1. In a standard square root spiral, the first right triangle has perpendicular sides of length 1 unit each. By the Pythagorean theorem, its hypotenuse is the square root of 2. The next right triangle is built using this hypotenuse as a base and adding a perpendicular side of 1 unit, making its hypotenRead more

    In a standard square root spiral, the first right triangle has perpendicular sides of length 1 unit each. By the Pythagorean theorem, its hypotenuse is the square root of 2. The next right triangle is built using this hypotenuse as a base and adding a perpendicular side of 1 unit, making its hypotenuse the square root of 3. This mathematical pattern continues sequentially for all successive right triangles, creating hypotenuse lengths of root 2, root 3, root 4, root 5, and so on.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/

    See less
    • 40
  2. We are given two equations: x + y + z = 0 and xy + yz + zx = 0. Using the algebraic identity, we know that (x + y + z) squared equals x2 + y2 + z2 + 2(xy + yz + zx). Substituting our given values into this identity results in 0 squared equals x2 + y2 + z2 + 2(0), which simplifies directly to x2 + y2Read more

    We are given two equations: x + y + z = 0 and xy + yz + zx = 0. Using the algebraic identity, we know that (x + y + z) squared equals x2 + y2 + z2 + 2(xy + yz + zx). Substituting our given values into this identity results in 0 squared equals x2 + y2 + z2 + 2(0), which simplifies directly to x2 + y2 + z2 = 0. Since the square of any real rational number is always non-negative, their sum can only equal zero if x, y, and z are all simultaneously zero.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/

    See less
    • 35
  3. The denominator of the rational number in its lowest form is given as 2 cubed times 5, which equals 8 times 5, or 40. A fraction with a denominator of the form 2 raised to m times 5 raised to n will always have a terminating decimal expansion. The number of decimal places is determined by the higherRead more

    The denominator of the rational number in its lowest form is given as 2 cubed times 5, which equals 8 times 5, or 40. A fraction with a denominator of the form 2 raised to m times 5 raised to n will always have a terminating decimal expansion. The number of decimal places is determined by the higher exponent of the prime factors 2 and 5. Since the exponent of 2 is 3, it will have exactly 3 decimal places.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/

    See less
    • 40
  4. To find 5 rational numbers between 1/6 and 2/5, we first convert them to have a common denominator. The least common multiple of 6 and 5 is 30, making the fractions 5/30 and 12/30. Since there is a sufficient gap between the numerators 5 and 12, we can directly choose five intermediate values. Thus,Read more

    To find 5 rational numbers between 1/6 and 2/5, we first convert them to have a common denominator. The least common multiple of 6 and 5 is 30, making the fractions 5/30 and 12/30. Since there is a sufficient gap between the numerators 5 and 12, we can directly choose five intermediate values. Thus, the five rational numbers are 6/30, 7/30, 8/30, 9/30, and 10/30.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/

    See less
    • 38
  5. To find 5 rational numbers between 2/5 and 3/5, we can scale up the fractions to have a larger common denominator. Multiplying the numerator and denominator of both fractions by 6 transforms them into 12/30 and 18/30. Now, we can easily select five consecutive numerators that lie between 12 and 18.Read more

    To find 5 rational numbers between 2/5 and 3/5, we can scale up the fractions to have a larger common denominator. Multiplying the numerator and denominator of both fractions by 6 transforms them into 12/30 and 18/30. Now, we can easily select five consecutive numerators that lie between 12 and 18. This gives the five required rational numbers: 13/30, 14/30, 15/30, 16/30, and 17/30.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/

    See less
    • 33