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  1. The inner playground is a square with an area of 40 times 40, which equals 1600 square metres. When a path of width s is added around all sides, the new outer boundary forms a larger square with a total side length of 40 + 2s metres. The total outer area is the square of (40 + 2s), which expands toRead more

    The inner playground is a square with an area of 40 times 40, which equals 1600 square metres. When a path of width s is added around all sides, the new outer boundary forms a larger square with a total side length of 40 + 2s metres. The total outer area is the square of (40 + 2s), which expands to 1600 + 160s + 4s square. Subtracting the inner playground area leaves the path area.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/

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  2. We find the dimensions by factoring the given cubic volume polynomial step by step. First, we pull out the common term 3p from all three components, yielding 3p times the trinomial s square - 5s + 4. Next, we apply the middle term splitting technique to this remaining quadratic expression. The middlRead more

    We find the dimensions by factoring the given cubic volume polynomial step by step. First, we pull out the common term 3p from all three components, yielding 3p times the trinomial s square – 5s + 4. Next, we apply the middle term splitting technique to this remaining quadratic expression. The middle coefficient minus 5 splits into minus 1 and minus 4, which solves into the final three dimensional factors.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/

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  3. To find three dimensional expressions for a cuboid volume, we must fully factorize the given expression into three parts. First, we isolate the highest common integer factor, which is 6. This gives 6 multiplied by the binomial quantity a square - 4b square. The bracketed part is a difference of squaRead more

    To find three dimensional expressions for a cuboid volume, we must fully factorize the given expression into three parts. First, we isolate the highest common integer factor, which is 6. This gives 6 multiplied by the binomial quantity a square – 4b square. The bracketed part is a difference of squares that expands into the linear components (a + 2b) and (a – 2b). These three distinct factors constitute the dimensions.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/

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  4. The area of the rectangle is given as a binomial with two perfect squares separated by a minus sign. We utilize the difference of squares algebraic identity to factor this expression. Here, the term 36s square is the square of 6s and the term 49t square is the square of 7t. Factoring gives the produRead more

    The area of the rectangle is given as a binomial with two perfect squares separated by a minus sign. We utilize the difference of squares algebraic identity to factor this expression. Here, the term 36s square is the square of 6s and the term 49t square is the square of 7t. Factoring gives the product of the sum and difference of these bases, yielding the length and breadth expressions.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/

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  5. To find the dimensions of the rectangle, we look for the factors of the given quadratic area polynomial. The expression matches the perfect square subtraction identity. The first term 25a square is the square of 5a and the last term 9b square is the square of 3b. The middle term minus 30ab representRead more

    To find the dimensions of the rectangle, we look for the factors of the given quadratic area polynomial. The expression matches the perfect square subtraction identity. The first term 25a square is the square of 5a and the last term 9b square is the square of 3b. The middle term minus 30ab represents minus two times 5a times 3b. Therefore, it factors completely into (5a – 3b) square, giving the possible side lengths.

     

    For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/

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