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SQUARES AND SQUARE ROOTS
Exercise 6.1
Q.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
Ans. (i)
The unit digit of 812 = 1
[12 = 1]
(ii)
The unit digit of 2722 = 4
[22 = 4]
(iii)
The unit digit of 7992 = 1
[92 = 81]
(iv)
The unit digit of 38532 = 9
[32 = 9]
(v)
The unit digit of 12342 = 6
[42 = 16]
(vi) The unit digit of 263872 = 9
[72 = 49]
(vii) The unit digit of 526982 = 4
[82 = 64]
(viii) The unit digit of 998802 = 0
[02 = 0]
(ix)
The unit digit of 12796 = 6
[62 = 36]
(x)
The unit digit of 55555 = 5
[52 = 25]
Q.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
Ans. (i) 1057 – it is not a perfect square as it ends with 7.
(ii) 23453 – it is not a perfect square as it ends with 3.
(iii) 7928 – it is not a perfect square as it ends with 8.
(iv) 222222 – it is not a perfect square as it ends with 2.
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(v) 64000 – it is not a perfect square as it does not have
even number of zeroes at the end.
(vi) 89722 – it is not a perfect square as it ends with 2.
(vii) 222000 – it is not a perfect square as it ends with an
odd number of zeroes.
(viii) 505050 – it is not a perfect square as it ends with an
odd number of zeroes.
Q.3. The squares of which of the following would be odd
numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Ans. (i), (iii) Since, 431 and 7779 are odd numbers, so their
squares will be odd numbers.
Q.4. Observe the following pattern and find the missing
digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 ……. 2 …….. 1
100000012 = …………………
Ans. Using the given pattern, we have
1000012 = 10000200001
100000012 = 100000020000001
Q.5. Observe the following pattern and supply the missing
numbers.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = ………………..
……….. 2 = 10203040504030201
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Ans. Using the given pattern, we have
10101012 = 1020304030201
1010101012 = 10203040504030201
Q.6. Using the given pattern, find the missing numbers.
12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + ___2 = 212
52 + ___2 + 302 = 312
62 + 72 + ___2 = ___2
Ans.
12 + 22 + 22
= 32
22 + 32 + 62
= 72
32 + 42 + 122 = 132
42 + 52 + 202 = 212
52 + 62 + 302 = 312
62 + 72 + 422 = 432
Q.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
Ans. (i) 1 + 3 + 5 + 7 + 9
Sum of first five odd numbers = (5)2 = 25
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
Sum of first ten odd numbers = (10)2 = 100
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Sum of first twelve odd numbers = (12)2 = 144
Q.8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
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Ans. (i) 49 = (7)2 = Sum of first 7 odd numbers
= 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = (11)2 = Sum of first 11 odd numbers
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Q.9. How many numbers lie between squares of the following
numbers?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100
Ans. (i) There are 2n numbers between the squares of the
numbers n and (n + 1)
Here n = 12 and n + 1 = 13
∴ 2 × 12 i.e., 24 numbers lie between 12 and 13.
(ii) Here n = 25 and n + 1 = 26
∴ Numbers between the squares of 25 and 26 are 2 × 25
i.e. 50
(iii) Here n = 99 and n + 1 = 100
∴ Numbers between the squares of 99 and 100 are 2 × 99
i.e. 198
Exercise 6.2
Q.1. Find the squares of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Ans. (i)
322 = (30 + 2)2 = (30 + 2) (30 + 2)
= 30(30 + 2) + 2 (30 + 2)
= 900 + 60 + 60 + 4
= 1024
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(ii)
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352 = (5 + 30)2 = (5 + 30) (5 + 30)
= 5(5 + 30) + 30 (5 + 30)
= 25 + 150 + 150 + 900
= 25 + 300 + 900
= 1225
(iii)
862 = (80 + 6)2 = (80 + 6) (80 + 6)
= 80(80 + 6) + 6 (80 + 6)
= 6400 + 480 + 480 + 36
= 7396
(iv)
932 = (90 + 3)2 = (90 + 3) (90 + 3)
= 90(90 + 3) + 3 (90 + 3)
= 8100 + 270 + 270 + 9
= 8649
(v)
712 = (70 + 1) (70 + 1)
= 70(70 + 1) + 1 (70 + 1)
= 4900 + 70 + 70 + 1
= 5041
(vi)
462 = (40 + 6)2 = (40 + 6) (40 + 6)
= 40(40 + 6) + 6 (40 + 6)
= 1600 + 240 + 240 + 36
= 2116
Q.2. Write a Pythagorean triplet whose one member is.
(i) 6
(ii) 14
(iii) 16
(iv) 18
Ans. For any natural number m > 1; (2m)2, (m2 – 1)2 and
(m2 + 1)2 forms a Pythagorean triplet.
(i)
Let 2m = 6 ------------- (i)
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⇒
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m = 3 (integer)
Also, m2 – 1 = 32 – 1 = 9 – 1 = 8 -------------- (ii)
And, m2 + 1 = 32 + 1 = 9 + 1 = 10 ------------- (iii)
From eqn. (i), (ii) and (iii), we get the required
triplet as 6, 8, 10.
(ii)
Let 2m = 14 ------------- (i)
⇒
m =
14
= 7 (integer)
2
∴ m2 – 1 = (7)2 – 1 = 49 – 1 = 48 ------------ (ii)
And, m2 + 1 = (7)2 + 1 = 49 + 1 = 50 ------------ (iii)
From eqn. (i), (ii) and (iii), we get the required
triplet as 14, 48, 50.
(iii)
Let 2m = 16 -------------- (i)
⇒
m = 8
(integer)
∴ m2 – 1 = (8)2 – 1 = 63 ------------- (ii)
And m2 + 1 = (8)2 + 1 = 65 -------------- (iii)
From eqn. (i), (ii) and (iii), we get the required
triplet as 16, 63, 65.
(iv)
Let 2m = 18 ------------ (i)
⇒
m = 9
(integer)
∴ m2 – 1 = (9)2 – 1 = 81 – 1 = 80 ------------- (ii)
And m2 + 1 = (9)2 + 1 = 81 + 1 = 82 ------------- (iii)
From eqn. (i), (ii) and (iii), we get the required
triplet, as 18, 80, 82.
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Exercise 6.3
Q.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
Ans. (i) As we know,
12 = 1 × 1 = 1;
One’s digit is 1.
And, 92 = 9 × 9 = 81;
One’s digit is 1.
Thus, the one’s digit of the square root of 9801 can
be 1 or 9.
(ii) As we know,
42 = 4 × 4 = 16;
One’s digit is 6
And, 62 = 6 × 6 = 36;
One’s digit is 6
Thus, the one’s digit of the square root of 99856 can be
4 or 6.
(iii) As discussed in part (i), the one’s digit of the square
root of 998001 can be 1 or 9.
(iv) As we know,
52 = 5 × 5 = 25
One’s digit is 5
Thus, the one’s digit of the square root of 657666025
can be 5.
Q.2. Without doing any calculation, find the numbers which
are surely not perfect squares.
(i) 153
(ii) 257
(iii) 408
(iv) 441
Ans. (i) 153 is not a perfect square because it ends with 3.
(ii) 257 is not a perfect square because it ends with 7.
(iii) 408 is not a perfect square because it ends with 8.
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Q.3. Find the square roots of 100 and 169 by the method of
repeated subtraction.
Ans. Method of repeated subtraction
For 169
For 100
Step (i)
169 – 1 = 168
Step (i) 100 – 1 = 99
Step (ii)
168 – 3 = 165
Step (ii) 99 – 3 = 96
Step (iii) 165 – 5 = 160
Step (iii) 96 – 5 = 91
Step (iv) 160 – 7 = 153
Step (iv) 91 – 7 = 84
Step (v)
153 – 9 = 144
–––––––
–––––––
–––––––
–––––––
–––––––
–––––––
Step (x) 9 – 19 = 0
Step (xiii) 25 – 25 = 0
From 100 and 169 we have subtracted successive odd
numbers starting from 1 and obtained 0 at 10th and 13th
steps respectively.
∴
169 = 13
and
100 = 10
Q.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
729 = 3 × 3 × 3 × 3 × 3 × 3
Ans. (i)
∴
729 = 3 × 3 × 3
= 27
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(ii)
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400 = 2 × 2 × 2 × 2 × 5 × 5
∴
400 = 2 × 2 × 5
= 20
(iii)
1764 = 2 × 2 × 3 × 3 × 7 × 7
∴
1764 = 2 × 3 × 7
= 42
(iv)
4096 = 2 × 2 × 2 × 2 × 2 × 2
×2×2×2×2×2×2
∴
4096 = 2 × 2 × 2 × 2 × 2 × 2
= 64
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(v)
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7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
∴
7744 = 2 × 2 × 2 × 11
= 8 × 11
= 88
(vi)
9604 = 2 × 2 × 7 × 7 × 7 × 7
∴
9604 = 2 × 7 × 7
= 98
(vii) 5929 = 7 × 7 × 11 × 11
∴
5929 = 11 × 7
= 77
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(viii) 9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
∴
9216 = 2 × 2 × 2 × 2 × 2 × 3
= 32 × 3
= 96
(ix)
529 = 23 × 23
∴
529 = 23
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
(x)
∴
8100 = 2 × 3 × 3 × 5
= 90
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Q.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
252 = 2 × 2 × 3 × 3 × 7
Ans. (i)
Here, 7 is unpaired
So, we will have to multiply 252 by 7 to
make it a perfect square.
∴ 252 × 7 = 2 × 2 × 3 × 3 × 7 × 7
⇒
1764 = 2 × 3 × 7
= 42
180 = 2 × 2 ×6 3 × 3 × 5
(ii)
Here, 5 is unpaired
So, we will have to multiply 180 by 5 to
make it a perfect square.
∴ 180 × 5 = 2 × 2 × 3 × 3 × 5 × 5
⇒
900 = 2 × 3 × 5
= 30
(iii)
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
Here, 7 is unpaired
So, we will have to multiply 1008 by 7 to
make it a perfect square.
∴ 1008 × 7 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
⇒
7056 = 2 × 2 × 3 × 7
= 84
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(iv)
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2028 = 2 × 2 × 3 × 13 × 13
Here, 3 is unpaired
So, we will have to multiply 2028 by 3 to
make it a perfect square.
∴ 2028 × 3 = 2 × 2 × 3 × 3 × 13 × 13
⇒
6084 = 2 × 3 × 13
= 78
(v)
1458 = 3 × 3 × 3 × 3 × 3 × 3 × 2
Here, 2 is unpaired
So, we will have to multiply 1458 by 2 to
make it a perfect square.
∴ 1458 × 2 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
⇒
2916 = 2 × 3 × 3 × 3
= 2 × 27
= 54
(vi)
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, 3 is unpaired
So, we will have to multiply 768 by 3 to
make it a perfect square.
∴ 768 × 3 = 2 × 2 × 2 × 2 × 2 × 2
×2×2×3×3
⇒
2304 = 2 × 2 × 2 × 2 × 3
= 48
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Q.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
252 = 2 × 2 × 3 × 3 × 7
Ans. (i)
Here, 7 is unpaired. If we divide 252 by 7,
we get a perfect square.
252 ÷ 7 = 36
= 2×2×3×3
Which is a perfect square.
Therefore, the required smallest number is 7.
Also, 36 = 2 × 3 = 6
(ii)
2925 = 3 × 3 × 5 × 5 × 13
Here, 13 is unpaired. If we divide
2925 by 13, we get a perfect square.
2925 ÷ 13 = 225
= 3×3×5×5
Which is a perfect square.
Therefore, the required smallest number is 13.
Also,
(iii)
225 = 3 × 5 = 15
396 = 2 × 2 × 3 × 3 × 11
Here, 11 is unpaired. If we divide
396 by 11, we get a perfect square.
396 ÷ 11 = 36
= 2×2×3×3
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Which is a perfect square.
Therefore, the required smallest number is 11.
Also, 36 = 2 × 3 = 6
(iv)
2645 = 5 × 23 × 23
Here, 5 is unpaired. If we divide 2645 by 5,
we get a perfect square.
2645 ÷ 5 = 529
= 23 × 23
Which is a perfect square.
Therefore, the required smallest number is 5.
Also,
(v)
529 = 23
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
Here, 7 is unpaired. If we divide 2800 by 7,
we get a perfect square.
2800 ÷ 7 = 400 = 2 × 2 × 2 × 2 × 5 × 5
Which is a perfect square.
Therefore, the required smallest number is 7.
Also,
(vi)
400 = 2 × 2 × 5 = 20
1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
Here, 5 is unpaired. If we divide 1620 by 5,
we get a perfect square.
1620 ÷ 5 = 324 = 2 × 2 × 3 × 3 × 3 × 3
Which is a perfect square.
Therefore, the required smallest number is 5.
Also,
324 = 2 × 3 × 3 = 18
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Q.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.
Ans. Let the number of students in the class VIII = x
Then, the rupees donated by each student = Rs x
According to question,
x × x = 2401
⇒
x =
=
∴
2401
7×7×7×7
x = 7 × 7 = 49
Hence, the number of students in the class is 49.
Q.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.
Ans. Let the number of plants in each row = x
Then, the number of rows = x
According to question,
x × x = 2025
⇒
x =
2025
⇒
x =
3×3×3×3×5×5
= 3×3×5
∴
x = 45
Hence, the number of rows and the numbers of plants in
each row is 45.
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Q.9. Find the smallest square number that is divisible by each
of the numbers 4, 9 and 10.
Ans. L.C.M. of 4, 9 and 10 = 2 × 2 × 3 × 3 × 5 = 180
Here 5 is unpaired. If we multiply 180 by 5, then
we will get a perfect square
Since,
180 × 5 = 2 × 2 × 3 × 3 × 5 × 5
Or,
900 = 2 × 2 × 3 × 3 × 5 × 5
⇒
900 = 2 × 3 × 5
= 30
Hence, required number is 900
Q.10. Find the smallest square number that is divisible by each
of the numbers 8, 15 and 20.
Ans. L.C.M. of 8, 15 and 20 = 2 × 2 × 2 × 3 × 5 = 120
Here, 2, 3 and 5 are unpaired. If we
multiply 120 by 2, 3 and 5
we will get a perfect square.
Since, 120 × 2 × 3 × 5 = 2 × 2 × 3 × 3 × 5 × 5
∴
3600 = 2 × 2 × 3 × 5
= 60
Hence, required number is 3600.
Exercise 6.4
Q.1. Find the square root of each of the following numbers by
Division Method.
(i) 2304 (ii) 4489
(iii) 3481
(iv) 529
(v) 3249 (vi) 1369
(vii) 5776
(viii) 7921
(ix) 576
(x) 1024
(xi) 3136
(xii) 900
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Ans. (i)
Therefore,
2304 = 48
Therefore,
4489 = 67
(ii)
(iii)
Therefore, 3481 = 59
(iv)
Therefore,
529 = 23
Therefore,
3249 = 57
(v)
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(vi)
Therefore, 1369 = 37
(vii)
Therefore,
5776 = 76
Therefore,
7921 = 89
Therefore,
576 = 24
(viii)
(ix)
(x)
Therefore, 1024 = 32
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(xi)
Therefore,
3136 = 56
Therefore,
900 = 30
(xii)
Q.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
Ans. We know that if a perfect square is of n-digits, then its
n
⎛ n + 1⎞
square root will have digits, if n is even or ⎜
⎟ , if n
2
⎝ 2 ⎠
is odd:
(i) Here, n = 2 (even)
So, number of digits in the square root of 64
=
n 2
= =1
2 2
(ii) Here n = 3 (odd)
So, number of digits in the square root of 144
=
n+1 3+1 4
=
= =2
2
2
2
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(iii) Here n = 4 (even)
So, number of digits in the square root of 4489
=
n 4
= =2
2 2
(iv) Here n = 5 (odd)
So, number of digits in the square root of 27225
=
n+1 5+1
=
=3
2
2
(v) Here n = 6 (even)
So, number of digits in the square root of 390625
=
n 6
= =3
2 2
Q.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
Ans. (i)
Therefore,
2.56 = 1.6
(ii)
Therefore,
21
7.29 = 2.7
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(iii)
Therefore,
51.84 = 7.2
Therefore,
42.25 = 6.5
(iv)
(v)
Therefore,
31.36 = 5.6
Q.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
Ans. (i) Here, the remainder is 2. It shows that 202 is
less than 402 by 2. This means if we subtract the
remainder from the given number, we will get
a perfect square.
Therefore, the required perfect square
is 402 – 2 = 400.
and,
400 = 20
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(ii)
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Here, the remainder is 53.
It shows that 442 is less than
1989 by 53. So, the number to be
subtracted from 1989 to make it a
perfect square is 53.
Therefore, required perfect square = 1989 – 53 = 1936
And,
(iii)
1936 = 44
Here, the remainder is 1.
It shows that 572 is less than 3250 by 1
So, the number to be subtracted from
3250 to make it a perfect square is 1.
Therefore, required perfect square = 3250 – 1 = 3249
and, 3249 = 57
(iv)
Here, the remainder is 41.
It shows that 282 is less than 825 by 41.
So, the number to be subtracted from
825 to make it a perfect square is 41.
Therefore, required perfect square = 825 – 41 = 784
and, 784 = 28
(v)
Here, the remainder is 31. It shows that
632 is less than 4000 by 31.
So, the number to be subtracted from
4000 to make it a perfect square is 31.
Therefore, required perfect square = 4000 – 31 = 3969
and, 3969 = 63
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Q.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
Ans. (i)
Here, the remainder is 41.
It shows that 222 is less
than 525 by 41.
⇒
222 < 525
So, next perfect square number is 232 = 529
Hence, the number to be added = 232 – 525
= 529 – 525 = 4
Square root of 529
= 529
= 23
(ii)
Here, the remainder is 69. It shows
that 412 is less than 1750 by 69.
⇒ 412 < 1750
So, next perfect square number is 422 = 1764
Hence, the number to be added = 422 – 1750
= 1764 – 1750 = 14
Square root of 1764 = 1764
= 42
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(iii) 252 :
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Here, we get remainder as 27.
It shows that 152 is less than 252 by 27.
⇒ 152 < 252
So, next perfect square number
is 162 = 256
Hence, the number to be added
= 162 – 252 = 256 – 252 = 4
Square root of 256 = 256
= 16
(iv) 1825 :
Here, we get remainder as 61. It shows
that 422 is less than 1825 by 61.
⇒ 422 < 1825
So, next perfect square number is
432 = 1849.
Hence, the number to be added
= 432 – 1825 = 1849 – 1825 = 24
Square root of 1849 = 1849 = 43
(v) 6412 :
Here, we get remainder as 12. It shows
that 802 is less than 6412 by 12.
⇒
802 < 6412
So, next perfect square number is
812 = 6561
Hence, the number to be added is 812 – 6412
= 6561 – 6412 = 149
Square root of 6561 = 6561 = 81
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Q.6. Find the length of the side of a square whose area
is 441 m2.
Ans. Let side of square = x metre
We know that area of square = side2
⇒
⇒
x2 = 441
x =
441 = 21 m
Hence, length of the side of the square is 21 m.
Q.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
Ans. (a) We know that
AC2 = AB2 + BC2 (By Pythagoras Theorem)
= (6)2 + (8)2
= 36 + 64
⇒ AC2 = 100
Or, AC = 10 cm
(b) We have
AC2 = AB2 + BC2
(By Pythagoras Theorem)
⇒ (13)2 = (AB)2 + (5)2
⇒ 169 = (AB)2 + 25
⇒ AB2 = 169 – 25 = 144
Or, AB = 12 cm
Q.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.
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Ans.
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Here, we get the remainder as 39.
It shows that 312 < 1000
So, next perfect square is 322 = 1024
So, the number to be added
= 322 – 1000 = 1024 – 1000 = 24
So, gardener needs 24 more plants.
Q.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.
Ans.
Here, we get the remainder as 16.
It show that 222 < 500
So, number to be subtracted from
500 to make it a perfect square = 16
Hence, 16 children would be left out in this
arrangement.
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