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Square & Square Roots
1. If a natural number m can be expressed as n², where n is also a natural number,
then m is a square number.
2. All square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
All numbers have numbers from 0 to 9 at their unit place and the number at the unit
place of square of that number behaves as per the number at unit place in the
original number. Following is the illustration of this property:
0²=0
1²=1
2²=4
3²=9
4²=16
5²=25
6²=36
7²=49
8²=64
9²=81
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4
9
6
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9
4
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unit’s
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3. Square numbers can only have even number of zeros at the end.
4. Square root is the inverse operation of square.
5. There are two integral square roots of a perfect square number.
Positive square root of a number is denoted by the symbol
For example, 3² = 9 gives
9 =3
Exercise 1
1. What will be the unit digit of the squares of the following numbers?
(i) 81
Answer: As 1² ends up having 1 as the digit at unit’s place so 81² will have 1 at
unit’s place.
(ii) 272
Asnwer: 2²=4
So, 272² will have 4 at unit’s place
(iii) 799
Answer: 9²=81
So, 799 will have 1 at unit’s place
(iv) 3853
Answer: 3²=9
So, 3853² will have 9 at unit’s place.
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(v) 1234
Answer: 4²=16
So, 1234² will have 6 at unit’s place
(vi) 26387
Answer: 7²=49
So, 26387² will have 9 at unit’s place
(vii) 52698
Answer: 8²=64
So, 52698² will have 4 at unit’s place
(viii) 99880
Answer: 0²=0
So, 99880² will have 0 at unit’s place
(ix) 12796
Answer: 6²=36
So, 12796² will have 6 at unit’s place
(x) 55555
Answer: 5²=25
So, 55555² will have 5 at unit’s place
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
Answer: (i), (ii), (iii), (iv), (vi) don’t have any of the 0, 1, 4, 5, 6, and 9 at unit’s
place, so they are not perfect squares.
(v), (vii) and (viii) don’t have even number of zeroes at the end so they are not
perfect squares.
3. The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Answer: (i) and (iii) will have odd numbers as their square, because an odd number
multiplied by another odd number always results in an odd number.
4. Observe the following pattern and find the missing digits.
11² = 121
101² = 10201
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1001² = 1002001
100001² = 1 0000 2 0000 1
10000001² = 100000020000001
Start with 1 followed as many zeroes as there are between the first and the last one,
followed by two again followed by as many zeroes and end with 1.
5. Observe the following pattern and supply the missing numbers.
11² = 1 2 1
101² = 1 0 2 0 1
10101² = 102030201
1010101² = 1020304030201
101010101² =102030405040201
Start with 1 followed by a zero and go up to as many number as there are number of
1s given, follow the same pattern in reverse order.
6. Using the given pattern, find the missing numbers.
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20²= 21²
5² + 6²+ 30² = 31²
6² + 7² + 42² = 43²
If the square of a number is added with square of its prime factors we get square of
a number which is 1 more than the original number.
7. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
Answer: 1 + 3 = 2² = 4
1 + 3 + 5 = 3² = 9
1 + 3 + 5 + 7 = 4² = 16
1 + 3 + 5 + 7 + 9 = 5² = 25
In other words this is a way of finding the sum of n odd numbers starting from 1.
Sum of n odd numbers starting from 1 = n²
(ii) 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19
Answer: There are 10 odd numbers in the given equation
So, sum =10²=100
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Answer: Sum = 12²=144
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8. (i) Express 49 as the sum of 7 odd numbers.
As you know 49=7²
So, 7² can be expressed as follows:
1+3+5+7+9+11+13
(ii) Express 121 as the sum of 11 odd numbers.
11²=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
Answer: 12²=144
13²=169
Now, 169-144=25
So, there are 25-1=24 numbers lying between 12² and 13²
(ii) 25 and 26
Answer: 25²=625
26²=676
Now, 676-625=51
So, there are 51-1=50 numbers lying between 25² and 26²
(iii) 99 and 100
Answer: 99²=9801
100²=10000
Now, 10000-9801=199
So, there are 199-1=198 numbers lying between 99² and 100²
Exercise 2
1. Find the square of the following numbers.
(i) 32
Answer: 32²=32 × 32= 1024
But above method can be tough to calculate. It is easier to calculate such values by
using algebraic identities.
So, 32²=(30+2) ²
Using (a+b) ² = a²+b²+2ab
We get (30+2) ²= 30²+2²+2 × 30 × 2
=900+4+120=1024
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(ii) 35
Answer: (35)²=(30+5) ²
=30²+5²+2 × 30 × 5
=900+25+300=1225
(iii) 86
Answer: 86²=(80+6) ²
=80²+6²+2 × 80 × 6
=6400+36+960=7396
(iv) 93
Answer=93²=(90+3²
=90²+3²+2 × 90 × 3
=8100+9+540
=8649
(v) 71
Answer: 71²=(70+1) ²
=70²+1²+2 × 70 × 1
=4900+1+140
=5040
(vi) 46
Answer: 46²=(40+6) ²
=40²+6²+2 × 40 × 6
=1600+36+480=2116
2. Write a Pythagorean triplet whose one member is.
(i) 6
Answer: As we know 2m, m²+1 and m²-1 form a Pythagorean triplet for any
number, m>1.
Let us assume 2m=6
Then m=3
⇒ m²+1=3²+1=10
⇒ m²-1=3²-1=8
Test: 6²+8²=36+64=100=10²
Hence, the triplet is 6, 8, and 10
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Note: Most of the Pythagorean triplets are in the ratio of 3:4:5
(ii) 14
Answer: Let us assume, 2m=14, then m=7
So, m²+1=7²+1=50
And, m²-1=7²-1=48
Test: 14²+48²=196+1304=2500=50²
Hence, the triplet is 14, 48, and 50
(iii) 16
Answer: Let us assume 2m=16, then m=8
So, m²+1=8²+1=65
And, m²-1=8²-1=63
Test: 16²+63²=256+3969=4225=65²
Hence, the triplet is 16, 63, and 65
(iv) 18
Answer: Let us assume 2m=18, then m=9
So, m²+1=9²+1=82
m²-1=9²-1=80
test: 18²+80²=6724=82²
Exercise 3
1. What could be the possible ‘one’s’ digits of the square root of each of the
following numbers?
(i) 9801
Answer: Since 1² and 9² give 1 at unit’s place, so these are the possible values of
unit digit of the square root.
(ii) 99856
Answer: 4²=16 and 6²=36, hence, 4 and 6 are possible
(iii) 998001
Answer: Same as question 1—(i)
(iv) 657666025
Answer: 5²=25, hence 5 is possible.
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2. Without doing any calculation, find the numbers which are surely not
perfect squares.
(i) 153 (ii) 257 (iii) 408 (iv) 441
Answer: Option 1 can be a perfect square, others can’t be perfect squares because
the unit digit of a perfect square can be only from 0, 1, 4, 5, 6, 9
3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Answer:
Repeated subtraction:
1. 100-1=99
2. 99-3=96
3. 96-5=91
4. 91-7=84
5. 84-9=75
6. 75-11=64
7. 64-13=51
8. 51-15=36
9. 36-17=19
10. 19-19=0
We get 0 at 10th step
So,
100 =10
1. 169-1=168
2. 168-3=165
3. 165-5=160
4. 160-7=153
5. 153-9=144
6. 144-11=133
7. 133-13=120
8. 120-15=105
9. 105-17=88
10. 88-19=69
11. 69-21=48
12. 48-23=25
13. 25-25=0
We get 0 at 13th step
So,
169 = 13
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4. Find the square roots of the following numbers by the Prime Factorisation
Method.
(i) 729
3 729
3 243
Answer:
3 81
3 27
39
3
⇒ 729= 3² × 3² × 3²
⇒ 729 = 3 × 3 × 3 = 27
(ii) 400
2 400
2 200
Answer:
2 100
2 50
5 25
5
⇒ 400 = 2² × 2² × 5²
⇒ 400 = 2 × 2 × 5 = 20
(iii) 1764
Answer: 1764= 2 × 882 = 2 × 2 × 441 = 2 × 2 × 3 × 147
= 2 × 2 × 3 × 3 × 49 = 2 × 2 × 3 × 3 × 7 × 7 = 2² × 3² × 7²
⇒
1764 = 2 × 3 × 7 = 42
(iv) 4096
Answer: 4096 = 2 × 2048
= 2 × 2 × 1024
= 2 × 2 × 2 × 512
= 2 × 2 × 2 × 2 × 256
= 2 × 2 × 2 × 2 × 2 × 128
= 2 × 2 × 2 × 2 × 2 × 2 × 64
= 2× 2× 2× 2× 2× 2× 8× 8
= 2² × 2² × 2² × 8²
⇒ 4096 = 2 × 2 × 2 × 8 = 64
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(v) 7744
Answer: 7744= 2 × 3872
= 2 × 2 × 1936
= 2 × 2 × 2 × 968
= 2 × 2 × 2 × 2 × 484
= 2 × 2 × 2 × 2 × 2 × 242
= 2 × 2 × 2 × 2 × 2 × 2 × 121
= 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
= 8² × 11²
⇒ 7744 = 88
(vi) 9604
Answer: 9604 = 4 × 2401
= 4 × 7 × 343
= 4 × 7 × 7 × 49
= 4× 7× 7× 7× 7
= 2² × 7² × 7²
⇒ 9604 = 98
(vii) 5929
Answer: 5929 = 11 × 539
= 11 × 11 × 49
= 11 × 11 × 7 × 7
= 11² × 7²
⇒ 5929 = 77
(viii) 9216
9216 = 9 × 1024
= 9 × 4 × 256
= 9 × 4 × 4 × 64
= 9× 4× 4× 8× 8
= 3² × 4² × 8²
⇒ 9216 = 96
Answer:
(ix) 529
529 = 23 × 23
= 23²
⇒ 529 = 23
Answer:
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(x) 8100
Answer: 8100 = 9 × 900
= 9 × 9 × 10 × 10
= 9² × 10²
⇒ 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
252 = 2 × 126
= 2 × 2 × 63
= 2 × 2 × 3× 3× 7
Answer:
Here, 2 and 3 are in pairs but 7 needs a pair, so 252 will become a perfect square
when multiplied by 7.
(ii) 180
180 = 2 × 2 × 45
= 2 × 2 × 3× 3× 5
Answer:
180 needs to be multiplied by 5 to become a perfect square.
(iii) 1008
1008 = 2 × 2 × 252
= 2 × 2 × 2 × 2 × 63
= 2 × 2 × 2 × 2 × 3× 3× 7
Answer:
1008 needs to be multiplied by 7 to become a perfect square
(iv) 2028
Answer: 2028 = 4 × 507
= 4 × 3 × 169
= 2 × 2 × 3 × 13 × 13
2028 needs to be multiplied by 3 to become a perfect square.
(v) 1458
1458 = 2 × 729
= 2 × 3 × 3 × 81
= 2 × 3× 3× 3× 3× 3× 3
Answer:
1458 needs to be multiplied by 2 to become a perfect square.
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(vi) 768
Answer: 768 = 2 × 2 × 192
= 2 × 2 × 2 × 2 × 48
= 2 × 2 × 2 × 2 × 2 × 2 × 12
= 2× 2× 2× 2× 2× 2× 2× 2× 3
768 needs to be multiplied by 3 to become a perfect square.
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
Answer: 252 = 2 × 2 × 63
= 2 × 2 × 3× 3× 7
252 needs to be divided by 7 to become a perfect square.
(ii) 2925
Answer: 2925 = 5 × 5 × 117
= 5 × 5 × 3 × 3 × 13
2925 needs to be divided by 13 to become a perfect square
(iii) 396
Answer: 396 = 2 × 2 × 99
= 2 × 2 × 3 × 3 × 11
396 needs to be divided by 11 to become a perfect square
(iv) 2645
Answer: 2645 = 5 × 529 5 × 23 × 23
2645 needs to be divided by 5 to become a perfect square.
(v) 2800
Answer: 2800 = 2 × 2 × 7 × 10 × 10
2800 needs to be divided by to become a perfect square.
(vi) 1620
Answer: 1620 = 2 × 2 × 405
= 2 × 2 × 3 × 3 × 45
= 2 × 2 × 3× 3× 3× 3× 5
1620 needs to be divided by 5 to become a perfect square
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7. The students of Class VIII of a school donated Rs 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.
Answer: We need to calculate the square root of 2401 to get the solution
2401 = 7 × 7 × 7 × 7
⇒ 2401 = 7 × 7 = 49
There are 49 students, each contributing 49 rupees
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.
2025 = 5 × 5 × 3 × 3 × 3 × 3
2025 = 5 × 3 × 3 = 45
Answer:
⇒
There are 45 rows with 45 plants in each of them.
9. Find the smallest square number that is divisible by each of the numbers
4, 9 and 10.
Answer: Let us find LCM of 4, 9 and 10
4 = 2× 2
9 = 3× 3
10 = 5 × 2
So, LCM = 2² × 3² × 5 = 180
Now the LCM gives us a clue that if 180 is multiplied by 5 then it will become a
perfect square.
The Required number = 180 × 5 = 900
10. Find the smallest square number that is divisible by each of the numbers
8, 15 and 20.
Answer:
8 = 2× 2× 2
15 = 3 × 5
20 = 2 × 2 × 5
So, LCM = 2 × 2 × 2 × 5 × 3 = 120
As 3 and 5 are not in pair in LCM’s factor so we need to multiply 120 by 5 and three
to make it a perfect square.
Required Number= 180 × 3 × 5 = 2700
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Exercise 4
1. Find the square root of each of the following numbers by Division
method.
(i) 2304
Answer:
48
4
4
2304
88
8
16
704
704
0
2304 = 48
(ii) 4489
Answer:
67
6
6
4489
127
7
36
889
889
0
4489 = 67
(iii) 3481
Answer:
59
5
5
3481
109
9
25
981
981
0
3481 = 59
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(iv) 529
Answer:
23
2
2
43
3
5 29
4
129
129
0
529 = 23
(v) 3249
Answer:
57
5
5
3249
107
7
25
749
749
0
3249 = 57
(vi) 1369
Answer:
37
3
1369
3
9
67
7
469
469
0
1369 = 37
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(vii) 5776
Answer:
76
7
7
5776
146
6
49
876
876
0
5776 = 76
(viii) 7921
Answer:
89
8
8
7921
169
9
64
1521
1521
0
7921 = 89
(ix) 576
Answer:
24
2
2
576
44
4
4
176
176
0
576 = 24
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(x) 1024
Answer:
32
3
3
1024
62
2
9
124
124
0
1024 = 32
(xi) 3136
Answer:
56
5
5
3136
106
6
25
636
636
0
3136 = 56
(xii) 900
Answer:
30
3
3
900
60
0
9
000
000
0
900 = 30
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2. Find the number of digits in the square root of each of the following
numbers (without any calculation).
(i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625
Answer:
If there are even number of digits in square then number of digits in square root =
n
2
If there are odd number of digits in square then number of digits in square root=
n+ 1
2
(i) 1, (ii) 2, (iii) 2, (iv) 3, (v) 3
3. Find the square root of the following decimal numbers.
(i) 2.56
Answer:
1
1
26
6
1.6
2.56
1
156
156
0
2.56 = 1.6
(ii) 7.29
Answer:
2
2
47
7
2.7
7.29
4
329
329
0
7.29 = 2.7
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(iii) 51.84
Answer:
7
7
142
2
7.2
51.84
49
284
284
0
51.84 = 7.2
(iv) 42.25
Answer:
6
6
125
5
6.5
42.25
36
625
625
0
42.25 = 65
(v) 31.36
Answer:
5
5
106
6
5.6
31.36
25
636
636
0
31.36 = 5.6
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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
Answer:
2
2
2
4
402
4
002
It is clear that if 2 is subtracted then we will get 400, which is a perfect square.
(ii) 1989
Answer:
4
4
8
1989
16
389
Here, 84X4=336 which is less than 389
And, 85X5=425, which is more than 389
Hence the required difference =389-336=53
1989-53=1936 is a perfect square.
(iii) 3250
Answer:
5
5
10
3250
25
750
Here, 107X7=749 is less than 750
108X8=864 is more than 750
Hence, the required difference = 750-749=1
3250-1=3249 is a perfect square.
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(iv) 825
Answer:
2
2
4
825
4
425
Here, 48X8=384 is less than 425
49X9=441 is more than 425
Hence, the required difference= 425-384=41
825-41=784 is a perfect square.
(v) 4000
Answer:
6
6
12
4000
36
400
Here, 123X3=369 is less than 400
124X4=496 is more than 400
Hence, the required difference = 400-369=31
4000-31=3969 is a perfect square.
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
Answer:
2
2
4
525
4
125
Here, 43X3=129 is more than 125
42X2=84 is less than 125
Hence, required addition= 129-125=4
525+4=529 is a perfect square.
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(ii) 1750
Answer:
4
1750
4
16
16
150
Here, 161X1=161 is 11 more than 150
So, 1750+11=1761 is a perfect square
(iii) 252
Answer:
1 252
1 1
2 152
Here, 25X5=125 is less than 152
26X6=156 is more than 152
Required difference= 156-152=4
So, 252+4=256 is a perfect square
(iv) 1825
Answer:
`
4
4
8
1825
16
225
Here, 82X2=164 is less than 225
83X3=249 is more than 225
Required difference= 249-225=24
So, 1825+24=1849 is a perfect square
(v) 6412
Answer:
8
8
16
6412
64
12
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Here, we need 161X1=161
Required difference=161-12=149
So, 6412+149=6561 is a perfect square
6. Find the length of the side of a square whose area is 441 m².
Answer: Area of Square = Side²
⇒ Side = Area
441 = 3 × 3 × 7 × 7
⇒ 441 = 3 × 7 = 21
7. In a right triangle ABC, ∠ B = 90°
.
(a) If AB = 6 cm, BC = 8 cm, find AC
Answer= AC²=AB²+BC²
=6²+8²=36+64=100
AC=
100 = 10
(b) If AC = 13 cm, BC = 5 cm, find AB
Answer: AB²=AC²-BC²
=13²-5²=169-25=144
AB = 144 = 12
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.
Answer:
3
3
6
1000
9
100
Here, 61X1=61 is less than 100
62X2=124 is more than 100
Hence, the required difference= 100-61=39
Min. number of plants required= 1000-39=961
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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.
Answer:
2
2
4
500
4
100
Here, 42X2=84 is less than 100
43X3=129 is more than 100
Hence, the required difference = 100-84=16
So, 16 children will be left out in the arrangement.
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