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October 11, 2024**Maths Formulas for Class 9:** Maths is one of the most important subjects for Class 9 students, and a clear understanding of the Maths Formulas for Class 9 is essential. Students must have outstanding Maths skills to contribute to great inventions in the field of Engineering, Science and Technology.

It takes a lot of practice, a deep comprehension of topics, and memorising formulas to do well in the CBSE Class 9 Maths exam. The Maths Formulas for Class 9 will help students easily score good marks in maths exams. Students who find difficulties in solving maths problems should go through the topic-wise Class 9 Maths formulas provided by Embibe. From this article, students can download the NCERT Class 9 Maths Formulas PDF, which will help them study better.

Mathematical formulas aren’t just for closing your eyes and learning. Conceptual clarity is crucial, as is understanding all the formulas of Math, implementing them, and analysing them. Before getting into the list of the formulae, let us check out the important chapters of Class 9 Maths for which formulas are needed:

- Numbers
- Polynomials
- Coordinate Geometry
- Algebra
- Triangles
- Areas of Parallelograms and Triangles
- Circles
- Heron’s Formula
- Surface Areas and Volumes
- Statistics
- Probability

**Download** – **Algebra Formulae for Class 9**

Here are some of the essential formulae for Class 9 polynomial identities and all Class 9 identities in Mathematics.

Any number that can be written in the form of p ⁄ q where p and q are integers and q ≠ 0 are rational numbers. Irrational numbers cannot be written in the p ⁄ q form.

- There is a unique real number that can be represented on a number line.
- If r is one such rational number and s is an irrational number, then (r + s), (r – s), (r × s) and (r ⁄ s) are irrational.
- For positive real numbers, the corresponding identities hold together:
- \(\sqrt{ab}\) = \(\sqrt{a} × \sqrt{b}\)
- \(\sqrt{\tfrac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\)
- \((\sqrt{a}+\sqrt{b})\times(\sqrt{a}-\sqrt{b})=a-b\)
- \((a+\sqrt{b})\times(a-\sqrt{b})=a^2-b\)
- \((\sqrt{a}+\sqrt{b})^2=a^2+2\sqrt{ab}+b\)

- If you want to rationalize the denominator of 1 ⁄ √ (a + b), then we have to multiply it by √(a – b) ⁄ √(a – b), where a and b are both integers.
- Suppose a is a real number (greater than 0) and p and q are the rational numbers.

A polynomial p(x) denoted for one variable ‘x’ comprises an algebraic expression in the form:

**p(x) = a**_{n}**x**^{n}** + a**_{n-1}**x**^{n-1}** + ….. + a**_{2}**x**^{2}** + a**_{1}**x + a**_{0} ; where a_{0}, a_{1}, a_{2}, …. a_{n} are constants where a_{n} ≠ 0

- Any real number; let’s say ‘a’ is considered to be the zero of a polynomial ‘p(x)’ if p(a) = 0. In this case, a is said to be the mysqladmin of the equation p(x) = 0.
- Every one variable linear polynomial will contain a unique zero, a real number which is a zero of the zero polynomial and a non-zero constant polynomial which does not have any zeros.
**Remainder Theorem:**If p(x) has the degree greater than or equal to 1 and p(x) when divided by the linear polynomial x – a will give the remainder as p(a).**Factor Theorem:**x – a will be the factor of the polynomial p(x), whenever p(a) = 0. The vice-versa also holds true every time.

Whenever you have to locate an object on a plane, you need two divide the plane into two perpendicular lines, thereby, making it a Cartesian Plane.

- The horizontal line is known as the x-axis, and the vertical line is called the y-axis.
- The coordinates of a point are in the form of (+, +) in the first quadrant, (–, +) in the second quadrant, (–, –) in the third quadrant and (+, –) in the fourth quadrant; where + and – denotes the positive and the negative real number respectively.
- The coordinates of the origin are (0, 0) and thereby it gets up to move in the positive and negative numbers.

Once the students have a hold over all Algebraic identities class 9, they will be able to solve all the Algebra related problems in their exams. Given below are Algebraic identities for class 9 which are considered very important Maths formulas for Class 9:

- (a + b)
^{2}= a^{2}+ 2ab + b^{2} - (a – b)
^{2}= a^{2}– 2ab + b^{2} - (a + b) (a – b) = a
^{2}-b^{2} - (x + a) (x + b) = x
^{2}+ (a + b) x + ab - (x + a) (x – b) = x
^{2}+ (a – b) x – ab - (x – a) (x + b) = x
^{2}+ (b – a) x – ab - (x – a) (x – b) = x
^{2}– (a + b) x + ab - (a + b)
^{3}= a^{3}+ b^{3}+ 3ab (a + b) - (a – b)
^{3}= a^{3}– b^{3}– 3ab (a – b) - (x + y + z)
^{2}= x^{2}+ y^{2}+ z^{2}+ 2xy +2yz + 2xz - (x + y – z)
^{2}= x^{2}+ y^{2}+ z^{2}+ 2xy – 2yz – 2xz - (x – y + z)
^{2}= x^{2}+ y^{2}+ z^{2}– 2xy – 2yz + 2xz - (x – y – z)
^{2}= x^{2}+ y^{2}+ z^{2}– 2xy + 2yz – 2xz - x
^{3}+ y^{3}+ z^{3}– 3xyz = (x + y + z) (x^{2}+ y^{2}+ z^{2}– xy – yz -xz) - x
^{2 }+ y^{2}= \(\frac{1}{2}\) [(x + y)^{2}+ (x – y)^{2}] - (x + a) (x + b) (x + c) = x
^{3 }+ (a + b + c)x^{2}+ (ab + bc + ca)x + abc - x
^{3}+ y^{3}= (x + y) (x^{2 }– xy + y^{2}) - x
^{3}– y^{3}= (x – y) (x^{2 }+ xy + y^{2}) - x
^{2}+ y^{2}+ z^{2}– xy – yz – zx = \(\frac{1}{2}\) [(x – y)^{2}+ (y – z)^{2}+ (z – x)^{2}]

A triangle is a closed geometrical figure formed by three sides and three angles.

- Two figures are congruent if they have the same shape and same size.
- If the two triangles ABC and DEF are congruent under the correspondence that A ↔ D, B ↔ E and C ↔ F, then symbolically, these can be expressed as ∆ ABC ≅ ∆ DEF.

Suppose ∆ ABC is a right-angled triangle with AB as the perpendicular, BC as the base and AC as the hypotenuse; then Pythagoras Theorem will be expressed as:

**(Hypotenuse)**^{2}** = (Perpendicular)**^{2}** + (Base)**^{2}

i.e. **(AC)**^{2}** = (AB)**^{2}** + (BC)**^{2}

A parallelogram is a type of quadrilateral that contains parallel opposite sides.

- Area of parallelogram = Base × Height
- Area of Triangle = \(\frac{1}{2}\) × Base × Height

A circle is a closed geometrical figure. All points on the boundary of a circle are equidistant from a fixed point inside the circle (called the centre).

- Area of a circle (of radius r) = π × r
^{2} - The diameter of the circle, d = 2 × r
- Circumference of the circle = 2 × π × r
- Sector angle of the circle, θ = (180 × l ) / (π × r )
- Area of the sector = (θ/2) × r
^{2}; where θ is the angle between the two radii - Area of the circular ring = π × (R
^{2}– r^{2}); where R – radius of the outer circle and r – radius of the inner circle

Heron’s Formula is used to calculate the area of a triangle whose all three sides are known. Let us suppose the length of the three sides is a, b and c.

**Step 1 –**Calculate the semi-perimeter, \(s=\frac{a+b+c}{2}\)**Step 2 –**Area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

In this section, students will find Class 9 surface area and volume formulas at one place. Below, LSA stands for Lateral/Curved Surface Area and TSA stands for Total Surface Area.

Name of the Solid Figure | Formulae |

Cuboid | LSA: 2h(l + b)TSA: 2(lb + bh + hl)Volume: l × b × hl = length, b = breadth, h = height |

Cube | LSA: 4a^{2}TSA: 6a^{2}Volume: a^{3}a = sides of a cube |

Right Circular Cylinder | LSA: 2(π × r × h)TSA: 2πr (r + h)Volume: π × r^{2} × hr = radius, h = height |

Right Pyramid | LSA: ½ × p × lTSA: LSA + Area of the baseVolume: ⅓ × Area of the base × hp = perimeter of the base, l = slant height, h = height |

Prism | LSA: p × hTSA: LSA × 2BVolume: B × hp = perimeter of the base, B = area of base, h = height |

Right Circular Cone | LSA: πrlTSA: π × r × (r + l)Volume: ⅓ × (πr^{2}h)r = radius, l = slant height, h = height |

Hemisphere | LSA: 2 × π × r^{2}TSA: 3 × π × r^{2}Volume: ⅔ × (πr^{3})r = radius |

Sphere | LSA: 4 × π × r^{2}TSA: 4 × π × r^{2}Volume: 4/3 × (πr^{3})r = radius |

Certain facts or figures which can be collected or transformed into some useful purpose are known as data. These data can be graphically represented to increase readability for people. Three measures of formulae to interpret the ungrouped data:

Category | Mathematical Formulae |

Mean, \(\bar{x}\) | \(\frac{\sum x}{n}\) x = Sum of the values; N = Number of values |

Standard Deviation, \(\sigma\) | \(\sigma= \sqrt{\frac{\sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}}{N-1}}\) x _{i} = Terms Given in the Data, x̄ = Mean, N = Total number of Terms |

Range, R | R = Largest data value – Smallest data value |

Variance, \(\sigma^2\) | \(\sigma^2\ = \frac{\sum x_{i}-\bar{x}}{N}\) x = Item given in the data, x̅ = Mean of the data, n = Total number of items |

Probability is the possibility of any event likely to happen. The probability of any event can only be from 0 to 1 with 0 being no chances and 1 being the possibility of that event happening.

\(Probability=\frac{Number\: of\: favourable\: outcomes}{Total\: Number\: of\: outcomes}\)

**Ans:** You can practice for Class 9 Maths questions at Embibe. Embibe offers you topic-wise questions which are available for.

**Ans :** Yes, for Class 9 NCERT Maths book is enough for students preparing for exams. Make sure you understand all the concepts and solve the questions diligently. Note that regular practice is a must.

**Ans :** Mathematics is a subject of logic. Therefore, it should be interpreted in the same way. You can learn these formulae by understanding them logically. Then, you can try solving the questions by implementing these formulae.

**Ans: **We have compiled Class 9 Maths formulas in this article so that students can understand them. These formulae are based on CBSE, ICSE, and other respective boards.

**Ans: **You can try to remember everything you are trying to learn in the form of a story. Sequencing will help you to memorise the formulas in a particular order. Also, make sure to understand the derivations of the formulas rather than rote learning. This way, you will be able to remember all formulas of Maths Class 9 for a long time.