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MATH CBSE 9TH SECOND CHAPTER

Introduction to Euclid’s Geometry with detailed explanations, examples, and solutions to help you excel..

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Example 1: Understanding Postulates

Question

Using Euclid’s postulates, prove that:
“A straight line can be drawn from any one point to any other point.”

Solution

  1. Postulate Referenced:
    • Euclid’s First Postulate states, “A straight line can be drawn from any one point to any other point.”
  2. Explanation:
    • Consider two points AA and BB in a plane.
    • According to the postulate, there exists one and only one straight line that passes through both AA and BB.
  3. Diagram:
    Draw a plane with two points labeled AA and BB, and connect them with a straight line.
    This represents the unique straight line as per the postulate.
  4. Conclusion:
    The statement is a direct application of Euclid’s First Postulate.

Example 2: Understanding Axioms

Question

Explain the axiom: “Things that are equal to the same thing are equal to one another.” Illustrate with an example.

Solution

  1. Axiom Statement:
    • This is Euclid’s First Axiom, which states that if two quantities are each equal to a third quantity, they are equal to each other.
  2. Explanation:
    • Let a=ba = b and b=cb = c.
    • Since both aa and cc are equal to bb, it follows that a=ca = c.
  3. Example:
    • Suppose AB=5 cmAB = 5 \, \text{cm}, BC=5 cmBC = 5 \, \text{cm}.
    • Since ABAB and BCBC are both equal to 5 cm, we conclude AB=BCAB = BC.
  4. Diagram:
    Draw a line segment divided into two parts ABAB and BCBC, both labeled 5 cm5 \, \text{cm}.

Example 3: Application of Euclid’s Fifth Postulate

Question

Explain the significance of Euclid’s Fifth Postulate and provide an example.

Solution

  1. Postulate Statement:
    • Euclid’s Fifth Postulate states, “If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side.”
  2. Explanation:
    • This is the basis for parallel and non-parallel lines.
    • If the sum of the interior angles is less than 180∘180^\circ, the lines will meet on that side.
  3. Example:
    • Consider two lines l1l_1 and l2l_2, and a transversal tt.
    • Measure the interior angles on the same side of the transversal.
      • If ∠1+∠2<180∘\angle 1 + \angle 2 < 180^\circ, the lines l1l_1 and l2l_2 will meet.
  4. Diagram:
    Draw two lines and a transversal, showing interior angles ∠1\angle 1 and ∠2\angle 2.

Example 4: Exam-Style Proof

Question

Prove that:
The sum of the angles in a triangle is 180∘180^\circ.

Solution

  1. Given: A triangle ABCABC.
  2. To Prove: ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ.
  3. Construction:
    • Draw a line DEDE parallel to BCBC through AA.
    • Extend ABAB and ACAC to meet DEDE.
  4. Proof:
    • By the Alternate Interior Angle Theorem, ∠CAB=∠DAB,∠ACB=∠EAC.\angle CAB = \angle DAB, \quad \angle ACB = \angle EAC.
    • Along the straight line DEDE, ∠DAB+∠BAC+∠EAC=180∘.\angle DAB + \angle BAC + \angle EAC = 180^\circ.
    • Substituting, ∠CAB+∠ABC+∠BCA=180∘.\angle CAB + \angle ABC + \angle BCA = 180^\circ.
  5. Conclusion: ∠A+∠B+∠C=180∘.\angle A + \angle B + \angle C = 180^\circ.
  6. Diagram:
    Draw triangle ABCABC with the parallel line DEDE and the angles marked.

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Polynomials, a critical evaluation and detailed explanation of its key concepts, with examples and problem-solving techniques

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Key Concepts in Polynomials

  1. Definition of a Polynomial: A polynomial is an algebraic expression consisting of variables, coefficients, and exponents, where the exponents are whole numbers. Examples:
    • Polynomial: 2×3−3×2+5x−72x^3 – 3x^2 + 5x – 7
    • Not a Polynomial: 2x−1+x2x^{-1} + \sqrt{x} (exponent is not a whole number).
  2. Degrees of a Polynomial:
    • The degree is the highest power of the variable in the polynomial.
    • Examples:
      • 5×3−2×2+45x^3 – 2x^2 + 4: Degree = 3.
      • 7y5+3y4+y7y^5 + 3y^4 + y: Degree = 5.
  3. Types of Polynomials Based on Degree:
    • Linear Polynomial: Degree = 1 (e.g., 3x+23x + 2).
    • Quadratic Polynomial: Degree = 2 (e.g., x2−4x+3x^2 – 4x + 3).
    • Cubic Polynomial: Degree = 3 (e.g., 2×3−x2+3x−52x^3 – x^2 + 3x – 5).
  4. Types of Polynomials Based on Terms:
    • Monomial: One term (e.g., 3x3x).
    • Binomial: Two terms (e.g., x2+2xx^2 + 2x).
    • Trinomial: Three terms (e.g., x3+x2+2x^3 + x^2 + 2).

Key Operations and Concepts

1. Addition, Subtraction, and Multiplication

  • Combine like terms while performing operations.

Example: Simplify (2×2+3x+4)+(x2−5x+6)(2x^2 + 3x + 4) + (x^2 – 5x + 6): =(2×2+x2)+(3x−5x)+(4+6).= (2x^2 + x^2) + (3x – 5x) + (4 + 6). =3×2−2x+10.= 3x^2 – 2x + 10.

2. Division of Polynomials

Division is performed using long division.

Example: Divide 2×3+3×2−x+52x^3 + 3x^2 – x + 5 by x+1x + 1.

  1. Divide 2x32x^3 by xx: Quotient = 2x22x^2.
  2. Multiply 2x22x^2 by x+1x + 1: 2×3+2x22x^3 + 2x^2.
  3. Subtract: (2×3+3×2−x+5)−(2×3+2×2)=x2−x+5(2x^3 + 3x^2 – x + 5) – (2x^3 + 2x^2) = x^2 – x + 5.
  4. Repeat until the degree of the remainder is less than the degree of the divisor.

The Remainder Theorem

If p(x)p(x) is divided by x−ax – a, the remainder is p(a)p(a).

Example: Find the remainder when p(x)=x3−3×2+4x−2p(x) = x^3 – 3x^2 + 4x – 2 is divided by x−2x – 2.

  1. Substitute x=2x = 2 into p(x)p(x): p(2)=23−3(22)+4(2)−2.p(2) = 2^3 – 3(2^2) + 4(2) – 2. =8−12+8−2=2.= 8 – 12 + 8 – 2 = 2.
  2. The remainder is 2.

The Factor Theorem

A polynomial p(x)p(x) has x−ax – a as a factor if p(a)=0p(a) = 0.

Example: Check if x−1x – 1 is a factor of p(x)=x3−4×2+3xp(x) = x^3 – 4x^2 + 3x.

  1. Substitute x=1x = 1 into p(x)p(x): p(1)=13−4(12)+3(1)=1−4+3=0.p(1) = 1^3 – 4(1^2) + 3(1) = 1 – 4 + 3 = 0.
  2. Since p(1)=0p(1) = 0, x−1x – 1 is a factor.

Zeros of a Polynomial

The zeros of a polynomial are the values of xx for which p(x)=0p(x) = 0.

Key Results:

  1. A polynomial of degree nn has at most nn zeros.
  2. Zeros can be real or complex numbers.

Example: Find the zeros of p(x)=x2−5x+6p(x) = x^2 – 5x + 6.

  1. Factorize: p(x)=(x−2)(x−3).p(x) = (x – 2)(x – 3).
  2. Zeros: x=2,x=3x = 2, x = 3.

Graphical Representation

The graph of a polynomial depends on its degree and coefficients:

  1. Linear Polynomials: Straight line.
  2. Quadratic Polynomials: Parabola.
  3. Cubic Polynomials: S-shaped curve.

Application Problems

Problem 1: Sum and Product of Zeros

If p(x)=ax2+bx+cp(x) = ax^2 + bx + c, the sum and product of zeros are: Sum of zeros=−ba,Product of zeros=ca.\text{Sum of zeros} = -\frac{b}{a}, \quad \text{Product of zeros} = \frac{c}{a}.

Example: For p(x)=2×2−5x+3p(x) = 2x^2 – 5x + 3:

  1. Sum of zeros: =−−52=52.= -\frac{-5}{2} = \frac{5}{2}.
  2. Product of zeros: =32.= \frac{3}{2}.

Problem 2: Verify Factor Theorem

Verify that x−2x – 2 is a factor of p(x)=x3−3×2+4x−8p(x) = x^3 – 3x^2 + 4x – 8.

  1. Substitute x=2x = 2 into p(x)p(x): p(2)=23−3(22)+4(2)−8=8−12+8−8=0.p(2) = 2^3 – 3(2^2) + 4(2) – 8 = 8 – 12 + 8 – 8 = 0.
  2. Since p(2)=0p(2) = 0, x−2x – 2 is a factor.

Critical Observations

  1. Importance of Remainder and Factor Theorem:
    • Simplifies finding roots and divisors of polynomials.
    • Basis for synthetic division.
  2. Role of Degree:
    • The degree dictates the behavior of polynomials (number of zeros, shape of graph).
  3. Graphical Interpretation:
    • Understanding graphs builds intuition about polynomial behavior.
  4. Applications:
    • Used in physics (trajectory of objects), economics (profit functions), and data fitting.

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