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Math Formulas for Class 9, 10, 11 & 12

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A comprehensive resource for all essential mathematical formulas, from basic algebra to advanced calculus. Perfect for board exams and competitive tests.

Class 9 Math Formulas (कक्षा 9 गणित सूत्र)

Number Systems (संख्या पद्धति)

Laws of Exponents for Real Numbers (घातांक के नियम)

  • $$a^m \cdot a^n = a^{m+n}$$
  • $$(a^m)^n = a^{mn}$$
  • $$\frac{a^m}{a^n} = a^{m-n}$$
  • $$a^m \cdot b^m = (ab)^m$$
  • $$a^0 = 1$$
  • $$a^{-n} = \frac{1}{a^n}$$
  • $$(\frac{a}{b})^n = \frac{a^n}{b^n}$$

Example: LCM & HCF

प्रश्न (Question): 36 और 48 का HCF तथा LCM निकालिए।

फ़ॉर्मूला (Formula): $$HCF \times LCM = a \times b$$

समाधान (Solution):

  1. Prime Factorization of 36: $$36 = 2^2 \times 3^2$$
  2. Prime Factorization of 48: $$48 = 2^4 \times 3^1$$
  3. HCF is the product of common factors with the lowest power:
    $$HCF = 2^2 \times 3^1 = 4 \times 3 = 12$$
  4. LCM is the product of all factors with the highest power:
    $$LCM = 2^4 \times 3^2 = 16 \times 9 = 144$$
  5. Verification: $$12 \times 144 = 1728$$ and $$36 \times 48 = 1728$$.

Final Answer: HCF = 12, LCM = 144

Polynomials & Algebraic Identities ( बहुपद और बीजगणितीय सर्वसमिकाएं)

Key Identities (मुख्य सर्वसमिकाएं)

  • $$(a + b)^2 = a^2 + 2ab + b^2$$
  • $$(a - b)^2 = a^2 - 2ab + b^2$$
  • $$a^2 - b^2 = (a - b)(a + b)$$
  • $$(x + a)(x + b) = x^2 + (a + b)x + ab$$
  • $$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$
  • $$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$$
  • $$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$$
  • $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$
  • $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
  • $$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$

Example: Factorization

प्रश्न (Question): $$x^2 + 5x + 6$$ को गुणनखंडों में लिखिए।

विधि (Method): Find two numbers whose product is the constant term (6) and whose sum is the middle coefficient (5).

समाधान (Solution):

  1. The constant term is 6. We need two numbers that multiply to 6.
  2. The middle coefficient is 5. The same two numbers must add up to 5.
    1. The numbers are 2 and 3 (since 2 × 3 = 6 and 2 + 3 = 5).
    2. Split the middle term: x2 + 2x + 3x + 6
    3. Factor by grouping: x(x + 2) + 3(x + 2)

    Final Answer: (x + 2)(x + 3)

Linear Equations (रैखिक समीकरण)

Example: Word Problem

प्रश्न (Question): एक व्यक्ति के पास 5-रु के सिक्के और 2-रु के सिक्के हैं। कुल सिक्कों की संख्या 20 है और कुल राशि ₹64 है। 5-रु के कितने सिक्के हैं?

विधि (Method): Set up and solve a pair of linear equations.

समाधान (Solution):

  1. Let 'x' be the number of ₹5 coins and 'y' be the number of ₹2 coins.
  2. Equation for total number of coins: $$x + y = 20 ---(1)$$
  3. Equation for total value: $$5x + 2y = 64 --- (2)$$
  4. From (1), we get $$y = 20 - x$$.
  5. Substitute this into (2): $$5x + 2(20 - x) = 64$$
  6. Solve for x: $$5x + 40 - 2x = 64 \rightarrow 3x = 24 \rightarrow x = 8$$.

Final Answer: There are 8 five-rupee coins.

Geometry: Lines, Angles & Triangles (ज्यामिति)

Area of a Triangle (त्रिभुज का क्षेत्रफल)

Standard Formula: $$Area = \frac{1}{2} \times \text{base} \times \text{height}$$

Example: Area of Triangle

प्रश्न (Question): बेस = 10 cm और ऊँचाई = 6 cm वाले त्रिभुज का क्षेत्रफल निकालिए।

फ़ॉर्मूला (Formula): $$Area = \frac{1}{2} \times \text{base} \times \text{height}$$

समाधान (Solution):

  1. Substitute the given values into the formula.
  2. $$Area = \frac{1}{2} \times 10 \text{ cm} \times 6 \text{ cm}$$
  3. $$Area = 5 \text{ cm} \times 6 \text{ cm} = 30 \text{ cm}^2$$

Final Answer: $$30 \text{ cm}^2$$

Heron's Formula (हीरोन का सूत्र)

For a triangle with side lengths a, b, and c:

First, calculate the semi-perimeter (s):

$$s = \frac{a + b + c}{2}$$

Then, the area is:

$$Area = \sqrt{s(s - a)(s - b)(s - c)}$$
Surface Area and Volume (पृष्ठीय क्षेत्रफल और आयतन)
  • Cube (घन): Surface Area = 6a2, Volume = a3 (where a = side)
  • Cuboid (घनाभ): Surface Area = 2(l × b + b × h + h × l), Volume = l × b × h
  • Cylinder (बेलन): Curved Surface Area = 2πr × h, Total Surface Area = 2πr(r + h), Volume = πr2 × h
  • Cone (शंकु): Curved Surface Area = $$\pi rl$$, Total Surface Area = $$\pi r(r + l)$$, Volume = $$\frac{1}{3}\pi r^2 h$$ (where $$l = \sqrt{r^2 + h^2}$$)
  • Sphere (गोला): Surface Area = $$4\pi r^2$$, Volume = $$\frac{4}{3}\pi r^3$$
  • Hemisphere (अर्धगोला): Curved Surface Area = $$2\pi r^2$$, Total Surface Area = $$3\pi r^2$$, Volume = $$\frac{2}{3}\pi r^3$$

Class 10 Math Formulas (कक्षा 10 गणित सूत्र)

Algebra: AP & Quadratic Equations (बीजगणित)

Arithmetic Progressions (A.P.) (समान्तर श्रेढ़ी)

n-th term (nवाँ पद): $$a_n = a + (n - 1)d$$

Sum of first n terms (n पदों का योग): $$S_n = \frac{n}{2} [2a + (n - 1)d]$$ OR $$S_n = \frac{n}{2} [a + l]$$

Quadratic Equations (द्विघात समीकरण)

Standard Form: $$ax^2 + bx + c = 0$$

Quadratic Formula (द्विघाती सूत्र): $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Example: Quadratic Roots

प्रश्न (Question): समीकरण $$x^2 - 5x + 6 = 0$$ के मूल निकालिए।

फ़ॉर्मूला (Formula): $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

समाधान (Solution):

  1. Identify a, b, and c: Here, $$a = 1, b = -5, c = 6$$.
  2. Calculate the discriminant (D): $$D = b^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1$$.
  3. Substitute into the formula: $$x = \frac{-(-5) \pm \sqrt{1}}{2(1)}$$
  4. Simplify: $$x = \frac{5 \pm 1}{2}$$.
  5. Find the two roots: $$x = \frac{5 + 1}{2} = 3$$ and $$x = \frac{5 - 1}{2} = 2$$.

Final Answer: The roots are x = 2 and x = 3.

Trigonometry (त्रिकोणमिति)

Basic Ratios and Identities (मूल अनुपात और सर्वसमिकाएं)

  • $$\sin^2\theta + \cos^2\theta = 1$$
  • $$1 + \tan^2\theta = \sec^2\theta$$
  • $$1 + \cot^2\theta = \csc^2\theta$$

Example: Basic Trigonometry

प्रश्न (Question): In a right-angled triangle, the side opposite to angle θ is 3 and the adjacent side is 4. Find the hypotenuse and the values of sinθ and cosθ.

फ़ॉर्मूला (Formula): Pythagoras Theorem: $$c = \sqrt{a^2 + b^2}$$. Also, $$\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$$, $$\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$$.

समाधान (Solution):

  1. Find the hypotenuse (c): $$c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$.
  2. Calculate sinθ: $$\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{5}$$.
  3. Calculate cosθ: $$\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5}$$.

Final Answer: Hypotenuse = 5, sinθ = 3/5, cosθ = 4/5.

Coordinate Geometry & Statistics (निर्देशांक ज्यामिति और सांख्यिकी)

Distance Formula (दूरी सूत्र)

$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Example: Distance Formula

प्रश्न (Question): बिंदु A(2, 3) और B(7, 11) के बीच की दूरी ज्ञात कीजिए।

फ़ॉर्मूला (Formula): $$Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

समाधान (Solution):

  1. Calculate the difference in x-coordinates: $$\Delta x = 7 - 2 = 5$$.
  2. Calculate the difference in y-coordinates: $$\Delta y = 11 - 3 = 8$$.
  3. Substitute into the formula: $$Distance = \sqrt{5^2 + 8^2}$$.
  4. Simplify: $$Distance = \sqrt{25 + 64} = \sqrt{89}$$.

Final Answer: $$\sqrt{89}$$ units.

Statistics (सांख्यिकी)

Mean (माध्य): $$\bar{x} = \frac{\Sigma x_i}{n}$$

Example: Mean

प्रश्न (Question): डेटा: 2, 4, 6, 8, 10। उनका औसत (mean) निकालिए।

फ़ॉर्मूला (Formula): $$\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}$$.

समाधान (Solution):

  1. Sum of observations: $$2 + 4 + 6 + 8 + 10 = 30$$.
  2. Number of observations: $$n = 5$$.
  3. Calculate the mean: $$\text{Mean} = \frac{30}{5} = 6$$.

Final Answer: The mean is 6.

Class 11 Math Formulas (कक्षा 11 गणित सूत्र)

Sets, Relations & Functions (समुच्चय, संबंध एवं फलन)

Functions (फलन)

A function f from a set A to a set B is a rule that assigns each element x in A to exactly one element y in B. We write f(x) = y.

Example: Evaluating a Function

प्रश्न (Question): If $$f(x) = 2x + 3$$, find the value of $$f(4)$$.

विधि (Method): Substitute the value of x into the function's definition.

समाधान (Solution):

  1. Replace 'x' with '4' in the expression $$2x + 3$$.
  2. $$f(4) = 2(4) + 3$$
  3. $$f(4) = 8 + 3 = 11$$

Final Answer: $$f(4) = 11$$

Sequences and Series (अनुक्रम तथा श्रेणी)

Arithmetic Progression (A.P.) (समान्तर श्रेढ़ी)

n-th term: $$a_n = a + (n - 1)d$$

Sum of n terms: $$S_n = \frac{n}{2} [2a + (n - 1)d]$$

Example: Arithmetic Progression

प्रश्न (Question): In an AP, the first term $$a=7$$ and the common difference $$d=3$$. Find the 10th term and the sum of the first 10 terms.

फ़ॉर्मूला (Formula): $$a_n = a + (n-1)d$$ and $$S_n = \frac{n}{2} [2a + (n-1)d]$$.

समाधान (Solution):

  1. Find the 10th term ($$a_{10}$$): $$a_{10} = 7 + (10 - 1) \times 3 = 7 + 27 = 34$$.
  2. Find the sum of the first 10 terms ($$S_{10}$$):
    $$S_{10} = \frac{10}{2} [2\times7 + (10 - 1)\times3]$$
  3. $$S_{10} = 5 [14 + 27] = 5 \times 41 = 205$$.

Final Answer: 10th term is 34, Sum is 205.

Advanced Trigonometry (उन्नत त्रिकोणमिति)

Sum and Difference Formulas

  • $$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$
  • $$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$

Example: Sum Formula

प्रश्न (Question): Find the value of $$\sin(75^\circ)$$ using $$\sin(45^\circ + 30^\circ)$$.

फ़ॉर्मूला (Formula): $$\sin(A + B) = \sin A \cos B + \cos A \sin B$$.

समाधान (Solution):

  1. Apply the formula with A=45° and B=30°.
  2. $$\sin(75^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ)$$
  3. Substitute known values: $$\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$$, $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$, $$\sin 30^\circ = \frac{1}{2}$$.
  4. $$= \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{1}{2}\right)$$
  5. $$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$$.

Final Answer: $$\frac{\sqrt{3} + 1}{2\sqrt{2}}$$

Calculus: Limits & Derivatives (कलन: सीमा और अवकलज)

Limits (सीमाएँ)

Example: Evaluating a Limit

प्रश्न (Question): Evaluate $$\lim_{x\to2} \frac{x^2 - 4}{x - 2}$$.

विधि (Method): Direct substitution gives 0/0, which is an indeterminate form. Factorize the numerator.

समाधान (Solution):

  1. Factor the numerator: $$x^2 - 4 = (x - 2)(x + 2)$$.
  2. Rewrite the expression: $$\frac{(x - 2)(x + 2)}{x - 2}$$.
  3. Cancel the common factor $$(x - 2)$$. The expression simplifies to $$x + 2$$ for x ≠ 2.
  4. Now substitute x = 2 into the simplified expression: $$2 + 2 = 4$$.

Final Answer: The limit is 4.

Standard Derivatives (मानक अवकलज)

  • $$\frac{d}{dx} (x^n) = nx^{n-1}$$
  • $$\frac{d}{dx} (\sin x) = \cos x$$
  • $$\frac{d}{dx} (\cos x) = -\sin x$$

Class 12 Math Formulas (कक्षा 12 गणित सूत्र)

Calculus: Differentiation & Applications (अवकलन और अनुप्रयोग)

Basic Differentiation (अवकलन)

Example: Basic Differentiation

प्रश्न (Question): Find dy/dx for $$y = x^3 - 5x^2 + 4x - 7$$.

फ़ॉर्मूला (Formula): Power rule: $$\frac{d}{dx} (x^n) = nx^{n-1}$$.

समाधान (Solution): Apply the power rule to each term.

  1. $$\frac{d}{dx}(x^3) = 3x^2$$
  2. $$\frac{d}{dx}(-5x^2) = -5 \times 2x = -10x$$
  3. $$\frac{d}{dx}(4x) = 4$$
  4. $$\frac{d}{dx}(-7) = 0$$ (Derivative of a constant is zero)

Final Answer: $$\frac{dy}{dx} = 3x^2 - 10x + 4$$

Application of Derivatives (अवकलज के अनुप्रयोग)

Example: Slope of Tangent

प्रश्न (Question): Find the slope of the tangent to the curve $$y = x^3$$ at the point where $$x = 2$$.

विधि (Method): The slope of the tangent at a point is the value of the derivative (y') at that point.

समाधान (Solution):

  1. Find the derivative of the function: If $$y = x^3$$, then $$y' = 3x^2$$.
  2. Evaluate the derivative at x = 2: $$y'(2) = 3 \times (2)^2$$.
  3. Calculate the result: $$3 \times 4 = 12$$.

Final Answer: The slope of the tangent is 12.

Calculus: Integration (कलन - समाकलन)

Standard Integrals (मानक समाकलन)

$$\int x^n \,dx = \frac{x^{n+1}}{n+1} + C$$

Example: Indefinite Integration

प्रश्न (Question): Find $$\int(6x^2 - 4x + 3) \,dx$$.

फ़ॉर्मूला (Formula): $$\int x^n \,dx = \frac{x^{n+1}}{n+1} + C$$.

समाधान (Solution): Integrate each term separately.

  1. $$\int 6x^2 \,dx = 6 \times \frac{x^3}{3} = 2x^3$$
  2. $$\int -4x \,dx = -4 \times \frac{x^2}{2} = -2x^2$$
  3. $$\int 3 \,dx = 3x$$
  4. Combine the results and add the constant of integration, C.

Final Answer: $$2x^3 - 2x^2 + 3x + C$$

Probability (प्रायिकता)

Binomial Distribution (द्विपद बंटन)

$$P(X=k) = {}^nC_k \, p^k q^{n-k}$$

Example: Binomial Probability

प्रश्न (Question): एक सिक्का 3 बार उछाला जाता है। सटीक दो बार ‘Heads’ आने की संभावना क्या है?

फ़ॉर्मूला (Formula): $$P(k \text{ successes}) = {}^nC_k \cdot p^k \cdot (1-p)^{n-k}$$. For a fair coin, $$p = 1/2$$.

समाधान (Solution):

  1. Identify the parameters: n = 3 (number of trials), k = 2 (number of successes), p = 1/2 (probability of getting a head).
  2. Calculate the combination $$^nC_k$$: $${}^3C_2 = \frac{3!}{2! \cdot 1!} = 3$$.
  3. Substitute into the formula: $$P(X=2) = 3 \times \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right)^1$$
  4. Calculate the result: $$P(X=2) = 3 \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{8}$$.

Final Answer: The probability is 3/8.

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