To find this square value using an identity shortcut, we express 41 as the sum of 40 and 1. We then apply the binomial identity where a is 40 and b is 1. The square of 40 is 1600 and the square of 1 is 1. The middle term is two times 40 times 1, which equals 80. Adding 1600, 80 and 1 together givesRead more
To find this square value using an identity shortcut, we express 41 as the sum of 40 and 1. We then apply the binomial identity where a is 40 and b is 1. The square of 40 is 1600 and the square of 1 is 1. The middle term is two times 40 times 1, which equals 80. Adding 1600, 80 and 1 together gives 1681.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
We evaluate this square by rewriting 27 as a subtraction expression, 30 - 3. This matches our subtraction square identity with a equal to 30 and b equal to 3. The square of 30 is 900 and the square of 3 is 9. The middle product term to subtract is two times 30 times 3, which equals 180. Computing 90Read more
We evaluate this square by rewriting 27 as a subtraction expression, 30 – 3. This matches our subtraction square identity with a equal to 30 and b equal to 3. The square of 30 is 900 and the square of 3 is 9. The middle product term to subtract is two times 30 times 3, which equals 180. Computing 900 minus 180 plus 9 yields 729.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
We find this large square value by splitting 135 into a three-part addition expression, 100 + 30 + 5. This allows us to apply the three-term identity where a is 100, b is 30 and c is 5. The three individual squares are 10000, 900 and 25. The double cross-products are 6000, 300 and 1000. Summing allRead more
We find this large square value by splitting 135 into a three-part addition expression, 100 + 30 + 5. This allows us to apply the three-term identity where a is 100, b is 30 and c is 5. The three individual squares are 10000, 900 and 25. The double cross-products are 6000, 300 and 1000. Summing all these evaluated values together gives 18225.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
To calculate this square without direct multiplication, we express 97 as the subtraction form 100 - 3. We apply the binomial identity where a is 100 and b is 3. The square of 100 is 10000 and the square of 3 is 9. The middle product term to subtract is two times 100 times 3, which equals 600. ComputRead more
To calculate this square without direct multiplication, we express 97 as the subtraction form 100 – 3. We apply the binomial identity where a is 100 and b is 3. The square of 100 is 10000 and the square of 3 is 9. The middle product term to subtract is two times 100 times 3, which equals 600. Computing 10000 minus 600 plus 9 results in 9409.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
To find this product using an identity approach, we can break down both numbers relative to base tens, writing them as the binomial product (20 - 2)(30 - 1). Applying the distributive expansion gives 20 times 30 minus 20 times 1 minus 2 times 30 plus 2 times 1. This evaluates to 600 minus 20 minus 6Read more
To find this product using an identity approach, we can break down both numbers relative to base tens, writing them as the binomial product (20 – 2)(30 – 1). Applying the distributive expansion gives 20 times 30 minus 20 times 1 minus 2 times 30 plus 2 times 1. This evaluates to 600 minus 20 minus 60 plus 2. Solving this basic arithmetic sequence yields 522.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
Select and use the identity that will help you to find the product without multiplying directly: (41) square
To find this square value using an identity shortcut, we express 41 as the sum of 40 and 1. We then apply the binomial identity where a is 40 and b is 1. The square of 40 is 1600 and the square of 1 is 1. The middle term is two times 40 times 1, which equals 80. Adding 1600, 80 and 1 together givesRead more
To find this square value using an identity shortcut, we express 41 as the sum of 40 and 1. We then apply the binomial identity where a is 40 and b is 1. The square of 40 is 1600 and the square of 1 is 1. The middle term is two times 40 times 1, which equals 80. Adding 1600, 80 and 1 together gives 1681.
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 lessSelect and use the identity that will help you to find the product without multiplying directly: (27) square
We evaluate this square by rewriting 27 as a subtraction expression, 30 - 3. This matches our subtraction square identity with a equal to 30 and b equal to 3. The square of 30 is 900 and the square of 3 is 9. The middle product term to subtract is two times 30 times 3, which equals 180. Computing 90Read more
We evaluate this square by rewriting 27 as a subtraction expression, 30 – 3. This matches our subtraction square identity with a equal to 30 and b equal to 3. The square of 30 is 900 and the square of 3 is 9. The middle product term to subtract is two times 30 times 3, which equals 180. Computing 900 minus 180 plus 9 yields 729.
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 lessSelect and use the identity that will help you to find the product without multiplying directly: (135) square
We find this large square value by splitting 135 into a three-part addition expression, 100 + 30 + 5. This allows us to apply the three-term identity where a is 100, b is 30 and c is 5. The three individual squares are 10000, 900 and 25. The double cross-products are 6000, 300 and 1000. Summing allRead more
We find this large square value by splitting 135 into a three-part addition expression, 100 + 30 + 5. This allows us to apply the three-term identity where a is 100, b is 30 and c is 5. The three individual squares are 10000, 900 and 25. The double cross-products are 6000, 300 and 1000. Summing all these evaluated values together gives 18225.
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 the product without multiplying directly: (97) square
To calculate this square without direct multiplication, we express 97 as the subtraction form 100 - 3. We apply the binomial identity where a is 100 and b is 3. The square of 100 is 10000 and the square of 3 is 9. The middle product term to subtract is two times 100 times 3, which equals 600. ComputRead more
To calculate this square without direct multiplication, we express 97 as the subtraction form 100 – 3. We apply the binomial identity where a is 100 and b is 3. The square of 100 is 10000 and the square of 3 is 9. The middle product term to subtract is two times 100 times 3, which equals 600. Computing 10000 minus 600 plus 9 results in 9409.
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 the product without multiplying directly: (18 x 29)
To find this product using an identity approach, we can break down both numbers relative to base tens, writing them as the binomial product (20 - 2)(30 - 1). Applying the distributive expansion gives 20 times 30 minus 20 times 1 minus 2 times 30 plus 2 times 1. This evaluates to 600 minus 20 minus 6Read more
To find this product using an identity approach, we can break down both numbers relative to base tens, writing them as the binomial product (20 – 2)(30 – 1). Applying the distributive expansion gives 20 times 30 minus 20 times 1 minus 2 times 30 plus 2 times 1. This evaluates to 600 minus 20 minus 60 plus 2. Solving this basic arithmetic sequence yields 522.
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