NCERT Solutions for Class 10 Maths

• 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 Applications of Trigonometry
• Chapter 10 Circle
• Chapter 11 Constructions
• Chapter 12 Areas related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability

There are 15 chapters in class 10 maths. These chapters lay a foundation for the chapters that will come in class 10. This pdf is accessible to everyone and they can use this pdf based on their convenience. Here below we are helping you with the overview of each and every chapter appearing in the textbook.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Real Numbers Class 10 has total of four exercises consists of 18 Problems. Prove Irrational, Problems based on Euclid’s division lemma, HCF and LCM and Divisibility are mostly asked topics in previous board exams. Other topics included are Fundamental Theorem of Arithmetic, important properties of positive integers, fraction to decimals and decimals to a fraction.

• Class 10 Maths Real Numbers Ex 1.1
• प्रश्नावली 1.1 का हल हिंदी में वास्तविक संख्याएँ
• Class 10 Maths Real Numbers Ex 1.2
• प्रश्नावली 1.2 का हल हिंदी में वास्तविक संख्याएँ
• Class 10 Maths Real Numbers Ex 1.3
• प्रश्नावली 1.3 का हल हिंदी में वास्तविक संख्याएँ
• Class 10 Maths Real Numbers Ex 1.4
• प्रश्नावली 1.4  का हल हिंदी में वास्तविक संख्याएँ
• Real Numbers Class 10 Extra Questions
• Real Numbers Class 10 Important Questions
• Real Numbers Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

Polynomials Class 10 has total of four exercises consists of 13 Questions. Problems related to finding polynomials, Properties zeros and coefficient, long division of polynomials, finding a quadratic polynomial, finding zeros of polynomials are scoring topics.

• Class 10 Maths Polynomials Ex 2.1
• प्रश्नावली 2.1 का हल हिंदी में बहुपद
• Class 10 Maths Polynomials Ex 2.2
• प्रश्नावली 2.2 का हल हिंदी में बहुपद
• Class 10 Maths Polynomials Ex 2.3
• प्रश्नावली 2.3 का हल हिंदी में बहुपद
• Class 10 Maths Polynomials Ex 2.4
• प्रश्नावली 2.4 का हल हिंदी में बहुपद
• Polynomials Class 10 Extra Questions
• Polynomials Class 10 Important Questions
• Polynomials Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Pair of Linear Equations Class 10 has total of seven exercises consists of 55 Problems. The problems will be based on concepts like linear equations in two variables, algebraic methods for solving linear equations, elimination method, cross-multiplication method Time and Work, Age, Boat Stream and equations reducible to a pair of linear equations these answers will give you ease in solving problems related to linear equations.

• Pair Of Linear Equations In Two Variables Class 10 Ex 3.1
• प्रश्नावली 3.1 का हल हिंदी में दो चरों में रैखिक समीकरणों का युग्म
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.2
• प्रश्नावली 3.2 का हल हिंदी में दो चरों में रैखिक समीकरणों का युग्म
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.3
• प्रश्नावली 3.3 का हल हिंदी में दो चरों में रैखिक समीकरणों का युग्म
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.4
• प्रश्नावली 3.4 का हल हिंदी में दो चरों में रैखिक समीकरणों का युग्म
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.5
• प्रश्नावली 3.5 का हल हिंदी में दो चरों में रैखिक समीकरणों का युग्म
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.6
• प्रश्नावली 3.6 का हल हिंदी में दो चरों में रैखिक समीकरणों का युग्म
• Pair Of Linear Equations In Two Variables Class 10 Ex 3.7
• प्रश्नावली 3.7 का हल हिंदी में दो चरों में रैखिक समीकरणों का युग्म
• Extra Questions for Class 10 Maths Linear Equations in Two Variables
• Pair of Linear Equations in Two Variables Class 10 Important Questions
• Pair of Linear equations in Two Variables Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Quadratic Equations Class 10  has total of four exercises consists of 24 Problems. The Questions are related to find roots of quadratic equations and convert world problem into quadratic equations are easily scoring topics in board exams.

• Class 10 Maths Quadratic Equations Ex 4.1
• प्रश्नावली 4.1 का हल हिंदी में द्विघातीय समीकरण
• Class 10 Maths Quadratic Equations Ex 4.2
• प्रश्नावली 4.2 का हल हिंदी में द्विघातीय समीकरण
• Class 10 Maths Quadratic Equations Ex 4.3
• प्रश्नावली 4.3 का हल हिंदी में द्विघातीय समीकरण
• Class 10 Maths Quadratic Equations Ex 4.4
• प्रश्नावली 4.4 का हल हिंदी में द्विघातीय समीकरण
• Quadratic Equations Class 10 Extra Questions
• Quadratic Equations Class 10 Important Questions
• Quadratic Equations Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Arithmetic Progressions Class 10 has total of four exercises consists of 49 Problems. find the nth terms and the sum of n consecutive terms are important topics in this chapter 5.

• Class 10 Maths Arithmetic Progressions Ex 5.1
• प्रश्नावली 5.1 का हल हिंदी में समान्तर श्रेणी
• Class 10 Maths Arithmetic Progressions Ex 5.2
• प्रश्नावली 5.2 का हल हिंदी में समान्तर श्रेणी
• Class 10 Maths Arithmetic Progressions Ex 5.3
• प्रश्नावली 5.3 का हल हिंदी में समान्तर श्रेणी
• Class 10 Maths Arithmetic Progressions Ex 5.4
• प्रश्नावली 5.4 का हल हिंदी में समान्तर श्रेणी
• Arithmetic Progressions Class 10 Extra Questions
• Arithmetic Progressions Class 10 Important Questions
• Arithmetic Progressions Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Triangles Class 10 has total of six exercises consists of 64 Problems. The Questions are based on properties of triangles and 9 important theorems which are important in scoring good marks in CBSE Class 10 Exams.

• Triangles Class 10 Maths Mind Map
• Triangles Class 10 Maths Ex 6.1
• प्रश्नावली 6.1 का हल हिंदी में त्रिभुज
• Triangles Class 10 Maths Ex 6.2
• प्रश्नावली 6.2 का हल हिंदी में त्रिभुज
• Triangles Class 10 Maths Ex 6.3
• प्रश्नावली 6.3 का हल हिंदी में त्रिभुज
• Triangles Class 10 Maths Ex 6.4
• प्रश्नावली 6.4 का हल हिंदी में त्रिभुज
• Triangles Class 10 Maths Ex 6.5
• प्रश्नावली 6.5 का हल हिंदी में त्रिभुज
• Triangles Class 10 Maths Ex 6.6
• प्रश्नावली 6.6 का हल हिंदी में त्रिभुज
• Extra Questions for Class 10 Maths Triangles
• Triangles Class 10 Notes Maths Chapter 6
• NCERT Exemplar Class 10 Maths Chapter 6 Triangles
• Important Questions for Class 10 Maths Chapter 6 Triangles
• Triangles Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Coordinate Geometry Class 10  has total of four exercises consists of 33 Problems. The Questions related to finding the distance between two points using their coordinates, Area of Triangle, Line divided in Ratio (Section Formula) are important models in class 10 boards.

• Class 10 Maths Coordinate Geometry Ex 7.1
• प्रश्नावली 7.1 का हल हिंदी में निर्देशांक ज्यामिति
• Class 10 Maths Coordinate Geometry Ex 7.2
• प्रश्नावली 7.2 का हल हिंदी में निर्देशांक ज्यामिति
• Class 10 Maths Coordinate Geometry Ex 7.3
• प्रश्नावली 7.3 का हल हिंदी में निर्देशांक ज्यामिति
• Class 10 Maths Coordinate Geometry Ex 7.4
• प्रश्नावली 7.4 का हल हिंदी में निर्देशांक ज्यामिति
• Coordinate Geometry Class 10 Extra Questions
• Coordinate Geometry Class 10 Important Questions
• Coordinate Geometry Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Introduction to Trigonometry Class 10  has total of four exercises consists of 27 Problems. The questions based on trigonometric ratios of specific angles, trigonometric identities and trigonometric ratios of complementary angles are the main topics you will learn in this chapter. Trigonometry Formulas plays important role in getting 100% marks in Board Exams.

• Class 10 Maths Introduction to Trigonometry Ex 8.1
• प्रश्नावली 8.1 का हल हिंदी में त्रिकोणमिति का परिचय
• Introduction to Trigonometry Class 10 Ex 8.2
• प्रश्नावली 8.2 का हल हिंदी में त्रिकोणमिति का परिचय
• Introduction to Trigonometry Class 10 Ex 8.3
• प्रश्नावली 8.3 का हल हिंदी में त्रिकोणमिति का परिचय
• Introduction to Trigonometry Class 10 Ex 8.4
• प्रश्नावली 8.4 का हल हिंदी में त्रिकोणमिति का परिचय
• Introduction to Trigonometry Class 10 Extra Questions
• Introduction to Trigonometry Class 10 Important Questions
• Introduction to Trigonometry Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Some Applications of Trigonometry Class 10  has one exercise consists of 16 Problems. In this chapter, you will be studying about real life applications of trigonometry and questions are based on the practical applications of trigonometry.

• Some Applications of Trigonometry Class 10 Ex 9.1
• प्रश्नावली 9.1 का हल हिंदी में त्रिकोणमिति के कुछ अनुप्रयोग
• Applications of Trigonometry Class 10 Extra Questions
• Some Applications of Trigonometry Class 10 Important Questions
• Some Applications of Trigonometry Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Circle Class 10  has total of two exercises consists of 17 Problems. Understand concepts such as tangent, secant, number tangents from a point to a circle and more.

• Circles Class 10 Ex 10.1
• प्रश्नावली 10.1 का हल हिंदी में वृत्त
• Circles Class 10 Ex 10.2
• प्रश्नावली 10.2 का हल हिंदी में वृत्त
• Circles Class 10 Extra Questions
• Circles Class 10 Important Questions
• Circles Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Constructions Class 10 has total of four exercises consists of 14 Problems. The Questions are based on drawing tangents and draw similar triangles are important topics.

• Constructions Class 10 Ex 11.1
• प्रश्नावली 11.1 का हल हिंदी में संरचना
• Constructions Class 10 Ex 11.2
• प्रश्नावली 11.2 का हल हिंदी में संरचना
• Extra Questions for Class 10 Maths Constructions
• Constructions Class 10 Important Questions
• Constructions Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles

Areas Related to Circles Class 10  has total of three exercises consists of  35 Problems. Solve problems related to ‘Perimeter and Area of a Circle’, ‘Areas of Combinations of Plane Figures’ and ‘Areas of Sector and Segment of a Circle’.

• Areas Related to Circles Class 10 Ex 12.1
• प्रश्नावली 12.1 का हल हिंदी में वृत्त से संबंधित क्षेत्र
• Areas Related to Circles Class 10 Ex 12.2
• प्रश्नावली 12.2 का हल हिंदी में वृत्त से संबंधित क्षेत्र
• Areas Related to Circles Class 10 Ex 12.3
• प्रश्नावली 12.3 का हल हिंदी में वृत्त से संबंधित क्षेत्र
• Areas Related to Circles Class 10 Extra Questions
• Areas Related to Circles Class 10 Important Questions
• Areas related to Circles Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

Surface Areas and Volumes Class 10  has total of five exercises consists of 36 Problems. In CBSE class 10 Maths, the ‘Surface Areas and Volumes’ chapter is a part of mensuration unit. The problems are based on finding areas and volumes of different solids such as cube, cuboid and cylinder, frustum, combination of solids.

• Surface Areas and Volumes Class 10 Ex 13.1
• प्रश्नावली 13.1 का हल हिंदी में सतह क्षेत्र और आयतन
• Surface Areas and Volumes Class 10 Ex 13.2
• प्रश्नावली 13.2 का हल हिंदी में सतह क्षेत्र और आयतन
• Surface Areas and Volumes Class 10 Ex 13.3
• प्रश्नावली 13.3 का हल हिंदी में सतह क्षेत्र और आयतन
• Surface Areas and Volumes Class 10 Ex 13.4
• प्रश्नावली 13.4 का हल हिंदी में सतह क्षेत्र और आयतन
• Surface Areas and Volumes Class 10 Ex 13.5
• प्रश्नावली 13.5 का हल हिंदी में सतह क्षेत्र और आयतन
• Surface Areas and Volumes Class 10 Extra Questions
• Surface Areas and Volumes Class 10 Important Questions
• Surface Areas and Volumes Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Statistics Class 10  has total of four exercises consists of 25 Problems. Problems related to find mean, mode or median of grouped data will be studied in this chapter. Solve questions by understanding the concept of cumulative frequency distribution.

• Statistics Class 10 Ex 14.1
• प्रश्नावली 14.1 का हल हिंदी में आंकड़े
• Statistics Class 10 Ex 14.2
• प्रश्नावली 14.2 का हल हिंदी में आंकड़े
• Statistics Class 10 Ex 14.3
• प्रश्नावली 14.3 का हल हिंदी में आंकड़े
• Statistics Class 10 Ex 14.4
• प्रश्नावली 14.4 का हल हिंदी में आंकड़े
• Statistics Class 10 Extra Questions
• Statistics Class 10 Important Questions
• Statistics Class 10 Notes

NCERT Solutions for Class 10 Maths Chapter 15 Probability

Probability Class 10 has total of two exercises consists of 30 Problems. Questions based on the concept of theoretical probability will be studied in this chapter.

• Probability Class 10 Ex 15.1
• प्रश्नावली 15.1 का हल हिंदी में प्रयकिता
• Probability Class 10 Ex 15.2
• प्रश्नावली 15.2 का हल हिंदी में प्रयकिता
• Probability Class 10 Extra Questions
• Probability Class 10 Important Questions
• Probability Class 10 Notes

NCERT Solutions of Class 10th Maths Book Chapter brief:

Students having trouble in solving tough Math problems can refer to these CBSE Maths Class 10 Solutions of NCERT for better guidance and for quick review. Solving these exercises in each chapter will assure positive results.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers – Term I

In Chapter 1 of Class 10, students will explore real numbers and irrational numbers. The chapter starts with the Euclid’s Division Lemma which states that “Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0≤r<b”. The Euclid’s Division algorithm is based on this lemma and is used to calculate the HCF of two positive integers. Then, the Fundamental Theorem of Arithmetic is defined which is used to find the LCM and HCF of two positive integers. After that, the concept of an irrational number, a rational number and decimal expansion of rational numbers are explained with the help of theorem.
 

Topics Covered in Class 10 Maths Chapter 1 Real Numbers for Term I:

Fundamental Theorem of Arithmetic – statements after reviewing work done earlier and after illustrating and motivating through examples. Decimal representation of rational numbers in terms of terminating/non-terminating recurring decimals.
Important Steps –
To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:
Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.
Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF. This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (c, d) denotes the HCF of c and d, etc.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials – Term I

In Polynomials, the chapter begins with the definition of degree of the polynomial, linear polynomial, quadratic polynomial and cubic polynomial. This chapter has a total of 4 exercises including an optional exercise. Exercise 2.1 includes the questions on finding the number of zeroes through a graph. It requires the understanding of Geometrical Meaning of the Zeroes of a Polynomial. Exercise 2.2 is based on the Relationship between Zeroes and Coefficients of a Polynomial where students have to find the zeros of a quadratic polynomial and in some of the questions they have to find the quadratic polynomial. In Exercise 2.3, the concept of division algorithm is defined and students will find the questions related to it. The optional exercise, 2.4 consists of the questions from all the concepts of Chapter 2.

Topics Covered in Class 10 Maths Chapter 2 Polynomials for Term I:

Zeroes of a polynomial. Relationship between zeroes and coefficients of quadratic polynomials only.
Important Steps –
We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees. Recall that arranging the terms in this order is called writing the polynomials in standard form.
Step 1: To obtain the first term of the quotient, divide the highest degree term of the dividend by the highest degree term of the divisor. Then carry out the division process.
Step 2: Now, to obtain the second term of the quotient, divide the highest degree term of the new dividend by the highest degree term of the divisor. Again, carry out the division process.
Step 3: Now, the degree of the remainder is less than the degree of the divisor. So, we cannot continue the division any further.
Here again, we see that Dividend = Divisor × Quotient + Remainder What we are applying here is an algorithm which is similar to Euclid’s division algorithm that you studied in Chapter 1.
This says that
If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).
This result is known as the Division Algorithm for polynomials.

NCERT Solutions of Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables – Term I

This chapter explains the concept of Pair of Linear Equations in Two Variables. This chapter has a total of 7 exercises, and in these exercises, different methods of solving the pair of linear equations are described. Exercise 3.1 describes how to represent a situation algebraically and graphically. Exercise 3.2 explains the methods of solving the pair of the linear equation through Graphical Method. Exercises 3.3, 3.4, 3.5 and 3.6 describe the Algebraic Method, Elimination Method, Cross-Multiplication Method, Substitution Method, respectively. Exercise 3.7 is an optional exercise which contains all types of questions. Students must practise these exercises to master the method of solving the linear equations.

Topics Covered in Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables for Term I:

Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically – by substitution and by elimination. Simple situational problems. Simple problems on equations reducible to linear equations.
Important Formulas –
The general form for a pair of linear equations in two variables x and y is
a₁ x + b₁ y + c₁ = 0
and a₂ x + b₂ y + c₂ = 0,
where a₁, b₁, c₁, a₂, b₂, c₂ are all real numbers and a₁² + b₁² ≠ 0, a₂² + b₂² ≠ 0.

NCERT Solutions of Class 10 Maths Chapter 4 Quadratic Equations – Term II

In this chapter, students will get to know the standard form of writing a Quadratic Equation. The chapter goes on to explain the method of solving the quadratic equation through the factorization method and completing the square method. The chapter ends with the topic on finding the nature of roots which states that, a quadratic equation ax² + bx + c = 0 has

1. Two distinct real roots, if b² – 4ac > 0
2. Two equal roots, if b² – 4ac = 0
3. No real roots, if b² – 4ac < 0

Topics Covered in Class 10 Maths Chapter 4 Quadratic Equations for Term II:

Standard form of a quadratic equation ax² + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by factorization, and by using quadratic formula. Relationship between discriminant and nature of roots. Situational problems based on quadratic equations related to day to day activities (problems on equations reducible to quadratic equations are excluded)
Important Formulas –
If b ² – 4ac > 0, we get two distinct real roots
`-\frac{b}{2a}+\frac{\sqrt{b^{2}-4ac}}{2a}\: and\: -\frac{b}{2a}-\frac{\sqrt{b^{2}-4ac}}{2a}`​​
If b ² – 4ac = 0, then
`x=-\frac{b}{2a}\pm0`
`x=-\frac{b}{2a} or-\frac{b}{2a}`​
So, the roots of the equation ax² + bx + c = 0 are both -b/2a.
Therefore, we say that the quadratic equation ax² + bx + c = 0 has two equal real roots in this case.
If b ² – 4ac < 0, then there is no real number whose square is b ² – 4ac. Therefore, there are no real roots for the given quadratic equation in this case.
Since b ² – 4ac determines whether the quadratic equation ax² + bx + c = 0 has real roots or not, b ² – 4ac is called the discriminant of this quadratic equation.
So, a quadratic equation ax² + bx + c = 0 has
(i) two distinct real roots, if b ² – 4ac > 0,
(ii) two equal real roots, if b ² – 4ac = 0,
(iii) no real roots, if b ² – 4ac < 0.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions – Term II

This chapter introduces students to a new topic that is Arithmetic Progression, i.e. AP. The chapter constitutes a total of 4 exercises. In Exercise 5.1, students will find the questions related to representing a situation in the form of AP, finding the first term and difference of an AP, finding out whether a series is AP or not. Exercise 5.2 includes the questions on finding out the nth term of an AP by using the following formula;
a_n = a + (n-1) d
The next exercise i.e., 5.3, contains the questions on finding the sum of first n terms of an AP. The last exercise includes higher-level questions based on AP to enhance students’ analytical and problem-solving skills.

Topics Covered in Class 10 Maths Chapter 5 Arithmetic Progressions for Term II:

Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems.
(Applications based on sum to n terms of an A.P. are excluded)
Important Formulas –
If a₁, a₂, a₃, a₄, a₅, a₆,_… are the terms of AP and d is the common difference between each term, then we can write the sequence as; a_, a+d, a+2d, a+3d, a+4d, a+5d,….,nth term… where a is the first term. Now, n^th term for arithmetic progression is given as;
n^th term = a + (n-1) d
Sum of the first n terms in Arithmetic Progression;
S_n = n/2 [2a + (n-1) d]

NCERT Solutions for Class 10 Maths Chapter 6 Triangles – Term I

In Chapter 6 of Class 10 CBSE Maths, students will study those figures which have the same shape but not necessarily the same size. The chapter Triangles starts with the concept of a similar and congruent figure. It further explains the condition for the similarity of two triangles and theorems related to the similarity of triangles. After that, the areas of similar triangles have been explained with a theorem. At the end of this chapter, the Pythagoras Theorem and converse of Pythagoras Theorem is described.

Topics Covered in Class 10 Maths Chapter 6 Triangles for Term I:

Definitions, examples, counter examples of similar triangles.
1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.
3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar.
4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar.
5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar.
6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other.
7. (Motivate) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.
9. (Motivate) In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angle opposite to the first side is a right angle.
Important Theorems –
Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Theorem 6.2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Theorem 6.3: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
Theorem 6.4: If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.
Theorem 6.5: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Theorem 6.6: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Theorem 6.7: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
Theorem 6.8: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Theorem 6.9: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry – Term I

In this chapter, students will learn how to find the distance between two points whose coordinates are given, and to find the area of the triangle formed by three given points. Along with this, students will also study how to find the coordinates of the point which divides a line segment joining two given points in a given ratio. For this purpose, students will get introduced to Distance Formula, Section Formula and Area of a Triangle in this chapter of Coordinate Geometry.

Topics Covered in Class 10 Maths Chapter 7 Coordinate Geometry for Term I:

LINES (In two-dimensions)
Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division)
Important Formulas –
Distance Formula
`PQ=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}`​
Section Formula
`m_{1}:m_{2}=(\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}})`
Area of Triangle = `\frac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry – Term I

This chapter will introduce students to Trigonometry. They will study some ratios of a right triangle with respect to its acute angles, called trigonometric ratios of the angles. The chapter also defines the trigonometric ratios for angles of 0⁰ and 90⁰. Further, students will also know how to calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities.

Topics Covered in Class 10 Maths Chapter 8 Introduction to Trigonometry for Term I:

Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined). Values of the trigonometric ratios of 30⁰, 45⁰ and 60⁰. Relationships between the ratios.
TRIGONOMETRIC IDENTITIES
Proof and applications of the identity sin²A + cos²A = 1. Only simple identities to be given
Important Formulas –
Trigonometry Maths formulas for Class 10 cover three major functions Sine, Cosine and Tangent for a right-angle triangle. Let a right-angled triangle ABC is right-angled at point B and have ∠θ.
`Sin\theta = \frac{  Side  opposite  to  angle  \theta }{Hypotenuse} = \frac{Perpendicular}{Hypotenuse} = \frac{P}{H}`
`Sin\theta = \frac{Side\opposite\to\angle\,\theta }{Hypotenuse} = \frac{Perpendicular}{Hypotenuse} = \frac{P}{H}`
Cos θ =
\frac{Adjacent\, side\, to\, angle\, \theta}{Hypotenuse}HypotenuseAdjacentsidetoangleθ​ =
\frac{Base}{Hypotenuse}HypotenuseBase​ = B/H
Tan θ =
\frac{Side\, opposite\, to\, angle\, \theta}{Adjacent\, side\, to\, angle\, \theta}AdjacentsidetoangleθSideoppositetoangleθ​ = P/B
`Sec \theta = \frac{1}{Cot \theta }`​
`Cot \theta = \frac{1}{Tan \theta }`​
`Cosec \theta = \frac{1}{Sin \theta }`​
`Tan \theta= \frac{Sin \theta }{Cos \theta }`​
Trigonometry Table

Trigonometric Ratios of Complementary Angles
sin (90° – A) = cos A,
cos (90° – A) = sin A,
tan (90° – A) = cot A,
cot (90° – A) = tan A,
sec (90° – A) = cosec A,
cosec (90° – A) = sec A
sin² A + cos² A = 1,
sec² A – tan² A = 1 for 0° ≤ A < 90°,
cosec² A = 1 + cot² A for 0° < A ≤ 90°

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry – Term II

This chapter is the continuation of the previous chapter as here the students will study the applications of trigonometry. It is used in geography, navigation, construction of maps, determining the position of an island in relation to the longitudes and latitudes. In this chapter, students will see how trigonometry is used for finding the heights and distances of various objects, without actually measuring them. They will get introduced to the term line of sight, angle of elevation, angle of depression.

Topics Covered in Class 10 Maths Chapter 9 Some Applications of Trigonometry for Term II:

HEIGHTS AND DISTANCES-Angle of elevation, Angle of Depression.
Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45°, 60°.
Important Points –
The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e., the case when we raise our head to look at the object.
The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e., the case when we lower our head to look at the point being viewed.
You would need to know the following:
(i) the distance DE at which the student is standing from the foot of the minar
(ii) the angle of elevation, ∠ BAC, of the top of the minar
(iii) the height AE of the student.
Assuming that the above three conditions are known, how can we determine the height of the minar?
In the figure, CD = CB + BD. Here, BD = AE, which is the height of the student.
To find BC, we will use trigonometric ratios of ∠ BAC or ∠ A.
In ∆ ABC, the side BC is the opposite side in relation to the known ∠ A. Our search narrows down to using either tan A or cot A, as these ratios involve AB and BC.
Therefore, tan A = BC/AB or cot A = AB/BC, which on solving would give us BC.
By adding AE to BC, you will get the height of the minar.

NCERT Solutions for Class 10 Maths Chapter 10 Circles – Term II

In earlier classes, students have studied about a circle and various terms related to a circle such as a chord, segment, arc, etc. In this chapter, students will study the different situations that arise when a circle and a line are given in a plane. So, they will get thorough with the concept of Tangent to a Circle and Number of Tangents from a Point on a Circle.

Topics Covered in Class 10 Maths Chapter 10 Circles for Term II:

Tangent to a circle at, point of contact
1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact.
2. (Prove) The lengths of tangents drawn from an external point to a circle are equal.
Important Theorems –
Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
Number of Tangents from a Point on a Circle
Case 1: There is no tangent to a circle passing through a point lying inside the circle.
Case 2: There is one and only one tangent to a circle passing through a point lying on the circle.
Case 3: There are exactly two tangents to a circle through a point lying outside the circle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions – Term II

This chapter consists of a total of 2 exercises. Whatever students have learned about construction in earlier classes will also help them. In Exercise 11.1, students will study how to divide a line segment, whereas in Exercise 11.2 they will study the construction of tangents to a circle. Methods and steps for construction are explained and also some examples are additionally given to make it clearer to the students.

Topics Covered in Class 10 Maths Chapter 11 Constructions for Term II:

1. Division of a line segment in a given ratio (internally).
2. Tangents to a circle from a point outside it.
Important Points –
Construction 11.1: To divide a line segment in a given ratio.
Construction 11.2: To construct a triangle similar to a given triangle as per given scale factor.
Construction 11.3: To construct the tangents to a circle from a point outside it.

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles – Term I

This chapter begins with the concepts of perimeter and area of a circle. Using this concept, the chapter further explains how to find the area of sector and segment of a circular region. Moreover, students will get clarity on finding the areas of some combinations of plane figures involving circles or their parts.

Topics Covered in Class 10 Maths Chapter 12 Areas Related to Circles for Term I:

Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60° and 90° only. Plane figures involving triangles, simple quadrilaterals and circle should be taken.)
Important Formulas –
circumference = 2πr
area of the circle = πr ²
Area of the sector of angle θ = (θ/360) × π r²
Length of an arc of a sector of angle θ = (θ/360) × 2 π r where r is the radius of the circle

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes – Term II

In Chapter 13, there are a total of 5 exercises. The first exercise consists of the questions based on finding the surface area of an object formed by combining any two of the basic solids, i.e cuboid, cone, cylinder, sphere and hemisphere. In Exercise, 13.2 questions are based on finding the volume of objects formed by combining any two of a cuboid, cone, cylinder, sphere and hemisphere. Exercise 13.3 deals with the questions in which a solid is converted from one shape to another. Exercise 13.4 is based on finding the volume, curved surface area and total surface area of a frustum of a cone. The last exercise is optional and has high-level questions based on all the topics of this chapter.

Topics Covered in Class 10 Maths Chapter 13 Surface Areas and Volumes for Term II:

1. Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinders/cones.
2. Problems involving converting one type of metallic solid into another and other mixed problems. (Problems with combination of not more than two different solids be taken).
Important Formulas –
TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere
Diameter of sphere = 2r
Surface area of sphere = 4 π r²
Volume of Sphere = 4/3 π r³
Curved surface area of Cylinder = 2 πrh
Area of two circular bases = 2 πr²
Total surface area of Cylinder = Circumference of Cylinder + Curved surface area of Cylinder = 2 πrh + 2 πr²
Volume of Cylinder = π r² h
Slant height of cone = l = √(r² + h²)
Curved surface area of cone = πrl
Total surface area of cone = πr (l + r)
Volume of cone = ⅓ π r² h
Perimeter of cuboid = 4(l + b +h)
Length of the longest diagonal of a cuboid = √(l² + b² + h²)
Total surface area of cuboid = 2(l×b + b×h + l×h)
Volume of Cuboid = l × b × h

NCERT Solutions for Class 10 Maths Chapter 14 Statistics – Term II

Here, students will learn the numerical representation of ungrouped data to grouped data and finding the Mean, Mode and Median. Also, the concept of cumulative frequency, cumulative frequency distribution and how to draw cumulative frequency curves will be explained.

Topics Covered in Class 10 Maths Chapter 14 Statistics for Term II:

Mean, median and mode of grouped data (bimodal situation to be avoided). Mean by Direct Method and Assumed Mean Method only.
Important Formulas –
The mean of the grouped data can be found by 3 methods.

1. Direct Method: x̅ =`\frac{\sum_{i=1}^{n}f_i x_i}{\sum_{i=1}^{n}f_i}`, where `\sum_{}^{}f_i x_i` is the sum of observations from value i = 1 to n And `\sum_{}^{}f_i` is the number of observations from value i = 1 to n
2. Assumed mean method : x̅ = 
   a+\frac{\sum_{i=1}^{n}f_i d_i}{\sum_{i=1}^{n}f_i}a+∑i=1n​fi​∑i=1n​fi​di​​
3. Step deviation method : x̅ = 
   a+\frac{\sum_{i=1}^{n}f_i u_i}{\sum_{i=1}^{n}f_i}\times ha+∑i=1n​fi​∑i=1n​fi​ui​​×h

The mode of grouped data:
Mode =
l+\frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times hl+2f1​–f0​–f2​f1​–f0​​×h
The median for a grouped data:
Median =
l+\frac{\frac{n}{2} – cf}{f} \times hl+f2n​–cf​×h

NCERT Solutions for Class 10 Maths Chapter 15 Probability – Term I

The last chapter deals with Probability. The chapter starts with the theoretical approach of probability. Subsequently, the chapter explains the difference between experimental probability and theoretical probability. There are various examples given to explain it in an effective way. So, before going through the exercise problems students must solve the examples of CBSE Maths first.

Topics Covered in Class 10 Maths Chapter 15 Probability for Term I:

Classical definition of probability. Simple problems on finding the probability of an event.
Important Formulas –

1. The theoretical probability (also called classical probability) of an event E, written as P(E), is defined as

P(E)= `\frac{Number\<space> of\, outcomes\\ favourable\: to\: E}{Number\: of\: all\: possible\: outcomes\: of\: the\: experiment}`
where we assume that the outcomes of the experiment are equally likely.

2. The probability of a sure event (or certain event) is 1.
3. The probability of an impossible event is 0.
4. The probability of an event E is a number P(E) such that 0 ≤ P (E) ≤ 1
5. An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.

1. as

P(E)= `\frac{Number  of  outcomes favourab le to E}{Number  of  all  possib le  outcomes  of  the  experiment}`