All Important Formula — Class 10 Maths Notes

All Important Formula · Class 10 Maths · 11 topics.

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Topics covered in All Important Formula

  1. 1.Important Formula of Real Numbers

    🧮 Formula:

    1. Euclid's Division Lemma:
      a=bq+ra = bq + r
      Where:

      • aa = Dividend
      • bb = Divisor
      • qq = Quotient
      • rr = Remainder ( 0≤r<b0 \leq r < b)

    2. HCF Calculation (Euclid’s Algorithm):
      To find the HCF of two numbers aa and bb:
      HCF(a,b)=HCF(b,r)\text{HCF}(a, b) = \text{HCF}(b, r) (Repeat until r=0r = 0)

    📖 Explanation:

    1. Euclid's Division Lemma:
      This formula is used to express any positive integer aa as a multiple of another integer bb, plus a remainder rr. The remainder is always smaller than bb.

    2. HCF (Highest Common Factor):

      • The HCF of two numbers is the largest number that divides both numbers exactly.
      • Euclid's Algorithm is a step-by-step method to find the HCF using the division lemma repeatedly until the remainder becomes 0.


    🔢 Step-by-Step Example:

    Let’s find the HCF of 56 and 12 using Euclid’s Algorithm.

    1. Step 1: Apply Euclid's Division Lemma to 56 and 12:

      56=12×4+856 = 12 \times 4 + 8

      (Here, q=4q = 4 and r=8r = 8)

    2. Step 2: Now, apply the lemma to 12 and 8:

      12=8×1+412 = 8 \times 1 + 4

      (Here, q=1q = 1 and r=4r = 4)

    3. Step 3: Next, apply the lemma to 8 and 4:

      8=4×2+08 = 4 \times 2 + 0

      (Here, r=0r = 0)

    Since the remainder is now 0, the divisor at this step, which is 4, is the HCF of 56 and 12.

    HCF = 4

  2. 2.Important Formula for Polynomials

    🧮 Formulas:

    1. Standard Form of a Polynomial:
      A polynomial P(x)P(x) in one variable xx is expressed as:
      P(x)=anxn+an−1xn−1+⋯+a1x+a0P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
      Where:

      • an,an−1,…,a0a_n, a_{n-1}, \ldots, a_0 are constants (coefficients).
      • nn is the degree of the polynomial (highest power of xx).

    2. Types of Polynomials:

      • Linear Polynomial: P(x)=ax+bP(x) = ax + b (Degree = 1)
      • Quadratic Polynomial: P(x)=ax2+bx+cP(x) = ax^2 + bx + c (Degree = 2)
      • Cubic Polynomial: P(x)=ax3+bx2+cx+dP(x) = ax^3 + bx^2 + cx + d (Degree = 3)

    3. Remainder Theorem:
      If P(x)P(x) is divided by (x−a)(x - a), the remainder is P(a)P(a).

    4. Factor Theorem:
      If P(a)=0P(a) = 0, then (x−a)(x - a) is a factor of P(x)P(x).

    5. Relationship Between Zeroes and Coefficients for Quadratic Polynomials:
      For ax2+bx+cax^2 + bx + c:

      • Sum of Zeroes: α+β=−ba\alpha + \beta = -\frac{b}{a}
      • Product of Zeroes: αβ=ca\alpha \beta = \frac{c}{a}


    📖 Explanation:

    1. Polynomials:
      A polynomial is an algebraic expression made up of variables, constants, and exponents, combined using addition, subtraction, and multiplication. The degree of the polynomial tells us the highest power of the variable.

    2. Types of Polynomials:

      • Linear: The highest power of xx is 1. Example: 3x+53x + 5.
      • Quadratic: The highest power of xx is 2. Example: x2+4x+4x^2 + 4x + 4.
      • Cubic: The highest power of xx is 3. Example: 2x3+3x2−12x^3 + 3x^2 - 1.

    3. Remainder and Factor Theorems:

      • Remainder Theorem: If you substitute x=ax = a into P(x)P(x), the result is the remainder when P(x)P(x) is divided by (x−a)(x - a).
      • Factor Theorem: If P(a)=0P(a) = 0, then (x−a)(x - a) is a factor of P(x)P(x).

    🔢 Step-by-Step Example:


    Let's find the remainder when P(x)=2x3−3x2+x+5P(x) = 2x^3 - 3x^2 + x + 5 is divided by (x−2)(x - 2).

    1. Step 1: By the Remainder Theorem, substitute x=2x = 2 in P(x)P(x).
      P(2)=2(2)3−3(2)2+(2)+5

      P(2) = 2(2)^3 - 3(2)^2 + (2) + 5

    2. Step 2: Calculate powers and multiply:
      P(2)=2×8−3×4+2+5

      P(2) = 2 \times 8 - 3 \times 4 + 2 + 5

    3. Step 3: Simplify:
      P(2)=16−12+2+5P(2) = 16 - 12 + 2 + 5
      P(2)=11P(2) = 11

    Remainder = 11

  3. 3.Important Formula of Pair of Linear Equation in Two Variables

    🧮 Key Formulas:

    1. General Form of a Pair of Linear Equations:

      a1x+b1y+c1=0a_1x + b_1y + c_1 = 0 a2x+b2y+c2=0a_2x + b_2y + c_2 = 0

      Where a1,b1,c1,a2,b2,c2a_1, b_1, c_1, a_2, b_2, c_2 are real numbers, and xx and yy are the variables.

    2. Solution of a Pair of Linear Equations:
      A solution is a pair of values (x,y)(x, y) that satisfies both equations simultaneously.

    3. Consistency of the System:

      • Consistent System (has a solution):
        a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2} → Unique Solution

      • Inconsistent System (no solution):
        a1a2=b1b2≠c1c2

        \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}
      • Dependent System (infinitely many solutions):
        a1a2=b1b2=c1c2

        \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

    4. Graphical Representation:

      • Two lines intersect at a point: Unique solution
      • Two lines are parallel: No solution
      • Two lines are coincident: Infinitely many solutions

    5. Algebraic Methods to Solve the Equations:

      • Substitution Method
      • Elimination Method
      • Cross-Multiplication Method: xb1c2−b2c1=yc1a2−c2a1=1a1b2−a2b1\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}

    📖 Explanation:

    1. Linear Equations in Two Variables:
      These are equations that can be written in the form ax+by+c=0ax + by + c = 0, where xx and yy are variables, and a,b,ca, b, c are constants. The graph of such an equation is a straight line.

    2. Pair of Linear Equations:
      When we have two linear equations in two variables, they form a system of equations. We solve this system to find values of xx and yy that satisfy both equations.

    3. Types of Solutions:

      • Unique Solution: The lines intersect at one point.
      • No Solution: The lines are parallel and never meet.
      • Infinitely Many Solutions: The lines are coincident (exactly overlapping).

    🔢 Step-by-Step Example:

    Solve the following pair of linear equations using the Elimination Method:

    1. 2x+3y=122x + 3y = 12
    2. x−2y=−3x - 2y = -3

    Step 1: Write the equations:

    1. 2x+3y=122x + 3y = 12
    2. x−2y=−3x - 2y = -3

    Step 2: Multiply the second equation by 2 to align the coefficients of xx:

    2(x−2y)=2(−3)2(x - 2y) = 2(-3) 2x−4y=−62x - 4y = -6

    Step 3: Subtract the second equation from the first:

    (2x+3y)−(2x−4y)=12−(−6)

    (2x + 3y) - (2x - 4y) = 12 - (-6) 2x+3y−2x+4y=18

    2x + 3y - 2x + 4y = 18 7y=187y = 18

    Step 4: Solve for yy:

    y=187y = \frac{18}{7}

    Step 5: Substitute y=187y = \frac{18}{7} into the second equation:

    x−2(187)=−3

    x - 2\left(\frac{18}{7}\right) = -3 x−367=−3x - \frac{36}{7} = -3

    Step 6: Solve for xx:

    x=−3+367x = -3 + \frac{36}{7} x=−21+367x = \frac{-21 + 36}{7} x=157x = \frac{15}{7}

    Solution:

    x=157,y=187x = \frac{15}{7}, \quad y = \frac{18}{7}


  4. 4.Important Formula for Quadratic Equation

    🧮 Formula:

    1. Standard Form of a Quadratic Equation:
      ax2+bx+c=0ax^2 + bx + c = 0
      Where:

      • a,b,a, b, and cc are real numbers and a≠0a \neq 0.
      • xx is the variable.

    2. Quadratic Formula (for solving quadratic equations):
      x=−b±b2−4ac2a

      x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    3. Discriminant:
      D=b2−4acD = b^2 - 4ac

      • If D>0D > 0: Two distinct real roots.
      • If D=0D = 0: One real root (roots are equal).
      • If D<0D < 0: No real roots (complex roots).

    4. Factorization Method:
      For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0:

      • Find two numbers whose product is a×ca \times c and sum is bb.
      • Factorize the middle term accordingly.

    📖 Explanation:

    1. What is a Quadratic Equation?
      A quadratic equation is a polynomial equation of degree 2. The highest power of xx is 2.

    2. Roots of a Quadratic Equation:
      The values of xx that satisfy the equation ax2+bx+c=0ax^2 + bx + c = 0 are called the roots of the quadratic equation.

    3. Methods to Solve Quadratic Equations:

      • Factorization Method
      • Quadratic Formula
      • Completing the Square

    4. Discriminant:
      The discriminant (DD) helps determine the nature of the roots:

      • Positive DD: The parabola intersects the xx-axis at two distinct points.
      • Zero DD: The parabola touches the xx-axis at exactly one point.
      • Negative DD: The parabola does not intersect the xx-axis.

    🔢 Step-by-Step Example:

    Solve the quadratic equation 2x2−7x+3=02x^2 - 7x + 3 = 0 using the Quadratic Formula.

    1. Step 1: Identify aa, bb, and cc:

      • a=2a = 2
      • b=−7b = -7
      • c=3

        c = 3

    2. Step 2: Apply the Quadratic Formula:
      x=−(−7)±(−7)2−4×2×32×2

      x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 2 \times 3}}{2 \times 2}

    3. Step 3: Simplify:
      x=7±49−244x = \frac{7 \pm \sqrt{49 - 24}}{4}
      x=7±254

      x = \frac{7 \pm \sqrt{25}}{4}

    4. Step 4: Calculate the Roots:
      x=7±54x = \frac{7 \pm 5}{4}

      • Root 1: x=7+54=124=3x = \frac{7 + 5}{4} = \frac{12}{4} = 3
      • Root 2: x=7−54=24=12x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}

    Solution: The roots are x=3x = 3 and x=12x = \frac{1}{2}.

  5. 5.Important Formula of Arithmetic Progression (AP)

    🧮 Key Formulas:

    1. General Form of an Arithmetic Progression:
      An AP is a sequence of numbers in which the difference between consecutive terms is constant.
      a,a+d,a+2d,a+3d,…a, a + d, a + 2d, a + 3d, \ldots

      Where:

      • aa = First term
      • dd = Common difference (constant difference between consecutive terms)

    2. nn-th Term of an AP:
      an=a+(n−1)da_n = a + (n-1)d
      Where:

      • ana_n = nn-th term
      • aa = First term
      • dd = Common difference
      • nn = Position of the term

    3. Sum of First nn Terms of an AP:
      Sn=n2(2a+(n−1)d)S_n = \frac{n}{2} \left(2a + (n-1)d\right)
      or
      Sn=n2(a+l)S_n = \frac{n}{2} \left(a + l\right)
      Where:

      • SnS_n = Sum of first nn terms
      • aa = First term
      • dd = Common difference
      • ll = Last term (or nn-th term)

    📖 Explanation:

    1. What is an Arithmetic Progression (AP)?
      An arithmetic progression (AP) is a sequence where each term increases (or decreases) by a fixed amount. The fixed amount is called the common difference (dd).

      Example: 2,5,8,11,14,…2, 5, 8, 11, 14, \ldots

      • Here, the common difference d=5−2=3d = 5 - 2 = 3.

    2. nn-th Term Formula:
      This formula helps you find the term at the nn-th position in the sequence.

    3. Sum of First nn Terms:
      This formula calculates the total sum of the first nn terms in the AP.

    🔢 Step-by-Step Example:

    1. Find the 10th term of the AP: 3,7,11,15,…3, 7, 11, 15, \ldots

      Step 1: Identify aa and dd:

      • a=3a = 3 (First term)
      • d=7−3=4d = 7 - 3 = 4 (Common difference)

      Step 2: Use the nn-th term formula:
      an=a+(n−1)da_n = a + (n-1)d

      Plug in n=10n = 10:
      a10=3+(10−1)×4a_{10} = 3 + (10-1) \times 4
      a10=3+9×4a_{10} = 3 + 9 \times 4
      a10=3+36a_{10} = 3 + 36
      a10=39a_{10} = 39

      10th term = 39

    2. Find the sum of the first 10 terms of the same AP.

      Step 1: Use the sum formula:
      Sn=n2(2a+(n−1)d)S_n = \frac{n}{2} \left(2a + (n-1)d\right)

      Plug in n=10n = 10, a=3a = 3, and d=4d = 4:
      S10=102(2×3+(10−1)×4)S_{10} = \frac{10}{2} \left(2 \times 3 + (10-1) \times 4\right)
      S10=5(6+9×4)S_{10} = 5 \left(6 + 9 \times 4\right)
      S10=5(6+36)S_{10} = 5 \left(6 + 36\right)
      S10=5×42S_{10} = 5 \times 42
      S10=210S_{10} = 210

      Sum of the first 10 terms = 210

  6. 6.Important Formula of Triangles

    🧮 Key Concepts and Formulas:

    1. Similar Triangles:
      Two triangles are said to be similar if:

      • Their corresponding angles are equal.
      • Their corresponding sides are in the same ratio.

    2. Criteria for Similarity of Triangles:

      • AA (Angle-Angle) Criterion: Two triangles are similar if two angles of one triangle are equal to two angles of another triangle.
      • SSS (Side-Side-Side) Criterion: Two triangles are similar if their corresponding sides are in the same ratio.
      • SAS (Side-Angle-Side) Criterion: Two triangles are similar if one angle of one triangle is equal to one angle of another triangle, and the sides containing these angles are in the same ratio.

    3. Basic Proportionality Theorem (Thales Theorem):
      If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

      In △ABC\triangle ABC, if DE∥BCDE \parallel BC, then:
      ADDB=AEEC

      \frac{AD}{DB} = \frac{AE}{EC}

    4. Pythagoras Theorem:
      In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

      For a right-angled triangle △ABC\triangle ABC where ∠B=90∘\angle B = 90^\circ:
      AC2=AB2+BC2AC^2 = AB^2 + BC^2

    5. Converse of Pythagoras Theorem:
      If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right-angled triangle.

    6. Area of Similar Triangles:
      The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

      If △ABC∼△DEF\triangle ABC \sim \triangle DEF, then:
      Area of △ABCArea of △DEF=(ABDE)2\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \left(\frac{AB}{DE}\right)^2

    📖 Explanation:

    1. Similar Triangles:

      • When two triangles have the same shape but not necessarily the same size, they are called similar triangles.
      • In similar triangles, the corresponding sides are proportional, and the corresponding angles are equal.

    2. Basic Proportionality Theorem (Thales Theorem):
      This theorem is useful to find unknown side lengths when a line parallel to one side of a triangle intersects the other two sides.

    3. Pythagoras Theorem:
      This theorem applies to right-angled triangles and is widely used in geometry and trigonometry.

    4. Converse of Pythagoras Theorem:
      It helps to determine if a triangle is right-angled by checking the sides.

    🔢 Step-by-Step Example:

    Example 1:
    In △ABC\triangle ABC, DEDE is drawn parallel to BCBC, and AD=4 cmAD = 4 \, \text{cm}, DB=6 cmDB = 6 \, \text{cm}, AE=5 cmAE = 5 \, \text{cm}. Find ECEC.


    Solution    :
    By the Basic Proportionality Theorem:

    ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}

    Plug in the values:

    46=5EC\frac{4}{6} = \frac{5}{EC}

    Cross-multiply:

    4×EC=6×5

    4 \times EC = 6 \times 5 4×EC=30

    4 \times EC = 30 EC=304=7.5 cmEC = \frac{30}{4} = 7.5 \, \text{cm}

    Answer: EC=7.5 cm

  7. 7.Impotant Formula of Trignometry

    🧮 Key Concepts and Formulas:

    1. Trigonometric Ratios:
      In a right-angled triangle, the following trigonometric ratios are defined for an angle θ\theta:

      • Sine (sin⁡θ\sin\theta):
        sin⁡θ=Opposite SideHypotenuse

        \sin\theta = \frac{\text{Opposite Side}}{\text{Hypotenuse}}

      • Cosine (cos⁡θ\cos\theta):
        cos⁡θ=Adjacent SideHypotenuse

        \cos\theta = \frac{\text{Adjacent Side}}{\text{Hypotenuse}}

      • Tangent (tan⁡θ\tan\theta):
        tan⁡θ=Opposite SideAdjacent Side

        \tan\theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}}

      • Cosecant (csc⁡θ\csc\theta):
        csc⁡θ=1sin⁡θ=HypotenuseOpposite Side


        \csc\theta = \frac{1}{\sin\theta} = \frac{\text{Hypotenuse}}{\text{Opposite Side}}

      • Secant (sec⁡θ\sec\theta):
        sec⁡θ=1cos⁡θ=HypotenuseAdjacent Side

        \sec\theta = \frac{1}{\cos\theta} = \frac{\text{Hypotenuse}}{\text{Adjacent Side}}

      • Cotangent (cot⁡θ\cot\theta):
        cot⁡θ=1tan⁡θ=Adjacent SideOpposite Side

        \cot\theta = \frac{1}{\tan\theta} = \frac{\text{Adjacent Side}}{\text{Opposite Side}}

    2. Trigonometric Ratios Table:


    1. Trigonometric Identities:

      • Pythagorean Identities:
        sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
        1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta
        1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta

    📖 Explanation:

    1. Trigonometric Ratios:
      These are ratios between the sides of a right-angled triangle with respect to an angle θ\theta. They are fundamental in solving problems involving heights and distances.

    2. Understanding the Ratios:

      • Hypotenuse: The side opposite the right angle (longest side).
      • Opposite Side: The side opposite the angle θ\theta.
      • Adjacent Side: The side next to the angle θ\theta (excluding the hypotenuse).

    3. Trigonometric Identities:
      These are equations that hold true for all values of θ\theta. They are useful for simplifying and solving trigonometric expressions.

    🔢 Step-by-Step Example:

    Example 1:
    In a right-angled triangle, if sin⁡θ=35\sin\theta = \frac{3}{5}, find cos⁡θ\cos\theta and tan⁡θ\tan\theta.

    1. Step 1: Given sin⁡θ=35\sin\theta = \frac{3}{5},

      • Opposite Side = 3
      • Hypotenuse = 5

    2. Step 2: Find the Adjacent Side using the Pythagoras theorem:

      Hypotenuse2=Opposite Side2+Adjacent Side2\text{Hypotenuse}^2 = \text{Opposite Side}^2 + \text{Adjacent Side}^2

      52=32+Adjacent Side25^2 = 3^2 + \text{Adjacent Side}^2
      25=9+Adjacent Side225 = 9 + \text{Adjacent Side}^2
      Adjacent Side2=16\text{Adjacent Side}^2 = 16
      Adjacent Side=4

      \text{Adjacent Side} = 4

    3. Step 3: Calculate cos⁡θ\cos\theta and tan⁡θ\tan\theta:

      • cos⁡θ=Adjacent SideHypotenuse=45\cos\theta = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{4}{5}
      • tan⁡θ=Opposite SideAdjacent Side=34\tan\theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{3}{4}

    Answer:

    • cos⁡θ=45

      \cos\theta = \frac{4}{5}
    • tan⁡θ=34
    • \tan\theta = \frac{3}{4}

  8. 8.Important Formula of Circles

    🧮 Key Concepts and Theorems

    1. Circle Definition:
      A circle is a set of points in a plane that are all equidistant from a fixed point called the center. The fixed distance is called the radius.

    2. Important Terms:

      • Radius: Distance from the center to any point on the circle.
      • Diameter: The longest chord of the circle, passing through the center. Diameter=2×Radius\text{Diameter} = 2 \times \text{Radius}.
      • Chord: A line segment connecting any two points on the circle.
      • Arc: A part of the circumference of a circle.
      • Sector: The region between two radii and the arc they form.
      • Tangent: A line that touches the circle at exactly one point.
      • Secant: A line that intersects the circle at two points.

    3. Tangent to a Circle:

      • A tangent to a circle is a straight line that touches the circle at only one point.
      • The point where the tangent touches the circle is called the point of contact.

    4. Theorem 1:
      The tangent to a circle is perpendicular to the radius at the point of contact.

      In △OPA\triangle OPA, if OPOP is the radius and APAP is the tangent, then:
      OP⊥AP

      OP \perp AP

    5. Theorem 2:
      The lengths of two tangents drawn from an external point to a circle are equal.

      If PAPA and PBPB are tangents from PP to the circle with center OO, then:
      PA=PBPA = PB

    📖 Explanation

    1. What is a Tangent?
      A tangent is a line that just touches the circle without crossing it. It touches the circle at a single point.

    2. Properties of a Tangent:

      • A tangent is always perpendicular to the radius at the point of contact.
      • From any point outside the circle, you can draw exactly two tangents to the circle, and these two tangents will be of equal length.

    3. Real-Life Examples of Circles and Tangents:

      • Bicycle Wheel: The ground touching the wheel at a single point is like a tangent.
      • Car Tires: The part of the tire that touches the road is an example of a tangent.

    🔢 Step-by-Step Example

    Problem:
    A circle has a radius of 5 cm. From a point PP located 13 cm away from the center OO, two tangents PAPA and PBPB are drawn to the circle. Find the length of each tangent.

    1. Given:

      • Radius r=5r = 5 cm
      • Distance OP=13OP = 13 cm

    2. Step 1: Apply the Pythagoras Theorem:
      In △OPA\triangle OPA, where:

      • OPOP is the hypotenuse = 13 cm
      • OAOA is the radius = 5 cm
      • PAPA is the tangent (to be found)

      By Pythagoras Theorem:
      OP2=OA2+PA2OP^2 = OA^2 + PA^2
      132=52+PA213^2 = 5^2 + PA^2
      169=25+PA2169 = 25 + PA^2
      PA2=169−25PA^2 = 169 - 25
      PA2=144PA^2 = 144
      PA=144=12 cmPA = \sqrt{144} = 12 \, \text{cm}

    Answer: Each tangent is 12 cm long.

  9. 9.Important Formula of Surface Areas & Volumes

    🧮 Key Formulas

    1. Surface Area and Volume of a Cube:

      • Surface Area:
        Surface Area=6a2\text{Surface Area} = 6a^2
      • Volume:
        Volume=a3\text{Volume} = a^3

      Where aa = Side of the cube

    2. Surface Area and Volume of a Cuboid:

      • Surface Area:
        Surface Area=2(lb+bh+hl)

        \text{Surface Area} = 2 (lb + bh + hl)
      • Volume:
        Volume=l×b×h\text{Volume} = l \times b \times h

      Where ll = Length, bb = Breadth, hh = Height

    3. Surface Area and Volume of a Cylinder:

      • Curved Surface Area (CSA):
        CSA=2πrh\text{CSA} = 2\pi rh
      • Total Surface Area (TSA):
        TSA=2πr(r+h)\text{TSA} = 2\pi r (r + h)
      • Volume:
        Volume=πr2h\text{Volume} = \pi r^2 h

      Where rr = Radius, hh = Height

    4. Surface Area and Volume of a Cone:

      • Curved Surface Area (CSA):
        CSA=πrl\text{CSA} = \pi rl
      • Total Surface Area (TSA):
        TSA=πr(r+l)\text{TSA} = \pi r (r + l)
      • Volume:
        Volume=13πr2h\text{Volume} = \frac{1}{3} \pi r^2 h

      Where rr = Radius, hh = Height, ll = Slant height (l=r2+h2l = \sqrt{r^2 + h^2})

    5. Surface Area and Volume of a Sphere:

      • Surface Area:
        Surface Area=4πr2\text{Surface Area} = 4\pi r^2
      • Volume:
        Volume=43πr3\text{Volume} = \frac{4}{3} \pi r^3

      Where rr = Radius

    6. Surface Area and Volume of a Hemisphere:

      • Curved Surface Area (CSA):
        CSA=2πr2\text{CSA} = 2\pi r^2
      • Total Surface Area (TSA):
        TSA=3πr2\text{TSA} = 3\pi r^2
      • Volume:
        Volume=23πr3\text{Volume} = \frac{2}{3} \pi r^3

      Where rr = Radius

    📖 Explanation

    1. Cube:

      • A cube has 6 equal square faces.
      • Example: Dice, ice cubes, and sugar cubes.

    2. Cuboid:

      • A cuboid has 6 rectangular faces.
      • Example: Bricks, matchboxes, and shoeboxes.

    3. Cylinder:

      • A cylinder has two circular faces and one curved surface.
      • Example: Water pipes, cans, and candles.

    4. Cone:

      • A cone has a circular base and a curved surface that meets at a point (vertex).
      • Example: Ice cream cones, party hats, and traffic cones.

    5. Sphere:

      • A sphere has a completely round surface with no edges.
      • Example: Balls, marbles, and planets.

    6. Hemisphere:

      • A hemisphere is half of a sphere.
      • Example: Bowls and domes.

    🔢 Step-by-Step Example

    Example 1:
    Find the surface area and volume of a cylinder with a radius of 7 cm and a height of 10 cm.

    1. Given:

      • Radius r=7r = 7 cm
      • Height h=10h = 10 cm

    2. Surface Area (TSA):
      TSA=2πr(r+h)\text{TSA} = 2\pi r (r + h)
      =2×3.14×7×(7+10)= 2 \times 3.14 \times 7 \times (7 + 10)
      =2×3.14×7×17= 2 \times 3.14 \times 7 \times 17
      =2×3.14×119= 2 \times 3.14 \times 119
      =747.46 cm2= 747.46 \, \text{cm}^2

    3. Volume:
      Volume=πr2h\text{Volume} = \pi r^2 h
      =3.14×72×10= 3.14 \times 7^2 \times 10
      =3.14×49×10= 3.14 \times 49 \times 10
      =3.14×490= 3.14 \times 490
      =1538.6 cm3= 1538.6 \, \text{cm}^3

    Answer:

    • Surface Area = 747.46 cm2747.46 \, \text{cm}^2
    • Volume = 1538.6 cm3
    • 1538.6 \, \text{सेमी}^3

  10. 10.Important Formula of Statistics

    🧮 Key Concepts and Formulas

    1. Mean (Average):
      The mean of a set of observations is the sum of all the observations divided by the total number of observations.

      Formula:
      Mean=Sum of observationsNumber of observations\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}}
      If the data is in a frequency table:
      Mean=∑fixi∑fi\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}
      Where:

      • fif_i = Frequency of the ii-th observation
      • xix_i = Value of the ii-th observation

    2. Median:
      The median is the middle value of a set of observations when arranged in ascending or descending order.

      • If the number of observations nn is odd:
        Median=Value of (n+12)-th observation\text{Median} = \text{Value of } \left(\frac{n + 1}{2}\right)\text{-th observation}
      • If nn is even:
        Median=Value of (n2)-th observation+Value of (n2+1)-th observation2

        \text{Median} = \frac{\text{Value of } \left(\frac{n}{2}\right)\text{-th observation} + \text{Value of } \left(\frac{n}{2} + 1\right)\text{-th observation}}{2}

    3. Mode:
      The mode is the observation that occurs most frequently in the dataset.

    4. Grouped Data Formulas:

      • Mean for Grouped Data:
        Mean=∑fixi∑fi

        \text{Mean} = \frac{\sum f_i x_i}{\sum f_i}

      • Median for Grouped Data:
        Median=l+(n2−Ff)×h\text{Median} = l + \left(\frac{\frac{n}{2} - F}{f}\right) \times h
        Where:

        • ll = Lower boundary of the median class
        • nn = Total frequency
        • FF = Cumulative frequency of the class before the median class
        • ff = Frequency of the median class
        • hh = Class width

      • Mode for Grouped Data:
        Mode=l+(f1−f02f1−f0−f2)×h\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h
        Where:

        • ll = Lower boundary of the modal class
        • f1f_1 = Frequency of the modal class
        • f0f_0 = Frequency of the class before the modal class
        • f2f_2 = Frequency of the class after the modal class
        • hh = Class width

    🔢 Step-by-Step Example

    Example: Calculate the mean for the following data:

    Observation xix_i 2 4 6 8 10
    Frequency fif_i 3 5 7 2 3
    1. Step 1: Calculate fixif_i x_i:


    1. Step 2: Sum of fif_i and fixif_i x_i:

      • ∑fi=3+5+7+2+3=20\sum f_i = 3 + 5 + 7 + 2 + 3 = 20
      • ∑fixi=6+20+42+16+30=114

        \sum f_i x_i = 6 + 20 + 42 + 16 + 30 = 114

    2. Step 3: Mean:

      Mean=∑fixi∑fi=11420=5.7\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{114}{20} = 5.7

    Answer: Mean = 5.7

  11. 11.Impotant Formula of Probability

    🧮 Key Concepts and Formulas

    1. Definition of Probability:
      Probability is a measure of the likelihood of an event happening. It ranges from 0 (impossible event) to 1 (certain event).


      Formula for Probability      :
      P(E)=Number of favorable outcomesTotal number of possible outcomesP(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

      Where:

      • P(E)P(E) = Probability of event EE
      • Favorable outcomes = Outcomes that satisfy the event EE
      • Total outcomes = Total possible outcomes in the experiment

    2. Types of Events:

      • Certain Event: An event that is guaranteed to happen. Probability = 1.
      • Impossible Event: An event that cannot happen. Probability = 0.
      • Equally Likely Events: Events that have the same chance of occurring.
      • Complementary Events: The probability of an event EE happening and not happening always sums up to 1.
        P(E)+P(Not E)=1

        P(E) + P(\text{Not } E) = 1

    3. Empirical Probability (Experimental Probability):
      Based on actual experiments or observations.
      P(E)=Number of times event E occursTotal number of trialsP(E) = \frac{\text{Number of times event } E \text{ occurs}}{\text{Total number of trials}}

    📖 Explanation

    1. What is Probability?
      Probability helps us measure how likely an event is to occur. It is widely used in games, weather predictions, sports, and decision-making.

    2. Example of Probability:

      • When you toss a fair coin, the possible outcomes are Head (H) and Tail (T).
      • The probability of getting a head:
        P(Head)=12

        P(\text{Head}) = \frac{1}{2}

    3. Complementary Events:

      • If the probability of rain tomorrow is 0.70.7, the probability that it will not rain is:
        P(Not Rain)=1−0.7=0.3P(\text{Not Rain}) = 1 - 0.7 = 0.3

    🔢 Step-by-Step Example

    Example:
    A bag contains 4 red balls, 5 blue balls, and 6 green balls. If one ball is drawn at random, what is the probability of drawing:

    1. A red ball

    2. A blue ball

    3. Step 1: Total Number of Balls:
      4+5+6=15

      4 + 5 + 6 = 15

    4. Step 2: Probability of a Red Ball:
      P(Red Ball)=Number of red ballsTotal number of balls=415

      P(\text{Red Ball}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{4}{15}

    5. Step 3: Probability of a Blue Ball:
      P(Blue Ball)=Number of blue ballsTotal number of balls=515=13P(\text{Blue Ball}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{5}{15} = \frac{1}{3}

    Answer:

    • Probability of drawing a red ball = 415

      \frac{4}{15}
    • Probability of drawing a blue ball = 13
    • \frac{1}{3}

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