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Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained: (i) 525 (ii) 1750 (iii) 252 (iv) 1825 (v) 6412

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NCERT Solutions for Class 8 Mathematics Chapter 6
Important NCERT Questions
Square and Square Roots Chapter 6 Exercise 6.4
NCERT Books for Session 2022-2023
CBSE Board
Questions No: 5

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1 Answer

  1. (i) 525
    Since remainder is 41. Therefore 22²
    Next perfect square number 23² = 529
    Hence, number to be added = 529 – 525 = 4
    ∴ 525+4=529
    Hence, the square root of 529 is 23.

    (ii) 1750
    Since remainder is 69. Therefore 41²<1750
    Next perfect square number 42²=1764
    Hence, number to be added = 1764 – 1750 = 14
    ∴ 1750+14=1764
    Hence, the square root of 1764 is 42.

    (iii) 252
    Since remainder is 27. Therefore 152 < 252
    Next perfect square number 162 = 256
    Hence, number to be added=256-252=4
    ∴ 252+4=256
    Hence, the square root of 256 is 16.

    (iv) 1825
    Since remainder is 61. Therefore 42²<1825
    Next perfect square number 43² = 1849
    Hence, number to be added = 1849 – 1825 = 24
    ∴ 1825 + 24 = 1849
    Hence, the square root of 1849 is 43.

    (v) 6412
    Since remainder is 12. Therefore 80²<6412
    Next perfect square number 81²=6561
    Hence, number to be added = 6561 – 6412 = 149
    ∴ 6412 + 149 = 6561
    Hence, the square root of 6561 is 81.

    Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-6/

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