NCERT Solutions for Class 8 Maths Chapter 5 - Square and Square Roots (Exercise 5.1)

Exercise 5.1

Q1. 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) Since, 1 × 1 = 1
∴ The unit’s digit of (81)² will be 1.
(ii) Since, 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.
(vii) Since, 8 × 8 = 64
∴ The unit’s digit of (52698)² will be 4.
(viii) Since 0 × 0 = 0
∴ The unit’s digit of (99880)² will be 0.
(ix) Since 6 × 6 = 36
∴ The unit’s digit of (12796)² will be 6.
(x) Since, 5 × 5 = 25
∴ The unit’s digit of (55555)² will be 5.

Q2. 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
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 3 (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
Has an odd number of zeros (3 zeros) → Not a perfect square.
∴ 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).
∴ 89722 is not a perfect square.
(viii) 222000
Has an odd number of zeros (3 zeros) → Not a perfect square.
∴  222000 is not a perfect square.
(viii) 505050
Has an odd number of zeros (1 zero) → Not a perfect square.
∴ 505050 can not be a perfect square.

Q3. The squares of which of the following would be odd numbers?
(i) 431 
(ii) 2826 
(iii) 7779 
(iv) 82004
Ans: 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.]
(iv) The square of 82004 is an even number.
       [∵ 82004 is an even number.]

Q4. Observe the following pattern and find the missing digits.
11² = 1 2 1
101² = 10201
1001² = 1002001
100001² = 1 … 2 …1
10000001² = …
Ans: We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and middle digit is 2. And the number of zeros between left most digits 1 and the middle digit 2 and right most digit 1 and the middle digit 2 is same as the number of zeros in the given number.
∴ (100001)² = 10000200001
(10000001)² = 100000020000001

Q5. Observe the following pattern and supply the missing number.
11² = 1 2 1
101² = 1 0 2 0 1
10101² = 102030201
1010101² = ………….
………….² = 10203040504030201
Ans: Observing the above, we have:
(i) (1010101)² = 1020304030201
(ii) 10203040504030201 = (101010101)²

Q6. 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?

Ans: Given, 1² + 2² + 2² = 3²
i.e 1² + 2² + (1×2 )² = ( 1² + 2² -1 × 2 )²
2² + 3² + 6² =7²
∴ 2² + 3² + (2×3 )² = (2² + 3² -2 × 3)²
3² + 4² + 12² = 13²
∴ 3² + 4² + (3×4 )² = (3² + 4² – 3 × 4)²
4² + 5² + (4×5 )² = (4² + 5² – 4 × 5)²
∴ 4² + 5² + 20² = 21²
5² + 6² + (5×6 )² = (5²+ 6² – 5 × 6)²
∴ 5² + 6² + 30² = 31²
6² + 7² + (6×7 )² = (6² + 7² – 6 × 7)²
∴ 6² + 7² + 42² = 43²

Q7. 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) The sum of first 5 odd numbers = 5²  = 25
(ii) The sum of first 10 odd numbers = 10²  = 100
(iii) The sum of first 12 odd numbers = 12² = 144

Q8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Ans:
(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

Q9. How many numbers lie between squares of the following numbers?
(i) 12 and 13    
(ii) 25 and 26     
(iii) 99 and 100
Solution: 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 5.2

Que 1. Find the square of the following numbers.
(i) 32 
(ii) 35 
(iii) 86 
(iv) 93 
(v) 71 
(vi) 46
Ans:

(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

Q2. Write a Pythagorean triplet whose one member is
(i) 6 
(ii) 14 
(iii) 16 
(iv) 18
Ans: 
(i) Let 2n = 6       ∴n = 3
Now, n² – 1 = 3² – 1 = 8
and n² + 1 = 3² + 1 = 10
Thus, the required Pythagorean triplet is 6, 8, 10.

(ii) Let 2n = 14        ∴ n = 7
Now, n² – 1 = 7² – 1 = 48
and n² + 1 = 7² + 1 = 50
Thus, the required Pythagorean triplet is 14, 48, 50.

(iii) Let 2n = 16      ∴n = 8
Now, n² – 1 = 8² – 1
= 64 – 1 = 63
and n² + 1 = 8² + 1
= 64 + 1 = 65
Thus, the required Pythagorean triplet is 16, 63, 65.

(iv) Let 2n = 18          ∴n = 9
Now, n² – 1 = 9² – 1
= 81 – 1 = 80
and n² + 1 = 9² + 1
= 81 + 1 = 82
Thus, the required Pythagorean triple is 18, 80, 82.