HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots Ex 6.1

Haryana State Board HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots Ex 6.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 6 Square and Square Roots Exercise 6.1

Question 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.
Solution:
(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 12796² = 6 [∵ 6² = 36]
(x) The unit digit of 55555² = 5 [∵ 5² = 25]

Question 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.
Solution:
(i) 1057 is not a perfect square.
[∵ Perfect square never end with 2, 3, 7 or 8 at unit’s place]
(ii) 23453 is not a perfect square because the number end with 3.
(iii) Since 7928 end with 8. So, it is not a perfect square.
(iv) Since 222222 end with 2. So, it is not a perfect square.
(v) 64000 because it do not have even number of zeros at the end.
(vi) 89722 is not a perfect square.
[∵ Perfect square never end with 2, 3, 7 or 8]
(vii) 222000 is not a perfect square as the number of zeros in the end is 3 (odd number).
(viii) 505050 is not a perfect square because number of zeros in the ends is 1 (odd number).

Question 3.
The squares of which of the following would be odd numbers ?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004.
Solution:
(i) 431 and (ii) 7779.
Because squares of odd number are always
odd.

Question 4.
Observe the following pattern and find the missing digits.
11² = 121
101² = 10201
1001² = 1002001
100001² = 1………….. 2 …………….. 1
10000001² = ………………..
Solution:
100001² = 10000200001
and 10000001² = 100000020000001
[Using pattern given in the question]

Question 5.
Observe the following pattern and supply the missing numbers.
11² = 121
101² = 10201
10101² = 102030201
1010101² = ………………..
………………..² = 10203040504030201
Solution:
1010101² = 1020304030201
and 101010101² = 10203040504030201
[Using pattern given in the question]

Question 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
To find pattern : Third number is related to first and second number. How ?
Fourth number is related to third number. How ?
Solution:
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
and 6² + 7² + 42² = 43²
∵ 1 × 2 = 2, 2 + 1 = 3
2 × 3 = 6, 6 + 1 = 7
3 × 4 = 12, 12 + 1 = 13
…………………………….
and so on.

Question 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.
Solution:
(i) 1 + 3 + 5 + 7 + 9
[Sum of first five odd numbers]
= 5² = 25.

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 [Sum of first ten odd numbers]
= 10² = 100.

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

Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of ii odd numbers.
Solution:
(i) 49 = (7)²
= Sum of first 7 odd numbers
= 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Similarly,
121 = (11)²
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Question 9.
How many numbers lie between squares of the following numbers ?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100.
Solution:
(i) Here, n = 12 and n + 1 = 13
∴ 2n numbers non perfect square numbers lie between square of n2 and (n + 1)²
Thus, 2 × 12 = 24
such numbers lie between 12² and 13².

(ii) Similarly,
2 × 25 = 50
numbers lie between 25² and 26².
2 × 99 = 198

and (iii) 2 × 99 = 198
numbers lie between 99² and 100².