Square Root – Class IX – Full Chapter

Perfect square number is an integer which is the product of two equal integers. Thus, 81 = 9 x 9 = (-9) x (-9) is a square number. Let us take another example, 144 = 12 x 12 = (-12) x (-12) is also a square number. We know that, there  are  two possibilities for a square root; one  positive and another  negative. When we say the Square Root, we always take the positive square root as a convention.

Given a number ‘a’, we say a number b is a Square Root  of ‘a’ if a =  b x b = b²

For any two numbers and b, √(a x b) = √a x √b, and √(a/b) = √a/√b (provided b ≠ 0)


Example 1: Find the  square root of 70.56

Solution: We write 70.56 in the form, 70.56 = 7056/100  and find the square root of 7056 and 100 separately using factorization. Observe 7056 =  16 x 441 = 16 x 9 x 49 = 4² x 3² x 7²

Hence √7056  =  4  x 3 x 7 = 84. We also have, √100 = 10. Thus, √(70.56)  = √(7056/100 ) = 84/10 =  8.4


Example 2:  Find the square  root of 0.017424

Solution: Write 0.017424 = 17424/1000000  . Observe, 17424 = 18 x 968 = 18 x 8 x 121 = 3² x 4² x 11² and 1000000 = 1000². Hence √17424  = 18  x 8 x 121  = 132. We also have, √1000000 = 1000. Thus, √(0.017424)  = √(17424/1000000 ) = 132/1000 =  0.132

The factorisation method of finding the square root of a perfect square may become more difficult whether number is too large. Here let us study another method called division method, which helps in finding the square root of positive decimal numbers. We shall see  that we can use this method for finding  the square root of a non-perfect  squares up to some decimal places.


Square Root Exercise 1.2.1

 

  1. Find the square root of the following numbers by the factorization method

(i) 82944

Solution:

82944 = 210  x 34

= (25)2   x (32)2

Square Root Exercise 1.2.1

√(82944) = √((25)2   x (32)) = 2x 32 = 288


(ii) 155236

Solution:

Square Root Exercise 1.2.1

155236 = (2)2 x  (197)2

√(155236) = √((2)2   x (197)) = 2 x 197 = 394


(iii) 19881

Solution:

3

19881 = (3)2 x  (47)2

√(19881) = √((3)2   x (47)) = 2 x 47 = 141


2. Find the square root of the following numbers.

(i) 184.96

Solution:

= (184.96X100)100 = (18496)100 

Square Root Exercise 1.2.1

((2³)²x(17)²)10²

√(184.96) = √ ((2³)²x(17)²⁄10²) = (2³)x(17)⁄10  
(136)10

= 13.6


(iii) 19.5364

Solution:

19.5364 = (19.5364×10000)10000
(195364)10000

Square Root Exercise 1.2.1

√(195364/10000) = √ ((2)²x(221)²⁄100²) = (2)x(221)⁄100  
(442)100

= 4.42


3. Exploration:

Find the squares of the numbers 9, 10, 99, 100, 1000, and 9999.Tabulate these numbers. How many digits are there in the squares of a numbers when it has even number of digits and odd number of digits?

Solution:

9² = 81; digits-2
10² = 100; digits-3
99² = 9801; digits-4
100² = 10000; digits-5
990² = 998001; digits-6
1000² = 1000000; digits-7
9999² = 99980001; digits-8

They follow the rule
2n if the number has even digits
2n-1 if the number has odd digits.


4. If a perfect square A has A digits, how many digits to you expect in √𝐀

Solution:
If a perfect square A has n digits then, √A has

(n)2 Digits if n is even
(n+1)2 Digits if n is odd.


1.1.2 Division method for finding the square root – Square Root

Division method  avoids this  lengthy and  tedious  factorisation.

Division method – Square Root

A method of finding the square root of a given number using division process, based on the identity (a + b)² = a² + (2a + b)b.  This also helps to find square root of non-perfect squares or decimal numbers to any degree of accuracy as required.

Example  3: Find the square root of 841.

Solution: We do it in several steps:

Step 1: Make several groups of pairs of digits starting from the units place and moving to its left. If the number is odd, the last group contains only one digit. Thus we write  841 as 8 41

Step 2: Find the largest number whose square is less than or equal to the number in the left most group. (We have 2² < 8 < 3²  and hence the required number here is 2.) Take this number both as divisor  and quotient, and the  left most group as dividend. Complete the division and get the remainder. (Here the divisor is 2, the dividend is 8, the quotient is 2 and the remainder is 4.)

Square Root

Step 3: Bring down the next group and write it next to the remainder . (Bring down 41 and write it next to the remainder 4, we get 441 as the new dividend)

Square Root

Step 4: Add the quotient and the divisor, and enter it with a blank space on its right. (Add 2 and 2 to get 4 and enter it with a blank space.)

Square Root

Step  5: Now guess the largest  possible digit to fill the blank space which will also be the new digit in the expected square root such that the product of the new divisor and the new quotient (that is the guest digit) does not exceed the new dividend.

Square Root

(Observe that 48 x 8 = 384 < 441 and 49 x 9 = 441. Thus the new digit will be 9.) Now enter the new digit to the right of the number on top and the new product below the new dividend and find the remainder. (Enter 9 after 2 on the top. Fill the blank space with 9 so that the new divisor is 49. Also write the new product 49 x 9 = 441 below the new dividend 441 and get the remainder 0.)

Step 6: Repeat the steps 3, 4, 5  till all the groups are exhausted. If the given number is perfect square, we get the last remainder 0. Now the number at the top is the required square root.

Here at the end of step 5, we see that all the groups are exhausted and the last remainder is 0. Hence √841 = 29.

Thus, 841 = 400 + 441 = 4 x 100 + 49 x 9 = 2² x 100 + (40 + 9) x 9. However, 29 = 20 + 9. Hence, 29² = (20 + 9)² = 20² + (2 x 20 x 9) + 9² = 2² x 100 + (2 x 20 + 9 ) x 9 = 2² x 100 + 49 x 9


Example 4:  Find the square root of 6889.

Solution: Let us follow the following steps:

Step 1: Group the following number as 68 89

Step 2: The largest number whose square does not exceed 68 is 8. Since 8² < 68 < 9². We take 8 both as divisor and quotient, and the group 68 as dividend. We get the remainder 4.

Square Root

Step 3: We bring down the next group 89 and write it to the right of the remainder 4 to get 489 as the new dividend.

Square Root

Step 4: Add the divisor 8 and the quotient 8 to get 16. Enter this with a blank space on its right.

Square Root

Step 5:  Observe that 162 x 2 = 324 and 163 x 3 = 489. Hence the new digit in the square root is 3. The new divisor is 163 and the quotient is 3 whose product gives 489. The remainder will be 0. This completes the division method.

Square Root

We thus get √6889 = 83. Here we split again,

6889 = 6400 + 489 = 80² + (163  x 3) = 80² + (2 x 80 + 3) x 3 = 80² + 2 x 80 x 3 + 3²

= (80 + 3)² = 83² . We can see that the reason for getting the divisors 8 and 163.


Example 5:  Find the square root of 96721

Solution: 

Square Root

We consolidate all the steps and write them in a single frame-work. First group 96721 as 9 67 21. Here 9 is the first group and 3² = 9. Hence 3 is the first digit in the square root. Take this both as divisor and quotient, and 9 as the dividend. The remainder is 0. Bring-down the next group 67. the new dividend is 67. Add the divisor and the quotient, we het 6. Since 61 x 1 = 61 < 67 and 62 x 2 = 124 > 67, the next digit in the quotient  is 1. The new quotient is 1 and the new divisor is 61.  We get the remainder S 67 – 61 = 6.

Bring-down the next group; the new dividend is 621. Add the divisor and the quotient, we have 6. Since 61 + 1 = 62. Now 621 x 1 = 621. Hence the next digit in the quotient  is 1. Take 1 as quotient and 621 as new divisor. We get 0 as the remainder. This completes the division method. Therefore, √96721 = 311.


Example 6: Find the square root of 2002225.

Solution:

Square Root

Thus, √(2002225) =  1415.

Suppose the given number is  not a perfect square. Then it  lies between to consecutive perfect squares. Thus we can add some integer to get the next perfect square or subtract some integer to get the previous perfect square.


Example 7:  Find the least number to be subtracted from 2011 to get a perfect square.

Solution:

As earlier, we follow the steps of finding the square root using division method. Observe 84 x 4 = 336 < 411 and 85 x 5 = 425 > 411. Thus 2011 = 442 + 75. Hence the least number to be subtracted from 2011 to get the perfect square is 75. The perfect square we get is 442.


Example 8: Find the least number to be added to 9300 to get a perfect square.

solution: We use the steps of finding the square root till all the  groups are exhausted.

The calculation shows that  9300 > 962 and 9300 = 962 + 84. Hence we have to look for the next  square, 972. Hence we have to add 109 to get the next perfect square.


Example 9: A square-field has area 3,32,929 m2. It has to be fenced using barbed wire. The barbed  wire  should go round the field 5 times. The barbed wire  is available  in bundles  of 100 m length. The vendor sells the wire only in complete bundles. How many bundles have to be bought?

Solution: First  we find the perimeter of square-field by finding its side length. We use division method.

We obtain √(332929) = 577. Hence the perimeter of the field is 577 x 4 = 2,308 m. The length of the required barbed wire is 5 x 2308 = 11,540 m.

We know, 115 x 110 = 11500 < 11540, and 116 x 100 = 11600 > 11540. Hence the number of bundles to be bought is 116.


Square Root Exercise 1.1.2

  1. Find the square root of the following numbers by division method:

1. 5329

Solution:

Square Root Exercise 1.1.2

Therefore √(5329) = 73


  1. 18769

Solution:

Square Root Exercise 1.1.2

Therefore √(18769) = 137


  1. 28224

Solution:

Square Root Exercise 1.1.2

Therefore √(28224) = 168


  1. 186624

Solution:

Square Root Exercise 1.1.2

Therefore √(186624) = 432


  1. Find the least number to be subtracted from the following numbers to get a perfect square.

1.6200

Solution:

Square Root Exercise 1.1.2

782 = 6084 < 6200 < 6241 = 792

So, we have to add 41 to get the next perfect square.


  1. 12675

Solution:

Square Root Exercise 1.1.2

1122 = 12544 < 12675 < 12769 = 1132

So, we have to add 94 to get the next perfect square.


  1. 88417

Solution:

Square Root Exercise 1.1.2

2972 = 88209 < 88417 < 88804 = 2982

So, we have to add 387 to get the next perfect square.


  1. 123456

Solution:

Square Root Exercise 1.1.2

3512 = 123201 < 123456 < 123904 = 3522

So, we have to add 448 to get the next perfect square.


  1. Find the least number to be subtracted from the following numbers to get a perfect square.

1)1234

Solution:

Square Root Exercise 1.1.2

352 = 1225 < 1234 < 1296 = 362

So, we have to subtract  9 to get the next perfect square.


  1. 4321

Solution:

Square Root Exercise 1.1.2

652 = 4225 < 4321 < 4356 = 662

So, we have to subtract  96 to get the next perfect square.


  1. 34567

Solution:

Square Root Exercise 1.1.2

1852 = 34225 < 34567 < 34596 = 1862

So, we have to subtract  342 to get the next perfect square.


  1. 109876

Solution:

Square Root Exercise 1.1.2

3312 = 109561 < 109876 < 110224 = 3322

So, we have to subtract  315 to get the next perfect square.


  1. Find the consecutive perfect squares between which the following numbers lie:

1)4567

Solution:

Square Root Exercise 1.1.2

672 = 4489 < 4567 < 4624 = 682

So, the square root of 4567 lies between 672 and 682.


  1. 56789

Solution:

Square Root Exercise 1.1.2

2382 = 56644 < 56789 < 57121 = 2392

So, the square root of 56789 lies between 2382 and 2392.


  1. 88888

Solution:

Square Root Exercise 1.1.2

2982 = 88804 < 88888 < 89401 = 2992

So, the square root of 88888 lies between 2982 and 2992.


  1. 123456

Solution:

Square Root Exercise 1.1.2

3512 = 123201 < 123456 < 123904 = 3522

So, the square root of 56789 lies between 3512 and 3522.


  1. A person has three rectangular plots of dimensions 112 m x 54 m, 84 m x 68 m and 140 m x 87 m at different places. He wants to sell all of them and buy a square plot of integral length of maximum possible area approximately equal to the sum of these plots. What would be the dimensions of such a square plot? How much area he may have to lose?

Solution:

The area of the 3 rectangular plots is

(112 x 54) m = 6048 sq. m.

(84 x 68) m = 5712 sq. m.

(140 x 87) m = 12180 sq. m.

Total area = 23940 sq. m.

Square Root Exercise 1.1.2

The area of the new square plot is 1542  = 23715 sq. m.

He will lose 224 m2.


1.1.3 Square root of a decimal number which is the square of another decimal number – Square Root

Consider the squares of some decimal number;

(1.2)2 = 1.44;

(2.13)2 = 4.5369;

(1.414)2 = 1.999396;

For a decimal number to be square of some other decimal number it should have even number of digits after the decimal point.


Example 10: Find the square root of 2.0164

Solution:

27.png

Therefore, square root of 2.0164 = 1.42


Note:

(1) Put a zero if necessary and make sure that you have even number of decimal points.

(2) Start from the rightmost digit and move leftward making of digits, as in the case of a whole number.


 Example 11: Find the square root of 0.009409

Solution:

Square Root

Therefore, square root of 0.009409 = 0.097


Square Root Exercise 1.1.3

  1. How many digits are there after the decimal points in the following?

(1) (3.16)2

Solution:

(3.16)2 = 9.9856

Hence, there are 4 digits after the decimal points.

  1. (1.234)2

Solution:

(1.234)2 = 1.522756

Hence, there are 6 digits after the decimal points.

  1. (0.0023)2

Solution:

(0.0023)2 = 0.00000529

Hence, there are 8 digits after the decimal points.

  1. (1.001)3

Solution:

(1.001)3 = 1.003003001

Hence, there are 9 digits after the decimal points.


  1. Given that the numbers below are all squares of some decimal numbers, how many digits do you expect in their square roots?

(a)84.8241

Solution:

The square root of 84.8241 will have 2 digits in it.

(b) 0.085849

Solution:

The square root of 0.085849 will have 3 digits in it.


© 1.844164

Solution:

The square root of 1.844164 will have 3 digits in it.


(d) 0.0089510521

Solution:

The square root of 0.0089510521 will have 5 digits in it.


  1. Find the square root of the following numbers using division method;

(1) 651.7809

Solution:

Square Root Exercise 1.1.3

Therefore, √(651.7809) = 25.53


(2) 0.431649

Solution:

Square Root Exercise 1.1.3

Therefore, √(0.431649) = 0.657


(3) 95.4529

Solution:

Square Root Exercise 1.1.3

Therefore, √(95.4529) = 9.77


(4) 73.393489

Solution:

Square Root Exercise 1.1.3

Therefore, √(73.393489) = 8.567


  1. A square garden has an are 24686.6944 m². A trench of one meter wide has to be dug along the boundary inside the garden. After digging the trench, what will be the area of the out garden?

Solution:

Area of the square garden is 24686.6944m2

Length of a side of the garden is √(24686.6644) = 157.12 m

If a trench of one meter is dug along the boundary inside the garden the length of a side of the garden will be (157.12 – 2) m = 155.12 m

Square Root Exercise 1.1.3

The area of the remaining garden is (155.12) = 24062.2144 m2


 


1.1.4 Moving closer to the square root of a non-perfect square – Square Root

Round off – Square Root

A process adopted for approximation. The accuracy of the approximation is measured using the number of decimal digits.

While rounding off we follow the following rule: if the last digit is more than or equal to 5, we round it off to the next multiple of 10; if the last digit is less than 5, we round it off to the previous multiple of 10. Rounding off is only a process of approximating the given number by close convenient number. Some times you get a number smaller than the given number and some times a larger number.

A necessary  condition for a decimal number to be the square root of another decimal number is that it should have even number of decimal digits.Writing any number of zeros after the last decimal digit does not affect the number; thus 2:346 = 2.34600000000000000000

Write some zeros at the end so that there are even number of digits after the decimal point. Then pair two digits at a time starting from right most digit and moving leftward. Thus 2.346 can be written as 2.23600000 and this can be paired as 2.34 60 00 00.


Example 12: Round off 12.341567 to 3 decimal places

Solution: We observe that 12.342 is nearer to 12.341567 than 12.341. Hence the answer is 12.342.


Example 13: Find the square root of 3 correct to 4 decimal places

Solution: Here we find the square root of 3 up to 5 decimal places and round it off to 4 decimal places.

Square Root

Here the fifth decimal digit is 5 and following our convention we round off 1.73205 to 1.7321. Hence 1.7321 is equal to √3 correct to 4 decimal places.


Example 14: Find the approximation of √(3.11) both from below and above 3 decimal places

Solution:

We find the square root of 3.11 up to 3 decimal places.

Square Root

Thus, the square root of √(3.11) to decimal places is 1.763.

Observe that (1.763)2 = 3.108169 < 3.11 < 3.111696 = (1.764)2. Thus, 1.763 is the approximation from below of 3.11 to 3 decimal places; and 1.764 is the approximation from above of 3.11 to 3 decimal places.


Square Root Exercise 1.1.4

  1. Round off the following numbers to 3 decimal places.
  1. 1.5678 ≈ 1.568
  2. 2.84671 ≈ 2.847
  3. 14.56789 ≈ 14.568
  4. 12.987564 ≈ 12.988
  5. 3.3333567 ≈ 3.333

  1. Find the square root of the following numbers correct to 3 decimal places.

(i) 12

Solution:

Square Root Exercise 1.1.4

√(12) = 3.4641 ≈ 3.464


(ii) 18

Solution:

Square Root Exercise 1.1.4

√(1.8) = 1.3416 ≈ 1.341


(iii) 133

Solution:

Square Root Exercise 1.1.4

√(133) = 11.5325 ≈ 11.533


(iv) 12.34

Solution:

Square Root Exercise 1.1.4

√(12.34) = 3.5128 ≈ 3.513


(v) 8.6666

Solution:

Square Root Exercise 1.1.4

√(8.6666) = 2.9439 ≈ 2.944


(vi) 234.234

Solution:

Square Root Exercise 1.1.4

√(234.234) = 15.3047 ≈ 15.305


  1. Find the square root of the following numbers correct to 4 decimal places.

(i) 13

Solution:

Square Root Exercise 1.1.4

√(13) = 3.60555 ≈ 3.605


(ii) 8.12

Solution:

Square Root Exercise 1.1.4

√(8.12) = 2.84956 ≈ 2.8496


(iii) 3333

Solution:

Square Root Exercise 1.1.4

√(3333) = 57.73214 ≈ 57.7321


  1. Find the approximation from below to 4 decimal places to the square root of the following numbers.

(i) 5

Solution:

46

√(5) = 2.236067

(2.236067)2  = 4.9996 < 5 < 5.000009 = (2.23607)2

2.236007 is the approximation from below of  √5 to 5 decimal places.


(ii) 8

Solution:

47

√8 = 2.828427

(2.82842) = 7.999959  <  8 <  8.000016  =  (2.828443)2

2.82842 is the approximation from below of  √8, correct to 5 decimal places.


  1. A square garden has area of 900m2. Additional land measuring equal area, surrounding it, has been added to it. If the resulting plot is also in the form of a square, what is its side correct to 3 decimal places?

Solution:

Ares of square garden is 900 m2. If additional land, measuring equal area to added to the garden, its area will be

(900+900)2 = 1800 m2

Side of the new garden is 1800 m.

Square Root Exercise 1.1.4

A = side x side = (side)2 = 42.43m


Glossary – Square Root

Perfect square – Square Root

A number of the form b x b = b²,where is an integer.

Square root – Square Root

If  a = b², then is  square root of ‘a’ and written as √a.

Division method – Square Root

A method of finding the square root of a given number using division process, based on the identity (a + b)² = a² + (2a + b)b.  This also helps to find square root of non-perfect squares or decimal numbers to any degree of accuracy as required.

Approximation – Square Root

A number which is very close to a given number, as close as one requires.

Round off – Square Root

A process adopted for approximation. The accuracy of the approximation is measured using the number of decimal digits.

Points to remember – Square Root

  • For any two numbers and b, √(a x b) = √a x √b, and √(a/b) = √a/√b (provided b ≠ 0)
  • A necessary  condition for a decimal number to be the square root of another decimal number is that it should have even number of decimal digits.
  • Writing any number of zeros after the last decimal digit does not affect the number; thus 2:346 = 2.34600000000000000000
  • Write some zeros at the end so that there are even number of digits after the decimal point. Then pair two digits at a time starting from right most digit and moving leftward. Thus 2.346 can be written as 2.23600000 and this can be paired as 2.34 60 00 00.
  • If you need to find the square root of a number correct to decimal places,take sufficient number of zeros at the end so that there are 2n digits after the decimal point. Then pair off and find the square root by division method.
  • If you want to find the square root of a number correct to  decimal places, first find the square root to  (n + 1) places and round off the last digit.
  • While rounding off we follow the following rule: if the last digit is more than or equal to 5, we round it off to the next multiple of 10; if the last digit is less than 5, we round it off to the previous multiple of 10.
  • Rounding off is only a process of approximating the given number by close convenient number. Some times you get a number smaller than the given number and some times a larger number.

 

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