# ESTIMATE SQUARE ROOTS

When we estimate the square root of a number, first we will find the two values between which the square root of the given number lies.

For example square root of the number 40 lies between 6 and 7.

Because square root of 36 is 6 and square root of 49 is 7. The given number lies between 36 and 49. So that we can say that the square root of the given number is between 6 and 7.

√1  =  1

√4  =  2

√9  =  3

√16  =  4

√25  =  5

√36  =  6

√49  =  7

√64  =  8

√81  =  9

√100  =  10

√121  =  11

√144  =  12

√169  =  13

√196  =  14

√225  =  15

√256  =  16

√289  =  17

√324  =  18

√361  =  19

√400  =  20

√441  =  21

√484  =  22

√529  =  23

√576  =  24

√625  =  25

√676  =  26

√729  =  27

√784  =  28

√841  =  29

√900  =  30

From the above square roots, we can come to know that

(i) If a perfect square has ‘n’ digits where n is even, its square root has n/2 digits.

(ii) If a perfect square has ‘n’ digits where n is odd, its square root has (n + 1)/2 digits.

Estimate the values of the following square roots to the nearest whole number :

Example 1 :

√80

Solution :

80 is not a perfect square.

Find the two perfect squares surrounding 80.

They are 64 and 81.

Then, we have

64 < 80 < 81

Taking square root,

√64 < √80 < √81

8 < √80 < 9

The value of √80 lies between 8 and 9.

In the inequality 64 < 80 < 81, since 80 is closer to 81, the approximate value of √80 is 9.

Example 2 :

√1000

Solution :

1000 is not a perfect square.

Find the two perfect squares surrounding 1000.

They are 961 and 1024.

Then, we have

961 < 1000 < 1024

Taking square root,

√961 < √1000 < √1024

31 < √1000 < 32

The value of √1000 lies between 31 and 32.

In the inequality 961 < 1000 < 1024, since 1000 is closer to 1024, the approximate value of √1000 is 32.

Example 3 :

√172

Solution :

172 is not a perfect square.

Find the two perfect squares surrounding 172.

They are 169 and 196.

Then, we have

169 < 172 < 196

Taking square root,

√169 < √172 < √196

13 < √172 < 14

The value of √172 lies between 13 and 14.

In the inequality 169 < 172 < 196, since 172 is closer to 169, the approximate value of √172 is 13.

Example 4 :

√5928

Solution :

5928 is not a perfect square.

Find the two perfect squares surrounding 5928.

They are 5776 and 5929.

Then, we have

5776 < 5928 < 5929

Taking square root,

√5776 < √5928 < √5929

76 < √5928 < 77

The value of √5928 lies between 76 and 77.

In the inequality 5776 < 5928 < 5929, since 5928 is closer to 5929, the approximate value of √5928 is 77.

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