius,
Angle,
13.Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are
Ans: Area of shaded region
Area of shaded region
Area of sector is given by,
Thus, Area of sector
Area of sector
Hence, area of the shaded region is
14. Find the area of the shaded region in figure, if ABCD is a square of side
Ans: From the given figure,
Area of shaded region = Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)
15. Find the area of the shaded region in figure, where a circular arc of radius
Ans: Area of the shaded region = Area of circle + Area of equilateral triangle
Area of a circle is given by,
Area of equilateral triangle is given by,
Hence, the area of the shaded region is
16. From each corner of a square of side
Ans: Given, a square of side
Also, a circle of diameter
Area of a square is given as
Area of a circle is given by,
Similarly, area of quadrant is given by
Thus,
Area of the remaining portion
Hence, area of the remaining portion is
17. In a circular table cover of radius
Ans: Given, circular table cover of radius,
Area of the design can be obtained by,
Area of design = Area of circle – Area of the equilateral triangle
Area of a circle is given by,
Area of equilateral triangle is given by,
Thus, area of design
So, we have
From the figure of equilateral triangle,
Substituting for
Area of design
Thus, area of the design is
18. In figure ABCD is a square of side
Ans: Given, a square of side
Thus, area of the shaded region = Area of the square – (
Area of a circle is given by,
19. The given figure depicts a racing track whose left and right ends are semi-circular. The difference between the two inner parallel line segments is
i. The distance around the track along its inner edge,
Ans: Given that, difference between the two inner parallel line segments is
Distance around the track along its inner edge
ii. The area of the track.
Ans: Area of track can be calculated as,
20. In 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 =
Ans: Given, radius of the bigger circle, OA or
Diameter of the smaller circle is the radius of the bigger circle.
Thus, Radius of the smaller circle,
Area of the circle is given by,
Area of the semicircle is given by,
Area of the shaded region can be obtained as:
Area of shaded region = Area of smaller circle + Area of semicircle
Area of shaded region
Thus, area of the shaded region is
21. The area of an equilateral triangle ABC is
Ans: Given, area of the equilateral triangle
Also, radius of each circle is half the length of the side of the equilateral triangle.
Area of the equilateral triangle is given by,
Let side of the equilateral triangle
Area of the shaded region can be obtained as:
Area of shaded region = Area of the equilateral triangle
We know, area of the sector of a circle
Thus,
Area of shaded region
Hence, area of the shaded region is
22. On a square handkerchief, nine circular designs each of radius
Ans: Given, radius of circular design is
Observing the figure, the side of the square is
The handkerchief consists of
Thus, area of the shaded region = Area of square
Area of shaded region
Hence, area of the shaded region is
23. In figure, OACB is a quadrant of a circle with centre O and radius
i. quadrant OACB
Ans: Given radius,
Area of the quadrant of a circle
Area of the quadrant
ii. shaded region
Ans: Given radius,
Area of the shaded region = Area of the quadrant
Area of the shaded region =
Hence, area of the shaded region is
24.In figure, a square OABC is inscribed in a quadrant OPBQ. If OA =
Ans: Given, a square
Taking the
Area of the shaded region = Area of quadrant
We know, area of the quadrant of a circle
Hence, area of the shaded region is
25. AB and CD respectively are the arcs of two concentric circles of radii
Ans: Given, two arcs
Also,
We know, area of the quadrant of a circle
Area of the shaded region = Area of arc
Hence, area of the shaded region is
26. In figure, ABC is a quadrant of a circle of radius
Ans: Given, a quadrant
Let us take,
We know, area of the quadrant of a circle
Area of the shaded region
Hence, area of the shaded region is
27.Calculate the area of the designed region in figure common between the two quadrants of circles of radius
Ans: Given, two quadrants each of radii
Let us consider the figure as:
Applying Pythagoras Theorem in the right- angled triangle
Draw a perpendicular,
Then
From the figure the right triangle AMB,
Area of
We know, area of the quadrant of a circle
Area of half shaded region = Area of quadrant
Thus, area of designed region can be obtained as, twice the area of half shaded region.
Hence, area of the designed region is
28. Find the circumference of a circle of diameter
Ans: Given, diameter of circle, d
Circumference of a circle is given by,
Since,
Thus, the circumference of the circle is,
The circumference of the circle is
29. The diameter of a circular pond is
Ans: Given, diameter of circular pond,
It is surrounded by a path of width
Equation for area of path
Hence, area of the path is
30. Find the area of the shaded region where ABCD is a square of side
Ans: Given, side of square,
Area of the shaded region can be calculated by subtracting area of
Area of shaded region = Area of square
Area of the given square
From the figure, we observe that the diameter of each circle is half the side of the square.
Thus, diameter of circle
Area of the circle
Hence, area of the shaded region is
31. The radius of a circle is
Ans: Given radius of the circle,
Area of a circle is given by,
Substituting for radius,
As the circle is divided into four parts of equal area, area of each circle would be,
Let
Thus the radius,
Radius of the largest of the three concentric circles is
32. OACB is a quadrant of a circle with centre O and radius
Ans: Given radius
Area of quadrant is given by,
From the figure it is clear that, angle
Area of
Area of shaded region = Area of quadrant
Thus, area of the shaded region is
33. A pendulum swings through on angle of
Ans: Given, pendulum swings at angle
Sketching the figure, we have:
In the below given figure let
AOB={{30}^{{}^\circ }}
Use
The length of the pendulum is
34. The cost of fencing a circular field at the rate of Rs.
Ans: As per the question the cost of fencing a circular field at the rate of Rs
With Rs.
Let
Circumference is given by,
Area of the field,
Rate = Rs.
Total cost of ploughing the field
Hence, the cost of ploughing the field is Rs.
35. Find the difference between the area of regular hexagonal plot each of whose side
Ans: Given, side of hexagonal plot
Area of equilateral triangle is given by
Area of hexagonal plot
Now,
Area of circular region,
Then the difference is,
36. In the given figure areas have been drawn of radius
Ans: Given, areas drawn with radius
Thus, required area is the area of the circle with radius
37. A wheel has diameter
Ans: Given, diameter of wheel,
Distance covered in one revolution = circumference
Circumference of the circle is given by
Distance covered in one revolution
Thus, for distance covered
Number of revolutions to complete
38. The given figure is a sector of a circle of radius
Ans: Given radius,
We know that circumference, i.e., perimeter of a sector of angle
Thus, the perimeter of a sector is
39. A car had two wipers which do not overlap. Each wiper has a blade of length
Ans: Given, radius of each wiper
Since, area of sector is given by,
Total area cleaned at each sweep of the blades
Thus, the total area cleaned at each sweep of the blades is
40. In the given figure arcs have been drawn with radii
Ans: Given, arc with radius
Area of sector is given by,
From the figure area of shaded region is,
Thus, area of shaded region is
41. The radii of two circles are
Ans: Given, radii of two circles as
Circumference of the circle is given by
C1 = circumference of the 1st circle
C2 = circumference of the 2nd circle
Circumference of the required circle
Thus, the radius of the required circle is
42. A car travels
Ans: Given, distance travelled by a wheel in 450 complete revolutions
Distance travelled in one revolution
Let
Circumference of the circular field is given by,
43. A sector is cut from a circle of diameter
Ans: Given, diameter,
Thus, radius can be obtained as,
Angle of sector
Area of the sector is given by,
Hence, area of the sector is,
44. In the given figure AOBCA represent a quadrant of area
Ans: Given, area of quadrant
From the figure,
Area of the quadrant of a circle
Area of the shaded region = Area of the quadrant
Area of the shaded region =
Hence, the area of the shaded region is
Short Answer Questions (3 Marks)
1. Find the area of the shaded region if
Ans: Given
Since QR is a diameter passing through the centre O of the circle
Angle of semi-circle,
Diameter of the circle
Thus radius,
Area of the semi-circle
Also, area of
From the given figure,
Area of the shaded region = Area of the semi-circle with O as centre and OQ as radius – area of
Hence, area of the shaded region can be calculated as,
Area of the shaded region
Area of the shaded region
2. The perimeter of a sector of a circle of radius
(i). The length of arc of the sector in cm.
Ans: Given radius,
Then
Now,
The length of the arc of the sector is
ii. The area of the sector in cm2 correct to the nearest
Ans: Area of sector OACB
Area of sector correct to nearest
3. Find the area of the shaded region in the given figure where ABCD is a square of side
Ans: Given, side of square ABCD
Area of square ABCD
Given that semicircle is drawn with side of square as diameter.
Diameter of semicircle
Radius of semicircle
Area of semi-circle
Area
Area of 4 semi-circles – Area of shaded region = Area of square ABCD
Area of shaded region
4. In the given figure
Ans: Given,
In
Area of shaded region = Area of sector OBPC – Area of
5. The radii of two circles are
Ans: Given, radii of two circles,
Area of a circle is given by
Let
Total area
Let R be the radius of the circle with area
Hence, required radius
6. A chord of a circle of radius
i. minor sector
Ans: Given, radius of circle,
Area of minor sector
Area of minor sector is
ii. major sector
Ans: Area of circle
Area of major sector = Area of circle – Area of minor sector
Area of major sector is
iii. minor segment
Ans: Area of minor segment = Area of minor sector OAB – Area of
Area of sector
Area of minor segment is
iv. major segment
Ans: Area of major segment = Area of the circle – Area of minor segment
Area of major segment is
7. In Akshita’s house, there is a flower pot. The sum of radii of circular top and bottom of a flowerpot is 140 cm and the difference of their circumference is
Ans: Given, sum of radii of circular top and bottom
Let radius of top
Circumference of top
Circumference of bottom
Difference of circumference
By the given conditions in the problem,
Diameter of top can be calculated as,
Radius of bottom can be obtained as,
8. A chord of a circle of radius
Ans: Given, radius of circle,
Area of a sector of a circle is given by,
Area of minor segment
Area of major segment = Area of the circle – area of minor segment
Area of minor segment is
9. A round table cover has six equal designs as shown in the figure. If the radius of the cover is
Ans: Given, radius of cover,
Area of equilateral triangle is given by,
From the figure area of one design
Total cost can be calculated as follows,
The total cost of making the design is
10. ABCD is a flower bed. If
Ans: Given,
Area of the flower bed (i.e., shaded portion)
Hence, area of the flower bed is
11. In a circle of radius
i. the length of the arc.
Ans: Given radius,
Length of arc
ii. area of the sector formed by the arc.
Ans: Area of the sector
iii. area of the segment formed by the corresponding chord
Ans: Area of segment formed by corresponding chord
In right triangles
Here
In right angled
Also
Using eq.
Area of segment formed by corresponding chord
12. A chord of a circle of radius
Ans: Given radius,
Area of minor sector can be obtained from
For, Area of
Draw
In right triangles
In right angled
Also,
And, Area of major segment
13. A chord of a circle of radius 12 cm subtends an angle of
Ans: Given radius,
Area of corresponding sector can be obtained by using the formula
For, Area of
Draw
In right triangles
In right angled
Also
Area of corresponding segment = Area of corresponding sector – Area of
14. Find the area of the shaded region in figure, if
Ans: Given radius,
Area of corresponding sector can be obtained by using the formula
For, Area of
Draw
In right triangles
In right angled
Also
Area of corresponding segment = Area of corresponding sector – Area of
Long Answer Questions (4 Marks)
1. Three horses are tethered with
Ans: Given, length of ropes at the three corners,
From the figure let,
Area which can be grazed by three horses
(
Sides of plot
Area grazed by the horses
2. In the given figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side
Ans: Given, side of square lawn
Area of the square lawn
Let
Again, area of sector
Substituting for
Area of
(Here,
Area of flower bed
Similarly, area of the other flowerbed
Total area of the lawn and the flowerbeds
3. An elastic belt is placed round the rein of a pulley of radius
Ans: Given, radius of rein of pulley,
Thus,
Length of the belt that is in contact with the rim of the pulley can be calculated by subtracting length of arc AB from Circumference of the rim.
Length of the belt
Length of the belt
Now, area of sector OAQB
Area of quadrilateral OAPB is given by,
Hence, shaded area can be obtained as,
Area of shaded region
Hence, area of shaded region is
Download Class 10 Maths Chapter 11 Important Questions PDF
Class 10 Maths Chapter 11 Important Questions Areas Related to Circles are given here based on the latest pattern of CBSE for 2024-2025. Students who are preparing for the board exams can practice Areas Related to Circles Important Questions to score full marks for the questions from this chapter. Along with the Important Questions for Class 10 Maths Chapter 11 related to Circles, we have also provided Extra Questions of Chapter 11 Class 10 Maths. Students can refer to the solutions whenever they get stuck while solving a problem. Also, in the end, some practice questions are provided for students to boost their preparation for the exam.
Areas Related to Circles chapter contains many formulas and concepts thus, it is important from the examination perspective, since most of the questions and objective-based questions will come in the exam from this chapter. Students can also refer to Chapter 11 Maths Class 10 Important Questions while preparing for their exam. NCERT Solution is also available at Vedantu to help the students in scoring full marks in the board exam.
Area related to circles includes the area of a circle, segment, sector, angle and length of a circle are provided in this pdf. In this chapter, we will discuss the concepts of the perimeter (also known as circumference) and area of a circle and apply this knowledge in finding the areas of two special parts of a circular region (or a circle) known as sector and segment. You can download Maths NCERT Solutions Class 10 and NCERT Solution for Class 10 Science to help you to revise the complete Syllabus and score more marks in your examinations.
Area of a Circle
The area is the region enclosed by the circumference of the circle. The area covered by one complete cycle of the radius of the circle on a two-dimensional plane is known as the area of that circle.
Area of the circle = πr2 square units.
Where r is the radius of the circle.
Area of Semicircle
To find the area of the semi-circle we need half of the area of the circle.
(Image will be uploaded soon)
The formula for the area of semicircle =
Where r is the radius of the circle.
Area of a Ring
(Image will be uploaded soon)
We know the area of circle πr2. Here the ring is having two radii inner and outer radius. Area of the ring i.e. the coloured part shown in the above figure is calculated by subtracting the area of the inner circle from the area of the bigger circle.
Area of the ring = πR2 - πr2 = π(R2 - r2)
Where R = radius of outer circle
r = radius of the inner circle
The Sector of a Circle
The sector of a circle is the region of a circle enclosed by an arc and two radii. The smaller area is called the minor sector and the larger area is called the major sector of the circle.
Areas of Sectors of a Circle
The area formed by an arc and the two radii joining the endpoints of the arc is known as a sector of a circle.
(Image will be uploaded soon)
In the above diagram, OACB is the minor sector. OADB is the major sector.
Minor Sector
The area including arc ∠AOB with point C is called Minor Sector. So, OACB is the minor sector. The area included in ∠AOB is the angle of the minor sector.
Area of the sector of angle
Major Sector
The area included in ∠AOB with point D is called the Major Sector. So OADB is the major sector. The angle of the major sector is 360° – ∠AOB.
If the degree measure of the angle at the center is 360, area of the sector = πr2.
Formula to find the area of Major Sector = πr2 - Area of the Minor Sector
Area of the circle can also be calculated by adding an area of a minor sector and area of a major sector.
Note: Area of Minor Sector + Area of Major Sector = Area of the Circle
To Calculate Length of an Arc of a Sector of Angle θ
An arc is the piece of the circumference of the circle so the length of an arc can be calculated as the θ part of the circumference.
Formula to calculate the length of an arc of a sector of angle
Areas of Segments of the Circle
The area made by an arc and a chord of a circle is called the segment of the Circle.
(Image will be uploaded soon)
Minor Segment
The minor segment is defined as the region bounded by the chord and the minor arc.
In the above diagram, AB is a chord of the circle. The area made by chord AB and arc X is known as the minor segment.
Formulas to calculate Area of Minor Segment = Area of Minor Sector – Area of ∆ABO
Major Segment
The major segment is defined as the region bounded by the chord and the major arc.
Formulas to calculate Area of Major Segment = πr2 - Area of Minor Segment
Note: Area of major segment + Area of minor segment = Area of the circle.
Areas of Combinations of Plane Figures
A plane figure is a geometric figure that has no thickness, it lies entirely on one plane. In this section, we will learn how to calculate the areas of some plane figures which are combinations of more than one plane figure.
To find the area of the composite area of the shapes, simply find the area of each shape and add them together. The order in which we calculate the areas does not matter and the commutative property states that it does not matter which order you add them in.
Following are the steps to find the area of combined figures:
Step I: First we divide the given combined figure into its simple geometrical shapes.
Step II: Then calculate the area of simple geometrical shapes separately.
Step III: Finally, to find the area of the combined figure we need to add or subtract these areas.
Below are some examples of combinations of plane figures:
(Image will be uploaded soon)
Above diagram is a combination of one rectangle and two semi-circles. So, to find an area of the figure, first, find an area of the rectangle and then add the area of two semi-circles.
Area of the given figure = Area of the rectangle + 2 x Area of semi-circles.
(Image will be uploaded soon)
Above diagram is the form of a ring.
Area of the required region =
(Image will be uploaded soon)
Above diagram is a combination of circle and square. To find the area of a given figure, first, calculate the area of the circle then subtract the area of the square.
Area of the required region = πr2 - a2
Important formula to remember:
Area of circle = πr2
Length of an arc of a sector of a circle with radius r and angle with degree measure θ is
.Area of a sector of a circle with radius r and angle with degrees measure θ is
Area of the segment of a circle = Area of the corresponding sector – Area of the corresponding triangle.
Conclusion:
Refer to this Important Questions For Class 10 Maths Chapter 11 for boards exam preparation. All the important concepts are available in the given PDF. The formula for the area of a circle also helps us calculate the area of circle sectors and segments as well. Download important questions of ch 11 Maths Class 10 from Vedantu site or app for more information.
Conclusion
Vedantu's offering of "Important Questions for CBSE Class 10 Maths Chapter 11 - Areas Related to Circles" is a valuable resource for students seeking to excel in their mathematics studies. The comprehensive collection of questions and solutions provides a deep understanding of the intricate concepts related to circles and their areas. By using this platform, students can gain confidence in tackling complex problems, enhance their problem-solving skills, and prepare effectively for their CBSE Class 10 examinations. Vedantu's commitment to quality education is evident in the well-structured and engaging content, making it an indispensable tool for students aiming to achieve academic success in mathematics.
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FAQs on Important Questions for CBSE Class 10 Maths Chapter 11 - Areas Related to Circles
1. How many sums are there in class 10 Maths Chapter 11 Areas Related to Circles?
Chapter 11 of Maths for class 10 teaches about the application of the chapter ‘Circles’. There are a total of 35 sums in all the exercises combined together from the chapter. All of these problems are unique in their own way and offer excellent practice and knowledge to students. At times, however, these can get very confusing. The Important questions by Vedantu for this chapter can help students here because of their simple language structure.
2. What students will study in chapter 11 Areas Related to Circles of class 10 Maths?
In this chapter, students learn the different applications of circles in terms of their areas. Concepts like the area of circles, their circumference, and the various parts of it, segment, sector, angle, and length, everything is discussed in this chapter. Important Questions for chapter 11 is specifically designed to focus on the important areas by providing students with questions so that they can practice with them and score good results for questions from this chapter by Vedantu.
3. What are the benefits of using important questions for chapter 10 of class 11 Maths?
Important Questions provided by Vedantu are good for students to learn ‘Areas Related to Circles’. They can access the questions free of cost by visiting the link-Important Questions for class 10 maths, and they can be assured of their quality content as each question is kept up to date and follows the latest curriculum. Moreover, the easily understandable structure provides them with a good practice experience for good marks. Since these questions are from the perspective of the Board exam paper, these are excellent sources for studying.
4. State the important formulae in chapter 11 of class 10 Maths.
Chapter 11 is named 'Areas Related to Circles' and this chapter is practically an application of the previous chapter in class 10 named 'Circles'. So, this chapter has some important formulae to solve various problems:
- Area of circle:
Area of a semicircle:
Area of quadrant:
Area of sector:
Area of Ring:
The perimeter of the sector of Circle: 2 Radius +
Length of arc = $\dfrac{2\pi r\theta}{360^\circ}
To know more students can download the vedantu app.
5. How to understand chapter 11 of class 10 Maths?
Circles are round figures without any sides or edges. Their applications are taught to students in chapter 11 of class 10 Maths where students learn how to solve different questions related to the areas of circles, the length and circumference, and a whole lot of other things. Since this chapter has a whole lot of things to learn, students can use Important Questions prepared for this chapter as it offers them an examination patterned structure, which is very beneficial.











