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):
We solve this multiplication by centering the numbers around the base value 40, transforming the problem into (40 - 6) multiplied by (40 + 3). This fits the identity formula matching the template (x + a)(x + b) = x square + (a + b)x + ab. Here, x is 40, a is minus 6 and b is 3. This gives 1600 minusRead more
We solve this multiplication by centering the numbers around the base value 40, transforming the problem into (40 – 6) multiplied by (40 + 3). This fits the identity formula matching the template (x + a)(x + b) = x square + (a + b)x + ab. Here, x is 40, a is minus 6 and b is 3. This gives 1600 minus 120 minus 18, resulting in 1462.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
To find the value using an algebraic shortcut, we rewrite 205 as the sum of 200 and 5. This lets us use the regular binomial identity where a equals 200 and b equals 5. The square of 200 is 40000 and the square of 5 is 25. The intermediate term is two times 200 times 5, which equals 2000. Adding 400Read more
To find the value using an algebraic shortcut, we rewrite 205 as the sum of 200 and 5. This lets us use the regular binomial identity where a equals 200 and b equals 5. The square of 200 is 40000 and the square of 5 is 25. The intermediate term is two times 200 times 5, which equals 2000. Adding 40000, 2000 and 25 together gives 42025.
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: (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 lessFind the product without multiplying directly: (34 x 43)
We solve this multiplication by centering the numbers around the base value 40, transforming the problem into (40 - 6) multiplied by (40 + 3). This fits the identity formula matching the template (x + a)(x + b) = x square + (a + b)x + ab. Here, x is 40, a is minus 6 and b is 3. This gives 1600 minusRead more
We solve this multiplication by centering the numbers around the base value 40, transforming the problem into (40 – 6) multiplied by (40 + 3). This fits the identity formula matching the template (x + a)(x + b) = x square + (a + b)x + ab. Here, x is 40, a is minus 6 and b is 3. This gives 1600 minus 120 minus 18, resulting in 1462.
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: (205) square
To find the value using an algebraic shortcut, we rewrite 205 as the sum of 200 and 5. This lets us use the regular binomial identity where a equals 200 and b equals 5. The square of 200 is 40000 and the square of 5 is 25. The intermediate term is two times 200 times 5, which equals 2000. Adding 400Read more
To find the value using an algebraic shortcut, we rewrite 205 as the sum of 200 and 5. This lets us use the regular binomial identity where a equals 200 and b equals 5. The square of 200 is 40000 and the square of 5 is 25. The intermediate term is two times 200 times 5, which equals 2000. Adding 40000, 2000 and 25 together gives 42025.
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