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  1. Asked: In:

    There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement?

    Yati

    Here, Number of children = 500 By getting the square root of this number, we get, In each row, the number of children is 22. And left out children are 16. 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/chapRead more

    Here, Number of children = 500
    By getting the square root of this number, we get,
    In each row, the number of children is 22.
    And left out children are 16.

    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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  2. Asked: In:

    A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and number of columns remain same. Find the minimum number of plants he needs more for this.

    Yati

    Here, plants = 1000 Since remainder is 39. Therefore 31² < 1000 Next perfect square number 32² = 1024 Hence, number to be added = 1024 – 1000 = 24 ∴ 1000 + 24 = 1024 Hence, the gardener required 24 more plants. Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video for more answers vist to: httpsRead more

    Here, plants = 1000
    Since remainder is 39. Therefore 31² < 1000
    Next perfect square number 32² = 1024
    Hence, number to be added = 1024 – 1000 = 24
    ∴ 1000 + 24 = 1024
    Hence, the gardener required 24 more plants.

    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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  3. Asked: In:

    In a right triangle ABC, ∠B = 90°. (i) If AB = 6 cm, BC = 8 cm, find AC. (ii) If AC = 13 cm, BC = 5 cm, find AB

    Yati

    (i) Using Pythagoras theorem, AC² = AB² + BC² ⇒ AC² = (6) ² + (8) ² ⇒ AC² = 36 + 84 = 100 ⇒ AC = 10 cm (ii) Using Pythagoras theorem, AC² = AB² + BC² ⇒ (13) ² = AB² + (5) ² ⇒ 169 = AB² + 25 ⇒ AB² = 169 – 25 ⇒ AB² = 144 ⇒ AB = 12 cm Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video for more answRead more

    (i) Using Pythagoras theorem,
    AC² = AB² + BC²
    ⇒ AC² = (6) ² + (8) ²
    ⇒ AC² = 36 + 84 = 100
    ⇒ AC = 10 cm
    (ii) Using Pythagoras theorem,
    AC² = AB² + BC²
    ⇒ (13) ² = AB² + (5) ²
    ⇒ 169 = AB² + 25
    ⇒ AB² = 169 – 25
    ⇒ AB² = 144
    ⇒ AB = 12 cm

    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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  4. Asked: In:

    Find the length of the side of a square whose area is 441 m²?

    Yati

    Let the length of side of a square be x meter. Area of square = (side)² = x²=441 ⇒ x=√441 = √3x3x7x7 = 3x7 ⇒ x = 21m Hence, the length of side of a square is 21 m. Class 8 Maths Chapter 6 Exercise 6.4 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/mRead more

    Let the length of side of a square be x meter.
    Area of square = (side)² = x²=441
    ⇒ x=√441 = √3x3x7x7 = 3×7
    ⇒ x = 21m
    Hence, the length of side of a square is 21 m.

    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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  6. Asked: In:

    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

    Yati

    (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 – 1Read more

    (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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