1. The total number of tiny squares in the image is 144, arranged in a 12 × 12 square. The prime factorisation of 144 is: 144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3² All exponents are even, so it confirms 144 is a perfect square. So, number of tiny squares = 144 Prime factorisation: 2⁴ × 3² Square root: 12Read more

    The total number of tiny squares in the image is 144, arranged in a 12 × 12 square. The prime factorisation of 144 is:
    144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3²
    All exponents are even, so it confirms 144 is a perfect square.
    So, number of tiny squares = 144
    Prime factorisation: 2⁴ × 3²
    Square root: 12

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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    • 92
  2. (i) 16² = 256, 17² = 289. So, numbers between them = 289 – 256 – 1 = 32. (ii) 99² = 9801, 100² = 10000. So, numbers between = 10000 – 9801 – 1 = 198. In general, numbers between n² and (n+1)² are 2n. Hence, between 16² and 17² → 32 numbers; between 99² and 100² → 198 numbers.   For more NCERT SRead more

    (i) 16² = 256, 17² = 289. So, numbers between them = 289 – 256 – 1 = 32.
    (ii) 99² = 9801, 100² = 10000. So, numbers between = 10000 – 9801 – 1 = 198.
    In general, numbers between n² and (n+1)² are 2n.
    Hence, between 16² and 17² → 32 numbers; between 99² and 100² → 198 numbers.

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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    • 8
  3. Prime factorisation of 9408 = 2⁶ × 3 × 7². For a perfect square, all powers must be even. The factor 3 is unpaired. Multiply 9408 by 3 → 28224 = 2⁶ × 3² × 7². Now, all powers are even. √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168. So, the smallest number to multiply is 3 and the square root of the result iRead more

    Prime factorisation of 9408 = 2⁶ × 3 × 7². For a perfect square, all powers must be even. The factor 3 is unpaired. Multiply 9408 by 3 → 28224 = 2⁶ × 3² × 7². Now, all powers are even. √28224 = 2³ × 3 × 7 = 8 × 3 × 7 = 168.
    So, the smallest number to multiply is 3 and the square root of the result is 168.

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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    • 90
  4. To find the smallest square divisible by 4, 9 and 10, we first compute their LCM: 4 = 2², 9 = 3², 10 = 2 × 5 ⇒ LCM = 2² × 3² × 5 = 180. Now find the smallest perfect square that includes all these prime factors in even powers. 900 = 2² × 3² × 5² = 30², which is a perfect square and divisible by 180.Read more

    To find the smallest square divisible by 4, 9 and 10, we first compute their LCM:
    4 = 2², 9 = 3², 10 = 2 × 5 ⇒ LCM = 2² × 3² × 5 = 180.
    Now find the smallest perfect square that includes all these prime factors in even powers. 900 = 2² × 3² × 5² = 30², which is a perfect square and divisible by 180.
    Answer: 900.

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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    • 6
  5. If the area of a square is 441 m², then the side is the square root of 441. Since 21 × 21 = 441, √441 = 21. So, the side of the square is 21 meters long. This follows the formula: Area = side × side → side = √Area This method helps in reverse-calculating side lengths from square areas.   For moRead more

    If the area of a square is 441 m², then the side is the square root of 441. Since 21 × 21 = 441, √441 = 21. So, the side of the square is 21 meters long. This follows the formula:
    Area = side × side → side = √Area
    This method helps in reverse-calculating side lengths from square areas.

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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