NCERT Solutions for Class 10 Maths Chapter 10 - Circles

NCERT Solutions for Class 10 Maths Chapter 10, "Circles", covers the basics of circles, including definitions of related terms and properties of circles. The chapter includes 16 exercise questions that require students to solve problems based on the length of the tangent, the number of common tangents, and proving theorems.

Exercise 10.1: This exercise deals with the basic concepts and properties of circles, including theorems related to tangents and chords.

Exercise 10.2: This exercise involves the application of these concepts in solving real-life problems related to circles, such as finding the angle of elevation or depression, the distance between two points on a circle, and more.

The solutions provided in both exercises aim to help students understand the concepts related to circles and apply them to solve problems.

Access NCERT Solutions for Class 10 Maths Chapter 10 – Circles

Exercise 10.1

1. How many tangents can a circle have?

Ans:

We know that a circle has an infinite number of points on its perimeter. Hence, there will be an infinite number of tangents on a circle.

2. Fill in the blanks: 

(i). A tangent to a circle intersects it in _________ point(s).

Ans:

A tangent to a circle intersects it in exactly one point(s).

(ii). A line intersecting a circle in two points is called a _______.

Ans:

A line intersecting a circle in two points is called a secant.

(iii). A circle can have _______ parallel tangents at the most.

Ans:

A circle can have two parallel tangents at the most.

(iv). The common point of a tangent to a circle and the circle is called ______.

Ans:

The common point of a tangent to a circle and the circle is called point of contact.

3. A tangent \[PQ\] at a point \[P\] of a circle of radius \[5\ \text{cm}\] meets a line through the centre \[O\] at a point \[Q\] so that \[OQ=12\ \text{cm}\]. 

Length of \[PQ\] is: (A) \[12\ \text{cm}\] (B) \[13\ \text{cm}\] (C) \[8.5\ \text{cm}\] (D) \[\sqrt{119}\].

Ans:

As, the tangent $PQ$ is base and radius is the height of $5\ \text{cm}$. Also, $OQ$ is the hypotenuse. Therefore, the length of $PQ$ will be –

$PQ=\sqrt{O{{Q}^{2}}-O{{P}^{2}}}$

\[\Rightarrow PQ=\sqrt{{{12}^{2}}-{{5}^{2}}}\]

\[\Rightarrow PQ=\sqrt{144-25}\]

\[\Rightarrow PQ=\sqrt{119}\]

Hence, option \[(D)\] is correct.

4. Draw a circle and two lines parallel to a given line such that one is tangent and the other, a secant to the circle.

Ans:

From the figure given above let us assume that a line \[l\] and a circle having centre \[O\], comprises a line \[PT\] which is parallel to the line \[l\] and is a tangent to the circle. Similarly, \[AB\] is a secant parallel to the line \[l\].

Exercise 10.2

1. From a point \[Q\], the length of the tangent to a circle is \[24\ \text{cm}\] and the distance of \[Q\] from the centre is \[25\ \text{cm}\]. The radius of the circle is 

(A) \[7\ \text{cm}\] (B) \[12\ \text{cm}\] (C) \[15\ \text{cm}\] (D) \[24.5\ \text{cm}\].

Ans:

From the figure given above, let us assume that \[O\] is the centre of the circle and given we have $PQ$ as the tangent to the circle of length $24\ \text{cm}$, and $OQ$ is of length $25\ \text{cm}$.

Now, by using the Pythagoras theorem we have –

$OP=\sqrt{O{{Q}^{2}}-P{{Q}^{2}}}$

$\Rightarrow OP=\sqrt{{{25}^{2}}-{{24}^{2}}}$

$\Rightarrow OP=\sqrt{625-576}$

$\Rightarrow OP=\sqrt{49}$

$\Rightarrow OP=7\ \text{cm}$.

Hence, option \[(A)\] is correct.

2. In the given figure, if \[TP\] and \[TQ\] are the two tangents to a circle with centre \[O\] so that \[\angle POQ=110{}^\circ \], then \[\angle PTQ\] is equal to 

(A) \[60{}^\circ \] (B) \[70{}^\circ \] (C) \[80{}^\circ \] (D) \[90{}^\circ \].

Ans:

As, from the given figure, we can observe that there are two tangents perpendicular to the radius of the circle as $TP$ and $TQ$. Since, they are perpendicular to the radius hence, we have $\angle OPT=90{}^\circ $ and 

$\angle OQT=90{}^\circ $.

Therefore, $\angle PTQ=360{}^\circ -\angle OPT-\angle OTQ-\angle POQ$

$\Rightarrow \angle PTQ=360{}^\circ -90{}^\circ -90{}^\circ -110{}^\circ $

$\Rightarrow \angle PTQ=360{}^\circ -290{}^\circ $

$\Rightarrow \angle PTQ=70{}^\circ $

Hence, option \[(B)\] is correct.

3. If tangents \[PA\] and \[PB\] from a point \[P\] to a circle with centre \[O\] are inclined to each other an angle of \[80{}^\circ \], then \[\angle POA\] is equal to 

(A) \[50{}^\circ \] (B) \[60{}^\circ \] (C) \[70{}^\circ \] (D) \[80{}^\circ \].

Ans:

As, we have two tangents $PA$ and $PB$ which will be perpendicular to the radius of the circle. Hence, we have $\angle PAO=90{}^\circ $ and 

$\angle PBO=90{}^\circ $. Since this is a quadrilateral, we will have the sum of all interior angles equal to $360{}^\circ $.

Therefore,

$\angle PAO+\angle PBO+\angle APB+\angle AOB=360{}^\circ $

$\Rightarrow 90{}^\circ +90{}^\circ +80{}^\circ +\angle AOB=360{}^\circ $

$\Rightarrow \angle AOB=360{}^\circ -260{}^\circ $

$\Rightarrow \angle AOB=100{}^\circ $.

Now, we know that $\angle POA$ is half of the $\angle AOB$.

Therefore,

$\angle POA=\frac{\angle AOB}{2}$

$\Rightarrow \angle POA=\frac{100{}^\circ }{2}$

$\Rightarrow \angle POA=50{}^\circ $.

Hence, option \[(A)\] is correct.

4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Ans:

As, from the figure given above, let us assume that \[O\] is the centre of the circle and \[AB\], \[CD\] are the two tangents at the ends of the diameter of the circle.

Now, we know that tangents are perpendicular to the radius of the circle.

Hence, \[\angle CQO=90{}^\circ \]

\[\angle DQO=90{}^\circ \]

\[\angle APO=90{}^\circ \]

\[\angle BPO=90{}^\circ \].

Therefore, we can say that \[\angle CPQ=\angle BQP\] because they are alternate angles. Similarly, \[\angle AQP=\angle QPD\].

Hence, if the interior alternate angles are equal then the lines \[AB\], \[CD\] should be parallel lines.

Hence, proved.

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Ans:

Let us assume that the line perpendicular at the point of contact to the tangent of a circle does not pass through the centre \[O\] but passes through a point \[Q\] as shown in the figure above.

Hence, we have the line \[PQ\bot AB\].

$\Rightarrow \angle QPB=90{}^\circ $

Also, we have $\angle OPB=90{}^\circ $.

After comparing both the equations we get –

$\Rightarrow \angle QPB=\angle OPB$

But, from the figure drawn above we can observe that this is not the case as $\angle QPB<\angle OPB$.

Therefore, we can conclude that $\angle QPB\ne \angle OPB$. They can only be equal when these two-line segments $QP$ and $OP$ will be equal. This implies that the line perpendicular at the point of contact to the tangent of a circle passes through the centre \[O\]. Hence, proved.

6. The length of a tangent from a point \[A\] at distance \[5\ \text{cm}\] from the centre of the circle is \[4\ \text{cm}\]. Find the radius of the circle.

Ans:

From the figure given above, let us assume that \[O\] is the centre of the circle and given we have $AB$ as the tangent to the circle of length $4\ \text{cm}$, and $OA$ is of length $5\ \text{cm}$.

Now, by using the Pythagoras theorem we have –

$OB=\sqrt{O{{A}^{2}}-A{{B}^{2}}}$

$\Rightarrow OB=\sqrt{{{5}^{2}}-{{4}^{2}}}$

$\Rightarrow OB=\sqrt{25-16}$

$\Rightarrow OB=\sqrt{9}$

$\Rightarrow OB=3\ \text{cm}$.

Therefore, the radius of the circle will be $3\ \text{cm}$.

7. Two concentric circles are of radii \[5\ \text{cm}\] and \[3\ \text{cm}\]. Find the length of the chord of the larger circle which touches the smaller circle.

Ans:

From the figure given above, we can observe that the line segment $PQ$ is the chord of the larger circle and a tangent to the smaller circle.

Therefore, \[OA\bot PQ\].

Now, we can observe that the \[\Delta OAP\] forms a right-angled triangle.

Hence, by applying Pythagoras theorem –

\[AP=\sqrt{O{{P}^{2}}-O{{A}^{2}}}\]

\[\Rightarrow AP=\sqrt{{{5}^{2}}-{{3}^{2}}}\]

\[\Rightarrow AP=\sqrt{25-9}\]

\[\Rightarrow AP=\sqrt{16}\]

\[\Rightarrow AP=4\ \text{cm}\].

Now, since the radius is perpendicular to the tangent therefore, we have \[AP=AQ\].

Hence, \[PQ=2AP\].

\[\Rightarrow PQ=2\times 4\]

\[\Rightarrow PQ=8\ \text{cm}\].

So, the length of the chord will be of \[8\ \text{cm}\].

8. A quadrilateral \[ABCD\] is drawn to circumscribe a circle. Prove that \[AB+CD=AD+BC\].

(Image will be uploaded soon)

Ans:

From the figure given we can observe that the sides of the quadrilateral acts as tangents to the circle. As, the tangents drawn from any external point will have the same length. Therefore, we have –

\[DR=DS\], 

\[CR=CQ\], 

\[BP=BQ\], and 

\[AP=AS\].

Now, we will add all the relations.

Hence, \[DR+CR+BP+AP=DS+CQ+BQ+AS\]

\[\Rightarrow \left( DR+CR \right)+\left( BP+AP \right)=\left( DS+AS \right)+\left( CQ+BQ \right)\]

\[\Rightarrow DC+AB=AD+BC\]

Hence, proved.

9. In the given figure, \[XY\] and \[X'Y'\] are two parallel tangents to a circle with centre \[O\] and another tangent \[AB\] with point of contact \[C\] intersecting \[XY\] at \[A\] and \[X'Y'\] at \[B\]. Prove that \[\angle AOB=90{}^\circ \].

Ans:

From the figure given we can observe that \[AB\], \[XY\] and \[X'Y'\] are tangents to the circle and will be perpendicular to its radius.

Now, let us consider two triangles \[\Delta OPA\] and \[\Delta OCA\], such that –

\[OP=OC\],

\[AP=AC\].

Hence, we have \[\Delta OPA\cong \Delta OCA\] by the SSS congruence rule.

\[\Rightarrow \angle POA=\angle COA\].

In the similar manner \[\Delta OQB\cong \Delta OCB\].

\[\Rightarrow \angle QOB=\angle COB\].

Now, we know that \[PQ\] is the diameter of the circle.

\[\Rightarrow \angle POA+\angle COA+\angle COB+\angle QOB=180{}^\circ \]

\[\Rightarrow 2\angle COA+2\angle COB=180{}^\circ \]

\[\Rightarrow \angle COA+\angle COB=90{}^\circ \]

\[\Rightarrow \angle AOB=90{}^\circ \].

Hence, proved.

10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Ans:

Let us assume a circle with centre at \[O\], which have two tangents \[PA\] and \[PB\] which are perpendicular to the radius of the circle as shown in the figure above.

Now, let us consider two triangles \[\Delta OAP\] and \[\Delta OBP\], such that –

\[PA=PB\], and 

\[OA=OB\].

Hence, we have \[\Delta OAP\cong \Delta OBP\] by the SSS congruence criteria.

Therefore, \[\angle OPA=\angle OPB\] and

\[\angle AOP=\angle BOP\].

This implies that \[\angle APB=2\angle OPA\] and

\[\angle AOB=2\angle AOP\].

Hence, in the right-angled triangle \[\Delta OAP\], we have –

\[\angle AOP+\angle OPA=90{}^\circ \]

\[2\angle AOP+2\angle OPA=180{}^\circ \]

\[\Rightarrow \angle AOB=180{}^\circ -\angle APB\]

\[\Rightarrow \angle AOP+\angle OPA=180{}^\circ \].

Hence, proved.

11. Prove that the parallelogram circumscribing a circle is a rhombus.

Ans:

(Image will be uploaded soon)

From the figure given above, we can observe that \[ABCD\] is a parallelogram whose sides are tangent to the circle. This implies that \[DR=DS\], 

\[CR=CQ\], 

\[BP=BQ\], and 

\[AP=AS\].

Now, we will add all the relations.

Hence, \[DR+CR+BP+AP=DS+CQ+BQ+AS\]

\[\Rightarrow\left( DR+CR \right)+\left( BP+AP \right)=\left( DS+AS \right)+\left( CQ+BQ \right)\]

\[\Rightarrow DC+AB=AD+BC\]

As, the sides of a parallelogram are always parallel and equal in length.

Therefore, \[AB=DC\] and

\[AD=BC\].

\[\Rightarrow 2AB=2BC\]

\[\Rightarrow AB=BC\]

Hence, all the sides of parallelogram are equal.

Therefore, we can conclude that it is a rhombus. Hence, proved.

12. A triangle \[ABC\] is drawn to circumscribe a circle of radius \[4\ \text{cm}\] such that the Segments \[BD\] and \[DC\] into which \[BC\] is divided by the point of contact \[D\] are of lengths \[8\ \text{cm}\] and \[6\ \text{cm}\] respectively. Find the sides \[AB\] and \[AC\].

Ans:

From the figure given above, we can observe that the sides of a triangle \[ABC\] are the tangents to the circle and are perpendicular to the radius of the circle.

Hence, in \[\Delta ABC\]

\[CF=CD=6\ \text{cm}\]

Similarly, \[BE=BD=8\ \text{cm}\] and

\[AE=AF=x\ \text{cm}\].

In \[\Delta ABE\],

\[AB=AE+BE\]

\[\Rightarrow AB=x+8\]

Similarly, \[BC=8+6=14\] and

\[CA=6+x\].

Therefore, we have \[2s=AB+BC+CA\]

\[\Rightarrow 2s=x+8+14+6+x\]

\[\Rightarrow 2s=2x+28\]

\[\Rightarrow s=x+14\].

Now, we know that the area of a triangle can be calculated by using the formula \[Area=\sqrt{s(s-a)(s-b)(s-c)}\].

Therefore,

\[Area\ \text{of}\ \Delta \text{ABC}=\sqrt{(14+x)(14+x-6-x)(14+x-8-x)}\]

\[\Rightarrow \sqrt{x(14+x)(8)(6)}\]

\[\Rightarrow 4\sqrt{3(14x+{{x}^{2}})}\].

Hence, \[Area\ \text{of}\ \Delta \text{OBC = }\frac{1}{2}\times OD\times BC\]

\[Area\ \text{of}\ \Delta \text{OBC =}28\]

\[\Rightarrow Area\ \text{of}\ \Delta \text{OCA = }\frac{1}{2}\times OF\times AC\]

\[Area\ \text{of}\ \Delta \text{OCA = 12+2x}\]

\[\Rightarrow Area\ \text{of}\ \Delta \text{OAB = }\frac{1}{2}\times OE\times AB\]

\[Area\ \text{of}\ \Delta \text{OAB = 16+2x}\]

Therefore, the total area of the triangle \[ABC\] will be –

\[=Area\ \Delta \text{OBC}+Area\ \Delta \text{OCA+}Area\ \Delta \text{OAB}\]

\[\Rightarrow 4\sqrt{3(14x+{{x}^{2}})}=28+12+2x+16x+2x\]

\[\Rightarrow \sqrt{3(14x+{{x}^{2}})}=14+x\]

\[\Rightarrow 3(14x+{{x}^{2}})={{(14+x)}^{2}}\]

Hence, after further solving we have –

\[(x+14)(x-7)=0\]

\[\Rightarrow x=-14\] or

\[\Rightarrow x=7\]

As, the side cannot be negative in nature, therefore, \[x=7\ \text{cm}\].

Hence, \[AB=7+8\]

\[\Rightarrow AB=15\ \text{cm}\] and

\[CA=6+7\]

\[\Rightarrow CA=13\ \text{cm}\].

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend Supplementary angles at the centre of the circle.

Ans:

(Image will be uploaded soon)

From the figure given we can observe that the sides of the quadrilateral acts as tangents to the circle. Therefore, we have –

\[\angle AOB+\angle COD=180{}^\circ \] and

\[\angle BOC+\angle DOA=180{}^\circ \]

Now, let us consider two triangles \[\Delta OAP\] and \[\Delta OAS\],

Hence, we have \[AP=AS\], and

\[OP=OS\]

\[\Rightarrow \Delta OAP\cong \Delta OAS\] by the SSS congruence criteria.

Thus, \[\angle POA=\angle AOS\]

\[\angle 1=\angle 8\]

Also,

\[\angle 2=\angle 3\],

\[\angle 4=\angle 5\],

\[\angle 6=\angle 7\]

Therefore, \[\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360{}^\circ \]

\[\Rightarrow \left( \angle 1+\angle 8 \right)+\left( \angle 2+\angle 3 \right)+\left( \angle 4+\angle 5 \right)+\left( \angle 6+\angle 7 \right)=360{}^\circ \]

\[\Rightarrow 2\angle 1+2\angle 2+2\angle 5+2\angle 6=360{}^\circ \]

\[\Rightarrow \left( \angle 1+\angle 2 \right)+\left( \angle 5+\angle 6 \right)=180{}^\circ \]

\[\Rightarrow \angle AOB+\angle COD=180{}^\circ \]

In the similar manner we can prove that \[\angle BOC+\angle DOA=180{}^\circ \].

Therefore, we have proved that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Overview of the Exercises Covered in Class 10 Maths Ch-10 Circles NCERT Solutions

The NCERT Solutions for Class 10 Maths Chapter 10 Circles comprise two exercises, Ex-10.1 and Ex-10.2. Solving the sums given in these two exercises is going to provide students with a deeper understanding of the concepts of circles, various theorems, corollaries, and applications of the formulas of Circles. 

Exercise 10.1: The first exercise of Class 10 Maths Chapter 10 Circles is a short one, consisting of only 4 sums. These sums will help students to assess their understanding of the properties of circles, including the concepts of tangents, secants, radii, diameters, chords, etc. There are fill-in-the-blank and short answer-type questions covered in Ex-10.1. Students will have to apply the Pythagoras theorem in solving one of the sums of this exercise, while the last sum is based on the construction of circles. 

Exercise 10.2: The first exercise consists of a total of 13 sums. Students will be able to get a good understanding of various concepts of angles, triangles, quadrilaterals, circles inscribed in quadrilaterals, circles circumscribing polygons, Pythagoras theorem, etc., when they solve the sums given in this exercise. Students will need to calculate the lengths and angles subtended by chords, tangents, etc., for some of the sums. The remaining sums of this exercise are mostly based on the deduction of equations using the application of various formulas and theorems of circles. The solutions to the sums given in NCERT Class 10 Maths Exercise 10.2 are very important from the exam point of view.

Overview of Class 10 Maths Chapter 10 Circles NCERT Solutions

Students of Class 10 have to prepare for their board exams and Mathematics is an important subject in their curriculum. One of the most significant chapters in the Class 10 Maths syllabus is known as Circles. It is an important chapter because it teaches students about basic geometry and much more. Students are able to learn some basic concepts related to circles such as the components of the circle, calculating the area of the circle, calculating the perimeter of the circle and much more.

They are further introduced to the different components of the circle. Tangents, the number of tangents from a particular point in the circle, and tangents to a circle are some of the other topics that are included in the chapter. Circles is definitely one of the most important chapters for Class 10 students. To give them a clear view of the chapter, Vedantu experts have come up with NCERT Solutions. Students can download the solutions and practice all the textbook problems along with other important questions that are commonly asked during the Class 10 board examination.

Circles is a chapter that has significant weightage in the Class 10 board exams. Thus, getting proper practice for the chapter is very important for students to complete their syllabus. Keeping that in mind, the experts at Vedantu have provided some help in the form of NCERT Solutions. These solutions are excellent study materials and will help students complete the chapter on time. Board exams are a stressful time for students and with these solutions in hand, they can have a boost o confidence and can score good marks.

NCERT Solutions for Class 10 Maths Chapter 10 Circles - PDF Download

Get Complete Details about Chapter 10 Maths Class 10 NCERT Solutions

You can opt for Chapter 10 - Circles NCERT Solutions for Class 10 Maths PDF for upcoming Exams  and also You can Find the Solutions of All the Maths Chapters below.

Importance of Circles in Mathematics

To put it in simple words, a circle is basically a round two-dimensional figure that consists of infinitely many points on its circumference that are all equidistant from the centre of the circle.

Circles are important in mathematics since they form a huge part of the geometry syllabus. There are several theorems related to the study of circles and their properties. This chapter on circles is an elaborate one and it is highly recommended that students go through all the topics covered in it and the NCERT solutions we have provided to get a good grasp on this concept. This will enable them to solve sums that are based on circles, and practising these questions will help them to score well in exams. 

NCERT Solutions for Class 10 Maths Chapter 10 Exercises

The expert teachers of Vedantu have solved each question as per NCERT Book guidelines in the pdf. If you practice it regularly, then you will better understand all the concepts of the chapter. When you solve more than 50-100 questions related to circles given in Chapter 10 Class 10 Maths NCERT Solutions, then you will become a perfectionist. Still, if you find yourself in any problem, our expert teachers will always help you. Along with this, you can take help from our online videos, lectures, topic lessons, other learning materials, etc.

A circle is a chapter in which students learned about the existence of tangents to a circle and their properties. In this chapter, all concepts are explained in detail, such as Tangent, Number of a tangent from a point on a circle, Tangents to a circle, and others. No doubt, Circle is a very important topic in board exams as well as competitions. Those students who have a goal to crack IIT, JEE, or any medical exam after their 12th, then they should not ignore this chapter at any condition. Many questions are asked from this topic.

With the help of NCERT Solutions for Class 10 Maths Chapter 10 Circles, you will understand the basic concepts. It includes lots of diagrams, figures, and lucid language, which is not easy to understand for a student without any guidance. Along with this, all questions are based on theorems, which are introduced in Maths NCERT Solutions Class 10 chapter 10. Students should know about them and understand the basic concept in detail before going to solve a numerical problem. These theorems are fundamental from an exam point of view.

It has been noticed that one or two questions based on these theorems are always asked in the board exam. That’s why our expert teacher has focused on the concept of theorems so that students haven’t faced any problem in the examination hall while solving questions. Apart from this, many activities are also given to practice in NCERT Solutions for Class 10 Maths Circles to build a foundation of the students. This way, students will acquire the concept of Geometry step by step.

Key Features for NCERT Solutions for Class 10 Maths Chapter 10 Circles

The key features of Vedantu’s Class 10 Maths NCERT Solutions for Chapter 10 Circles are listed below.

• Solved by highly experienced teachers 

• Easy to understand

• In compliance with the latest CBSE guidelines for Class 10 Term 2

• Detailed step-wise explanations for each sum given 

• Available in free downloadable PDF

• Best guide for self-study

Benefits of Class 10 NCERT Solutions Maths Chapter 10

Why Should Students Choose Vedantu?

As you all know, how vital is Class 10th Maths Chapter 10 (circles) for the board exam. There is no single-board exam until a date in which a question has not been asked from the circle chapter. The expert teachers of Vedantu have great experience in their respective subjects and use it while designing NCERT Maths Circle Solutions.

They include every question in NCERT solutions, which are asked in previous year's board examinations. Different easy methods have solved all problems by using easy concepts. Students may download these solutions in PDF form, which can be used offline too. Also, you can download the Vedantu app, where you can analyze your preparation by solving online test series.

NCERT Class 10 Maths Chapter 10 Circles is an easy and scoring chapter. Students will get a good grasp of the concepts and extensive applications of the properties and theorems of circles by solving the sums given in this exercise. One of the notable theorems that students must have prior knowledge of to solve the sums of this chapter is the Pythagoras theorem. They can download and refer to the NCERT Solutions for Class 10 Chapter 10 Circles for free and prepare for CBSE Term 2 Maths examination.

NCERT Solutions for Class 10 Maths - Other Chapters

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Download Class 10 Maths Chapter 10 NCERT Solutions to Learn

Get Vedantu’s NCERT Solutions for Class 10 Maths Chapter 10 Circles and learn all the concepts of the chapter. Solve the questions on your own and compare the answers to the solutions provided by experts. This way, you will be able to identify the areas for improvement and rectify your own errors. This will help you score good marks in the exams.