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For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained: (i) 252 (ii) 180(iii) 1008 (iv) 2028 (v) 1458 (vi) 768

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NCERT Solutions for Class 8 Mathematics Chapter 6
Important NCERT Questions
Square and Square Roots Chapter 6 Exercise 6.3
NCERT Books for Session 2022-2023
CBSE Board
Questions No: 5

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1 Answer

  1. (i) 252 = 2 x 2 x 3 x 3 x 7
    Here, prime factor 7 has no pair. Therefore 252 must be
    multiplied by 7 to make it a perfect square.
    ∴ 252 x 7 = 1764
    and √1764 = 2x3x7=42

    (ii) 180 = 2 x 2 x 3 x 3 x 5
    Here, prime factor 5 has no pair. Therefore 180 must be
    multiplied by 5 to make it a perfect square.
    ∴ 180 x 5 = 900
    and √900= 2x3x5=30

    (iii) 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7
    Here, prime factor 7 has no pair. Therefore 1008 must be
    multiplied by 7 to make it a perfect square.
    ∴ 1008 x 7 = 7056
    and √7056 = 2x2x3x7 =84

    (iv) 2028 = 2 x 2 x 3 x 13 x 13
    Here, prime factor 3 has no pair. Therefore 2028 must be
    multiplied by 3 to make it a perfect square.
    ∴ 2028 x 3 = 6084
    And √6080 = √2x2x3x3x13x13 =78

    (v) 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3
    Here, prime factor 2 has no pair. Therefore 1458 must be
    multiplied by 2 to make it a perfect square.
    ∴ 1458 x 2 = 2916
    and √2916 = 2x3x3x3 = 54

    (vi) 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
    Here, prime factor 3 has no pair. Therefore 768 must be
    multiplied by 3 to make it a perfect square.
    ∴ 768 x 3 = 2304
    and √2304 = 2x2x2x2x3 = 48

    Class 8 Maths Chapter 6 Exercise 6.3 Solution in Video

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

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