CBSE Class 10

On this page, you will find Triangles Class 10 Notes Maths Chapter 6 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 6 Triangles will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 6 Notes Triangles

Triangles Class 10 Notes Understanding the Lesson

In X standard we have learnt about congruent figures.

Congruent figure: Those two geometric figures having the same shape and size are known as congruent figures.

Rules of Congruency

1. SAS (Side-Angle-Side): Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

2. ASA (Angle-Side-Angle): Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.

3. AAS (Angle-Angle-Side): Two triangles are Congruent if any two pairs of angles and a pair of corresponding sides are equal.

4. SSS (Side-Side-Side): If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

5. RHS (Right angle-Hypotenuse-Side: In two right angle triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Note: All congruent figures or triangles are similar.

Similar Figure: Two figure which are of same shape (but not necessarily the same size) are called similar figures. For example,

• All line segments are similar.
• All circles are similar.
• Two or more squares are similar.
• Two or more equilateral triangles are similar.

Note:

• All rectangles are not similar.
• All triangles are not similar.

Similar Polygons: Two polygons with the same number of sides are similar, if (1) their corresponding angles are equal. (2) their corresponding sides in the same ratio.

Similarity of Triangles

Two triangles are similar if

• Their corresponding angles are equal; and
• Their corresponding sides are in the same ratio

Famous Greek mathematician Militus Thales gives the relation to the two equiangular triangle is known as BPT or Thales theorem.

Equiangular triangles: If corresponding angles of two triangles are equal then they are equiangular triangles

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.

Given: A AABC in which a line DE || BC intersects the other two sides AB and AC at D and E respectively.
To Prove that \(\frac{A D}{D B}=\frac{A E}{E C}\)
Construction: Join BE and CD and draw DM ⊥ AC and EN ⊥ AB

(Because both are on the same base DE and between the same parallels BC and DE) from eqn (1), (2) and (3) AD AE

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.

AD AE Given: In ΔABC, \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
To prove: DE || BC

Construction: Let us suppose that DE is not parallel to BC, so we draw a line DF || BC
Proof: DF || BC
Therefore by Basic Proportionality Theorem,

which is not possible. We come at the contradiction. So our supposition was wrong, it is only possible, if point F will coincide the point E.
Therefore DE || BC.

Criteria for Similarity of Triangles

In previous section, we have studied that two triangles are similar, if (I) their corresponding angles are similar (II) their corresponding sides are proportional (or are in the same ratio).

Theorem 6.3: AAA Criterion: 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.

In ΔABC and ΔDEF,

Remark : AA Similarity Creterian: If two angles of a triangle are equal to two angles of another triangle, then their corresponding angles are equal and the triangles are similar.

In ΔABC and ΔDEF
∠A =∠D, ∠C = ∠F then ΔABC ~ ΔDEF (by AA similarity)

Theorem 6.4: SSS Similarity Criterion: If the corresponding sides of two triangles are proportional (i.e. in the same ratio), then their corresponding angles are equal and the triangles are similar.

In ΔABC and ΔDEF
\(\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F} DF \)then ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
Hence ΔABC ~ ΔDEF

Theorem 6.5: SAS Similarity Criterion: If one angle of a triangle is equal to one angle of the other and the sides including these angles are proportional, the triangles are similar.

In ΔABC and ΔDEF
∠BAC = ∠EDF
\(\frac{A B}{D E}=\frac{A C}{D F}\)
Hence ΔABC ~ ΔDEF

Areas of Similar Triangles
We have study in two similar triangles. Ratio of the corresponding sides of two similar triangles is same.

Theorem 6.6: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: ΔABC ~ ΔPQR

Theorem 6.7: If a perpendicular is drawn from the verities 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.

• ΔADB ~ ΔABC
• ΔBDC ~ ΔABC
• ΔADB ~ ΔBDC

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.

Given: ABC is a right angle triangle which is right angled at B.
To prove:  AC² = AB² + BC²
Construction: Draw BD ⊥  AC
Proof:  ΔADB ~ ΔABC
(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)

From equation (2) and (3)
Adding eqn (1) and (2)
(If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse than triangles on both sides of the perpendicular are similar to the whole triangle and to each other)
AB² + BC² = AD . AC + CD . AC
⇒ AB² + BC² = AC (AD + CD)
⇒ AB² + BC² = AC x AC
⇒ AB² + BC² = AC²
⇒ AC² = AB² + BC²

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.

Given: We have ΔABC in which
AC² = AB² + BC²
To prove: ∠ABC = 90°
Construction: Construct a ΔDEF right angled at E such that EF = BC and DE = AB

Proof: In ΔDEF
DF² = EF² + DE² (Pythagoras theorem) (given)
DF² = BC² + AB²…(1) (by construction)
But AC² = BC² + AB²…(2)

From eqn (1) and (2)
AC² = DF²
⇒ AC = DF… (3)

In ΔABC and ΔDEF
AB = DE(by construction)
BC = EF(by construction)
AC = DF(proved above in eqn (3))
ΔABC ≅ ΔDEF(by SSS congruence)
⇒ ∠ABC = ∠DEF (by CPCT)
Therefore ∠ABC = 90°

On this page, you will find Arithmetic Progressions Class 10 Notes Maths Chapter 5 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 5 Arithmetic Progressions will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 5 Notes Arithmetic Progressions

Arithmetic Progressions Class 10 Notes Understanding the Lesson

We have observed many things in our daily life, follow a certain pattern.
(a) 1, 4, 7, 10, 13, 16, …….
(b) 15, 10, 5, 0, -5, -10,………….
(c) 1,\(\frac{1}{2}\),0,\(-\frac{1}{2}\)………………
These patterns are generally known as sequence. Two such sequences are arithmetic and geometric sequences. Let us investigate the Arithmetic sequence.

1. Sequence: A sequence is a ordered list of numbers.
Terms: The various numbers occurring in a sequence are called its terms. Terms of sequence are denoted by a₁ a₂, a₃, …………… a_n.

2. Arithmetic Progression: An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms are equal.

3. Common difference: The difference between two consecutive terms of an arithmetic progression is called common difference.
d = a₂ – a₁
d  = a₃ – a₂
d = a₄ – a₃
……………..
……………..
d = a_n – a_n-1

4. Finite Arithmetic Progression: A sequence which has finite or definite number of terms is called finite sequence.
Example, (1, 3, 5, 7, 9)… which has 5 terms.

5. Infinite Arithmetic Progression: A sequence which has indefinite or infinite number of terms is called infinite arithmetic progression.
Example, 1, 2, 3, 4, 5, …

In general, arithmetic progression can be written as a, a + d, a + 2d, where a is the first term and d is called the common difference i.e. difference between two consecutive terms.

6. General form of an AP: Let a be the first term and d is the common difference then the AP is

Here
a₁ = a (we take) (a is first term of AP)
a₂ = a₁ + d = a + d
a₃ = a₂ + d = a + d + d = a + 2d
a₄ = a₃+ d = a + 2d + d = a + 3d
…………….
……………
a_n = a + (n – 1) d
i.e. AP is a, a + d, a + 2d, a + 3d,………… , a + (n – 1)d.
n^th term of AP = a + (n -1)d
Note: Common difference of AP can be positive, negative or zero.

1. n^th term or General term of an AP
n^th term of an AP = a + (n – 1) d where
a → first term of the AP
n → number of terms
d→common difference of an AP.

2. n^th term of an AP from the end: Let us consider an AP where first term a and common difference is If m is number of terms in the AP. then
n^th term from the end = [m – n + 1]^th term from the beginning.
n^th term from the end = a + (m-n +1 – 1)d – a + (m – n) d
It  l is the last term of the AP, then n^th term from the end is the n^th term of an AP where first term is l and common difference is – d
n^th term from the end – 1 + (n – 1) (-d)
= 1 – (n – 1) d

Sum of first n terms of an AP
Let S_n denote the sum of first n terms of an AP
S_n = a + a + d + a + 2d + a + 3d …. + a + (n – 1)d ……….. (1)
Rewriting the terms in reverse order.
S_n = a + (n – 1)    + a + (n – 2)d + a + (n-3)d + ………….+a ……… (2)
Adding equations (1) and (2)
2S_n = [2a + (n – 1)d] + [2a + (n-1)d] + … + [2a + (n – 1)d]
2S_n = n[2a + (n – 1)d]
S_n=\(\frac{n}{2}\)[2a+(n-1)d]
We can Write
S_n=\(\frac{n}{2}\)[a+a+(n-1)d] [l=a+(n-1)d]
S_n=\(\frac{n}{2}\)[a+l]

Note:
(i) The Tith term of an AP = S_n – S_n-1 or a_n = S_n+1 – S_n
Sum of first n positive integer
\(S_{n}=\frac{n(n+1)}{2}\)
(iii) Sum of n odd positive integer = n²
(iv) Sum of n even positive integer = n(n + 1)

On this page, you will find Quadratic Equations Class 10 Notes Maths Chapter 4 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 4 Quadratic Equations will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 4 Notes Quadratic Equations

Quadratic Equations Class 10 Notes Understanding the Lesson

1. Quadratic Equation: A quadratic equation in the variable x is of the form ax² + bx + c = 0, where a, b, c are real number and a ≠ 0.

2. Roots (or zeroes of a quadratic equation): A real number a is called the root of the quadratic equation
ax² + bx + c = 0 if aα² + bα + c = 0.

Alternatively, any equation of the form p(x) = 0, where p(x) is a quadratic polynomial is a quadratic equation and if p(α) = 0 for any real number a; the a is said to be the root (or zero) of p(x).

Solution of a quadratic equation by factorization
Finding the roots of a quadratic equation by the method of factorization means finding out the linear factors of the quadratic equation and equating it to zero, the roots can be found. i.e. ax² + bx + c = 0
(Ax + B) (Cr + D) = 0
where A, B, C and D are real numbers, A, C≠ 0.
We get Ax + B = 0 or Cx + D = 0
x =\(-\frac{B}{A}\) or x =\(-\frac{D}{C}\)
x =\(-\frac{\mathrm{B}}{\mathrm{A}},-\frac{\mathrm{D}}{\mathrm{C}}\) are the two roots of quadratic equation.

Solution of a quadratic equation by completing the square
For given quadratic equation ax²+ bx + c = 0
Divide the equation by a, so that the coefficient of x² becomes 1.
\(x^{2}+\frac{b}{a} x+\frac{c}{a}=0\)

Adding and subtracting \(\left(\frac{b}{2 a}\right)^{2}\) i.e., square of the half of the coefficient of x.
This formula is known as quadratic formula.
If α and β are roots of the given equation, then

ax² + bx + c = 0,
a ≠ 0, a, b, c ∈ R

Discriminant D = b² – 4ac

Condition exists  Nature of roots
(i) b² – 4ac > 0    Real and unequal
(ii) b² – 4ac = 0   Real and equal
(iii) b² – 4ac < 0  No real roots

On this page, you will find Pair of Linear Equations in Two Variables Class 10 Notes Maths Chapter 3 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 3 Pair of Linear Equations in Two Variables will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 3 Notes Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables Class 10 Notes Understanding the Lesson

Two linear equations in the same two variables are called a pair of linear equations in two variables. The most general form of a pair of linear equation is
a₁x + b₁y + c₁ = 0
a₂x+ b₂y + c₂ = 0
where a₁,a₂, b_1,b₂, c₁ c₂ are real numbers. For the pair of linear equations, the following situations can arise:
(i) \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) In this case, the pair of linear equations is consistent.

(ii) \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) The pair of linear equations in inconsistent.

(iii) \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\) The pair of linear equations is consistent.

2. A pair of linear equations in two variables, can be represented, and solved by the

• Graphical method
• Algebraic method

3. Graphical Method: The graph of a pair of linear equations in two variables is represented by two lines, following three possibilities can occur.

• Two lines intersect at one point, then that point gives the unique solution of the two equations and the pair of equations is consistent.
• Two lines will not intersect, i.e. they are parallel, the pair of linear equations is inconsistent and the pair of equations will have no solution.
  
• The graph will be a pair of coincident lines. Each point on the lines will be a solution, so the pair of equations will have infinitely many solution and is consistent.

4. Algebraic Method: A pair of linear equations can be solved by any of the following three methods:

• Substitution method
• Elimination method
• Cross-multiplication method

5. Graphical Method of Solution of a pair of Linear Equations:

If the lines represented by the pair of linear equations in two variables are given by
a₁x + b₁y + c₁ = 0
a₂x+ b₂y + c₂ = 0

Following are the cases:

(i) If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) then the lines are intersecting lines and intersect at one point. In this case, the pair of  linear equations in consistent.

(ii) If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\), then the lines are coincident. In this case, the pair of linear equation is consistent  (dependent)

(iii) If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) then the lines are parallel to each other. In this case, the pair of linear equations inconsistent.

On this page, you will find Polynomials Class 10 Notes Maths Chapter 2 Pdf free download. CBSE NCERT Class 10 Maths Notes Chapter 2 Polynomials will seemingly help them to revise the important concepts in less time.

CBSE Class 10 Maths Chapter 2 Notes Polynomials

Polynomials Class 10 Notes Understanding the Lesson

1. The value of the polynomial p(x) at x = a is p(a).

2. Zeroes of the polynomial p(x) can be find by equating p(x) to zero and solving the equation for

3. If for p(x) = ax² + bx + c = 0, a ≠ 0; α and β are the zeroes, then

4. If for p(x) = ax³ + bx² + cx + d = 0; a ≠ 0; α, β,γ are the zeroes, then
α + β + γ = \(\frac{-b}{a}\)
αβ+βγ + αγ = \(\frac{c}{a}\)
αβγ =\(\frac{-d}{a}\)

5. If α and β are the zeroes; then quadratic polynomial will be given by K[x² – Sx + P]
where
S = α +β
P = αβ
K (≠0) is real.

6. The cubic polynomial with zeroes α, β and γ is given by
K[x³ – S₁x² + S₁x² S₃]
where
S₁ = α + β + γ
S₂ = αβ + βγ + αγ
S₃= αβγ
K(≠ 0) is real.

Degree of a Polynomial:

1. The degree of a polynomial p(x) in x is the highest power of x in p(x)

Note: Expressions like \(\frac{1}{\sqrt{x}}, \frac{1}{x^{2}+1}, \sqrt{x+2}\)

2. (i) Polynomial with degree 1, i.e., polynomial of the form ax + b; a ≠ 0 is called linear polynomial.
(ii) Polynomial with degree 2, i.e., polynomial of the form ax2 + bx + c; a ≠ 0 is called quadratic polynomial.
(iii) Polynomial with degree 3, i.e. polynomial of the form ax³ + bx² + cx + d ; a ≠ 0 is called cubic polynomial.
(iv) Polynomial with degree 4, i.e. polynomial of the form ax⁴ + bx³ + cx² + dx + e; a ≠ 0 is called biquadratic polynomial.

Geometrical Meaning of the Zeroes of a Polynomial:

1. For any polynomial y = f(x), the number of points on which the graph of y = f(x) intersects at x-axis is called the number of the zeroes of the polynomial and the x-coordinates of these points are called the zeroes of the polynomial y = f(x).

2. Polynomial with degree ‘n’ has maximum ‘n’ number of zeroes. A constant polynomial has no zeroes.

3. Geometrical representation of a linear polynomial is always a straight line.

4. Geometrical representation of a quadratic polynomial is the graph of the shape either open upwards like ‘∪’ or open downwards like ‘∩’ according to a > 0 or a < 0. These curves are called Parabola.

Relationship Between Zeroes and Coefficient of a Quadratic Polynomial:

1. If α and β are the zeroes of the quadratic polynomial  p(x) = ax² + bx + c a≠0 then

Relationship Between Zeroes and Coefficient of a Cubic Polynomial

1. If α, β and γ are the zeroes of the cubic polynomial

2. A quadratic polynomial p(x) with zeroes α and β is given by
p(x) = K[x² – (α + β)x + αβ]
where K(≠0) is real.

3. A cubic polynomial p(x) with α, β and γ as zeroes is given by
p(x) = K[x³ – (α + β + γ)x² + (αβ +βγ + αγ)x – αβγ
where K(≠0) is real.

Division Algorithm for Polynomials:
If p(x) and g(x) are any two polynomials where g(x) ≠ 0. Then on dividing p(x) by g(x), we find other two polynomials q(x) and r(x) such that
p(x) = g(x) x q(x) + r(x);
where deg. of r(x) < deg. of gix)
or Dividend = Divisor x Quotient + Remainder

Note:

• If r(x) = 0, then g(x) will be a factor of p(x) otherwise not.
• If any real number ‘a’ is a zero of the polynomial p(x), then (x – a) will be a factor of p(x).