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NCERT Solutions for Class 8 Maths chapter 6 Square and Square Root
class 8th maths

NCERT Solutions for Class 8 Maths chapter 6 Square and Square Root
class 8th maths
Ex 6.1

 

Exercise 6.1

1. What will be the unit digit of the squares of the following numbers?

(i) 81	(ii) 272	(iii) 799	(iv) 3853
(v) 1234	(vi) 26387	(vii) 52698	(viii) 99880
(ix) 12796	(x) 55555

Solutions :
(i) 81
One’s in number 81 is = 1
∴ 1 × 1 = 1
So, unit digit of the square of 81 is = 1

(ii) 272
One’s in number 81 is = 2
∴ 2 × 2 = 4
So, unit digit of the square of 272 is = 4

(iii) 799
One’s in number 799 is = 9
∴ 9 × 9 = 81
So, unit digit of the square of 799 is = 1

(vi) 3853
One’s in number 3853 is = 3
∴ 3 × 3 = 9
So, unit digit of the square of 3853 is = 9

(v) 1234
One’s in number 1234 is = 4
∴ 4 × 4 = 16
One’s in number 1234 is = 6

(vi) 26387
One’s in number 26387 is = 7
∴ 7 × 7 = 49
So, unit digit of the square of 26387 is = 9

(vii) 52698
One’s in number 52698 is = 8
∴ 8 × 8 = 64
So, unit digit of the square of 52698 is = 4

(viii) 99880
One’s in number 99880 is = 0
∴ 0 × 0 = 0
So, unit digit of the square of 99880 is = 0

(xi) 12796
One’s in number 12796 is = 6
∴ 6 × 6 = 36
So, unit digit of the square of 12796 is = 6

(x) 55555
One’s in number 55555 is = 5
∴ 5 × 5 = 25
So, unit digit of the square of 55555 is = 5

2. The following numbers are obviously not perfect squares. Give reason.

(i) 1057	(ii) 23453	(iii) 7928	(iv) 222222
(v) 64000	(vi) 89722	(vii) 222000	(viii) 505050

Solutions :
Numbers whose units digit is 2, 3, 7 or 8 are not perfect square numbers.
Even if there is an odd numbers of zero at the end of the number, that number will not be a perfect square number. 

(i) 1057
The unit’s digit of the number 1057 is 7, so it is not a perfect square number.

(ii) 23453
The unit’s digit of the number 23453 is 3, so it is not a perfect square number.

(iii) 7928
The unit’s digit of the number 7928 is 8, so it is not a perfect square number.

(iv) 222222
The unit’s digit of the number 222222 is 2, so it is not a perfect square number.

(v) 64000
The number 64000 has odd numbers of zeros at the end, so it is not a perfect square.

(vi) 89722
The unit’s digit of the number 89722 is 2, so it is not a perfect square number.

(vii) 222000
The number 222000 has odd numbers of zeros at the end, so it is not a perfect square.

(viii) 505050
The number 505050 has odd numbers of zeros at the end, so it is not a perfect square.

3. 3. The squares of which of the following would be odd numbers?

(i) 431

(ii) 2826

(iii) 7779

(vi) 82004

Solutions : 
The square of odd numbers is an odd number.

(i) 431
Number 431 is an odd number, so its square will also be an odd number.

(ii) 2826
The number 2826 is an even number, so its square will also be an even number.

(iii) 7779
The number 7779 is an even number, so its square will also be an even number.

(iv) 82004
The number 82004 is an even number, so its square will also be an even number.

4. Observe the following pattern and find the missing digits.

11²	=	121
101²	=	10201
1001²	=	1002001
100001²	=	1………..2……….1
10000001²	=	………………………………

Solutions :

11²	=	121
101²	=	10201
1001²	=	1002001
100001²	=	10000200001
10000001²	=	100000020000001

5. Observe the following pattern and supply the missing numbers.

11²	=	121
101²	=	10201
10101²	=	102030201
1010101²	=	………………………..
…………………²	=	10203040504030201

Solutions :

11²	=	121
101²	=	10201
10101²	=	102030201
1010101²	=	1020304030201
101010101²	=	10203040504030201

6. Using the given pattern, find the missing numbers.
1^² + 2^² + 2^² = 3^²
2^² + 3^² + 6^² = 7^²
3^² + 4^² + 12^² = 13^²
4^² + 5^² + __² = 21²
5^² + __² + 30^² = 31^²
6^² + 7^² + __² = __²
Solutions :
The second number in the given pattern is obtained by adding 1 to the first number. When the first number is multiplied by the second number, the third number is obtained and the number after the equal sign is obtained by adding one to the third number.

1^² + 2^² + 2^² = 3^²
2^² + 3^² + 6^² = 7^²
3^² + 4^² + 12^² = 13^²
4^² + 5^² + 20² = 21²
5^² + 6² + 30^² = 31^²
6^² + 7^² + 42² = 43²

7. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solutions :
The sum of the first n odd natural numbers is n².
(i) 1 + 3 + 5 + 7 + 9
Sum of first 5 odd natural numbers = 5²
= 25

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
Sum of first 10 odd natural numbers = 10²
= 100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Sum of first 12 odd natural numbers = 12²
= 144

8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
हल :
(i) Express 49 as the sum of 7 odd numbers.
The sum of the first n odd natural numbers is n².
1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Express 121 as the sum of 11 odd numbers.
The sum of the first n odd natural numbers is n².
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9. How many numbers lie between squares of the following numbers?

(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100

Solutions :
Numbers between squares of any two numbers = difference of square of two numbers − 1
(i) 12 and 13
12² = 144
13² = 169
difference of square of both = 169 − 144
= 25
Now, one less than 25 = 25 − 1 = 24
So, There are 24 numbers between the squares of 12 and 13.

(ii) 25 and 26 
25² = 625
26² = 676
difference of square of both = 676 − 625
= 51
Now, one less than 51 = 51 − 1 = 50
So, There are 50 numbers between the squares of 25 and 26.

(iii) 99 and 100
99² = 9801
100² = 10000
difference of square of both = 10000 − 9801
= 199
Now, one less than 199 = 199 − 1 = 198
So, There are 198 numbers between the squares of 99 and 100.

NCERT Solutions for Class 8 Maths chapter 6
class 8th math
Ex 6.2

Exercise 6.2

1. Find the square of the following numbers.

(i) 32	(ii) 35	(iii) 86	(iv) 93
(v) 71	(vi) 46

Solutions : 
(i) 32
identity (a + b)² = a² + 2ab + b²
(30 + 2)² = 30² + 2 × 30 × 2 + 2²
32² = 900 + 120 + 4
32² = 1024
So the square of 32 is 1024.

(ii) 35
identity (a + b)² = a² + 2ab + b²
(35 + 4)² = 30² + 2 × 30 × 5 + 5²
35² = 900 + 300 + 25
35² = 1225
So the square of 35 is 1225.

(iii) 86
identity (a + b)² = a² + 2ab + b²
(80 + 6)² = 80² + 2 × 80 × 6 + 6²
86² = 6400 + 960 + 36
86² = 7396
So the square of 86 is 7396.

(iv) 93
identity (a + b)² = a² + 2ab + b²
(90 + 3)² = 90² + 2 × 90 × 3 + 3²
93² = 8100 + 540 + 9
93² = 8649
So the square of 93 is 8649.

(v) 71
identity (a + b)² = a² + 2ab + b²
(70 + 1)² = 70² + 2 × 70 × 1 + 1²
71² = 4900 + 140 + 1
71² = 5041
So the square of 71 is 5041.

(vi) 46
identity (a + b)² = a² + 2ab + b²
(40 + 6)² = 40² + 2 × 40 × 6 + 6²
40² = 1600 + 480 + 36
40² = 5041
So the square of 46 is 5041.

2. Write a Pythagorean triplet whose one member is.

(i) 6

(ii) 14

(iii) 16

(iv) 18

Solutions :
From the simple form 2m, m² – 1, m² + 1 we can find the Pythagorean triplet.
(i) 6 
If we take 2m = 6 then –
\(\displaystyle m=\frac{{6}}{2}\)
m = 3
which is the integer value for m.
So, m² – 1
put the value of m 
= 3² – 1
= 9 – 1
= 8
same as, put the value of m in m² + 1 
3² + 1
= 9 + 1
= 10
So the Pythagorean triples are 6, 8 and 10.

(ii) 14 
If we take 2m = 14 then
\(\displaystyle m=\frac{{14}}{2}\)
m = 7
which is the integer value for m.
So, m² – 1
put the value of m
= 7² – 1
= 49 – 1
= 48
same as, put the value of m in m² + 1
7² + 1
= 49 + 1
= 50
So the Pythagorean triples are 14, 48 and 50.

(iii) 16 
If we take 2m = 16 then
\(\displaystyle m=\frac{{16}}{2}\)
m = 8
which is the integer value for m.
So, m² – 1
put the value of m
= 8² – 1
= 64 – 1
= 63
same as, put the value of m in m² + 1
8² + 1
= 64 + 1
= 65
So the Pythagorean triples are 16, 63 and 65.

(iv) 18 
If we take 2m = 18 then
\(\displaystyle m=\frac{{18}}{2}\)
m = 9
which is the integer value for m.
So, m² – 1
put the value of m
= 9² – 1
= 81 – 1
= 80
same as, put the value of m in m² + 1
9² + 1
= 81 + 1
= 82
So the Pythagorean triples are 9, 80 and 82.

NCERT Solutions for Class 8 Maths chapter 6
class 8th math
Ex 6.3

Exercise 6.3

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801

(ii) 99856

(iii) 998001

(iv) 657666025

Solutions : 
(i) 9801
The unit’s digit in the number 9801 is 1.
So the units digit 1 comes only when 1 is multiplied by 1 or 9 by 9.
So the unit’s digit in the square root of the number 9801 will be 1 or 9.

(ii) 99856
The unit’s digit in the number 99856 is 6.
So the units digit 6 comes only when 4 is multiplied by 4 or 6 by 6.
So the unit’s digit in the square root of the number 99856 will be 4 or 6.

(iii) 998001
The unit’s digit in the number 998001 is 1.
So the units digit 1 comes only when 1 is multiplied by 1 or 9 by 9.
So the unit’s digit in the square root of the number 998001 will be 1 or 9.

(iv) 657666025
The unit’s digit in the number 657666025 is 5.
So the unit’s digit 5 comes only when 5 is multiplied by 5.
Hence the unit’s digit in the square root of the number 657666025 will be 5.

3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Solutions : 
100 square root by the method of repeated subtraction – 
As we know that the sum of the first n odd natural numbers is n².

100 – 1 = 99	99 – 3 = 96	96 – 5 = 91	91 – 7 = 84
85 – 9 = 75	76 – 11 = 64	66 – 13 = 51	53 – 15 = 36
38 – 17 = 19	19 – 19 = 0

Subtracting consecutive odd numbers from the number 100, 0 comes in the 10th term.
So the square root of 100 will be 10.
\(\displaystyle \sqrt{{100}}=10\)

169 square root by the method of repeated subtraction – 
As we know that the sum of the first n odd natural numbers is n².

169 – 1 = 168	168 – 3 = 165	165 – 5 = 160	160 – 7 = 153
153 – 9 = 144	144 – 11 = 133	133 – 13 = 120	120 – 15 = 105
105 – 17 = 88	88 – 19 = 69	69 – 21 = 48	48 – 23 = 25
25 – 25 = 0

Subtracting consecutive odd numbers from the number 169, 0 comes in the 13th term.
So the square root of 169 will be 13.
\(\displaystyle \sqrt{{169}}=13\)

4. Find the square roots of the following numbers by the Prime Factorisation Method.

(i) 729	(ii) 400	(iii) 1764	(iv) 4096
(v) 7744	(vi) 9604	(vii) 5929	(viii) 9216
(ix) 529	(x) 8100

Solutions : 
(i) 729
\(\displaystyle \begin{array}{l}\underline{{3\left| {729} \right.}}\\\underline{{3\left| {243} \right.}}\\\underline{{3\left| {81} \right.}}\\\underline{{\left. 3 \right|27}}\\\underline{{3\left| 9 \right.}}\\\underline{{3\left| 3 \right.}}\\\left. {} \right|1\end{array}\)
prime factors of the number 729 = 3 × 3 × 3 × 3 × 3 × 3
On creating a pair of prime factors = 3 × 3 × 3 × 3 × 3 × 3 
= 3 × 3 × 3
= 27
So, \(\displaystyle \sqrt{{729}}=27\)

(ii) 400 
\(\displaystyle \begin{array}{l}\underline{{2\left| {400} \right.}}\\\underline{{2\left| {200} \right.}}\\\underline{{2\left| {100} \right.}}\\\underline{{\left. 2 \right|50}}\\\underline{{5\left| {25} \right.}}\\\underline{{5\left| 5 \right.}}\\\left. {} \right|1\end{array}\)
prime factors of the number 400 = 2 × 2 × 2 × 2 × 5 × 5
On creating a pair of prime factors = 2 × 2 × 2 × 2 × 5 × 5 
= 2 × 2 × 5
= 20
So, \(\displaystyle \sqrt{{400}}=20\)

(iii) 1764
\(\displaystyle \begin{array}{l}\underline{{2\left| {1764} \right.}}\\\underline{{2\left| {882} \right.}}\\\underline{{3\left| {441} \right.}}\\\underline{{\left. 3 \right|147}}\\\underline{{7\left| {49} \right.}}\\\underline{{7\left| 7 \right.}}\\\left. {} \right|1\end{array}\)
prime factors of the number 1764 = 2 × 2 × 3 × 3 × 7 × 7
On creating a pair of prime factors = 2 × 2 × 3 × 3 × 7 × 7
= 2 × 3 × 7
= 42
So, \(\displaystyle \sqrt{{1764}}=42\)

(iv) 4096
\(\displaystyle \begin{array}{l}\underline{{2\left| {4096} \right.}}\\\underline{{2\left| {2048} \right.}}\\\underline{{2\left| {1024} \right.}}\\\underline{{2\left| {512} \right.}}\\\underline{{2\left| {256} \right.}}\\\underline{{2\left| {128} \right.}}\\\underline{{2\left| {64} \right.}}\\\underline{{2\left| {32} \right.}}\\\underline{{2\left| {16} \right.}}\\\underline{{2\left| 8 \right.}}\\\underline{{2\left| 4 \right.}}\\\underline{{2\left| 2 \right.}}\\\left. {} \right|1\end{array}\)
prime factors of the number 4096= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
On creating a pair of prime factors = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2 × 2 × 2 × 2 × 2 × 2
= 64
So, \(\displaystyle \sqrt{{4096}}=64\)

(v) 7744
\(\displaystyle \begin{array}{l}\,\,\underline{{2\left| {7744} \right.}}\\\,\,\underline{{2\left| {3872} \right.}}\\\,\,\underline{{2\left| {1936} \right.}}\\\,\,\underline{{2\left| {968} \right.}}\\\underline{{\,\,2\left| {484} \right.}}\\\underline{{\,\,2\left| {242} \right.}}\\\underline{{11\left| {121} \right.}}\\\underline{{11\left| {11} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)
prime factors of the number 7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
On creating a pair of prime factors = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
= 2 × 2 × 2 × 11
= 88
So, \(\displaystyle \sqrt{{7744}}=88\)

(vi) 9604
\(\displaystyle \begin{array}{l}\underline{{2\left| {9604} \right.}}\\\underline{{2\left| {4802} \right.}}\\\underline{{7\left| {2401} \right.}}\\\underline{{7\left| {343} \right.}}\\\underline{{7\left| {49} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factors of the number 9604 = 2 × 2 × 7 × 7 × 7 × 7
On creating a pair of prime factors = 2 × 2 × 7 × 7 × 7 × 7
= 2 × 7 × 7
= 98
So, \(\displaystyle \sqrt{{9604}}=98\)

(vii) 5929
\(\displaystyle \begin{array}{l}\underline{{\,\,7\left| {5929} \right.}}\\\underline{{\,\,7\left| {847} \right.}}\\\underline{{11\left| {121} \right.}}\\\underline{{11\left| {11} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)
prime factors of the number 5929 = 7 × 7 × 11 × 11
On creating a pair of prime factors = 7 × 7 × 11 × 11
= 7 × 11
= 77
So, \(\displaystyle \sqrt{{5929}}=77\)

(viii) 9216
\(\displaystyle \begin{array}{l}\underline{{2\left| {9216} \right.}}\\\underline{{2\left| {4608} \right.}}\\\underline{{2\left| {2304} \right.}}\\\underline{{2\left| {1152} \right.}}\\\underline{{2\left| {576} \right.}}\\\underline{{2\left| {288} \right.}}\\\underline{{2\left| {144} \right.}}\\\underline{{2\left| {72} \right.}}\\\underline{{2\left| {36} \right.}}\\\underline{{2\left| {18} \right.}}\\\underline{{3\left| 9 \right.}}\\\underline{{3\left| 3 \right.}}\\\left. {} \right|1\end{array}\)
prime factors of the number 9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
On creating a pair of prime factors = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2 × 2 × 2 × 2 × 2 × 3
= 96
So, \(\displaystyle \sqrt{{9216}}=96\)

(ix) 529
\(\displaystyle \begin{array}{l}\underline{{23\left| {529} \right.}}\\\underline{{23\left| {23} \right.}}\\\left. {\,\,\,\,\,} \right|1\end{array}\)
prime factors of the number 529 = 23 × 23
On creating a pair of prime factors = 23 × 23
= 23
So, \(\displaystyle \sqrt{{529}}=23\)

(x) 8100
\(\displaystyle \begin{array}{l}\underline{{2\left| {8100} \right.}}\\\underline{{2\left| {4050} \right.}}\\\underline{{3\left| {2025} \right.}}\\\underline{{3\left| {675} \right.}}\\\underline{{3\left| {225} \right.}}\\\underline{{3\left| {75} \right.}}\\\underline{{5\left| {25} \right.}}\\\underline{{5\left| 5 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factors of the number 8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
On creating a pair of prime factors = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
= 2 × 3 × 3 × 5
So, \(\displaystyle \sqrt{{8100}}=90\)

5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

(i) 252	(ii) 180	(iii) 1008	(iv) 2028
(v) 1458	(vi) 768

Solutions : 
(i) 252
\(\displaystyle \begin{array}{l}\underline{{2\left| {252} \right.}}\\\underline{{2\left| {126} \right.}}\\\underline{{3\left| {63} \right.}}\\\underline{{3\left| {21} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number 252 = 2 × 2 × 3 × 3 × 7
On creating a pair of prime factors = 2 × 2 × 3 × 3 × 7
As the prime factor 7 has no pair, 252 is not a perfect square.
If 7 gets a pair then the number will become perfect square. So, we multiply 252 by 7 to get,
= 252 × 7
= 1764 is a perfect square.
252 = 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore,
= 2 × 3 × 7
Thus the required smallest multiple of 252 is 1764 which is a perfect square.
\(\displaystyle \sqrt{{1764}}=42\)

(ii) 180
\(\displaystyle \begin{array}{l}\underline{{2\left| {180} \right.}}\\\underline{{2\left| {90} \right.}}\\\underline{{3\left| {45} \right.}}\\\underline{{3\left| {15} \right.}}\\\underline{{5\left| 5 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number = 2 × 2 × 3 × 3 × 5
On creating a pair of prime factors = 2 × 2 × 3 × 3 × 5
As the prime factor 5 has no pair, 180 is not a perfect square.
If 5 gets a pair then the number will become perfect square. So, we multiply 180 by 7 to get,
= 180 × 5
= 900 is a perfect square.
= 2 × 2 × 3 × 3 × 5 × 5
Now each prime factor is in a pair. Therefore,
= 2 × 3 × 5
Thus the required smallest multiple of 180 is 900 which is a perfect square.
\(\displaystyle \sqrt{{900}}=30\)

(iii) 1008
\(\displaystyle \begin{array}{l}\underline{{2\left| {1008} \right.}}\\\underline{{2\left| {504} \right.}}\\\underline{{2\left| {252} \right.}}\\\underline{{2\left| {126} \right.}}\\\underline{{3\left| {63} \right.}}\\\underline{{3\left| {21} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number = 2 × 2 × 2 × 2 × 3 × 3 × 7
On creating a pair of prime factors = 2 × 2 × 2 × 2 × 3 × 3 × 7
As the prime factor 7 has no pair, 1008 is not a perfect square.
If 7 gets a pair then the number will become perfect square. So, we multiply 1008 by 7 to get,
= 1008 × 7
= 7056 is a perfect square.
= 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore,
= 2 × 2 × 3 × 7
Thus the required smallest multiple of 1008 is 7056 which is a perfect square. \(\displaystyle \sqrt{{7056}}=84\)

(iv) 2028
\(\displaystyle \begin{array}{l}\,\,\underline{{2\left| {2028} \right.}}\\\,\,\underline{{2\left| {1014} \right.}}\\\,\,\underline{{3\left| {507} \right.}}\\\underline{{13\left| {169} \right.}}\\\underline{{13\left| {13} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number = 2 × 2 × 3 × 13 × 13
On creating a pair of prime factors = 2 × 2 × 3 × 13 × 13
If 3 gets a pair then the number will become perfect square. So, we multiply 2028 by 3 to get
= 2028 × 3
= 6084 is a perfect square.
= 2 × 2 × 3 × 3 × 13 × 13
Now each prime factor is in a pair. Therefore,
= 2 × 3 × 13
Thus the required smallest multiple of 2028 is 6084 which is a perfect square.
\(\displaystyle \sqrt{{6084}}=78\)

(v) 1458
\(\displaystyle \begin{array}{l}\underline{{2\left| {1458} \right.}}\\\underline{{3\left| {729} \right.}}\\\underline{{3\left| {243} \right.}}\\\underline{{3\left| {81} \right.}}\\\underline{{3\left| {27} \right.}}\\\underline{{3\left| 9 \right.}}\\\underline{{3\left| 3 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number = 2 × 3 × 3 × 3 × 3 × 3 × 3
On creating a pair of prime factors = 2 × 3 × 3 × 3 × 3 × 3 × 3
If 2 gets a pair then the number will become perfect square. So, we multiply 1458 by 2 to get
= 1458 × 2
= 2916 is a perfect square.
= 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Now each prime factor is in a pair. Therefore,
= 2 × 3 × 3 × 3
Thus the required smallest multiple of 1458 is 2916 which is a perfect square.
\(\displaystyle \sqrt{{2916}}=54\)

(vi) 768
\(\displaystyle \begin{array}{l}\underline{{2\left| {768} \right.}}\\\underline{{2\left| {384} \right.}}\\\underline{{2\left| {192} \right.}}\\\underline{{2\left| {96} \right.}}\\\underline{{2\left| {48} \right.}}\\\underline{{2\left| {24} \right.}}\\\underline{{2\left| {12} \right.}}\\\underline{{2\left| 6 \right.}}\\\underline{{3\left| 3 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
On creating a pair of prime factors = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
If 3 gets a pair then the number will become perfect square. So, we multiply 768 by 3 to get
= 768 × 3
= 2304 is a perfect square.
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
Now each prime factor is in a pair. Therefore,
= 2 × 2 × 2 × 2 × 3
Thus the required smallest multiple of 768 is 2304 which is a perfect square.
\(\displaystyle \sqrt{{768}}=48\)

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.

(i) 252	(ii) 2925	(iii) 396	(iv) 2645
(v) 2800	(vi) 1620

Solutions :
(i) 252
\(\displaystyle \begin{array}{l}\underline{{2\left| {252} \right.}}\\\underline{{2\left| {126} \right.}}\\\underline{{3\left| {63} \right.}}\\\underline{{3\left| {21} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\) prime factor of the number 252 = 2 × 2 × 3 × 3 × 7
On creating a pair of prime factors = 2 × 2 × 3 × 3 × 7
If we divide 252 by the factor 7, then
= 252 ÷ 7
= 36 is a perfect square.
Therefore, the smallest number be divide is 7.
prime factor of new number 36 = 2 × 2 × 3 × 3
= 2 × 3
= 6
Thus, square root of new number is \(\displaystyle \sqrt{{36}}=6\)

(ii) 2925
\(\displaystyle \begin{array}{l}\underline{{\,\,3\left| {2925} \right.}}\\\underline{{\,\,3\left| {975} \right.}}\\\underline{{\,\,5\left| {325} \right.}}\\\underline{{\,\,5\left| {65} \right.}}\\\underline{{13\left| {13} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number 2925 = 3 × 3 × 5 × 5 × 13
On creating a pair of prime factors = 3 × 3 × 5 × 5 × 13
If we divide 2925 by the factor 13, then
= 2925 ÷ 13
= 225 is a perfect square.
Therefore, the smallest number be divide is 13.
prime factor of new number 225 = 3 × 3 × 5 × 5
= 3 × 5
= 15
Thus, square root of new number is \(\displaystyle \sqrt{{225}}=15\)

(iii) 396
\(\displaystyle \begin{array}{l}\underline{{\,\,2\left| {396} \right.}}\\\underline{{\,\,2\left| {198} \right.}}\\\underline{{\,\,3\left| {99} \right.}}\\\underline{{\,\,3\left| {33} \right.}}\\\underline{{11\left| {11} \right.}}\\\,\,\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number 396 = 2 × 2 × 3 × 3 × 11
On creating a pair of prime factors = 2 × 2 × 3 × 3 × 11
If we divide 396 by the factor 11, then
= 396 ÷ 11
= 36 is a perfect square.
Therefore, the smallest number be divide is 11.
prime factor of new number 36 = 2 × 2 × 3 × 3
= 2 × 3
= 6
Thus, square root of new number is \(\displaystyle \sqrt{{36}}=6\)

(iv) 2645
\(\displaystyle \begin{array}{l}\underline{{\,\,\,5\left| {2645} \right.}}\\\underline{{23\left| {529} \right.}}\\\underline{{23\left| {23} \right.}}\\\,\,\,\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number 2645 = 5 × 23 × 23
On creating a pair of prime factors = 5 × 23 × 23
If we divide 2645 by the factor 5, then
= 2645 ÷ 5
= 529 is a perfect square.
Therefore, the smallest number be divide is 5.
prime factor of new number 529 = 23 × 23
= 23
Thus, square root of new number is \(\displaystyle \sqrt{{529}}=23\)

(v) 2800
\(\displaystyle \begin{array}{l}\underline{{2\left| {2800} \right.}}\\\underline{{2\left| {1400} \right.}}\\\underline{{2\left| {700} \right.}}\\\underline{{2\left| {350} \right.}}\\\underline{{5\left| {175} \right.}}\\\underline{{5\left| {35} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number 2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
On creating a pair of prime factors = 2 × 2 × 2 × 2 × 5 × 5 × 7
If we divide 2800 by the factor 7, then
= 2800 ÷ 7
= 400 is a perfect square.
Therefore, the smallest number be divide is 13.
prime factor of new number 400 = 2 × 2 × 2 × 2 × 5 × 5
= 2 × 2 × 5
Thus, square root of new number is \(\displaystyle \sqrt{{400}}=20\)

(vi) 1620
\(\displaystyle \begin{array}{l}\underline{{2\left| {1620} \right.}}\\\underline{{2\left| {810} \right.}}\\\underline{{3\left| {405} \right.}}\\\underline{{3\left| {135} \right.}}\\\underline{{3\left| {45} \right.}}\\\underline{{3\left| {15} \right.}}\\\underline{{5\left| 5 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
prime factor of the number 1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
On creating a pair of prime factors = 2 × 2 × 3 × 3 × 3 × 3 × 5
If we divide 1620 by the factor 5, then
= 1620 ÷ 5
= 324 is a perfect square.
Therefore, the smallest number be divide is 13.
prime factor of new number 324 = 2 × 2 × 3 × 3 × 3 × 3
= 2 × 3 × 3
Thus, square root of new number is \(\displaystyle \sqrt{{324}}=18\)

7. The students of Class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solutions :
Let the number of students in the class = x
And each student deposited in the Prime Minister’s National Fund = ₹ x
The total amount donated to the Prime Minister’s National Fund = 2401
According to Question –
x × x = 2401
x² = 2401
\(\displaystyle x=\sqrt{{2401}}\)
\(\displaystyle \begin{array}{l}\underline{{7\left| {2401} \right.}}\\\underline{{7\left| {343} \right.}}\\\underline{{7\left| {49} \right.}}\\\underline{{7\left| 7 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
\(\displaystyle x=\sqrt{{\underline{{7\times 7}}\times \underline{{7\times 7}}}}\)
x = 7 × 7
x = ₹ 49
So there were 49 students in the school and each student donated ₹ 49.

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solutions :
Let the number of rows = x
Number of plants in each row = x
Total number of plants = 2025
According to Question –
x × x = 2025
x² = 2025
\(\displaystyle x=\sqrt{{2025}}\)
\(\displaystyle \begin{array}{l}\underline{{5\left| {2025} \right.}}\\\underline{{5\left| {405} \right.}}\\\underline{{3\left| {81} \right.}}\\\underline{{3\left| {27} \right.}}\\\underline{{3\left| 9 \right.}}\\\underline{{3\left| 3 \right.}}\\\,\,\,\left| 1 \right.\end{array}\)
\(\displaystyle x=\sqrt{{\underline{{5\times 5}}\times \underline{{3\times 3}}\times \underline{{3\times 3}}}}\)
x = 5 × 3 × 3
x = 45
So the number of rows will be 45 and the number of plants in each row will be 45.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution :
This can be solved in two steps. First we will find the smallest common multiple. And after that find the required class number.
The smallest number into which 4, 9 and 10 will be divided is its L.C.M Is.
\(\displaystyle \begin{array}{l}\underline{{2\left| {4,9,10} \right.}}\\\underline{{2\left| {2,9,5} \right.}}\\\underline{{3\left| {1,9,5} \right.}}\\\underline{{3\left| {1,3,5} \right.}}\\\underline{{5\left| {1,1,5} \right.}}\\\,\,\,\left| {1,1,1} \right.\end{array}\)
L.C.M of numbers 4, 9 and 10. S. = 180 है।
prime factors of LCM = 2 × 2 × 3 × 3 × 5
In prime factorization, we see that pairs of 5 are not formed. Hence 180 is not a perfect square number.
If 5 gets a pair then the number will become perfect square. So, we multiply 180 by 5 to get,
So, if we multiply 180 by 5 then we get 900, which is a perfect square number.

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
वह सबसे छोटी वर्ग संख्या ज्ञात कीजिए जो प्रत्येक 8, 15 और 20 से विभाजित हो जाए।
Solution : 
This can be solved in two steps. First we will find the smallest common multiple. And after that find the required class number.
The smallest number into which 8, 15 and 20 will be divided is its L.C.M’s.
\(\displaystyle \begin{array}{l}\underline{{2\left| {8,15,20} \right.}}\\\underline{{2\left| {4,9,10} \right.}}\\\underline{{2\left| {2,9,5} \right.}}\\\underline{{3\left| {1,9,5} \right.}}\\\underline{{3\left| {1,3,5} \right.}}\\\underline{{5\left| {1,1,5} \right.}}\\\,\,\,\left| {1,1,1} \right.\end{array}\)
The L.C.M of numbers 8, 15 and 20. = 360 है।
prime factor of L.C.M 360 are = 2 × 2 × 2 × 3 × 3 × 5
In prime factorization, we see that pairs of 2, 5 are not formed. Hence 360 is not a perfect square number.
If 2 and 5 gets a pair then the number will become perfect square. So, we multiply 360 by 10 to get,
So, if we multiply 360 by 10 then we get 3600, which is a perfect square number.

 

NCERT Solutions for Class 8 Maths chapter 6
class 8th math
Ex 6.4

1. Find the square root of each of the following numbers by Division method.
(i) 2304
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,} \right|\,\,\,\,\,\,48\,\,\,}}\\\left. {\,\,\,4} \right|\,\,\,\,\,\underline{{23}}\underline{{04}}\\\underline{{\left. {+4} \right|-16\,\downarrow \,}}\\\left. {\,88} \right|\,\,\,\,\,\,\,\,704\\\underline{{\left. {\,\,\,\,8} \right|\,\,\,-\,\,704}}\\\left. {\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\) Step 1 Starting with the unit digit, put a bar on each pair. Thus the number is written as 5284. \(\displaystyle \overline{{23}}\overline{{04}}\) 
Step 2 The first pair on the left is 23. Now we have to find a number which when multiplied by it will either give 23 or less than 23.
as, 4² = 4 × 4 = 16
5² = 5 × 5 = 25
So we take the number 4.
Step 3 Take this number as the divisor and quotient with the dividend 23 at the bottom of the leftmost bar. Divide and find the remainder.
Step 4 Write the number in the middle of the next bar to the right of the remainder. So the next composite number will be 704.
Step 5 Double the denominator and write it with a space on the right.
Step 6 To fill in the blanks guess the largest possible digit which will be the new digit in the quotient and multiplying the new divisor by the new quotient, the product will be less than or equal to the quotient.
In this situation-
89 × 9 = 801
88 × 8 = 704
Since 88 × 8 = 704, So to get the remainder, the new digit 8 is chosen.
Step 7 Because the remainder is 0, and there are no digits left in the given number.
So, \(\displaystyle \sqrt{{2304}}=48\)

(ii) 4489
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,67\,\,\,\,\,}}\\\left. {\,\,\,\,\,\,6} \right|\,\,\,\,\,\overline{{44}}\overline{{89}}\\\underline{{\left. {+\,\,\,6} \right|-36\,\,\downarrow \,}}\\\left. {127} \right|\,\,\,\,\,\,\,\,\,889\\\underline{{\left. {\,\,\,\,\,7} \right|\,\,\,\,\,-\,889}}\\\left. {\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{4489}}=67\)

(iii) 3481
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,59\,\,\,\,\,}}\\\left. {\,\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{34}}\overline{{81}}\\\underline{{\left. {+\,\,\,5} \right|-25\,\,\downarrow \,}}\\\left. {109} \right|\,\,\,\,\,\,\,\,981\\\underline{{\left. {\,\,\,\,\,9} \right|\,\,\,\,\,-\,981}}\\\left. {\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{3481}}=59\)

(iv) 529
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,23\,\,\,\,\,}}\\\left. {\,\,\,\,\,\,2} \right|\,\,\,\,\,\overline{5}\overline{{29}}\\\underline{{\left. {+\,\,\,2} \right|-4\,\,\downarrow \,}}\\\left. {\,\,\,43} \right|\,\,\,\,\,\,129\\\underline{{\left. {\,\,\,\,\,\,3} \right|\,\,-\,129}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{529}}=23\)

(v) 3249
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,57\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\overline{{32}}\overline{{49}}\\\underline{{\left. {+\,\,5} \right|-25\,\,\downarrow \,}}\\\left. {107} \right|\,\,\,\,\,\,749\\\underline{{\left. {\,\,\,\,\,7} \right|\,\,-\,749}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{3249}}=57\)

(vi) 1369
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,37\,\,\,\,\,}}\\\left. {\,\,\,\,\,3} \right|\,\,\,\overline{{13}}\overline{{69}}\\\underline{{\left. {+\,\,3} \right|-9\,\,\downarrow \,}}\\\left. {\,\,67} \right|\,\,\,\,\,469\\\underline{{\left. {\,\,\,\,\,7} \right|-469}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{1369}}=37\)

(vii) 5776
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,76\,\,\,\,\,}}\\\left. {\,\,\,\,\,7} \right|\,\,\,\,\,\overline{{57}}\overline{{76}}\\\underline{{\left. {+\,\,7} \right|-49\,\,\downarrow \,}}\\\left. {146} \right|\,\,\,\,\,\,\,\,876\\\underline{{\left. {\,\,\,\,\,6} \right|\,\,\,\,-876}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{5776}}=76\)

(viii) 7921
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,89\,\,\,\,\,}}\\\left. {\,\,\,\,\,8} \right|\,\,\,\,\,\overline{{79}}\overline{{21}}\\\underline{{\left. {+\,\,8} \right|-64\,\,\downarrow \,}}\\\left. {169} \right|\,\,\,\,\,1521\\\underline{{\left. {\,\,\,\,\,9} \right|\,-1521}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{7921}}=89\)

(ix) 576
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,24\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{5}\overline{{76}}\\\underline{{\left. {+\,\,2} \right|-4\,\,\downarrow \,}}\\\left. {\,\,44} \right|\,\,\,\,\,176\\\underline{{\left. {\,\,\,\,\,4} \right|\,-176}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{576}}=24\)

(x) 1024
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,32\,\,\,\,\,}}\\\left. {\,\,\,\,\,3} \right|\,\,\,\,\,\overline{{10}}\overline{{24}}\\\underline{{\left. {+\,\,3} \right|\,\,\,\,-9\,\,\downarrow \,}}\\\left. {\,\,62} \right|\,\,\,\,\,\,\,\,124\\\underline{{\left. {\,\,\,\,\,2} \right|\,\,\,\,-124}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{1024}}=32\)

(xi) 3136
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,56\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{31}}\overline{{36}}\\\underline{{\left. {+\,\,5} \right|\,-25\,\,\downarrow \,}}\\\left. {106} \right|\,\,\,\,\,\,636\\\underline{{\left. {\,\,\,\,\,6} \right|\,\,-636}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{3136}}=56\)

(xii) 900
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,30\,\,\,\,\,}}\\\left. {\,\,\,\,\,3} \right|\,\,\,\,\,\overline{9}\overline{{00}}\\\underline{{\left. {+\,\,3} \right|\,\,-9\,\,\downarrow \,}}\\\left. {\,\,60} \right|\,\,\,\,\,\,000\\\underline{{\left. {\,\,\,\,\,0} \right|\,\,-000}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{900}}=30\)

3. Find the square root of the following decimal numbers.

(i) 2.56	(ii) 7.29	(iii) 51.84	(iv) 42.25
(v) 31.36

Solutions : 
(i) 2.56
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,1.6\,\,\,\,\,}}\\\left. {\,\,\,\,\,1} \right|\,\,\,\,\,\overline{2}.\overline{{56}}\\\underline{{\left. {+\,\,1} \right|\,\,-1\,\,\downarrow \,}}\\\,\,\left. {26} \right|\,\,\,\,\,\,\,\,156\\\underline{{\left. {\,\,\,\,\,6} \right|\,\,-156}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{2.56}}=1.6\)

(ii) 7.29
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,2.7\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{7}.\overline{{29}}\\\underline{{\left. {+\,\,2} \right|\,\,-4\,\,\downarrow \,}}\\\,\,\left. {47} \right|\,\,\,\,\,\,329\\\underline{{\left. {\,\,\,\,\,7} \right|\,\,-329}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{7.29}}=2.7\)

(iii) 51.84
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,7.2\,\,\,\,\,}}\\\left. {\,\,\,\,\,7} \right|\,\,\,\,\,\overline{{51}}.\overline{{84}}\\\underline{{\left. {+\,\,7} \right|\,-49\,\,\downarrow \,}}\\\left. {142} \right|\,\,\,\,\,\,\,284\\\underline{{\left. {\,\,\,\,\,2} \right|\,\,\,-284}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{51.84}}=7.2\)

(iv) 42.25
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,6.5\,\,\,\,\,}}\\\left. {\,\,\,\,\,6} \right|\,\,\,\,\,\overline{{42}}.\overline{{25}}\\\underline{{\left. {+\,\,6} \right|\,-36\,\,\downarrow \,}}\\\left. {125} \right|\,\,\,\,\,\,\,\,\,625\\\underline{{\left. {\,\,\,\,\,5} \right|\,\,\,\,\,-625}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{42.25}}=6.5\)

(v) 31.36
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,5.6\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{31}}.\overline{{36}}\\\underline{{\left. {+\,\,5} \right|\,\,-25\,\,\downarrow \,}}\\\left. {106} \right|\,\,\,\,\,\,\,\,636\\\underline{{\left. {\,\,\,\,\,6} \right|\,\,\,\,\,-636}}\\\left. {\,\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{31.36}}=5.6\)

4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

(i) 402	(ii) 1989	(iii) 3250	(iv) 825
(v) 4000

Solutions : 
(i) 402
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,2\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{4}\overline{{02}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,4\,\,\,} \right|\,\,\,\,\,\,\,\,02\end{array}\)
The remainder is 2, so subtracting 2 from 402 gives the number 400.
which is a perfect square number. Whose square root will be 20.
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,20\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{4}\overline{{00}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,40} \right|\,\,\,\,\,\,\,\,00\\\underline{{\,\,\,\,\left. {\,\,\,} \right|\,\,\,\,\,\,\,\,00}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \,\,\,\end{array}\)
\(\displaystyle \sqrt{{400}}=20\)

(ii) 1989
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,4\,\,\,\,\,}}\\\left. {\,\,\,\,\,4} \right|\,\,\,\,\,\overline{{19}}\overline{{89}}\\\underline{{\left. {+\,\,4} \right|\,-16\,\,\downarrow \,}}\\\left. {\,\,\,84} \right|\,\,\,\,\,\,\,389\\\underline{{\,\,\,\,\left. {\,\,4} \right|\,\,\,\,\,\,\,336}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,53\end{array}\) 
The remainder is 53, so subtracting 53 from 1989 gives the number 1936.
which is a perfect square number. Whose square root will be 44.
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,44\,\,\,\,\,}}\\\left. {\,\,\,\,\,4} \right|\,\,\,\,\,\overline{{19}}\overline{{36}}\\\underline{{\left. {+\,\,4} \right|\,-16\,\,\downarrow \,}}\\\left. {\,\,\,84} \right|\,\,\,\,\,\,\,336\\\underline{{\,\,\,\,\left. {\,\,4} \right|\,\,\,\,\,\,\,336}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,\times \end{array}\)
\(\displaystyle \sqrt{{1936}}=44\)

(iii) 3250
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,57\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{32}}\overline{{50}}\\\underline{{\left. {+\,\,5} \right|\,-25\,\,\downarrow \,}}\\\left. {107} \right|\,\,\,\,\,\,\,750\\\underline{{\,\,\,\,\left. {\,7} \right|\,\,\,\,\,\,\,749}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,1\end{array}\)
The remainder is 1, so subtracting 1 from 3250 gives the number 3249.
which is a perfect square number. whose square root will be 57.
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,57\,\,\,\,\,}}\\\left. {\,\,\,\,\,5} \right|\,\,\,\,\,\overline{{32}}\overline{{49}}\\\underline{{\left. {+\,\,5} \right|\,-25\,\,\downarrow \,}}\\\left. {107} \right|\,\,\,\,\,\,\,749\\\underline{{\,\,\,\,\left. {\,7} \right|\,\,\,\,\,\,\,749}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\,\times \end{array}\)
\(\displaystyle \sqrt{{3249}}=57\)

(iv) 825
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,28\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{8}\overline{{25}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,48} \right|\,\,\,\,\,425\\\underline{{\,\,\,\,\left. {\,8} \right|\,\,\,\,\,384}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,41\end{array}\)
The remainder is 41, so subtracting 41 from 825 gives the number 784.
which is a perfect square number. Whose square root will be 28.
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,28\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{7}\overline{{84}}\\\underline{{\left. {+\,\,2} \right|\,-3\,\,\downarrow \,}}\\\left. {\,\,48} \right|\,\,\,\,\,384\\\underline{{\,\,\,\,\left. {\,8} \right|\,\,\,\,\,384}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\times \end{array}\)
\(\displaystyle \sqrt{{784}}=28\)

(v) 4000
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,63\,\,\,\,\,}}\\\left. {\,\,\,\,\,6} \right|\,\,\,\,\,\overline{{40}}\overline{{00}}\\\underline{{\left. {+\,\,6} \right|\,-36\,\,\downarrow \,}}\\\left. {123} \right|\,\,\,\,\,\,\,400\\\underline{{\,\,\,\,\left. {\,3} \right|\,\,\,\,\,\,\,369}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,31\end{array}\)
The remainder is 31, so subtracting 31 from 4000 gives the number 3969.
which is a perfect square number. Whose square root will be 63.

5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.

(i) 525	(ii) 1750	(iii) 252	(iv) 1825
(v) 6412

Solutions : 
(i) 525
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,22\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{5}\overline{{25}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,42} \right|\,\,\,\,\,125\\\underline{{\,\,\,\,\left. {\,2} \right|\,\,\,\,\,\,\,84}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,41\end{array}\)
On doing the square root of the number 525 by the division method, a remainder of 41 is left. which shows that
22² < 525
next whole number 23² = 529
so the required number = 23² – 525
529 – 525 = 4
So adding 4 to the number 525 will make a perfect square number whose square root will be 23.

(ii) 1750
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,41\,\,\,\,\,}}\\\left. {\,\,\,\,\,4} \right|\,\,\,\,\,\overline{{17}}\overline{{50}}\\\underline{{\left. {+\,\,4} \right|-16\,\,\downarrow \,}}\\\left. {\,\,\,81} \right|\,\,\,\,\,\,\,\,150\\\underline{{\,\,\,\,\,\left. 1 \right|\,\,\,\,\,\,\,\,\,\,\,81}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,69\end{array}\)
On doing the square root of the number 1750 by the division method, a remainder of 69 is left. which shows that
41² < 1750
next whole number 42² = 1764
so the required number = 42² – 1750
1764 – 1750 = 14
So adding 14 to the number 1750 will make a perfect square number whose square root will be 42.

(iii) 252
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,15\,\,\,\,\,}}\\\left. {\,\,\,\,\,1} \right|\,\,\,\,\,\overline{2}\overline{{52}}\\\underline{{\left. {+\,\,1} \right|\,\,-1\,\,\downarrow \,}}\\\left. {\,\,25} \right|\,\,\,\,\,\,152\\\underline{{\,\,\,\,\,\left. 5 \right|\,\,\,\,\,\,125}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,27\end{array}\)
On doing the square root of the number 252 by the division method, a remainder of 27 is left. which shows that
15² < 252
next whole number 16² = 256
so the required number = 16² – 252
256 – 252 = 4
So adding 4 to the number 252 will make a perfect square number whose square root will be 16.

(iv) 1825
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,42\,\,\,\,\,}}\\\left. {\,\,\,\,\,4} \right|\,\,\,\,\,\overline{{18}}\overline{{25}}\\\underline{{\left. {+\,\,4} \right|\,\,-16\,\,\downarrow \,}}\\\left. {\,\,82} \right|\,\,\,\,\,\,\,\,225\\\underline{{\,\,\,\,\,\left. 2 \right|\,\,\,\,\,\,\,\,164}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,61\end{array}\)
On doing the square root of the number 1825 by the division method, a remainder of 61 is left. which shows that
42² < 1825
next whole number 43² = 1849
so the required number = 43² – 1825
1849 – 1825 = 24
Therefore, adding 24 to the number 1825 will become a perfect square number whose square root will be 43.

(v) 6412
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,80\,\,\,\,\,}}\\\left. {\,\,\,\,\,8} \right|\,\,\,\,\,\overline{{64}}\overline{{12}}\\\underline{{\left. {+\,\,8} \right|\,\,-64\,\,\downarrow \,}}\\\left. {160} \right|\,\,\,\,\,\,\,\,012\\\underline{{\,\,\,\,\,\left. 0 \right|\,\,\,\,\,\,\,\,000}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,12\end{array}\)
On doing the square root of the number 6412 by the division method, a remainder of 12 is left. which shows that
80² < 6412
next whole number 81² = 6561
so the required number = 81² – 6412
6561 – 6412 = 149
So adding 149 to the number 6412 will make a perfect square number whose square root will be 81.

6. Find the length of the side of a square whose area is 441 m².
Solutions :
Let the side of the square be x.
area of square = 441 m²
side × side = 441
x × x = 441
x² = 441
\(\displaystyle x=\sqrt{{441}}\)
On finding the square root by division method –
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,21\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{4}\overline{{41}}\\\underline{{\left. {+\,\,2} \right|\,\,-4\,\,\downarrow \,}}\\\left. {\,\,41} \right|\,\,\,\,\,\,041\\\underline{{\,\,\,\,\,\left. 1 \right|\,\,\,\,\,\,\,\,\,\,41}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,\times \end{array}\)
So the side of the square will be x = 21 m

7. In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC
(b) If AC = 13 cm, BC = 5 cm, find AB
Solutions : 
(a) If AB = 6 cm, BC = 8 cm, find AC
we know that
(hypotenuse)² = (base)² + (perpendicular)²
(AC)² = (AB)² + (BC)²
Putting the value in the formula –
(AC)² = (6)² + (8)²
(AC)² = 36 + 64
(AC)² = 100
\(\displaystyle AC=\sqrt{{100}}\)
AC = 10 cm

(b) If AC = 13 cm, BC = 5 cm, find AB
we know that
(hypotenuse)² = (base)² + (perpendicular)²
(AB)² = (AC)² – (BC)²
(AB)² = (13)² – (5)²
(AB)² = 169 – 25
(AB)² = 144
\(\displaystyle AB=\sqrt{{144}}\)
AB = 12 cm

8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain same. Find the minimum number of plants he needs more for this.
Solution :
Let the number of rows = x
and the number of plants in each row = x
According to Question –
x × x = 1000
x² = 1000
\(\displaystyle x=\sqrt{{1000}}\)
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,31\,\,\,\,\,}}\\\left. {\,\,\,\,\,3} \right|\,\,\,\,\,\overline{{10}}\overline{{00}}\\\underline{{\left. {+\,\,3} \right|\,\,\,\,-9\,\,\downarrow \,}}\\\left. {\,\,91} \right|\,\,\,\,\,\,\,\,100\\\underline{{\,\,\,\,\,\left. 1 \right|\,\,\,\,\,\,\,\,\,\,91}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,9\end{array}\)
On doing the square root of the number 1000 by the division method, a remainder of 9 is left. which shows that
31² < 1000
next whole number 32² = 1024
so the required number = 32² – 1000
1024 – 1000 = 24
So adding 24 to the number 1024 will become a perfect square number.
The gardener will need 24 plants.

9. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement.
Solution :
Let the number of rows = x
and number of columns = x
According to Question –
x × x = 500
x² = 500
\(\displaystyle \begin{array}{l}\underline{{\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,22\,\,\,\,\,}}\\\left. {\,\,\,\,\,2} \right|\,\,\,\,\,\overline{5}\overline{{00}}\\\underline{{\left. {+\,\,2} \right|\,-4\,\,\downarrow \,}}\\\left. {\,\,42} \right|\,\,\,\,\,\,100\\\underline{{\,\,\,\,\,\left. 2 \right|\,\,\,\,\,\,\,84}}\\\left. {\,\,\,\,\,\,\,} \right|\,\,\,\,\,\,\,\,\,\,16\end{array}\)
The remainder is 16, so subtracting 16 from 500 gives the number 484.
which is a perfect square number. Whose square root will be 22.
So 16 students have to go out to make arrangements.