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Find possible expressions for the length and breadth of the rectangle whose area is 36s square – 49t square
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/
See lessFind possible expressions for the length and breadth of the rectangle whose area is 25a square – 30ab + 9b square
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/
See lessFactor using suitable identities: 9a square + 4b square + c square – 12ab + 6ac – 4bc
To factor this six-term polynomial, we look at the signs of the products to determine which base is negative. The perfect square bases are 3a, 2b and c. We notice that the terms 12ab and 4bc are negative, while 6ac is positive. Since the negative terms both contain the variable b, the base associateRead more
To factor this six-term polynomial, we look at the signs of the products to determine which base is negative. The perfect square bases are 3a, 2b and c. We notice that the terms 12ab and 4bc are negative, while 6ac is positive. Since the negative terms both contain the variable b, the base associated with b must carry the negative sign. This results in the factor (3a – 2b + c) square.
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/
See lessExpand using the identity (a + b + c) square: (p + 3q + 7r) square
We use the three-term square identity to expand the given expression. By setting a equal to p, b equal to 3q and c equal to 7r, we square each term to get p square, 9q square and 49r square. Then we calculate the three double cross-product terms: two times p times 3q gives 6pq; two times 3q times 7rRead more
We use the three-term square identity to expand the given expression. By setting a equal to p, b equal to 3q and c equal to 7r, we square each term to get p square, 9q square and 49r square. Then we calculate the three double cross-product terms: two times p times 3q gives 6pq; two times 3q times 7r gives 42qr; two times 7r times p gives 14rp. Combining them gives the full expansion.
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/
See lessExpand using the identity (a + b + c) square: (3x – 2y + 4z) square
We expand this expression using the three-term standard identity. Here, our terms are 3x, minus 2y and 4z. Squaring these individual parts results in 9x square, 4y square and 16z square. When calculating the paired products, any term multiplied by minus 2y becomes negative, resulting in minus 12xy aRead more
We expand this expression using the three-term standard identity. Here, our terms are 3x, minus 2y and 4z. Squaring these individual parts results in 9x square, 4y square and 16z square. When calculating the paired products, any term multiplied by minus 2y becomes negative, resulting in minus 12xy and minus 16yz. The final product two times 4z times 3x stays positive at 24zx.
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/
See less