CHAPTER – 6
SQUARES AND SQUARE ROOTS
Points to Remember:
t Numbers like 1, 4, 9,
16 , 25, ……………………….are known as square numbers.
t All square numbers end
with 0, 1, 4, 5, 6, or 9 at unit’s place.
t Positive square root
of a number is denoted by
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
Sol: (i) ∵ 1 × 1 = I
(ii)
2 × 2 = 4
The
unit’s digits of (272)2 will be 4.
(iii)
Since, 9 × 9 = 81
The
unit’s digit of (799)2 will be 1.
(iv)
Since, 3 × 3 = 9
The
unit’s digit of (3853)2 will be 9.
(v)
Since, 4 × 4 = 16
The
unit’s digit of (1234)2 will be 6.
(vi)
Since 7 × 7 = 49
The
unit’s digit of (26387)2 will be 9.
(v)
Since, 8 × 8 = 64
The
unit’s digit of (52698)2 will be 4.
(vi)
Since 0 × 0 = 0
The
unit’s digit of (99880)2 will be O.
(vii)
Since 6 × 6 = 36
The
unit’s digit of (12796)2 will be 6.
(x)
Since, 5 × 5 = 25
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
Sol: (i) 1057
Since,
the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9)
∴1057 is not a perfect square.
(ii) 23453
Since,
the ending digit is 7 (which is not one of 0, 1, 4, 5, 6 or 9).
∴23453 is not a perfect square.
(iii) 7928
Since,
the ending digit is 8 (which is not one of 0, 1, 4, 5, 6 or 9).
∴7928 is not a perfect square.
(iv) 222222
Since,
the ending digit is 2 (which is not one of 0, 1, 4, 5, 6 or 9).
∴222222 is not a perfect square.
(v) 64000
Since,
the number of zeros is odd.
∴64000 is not a perfect square.
(vi) 89722
Since,
the ending digits is 2 (which is not one of 0, 1, 4, 5, 6 or 9).
(vii) 222000
Since,
the number of zeros is odd.
∴222000 is not a perfect square.
(viii) 505050
The
unit’s digit is odd zero.
∴505050 cannot be a perfect square.
3. The squares of which of the
following would be odd numbers?
(i)
431 (ii) 2826 (iii) 7779
(iv) 82004
Sol: Since the square
of an odd natural number is odd and that of an even number
is an even number.
(i)
The square of 431 is an odd number.
[∵ 431 is an odd number.]
(ii)
The square of 2826 is an even number.
[∵ 2826 is an even number.]
(iii)
The square of 7779 is an odd number.
[∵ 7779 is an odd number.]
(v)
The square of 82004 is an even number.
[∵ 82004 is an even number.]
4. Observe the following
pattern and find the missing digits.
112 =121
1012 =10201
10012 =1002001
1000012 =1.............2
............. 1
100000012 =............
Sol: Observing the
above pattern, we have
(i)
(100001)2 - 10000200001
(ii)
(10000001)2 = 100000020000001
5. Observe
the, following pattern and supply the missing number.
112 =
121
1012 =
10201
101012 =
102030201
10101012 =
..............2 =
10203040504030201
Sol: Observing the
above, we have
(i)
(1010101)2 = 1020304030201
(ii)
10203040504030201 = (101010101)2
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
Note:
To find pattern:
Third
number is related to first and second number. How?
Fourth
number is related to third number. How?
Sol: The missing
numbers are
(i)
42 + 52 + 202 = 212
(ii) 52 +
22 + 302 = 312
(iii)
62 + 72 + 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
Sol: (i) The sum of
first 5 odd = 52
=
25
(ii)
The sum of first 10 odd numbers = 102
=
100
(iii)
The sum of first 12 odd numbers = 122
=
144
8. (i) Express 49 as the sum of
7 odd numbers.
(ii)
Express 121 as the sum of 11 odd numbers.
Sol: (i) 49 = 72 =
Sum of first 7 odd numbers
=
1 + 3 + 5 + 7 + 9 + 11 + 13
(ii)
121 = 112 = Sum of first 11 odd numbers
=
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
Sol: Since between n2 and
(n + 1)2, there are 2n non-square numbers.
∴ (i) Between 122 and 132,
there are 2 × 12, i.e. 24 numbers
(ii)
Between 252 and 262, there are 2 × 25, i.e. 50
numbers
(iii)
Between 992 and 1002, there are 2 × 99, i.e. 198
numbers
EXERCISE 6.2
1. Find the square of the
following numbers.
(i)
32 (ii) 35 (iii) 86 (iv) 93
(v) 71 (vi) 46
Sol: (i) (32)2 = (30 + 2)2
=
302 + 2(30)(2) + (2)2
=
900 + 120 + 4 = 1024
(ii) (35)2 = (30 + 5)2
=
(30)2 + 2(30)(5) + (5)2
=
900 + 300 + 25
=
1200 + 25 = 1225
Second
method
352 =
3 × (3 + 1) × 100 + 25
=
3 × 4 × 100 + 25
=
1200 + 25 = 1225
(iii)
(86)2 = (80 + 6)2
=
(80)2 + 2(80)(6) + (6)2
=
6400 + 960 + 36 = 7396
(iv)
(93)2 = (90 + 3)2
=
(90)2 + 2(90)(3) + (3)2
=
8100 + 540 + 9 - 8649
(v)
(71)2 = (70 + 1)2
=
(70)2 + 2(70)(1) + (1)2
=
4900 + 140 + 1 = 5041
(vi)
(46)2 = (40 + 6)2
=
(40)2 + 2(40)(6) + (6)2
=
1600 + 480 + 36 = 2116
(i)
6 (ii) 14 (iii) 16
(iv) 18
Sol: (i) Let 2m
=6 ∴m=3
and
m2 + 1 = 32 + 1 = 10
Thus, the required Pythagorean triplet
is 6, 8, 10.
(ii)
Let 2m = 14
∴m = 7
Now,
m2 – 1 = 72 – 1 = 48
and
m2 + 1 = 72 + 1 = 50
Thus, the required Pythagorean triplet is 14,
48, 50.
(iii)
Let 2m = 16
m = 8
Now,
m2 – 1 = 82 – 1
=
64 – 1 = 63
and
m2 + 1 = 82 + 1
=64
+ 1 = 65
∴ The required Pythagorean triplet is 16, 63,
65.
(iv)
Let 2m = 18 m = 9
Now,
m2 – 1 = 92 – 1
=81
– 1 = 80
and
m2 + 1 = 92 + 1
=
81 + 1 = 82
∴ The required Pythagorean triple is 18, 80,
82.
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
Sol: The possible
digit at one’s place of the square root of:
(i)
9801 can be 1 or 9.
[∵ 1 × 1 = 1 and 9 × 9 = 81]
(ii)
99856 can be 4 or 6.
[∵ 4 × 4 = 16 and 6 × 6 = 36]
(iii)
998001 can be 1 or 9.
(iv)
657666025 can be 5.
[∵ 5 × 5 = 25]
2. Without doing any
calculation, find the numbers which are surely not perfect squares.
(i) 152 (ii) 257
(iii) 408 (iv) 441
Sol: We know that the ending digit of perfect
square is 0, 1, 4, 5, 6, and 9.
∴ A number ending in 2, 3, 7 or 8 can never be
a perfect square.
(i) 153, cannot
be a perfect square.
(ii)
257, cannot be a perfect square.
(iii)
408, cannot be a perfect square.
(iv)
441, can he a perfect square.
Thus, (1)
153, (ii) 257 and (iii) 408 are surely not perfect squares.
3. Find the square roots of.
100 and 169 by the method of repeated subtraction.
4. Find the square roots of the
following numbers by the Prime Factorization
Method.
(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929(viii) 9216
(ix) 529
(x) 8100
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
Sol: (i) We have
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.
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
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.
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.
Sol: Let the number of
students = x
Each
student donated Rs x.
Total
amount donated by the class = Rs x × x = Rs x2
Thus,
x2 = 2401
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 row and the number
of plants in each row.
Soln: Let the number
of rows = x
∴ Number of plants is a row = x
So,
the number of plants to be planted = x × x = x2
4, 9 and 10
We know that LCM is
the smallest number divisible by all its factors.
Since,
LCM of 4, 9 and 10 = 2 × 2 × 9 × 5 = 180
But
180 is not a perfect square.
Again,
∴ 180 = 2 × 2 × 3 × 3 × 5
it has 5 as unpaired
so multiply by 5 on both sides
180 x 5 = 2 x 2 x 3 x 3 x 5 x 5
= 900
∴ 900 is a perfect square.
Thus,
the required number = 900
10. Find the smallest
square number that is divisible by each of the numbers
8, 15 and 20
Sol: The smallest
number divisible by 8, 15 and 20 is their LCM.
We
have
LCM
= 2 × 2 × 5 × 2 × 3 = 120
But
120 is not a square number.
Now,
to make it a perfect square, we have
120
= 2 × 2 × 2 × 3 × 5
or
[120] × 2 × 3 × 5 = [2 × 2 × 2 × 3 × 5] × 2 × 3 × 5
or
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
All
factors of 3600 are paired. Therefore, 3600 is a perfect squared.
∴ The required number = 3600.
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