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Find the square of 78 using a suitable identity
We evaluate this calculation by changing the number 78 into the expression 80 - 2. This lets us use the standard subtraction square identity where a is set to 80 and b is set to 2. The square of 80 is 6400, and the square of 2 is 4. The product term to subtract is two multiplied by 80 multiplied byRead more
We evaluate this calculation by changing the number 78 into the expression 80 – 2. This lets us use the standard subtraction square identity where a is set to 80 and b is set to 2. The square of 80 is 6400, and the square of 2 is 4. The product term to subtract is two multiplied by 80 multiplied by 2, which is 320. Computing 6400 minus 320 plus 4 yields 6084.
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 square of 198 using a suitable identity
To find this square value quickly, we convert 198 into 200 - 2. We apply the identity by substituting the values where a equals 200 and b equals 2. Squaring the first term gives 40000, and squaring the second term gives 4. The intermediate term to subtract is calculated as two times 200 times 2, whiRead more
To find this square value quickly, we convert 198 into 200 – 2. We apply the identity by substituting the values where a equals 200 and b equals 2. Squaring the first term gives 40000, and squaring the second term gives 4. The intermediate term to subtract is calculated as two times 200 times 2, which equals 800. Subtracting 800 from 40000 and adding 4 results in 39204.
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 square of 214 using a suitable identity
We evaluate the square of 214 by breaking it down into the three-term addition expression 200 + 10 + 4. This fits the identity formula where a is 200, b is 10, and c is 4. The squares are 40000, 100, and 16. The cross-products are two times each pair combination, which gives 4000, 1600, and 80. AddiRead more
We evaluate the square of 214 by breaking it down into the three-term addition expression 200 + 10 + 4. This fits the identity formula where a is 200, b is 10, and c is 4. The squares are 40000, 100, and 16. The cross-products are two times each pair combination, which gives 4000, 1600, and 80. Adding all these numerical values together gives 45796.
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 square of 1104 using a suitable identity
For this larger number, we use the identity by rewriting 1104 as the sum of 1000, 100, and 4. By assigning these values to a, b, and c in our expansion formula, we calculate the squares as 1000000, 10000, and 16. The double-product terms equal 200000, 8000, and 800. Combining all six evaluated partsRead more
For this larger number, we use the identity by rewriting 1104 as the sum of 1000, 100, and 4. By assigning these values to a, b, and c in our expansion formula, we calculate the squares as 1000000, 10000, and 16. The double-product terms equal 200000, 8000, and 800. Combining all six evaluated parts results in the final output of 1218816.
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 square of 1120 using a suitable identity
To calculate this square cleanly, we break down 1120 into the addition form 1100 + 20. We apply our regular binomial square identity where a equals 1100 and b equals 20. The square of 1100 is 1210000, and the square of 20 is 400. The middle product term is computed as two multiplied by 1100 multipliRead more
To calculate this square cleanly, we break down 1120 into the addition form 1100 + 20. We apply our regular binomial square identity where a equals 1100 and b equals 20. The square of 1100 is 1210000, and the square of 20 is 400. The middle product term is computed as two multiplied by 1100 multiplied by 20, which equals 44000. Summing these values gives 1254400.
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