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CBSE Tuition CBSE/Class 10/Mathematics

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A Mathematician With 2-3 Years of Experience Teaching

By Checking the Digit at the Units Place. There will be a pattern.
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Easy methods and tricks are not valid all the time, so basic method are always applicable.
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Yes, there are. I am trying to explain step by step.Say we want the square of n = 67. First of all, the square of 7 is 49. Write 9 in the answer place and keep 4 for carrying over process.∴n^2 = 9Then, see that 6 and 7 form the number 67. Their product is 42. The rule says you have to double it.... read more

Yes, there are. I am trying to explain step by step.
Say we want the square of n = 67. First of all, the square of 7 is 49. Write 9 in the answer place and keep 4 for carrying over process.
∴n^2 = 9
Then, see that 6 and 7 form the number 67. Their product is 42. The rule says you have to double it. Hence you get double of 42 as 84. Also, you have 4 in your hand for carrying over. Together you get 84+4=88. Write the unit digit eight beside 9 and keep 10s digit 8 for next round carry-over process.
∴ n^2 = 89
Finally, the 10s digit is 6. Square it, and you will get 36. Plus you have 8 in your hand. Together we get 36+8, i.e. 44. Now write it beside 89, and we are done.
∴ 67^2 = 4489.
Try your self: 39^2, 23^2, 52^2.
 
Verify your answer using a calculator.
 
For squaring three digits or more, there is procedure indeed, but I can't explain them here within short place. 
 
Although I can show you some other special technique for squaring 2 digit numbers of special types. 
 
1. If the number ends with 5: 
Let's say we need to square 85. First, write 25 straight forward in answer place. 
85^2 = 25
 
Then, note that after ignoring five we are left with the number 8 in the given number 85. The next number of 8 is 9. The rule says to make the product which is in our present case is 8 x 9, i.e. 72.
 
Write it down, and we are done.
85^2 = 7225.
Similarly, 95^2 = 9025 because 9 x 10 = 90 and 25 already is written etc.
115^2 = 13225 because 11 x 12 = 132. Etc
 
2. If the number is near to 100.
Let's say we need to square 93. 
 
The difference between 93 and 100 is 7. The rule says to write down the square of 7 first, which is here 49.
Next, decrease 93 by 7. Which gives 86. Hence the final answer becomes 
93^2=8649.
Similarly, 95^2 is 9025 because the difference between 95 and 100 is 5. The square of 5 is 25, and if we decrease 95 by 5, we shall get 90
 Together we thus obtain 
95^2= 9025.
Notice we have achieved the same already before by using the general squaring technique explained above.
 
Caution. Suppose we want to square 98. This time doesn't be tempted to write 964. That's not correct. While you are squaring the difference between 100 and the given number, make sure the answer occupies exactly two places only. If not, use 0 to fill up the lattice. Hence 98^2 is not 964 but in 9604.
 
Similarly, for 88^2, first, write down the square of the difference which is 12^2, i.e. 144. But this time it is a three-digit answer. So keep the last two digits and carry over the excess digit which is 1 in this present case. Now decrease 88 by 12, and we get 76. Added by carried over one we get 77. Hence 88^2 = 7744.
 
There are more so many special rules. But we can discuss them some other days.
 
Take care.

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Tutor with 2 Years of Experience

Yes there are many methods to find out the square of a number. For 2 digit numnber, you can simply multiply. It is easier only. For 3 digit or more you cancalculate by vadic multiplication method that will be easier.. Thank you.
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Mathematics Tutor

Yes. Every number is special. you can use algebraic identity to find to the square of any number. First we see about the number end with 9.. Eg:1) 99=100-1 (100-1)^2=100^2-2(100)(1)+1^2 =10000-200+1 =9800+1 =9801. Eg:2) 29=30-1 (30-1)^2=30^2-2(30)(2)+1^2 ... read more

Yes. Every number is special.

you can use algebraic identity to find to the square of any number.

 

First we see about the number end with 9..

Eg:1) 99=100-1

(100-1)^2=100^2-2(100)(1)+1^2

                =10000-200+1

                =9800+1

                =9801.

Eg:2) 29=30-1

(30-1)^2=30^2-2(30)(2)+1^2

               =900-120+1

               =780+1

               =781

If you practice this directly, start from 2nd step.

 

 

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Have 25 Years Experience in Teaching

For numbers that end with 5 fix the last 2 digits as 25. And the number to be fixed before, is the product of the 1st digit and it's consecutive number. Example: 45^2= last 2 digits 25. Previous number is the product of 4 and 4+1 , that is, 4×5 =20. So 45^2 =2025
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Yes, easily you can find square upto 5 digits square using vedic maths. For example (ab)^2 = a^2|2ab|b^2 (76)^2 = 7^2|2.7.6|6^2 = 49|84|36 = (49+8)|(84+3)|6 = 5776
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It Professional With 5+years of Experience in the Field of Software Development

This method uses the identity (a + b) 2 = a 2 + 2ab + b 2 Step 1 : Find 572 Here a = 5 and b =7 Column I Column II Column III a2 2 x a x b b2 52= 25 2 x 5 x 7 =70 72= 49 Step II: Underline the digit of b 2 ( in column III) and add its tens digit, if any, to 2 x a x b (in... read more


This method uses the identity (a + b) 2 = a 2 + 2ab + b 2 
Step 1 : Find 572 
Here a = 5 and b =7 

Column I
Column II
Column III
a2
2 x a x b
b2
52= 25
2 x 5 x 7 =70
72= 49


Step II: Underline the digit of b 2 ( in column III) and add its tens digit, if any, to 2 x a x b (in column III)

Column I
Column II
Column III
a2
2 x a x b
b2
25
70 + 4 = 74
49


Step III: Underline the digit in column II and add the number formed by tens and other digit, if any, to a 2 in column I. 

Column I
Column II
Column III
a2
2 x a x b
b2
25 + 7
70 + 4 = 74
49


Step IV: under the number in column I 

Column I
Column II
Column III
a2
2 x a x b
b2
32
74
49


Write the underlined digits from the unit digit.
Therefore, 57 2 = 3,249 . 

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You have to only remember square of number upto 20 , and this will be enough
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Teacher

Practice practice and practice
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