Class 8 Maths Chapter 5 Ex 5.1 Squares and Square Roots

The square of a number is obtained by multiplying the number by itself.
Example: The square of 4 is 4² = 4 x 4 = 16.

Properties of Square Numbers

1. Sum of Consecutive Odd Numbers:

A square number is the sum of consecutive odd numbers starting from 1.
Example: 4² = 16 and 16 = 1 + 3 + 5 + 7.

2. Ending Digits:

Square numbers have specific ending digits. They never end in 2, 3, 7, or 8.
Example: Squares end in 0, 1, 4, 5, 6, or 9.

3. Square Ending in 6:

If a square number ends in 6, the number whose square it is will have either 4 or 6 in the unit’s place.
Example: 14² = 196 and 16² = 256.

4. Squares of Numbers with 1 or 9 in Unit’s Place:

If a number has 1 or 9 in the units place, its square ends in 1.
Example: 11² = 121 and 19² = 361.

NCERT Solutions for Class 8 Maths Exercise 5.1 Chapter 5 Squares and Square Roots

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

Hint: The unit digit of a square depends only on the unit digit of the original number.

(i) 81
Unit digit of 81 = 1
Square of unit digit = 1² = 1

(ii) 272
Unit digit of 272 = 2
Square of unit digit = 2² = 4

(iii) 799
Unit digit of 799 = 9
Square of unit digit = 9² = 81
Unit digit of the square of the unit digit is 1

(iv) 3853
Unit digit of 3853 = 3
Square of unit digit = 3² = 9

(v) 1234
Unit digit of 1234 = 4
Square of unit digit = 4² = 16
Unit digit of the square of the unit digit is 6

(vi) 26387
Unit digit 26387 = 7
Square of unit digit = 7² = 49
Unit digit of the square of the unit digit is 9

(vii) 52698
Unit digit of 52698 = 8
Square of unit digit = 8² = 64
Unit digit of the square of the unit digit is 4

(viii) 99880
Unit digit of 99880 = 0
Square of unit digit = 0² = 0

(ix) 12796
Unit digit of 12796 = 6
Square of unit digit = 6² =36
Unit digit of the square of the unit digit is 6

(x) 55555
Unit digit of 55555 = 5
Square of unit digit = 5² = 25
Unit digit of the square of the unit digit is 5

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

Hint: A perfect square never ends in 2, 3, 7, or 8. In addition is a number end with odd number of zeros, the number cannot be a perfect square

(i) 1057
Given number 1057 ends with 7
The number has end digit in 2, 3, 7, or 8
1057 is not a perfect square.

(ii) 23453
Given number 23453 ends with 3
The number has end digit in 2, 3, 7, or 8
23453 is not a perfect square.

(iii) 7928
Given number 7928 ends with 8
The number has end digit in 2, 3, 7, or 8
7928 is not a perfect square.

(iv) 222222
Given number 222222 ends with 2
The number has end digit in 2, 3, 7, or 8
222222 is not a perfect square.

(v) 64000
Given number 64000 ends with 000 (last 3 digits)
The number ends with odd number of zeroes due to which it cannot be a perfect square

(vi) 89722
Given number 89722 ends with 2
The number has end digit in 2, 3, 7, or 8
89722 is not a perfect square.

(vii) 222000
Given number 222000 ends with 000 (last 3 digits)
The number ends with odd number of zeroes due to which it cannot be a perfect square

(viii) 505050
Given number 505050 ends with 0
The number ends with odd number of zeroes due to which it 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

Solution

Hint: The square of an odd number is always odd.

(i) 431
431 is a Odd number
The square of 431 will be odd.

(ii) 2826
2826 is an Even number
The square of 2826 will be even.

(iii) 7779
7779 is a Odd number
The square of 7779 will be odd.

(iv) 82004
82004 is an Even number
The square of 82004 will be even.

Squares of 431 and 7779 will be odd numbers.

4. Observe the following pattern and find the missing digits.

11² = 121
101² = 10201
1001² = 1002001
100001² = 1 ……… 2 ……… 1
10000001² = ………………………

Solution

Pattern: Squaring numbers with only 1s.

11² = 121
101² = 10201
1001² = 1002001
100001² = 100020001 (Explanation: 100001² is 1 followed by double the number of zeros in 100001, then 2, then the same number of zeros, and ends in 1.)
10000001² = 1000002000001 (Explanation: Same pattern, 10000001² is 1 followed by double the number of zeros in 10000001, then 2, then the same number of zeros, and ends in 1.)

5. Observe the following pattern and supply the missing numbers.

11² = 1 2 1
101² = 1 0 2 0 1
10101² = 102030201
1010101² = ………………………
…………² = 10203040504030201

Solution

Pattern: Squaring numbers with alternating 0s and 1s.

11² = 121
101² = 10201
10101² = 102030201
1010101² = 1020304030201 (Explanation: Each square increases the sequence in the middle by the next odd number (3, 5, 7, etc.))
101010101² = 10203040504030201

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

Solution

Pattern: Squaring and adding numbers.

1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20² = 21² (Explanation: Square each number and add. The sum of squares on the left equals the square of a single number on the right.)
5² + 6² + 30² = 31² (Explanation: 25 + 36 + 900 = 961, and 31² = 961.)
6² + 7² + 42² = 50² (Explanation: 36 + 49 + 1764 = 2500, and 50² = 2500.)

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

Hint: Use the formula for the sum of the first n odd numbers: Sum = n².

(i) 1 + 3 + 5 + 7 + 9
The expression is sum of consecutive odd numbers
There are 5 terms in the expression.
Therefore n = 5
Sum of first 5 odd numbers = n² = 5² = 25

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
The expression is the sum of consecutive odd numbers.
There are 10 terms in the expression.
Therefore, n = 10
Sum of the first 10 odd numbers = n² = 10² = 100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
The expression is the sum of consecutive odd numbers.
There are 12 terms in the expression.
Therefore, n = 12
Sum of the first 12 odd numbers = n² = 12² = 144

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

Solution

(i) Express 49 as the sum of 7 odd numbers
49 = 7² = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Express 121 as the sum of 11 odd numbers.
121 = 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 (ii) 25 and 26 (iii) 99 and 100

Solution

Hint: The difference between consecutive square numbers is given by the formula (n+1)² – n².

(i) 12 and 13
12 and 13 are 2 consecutive numbers
n = 12 and n + 1 = 13
Total Numbers between n² and (n+1)² is 2n (basic algebra method)
Total numbers between squares of 12 and 13 are = 2n = 2 × 12 = 24

Method 2:
13² – 12² = 169 – 144 = 25. So, 24 numbers lie between.

(ii) 25 and 26
25 and 26 are 2 consecutive numbers
n = 25 and n + 1 = 26
Total Numbers between n² and (n+1)² is 2n (basic algebra method)
Total numbers between squares of 25 and 26 are = 2n = 2 × 25 = 50

Method 2:
26² – 25² = 676 – 625 = 51. So, 50 numbers lie between.

(iii) 99 and 100
99 and 100 are 2 consecutive numbers
n = 99 and n + 1 = 100
Total Numbers between n² and (n+1)² is 2n (basic algebra method)
Total numbers between squares of 99 and 100 are = 2n = 2 × 99 = 198

Method 2:
100² – 99² = 10000 – 9801 = 199. So, 198 numbers lie between.

Practice Worksheet with Challenging Questions For Class 8 Exercise 5.1 Squares and Square Roots

Questions

1. What will be the unit digit of the squares of the following numbers? (i) 123 (ii) 4567 (iii) 89012
2. Express 36 as the sum of six odd numbers.
3. How many numbers lie between the squares of 32 and 33?
4. Find the sum without adding: (i) 2 + 4 + 6 + 8 + 10 (ii) 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20
5. Using the pattern, find the missing number: 12 + 42 + 32 = 52; 22 + 62 + 52 = 82; 32 + 82 + _2 = 112
6. Find the missing digits in the pattern: 1000000012 = ………………………
7. Express 64 as the sum of eight odd numbers.
8. Is 707070707 a perfect square? Justify your answer.
9. Identify if the following numbers are perfect squares without calculation: (i) 50005000 (ii) 44444444
10. How many numbers lie between the squares of 50 and 51?
11. Find the unit digit of the square of the number 987654321.
12. Express 81 as the sum of nine odd numbers.
13. Find the sum without adding: 1 + 5 + 9 + 13 + … up to 20 terms.
14. Using the pattern, find the missing numbers: 52 + 72 + 112 = 132; 62 + 82 + _2 = 152; 72 + 102 + _2 = 172
15. Determine if 12345678987654321 is a perfect square.

Answers

1. (i) 9 (ii) 9 (iii) 4 (Only the unit digit affects the unit digit of the square)
2. 1 + 3 + 5 + 7 + 9 + 11 (Start with the smallest odd numbers)
3. 63 numbers (Difference between squares of consecutive numbers)
4. (i) 30 (ii) 110 (Sum of the first n even numbers)
5. 72 (Pythagorean triples)
6. 1000000000000000000000001 (Observe the pattern in the number of zeros)
7. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 (Use consecutive odd numbers)
8. No, the unit digit of a square cannot be 7 (Look at the unit digit)
9. (i) No (ii) Yes (Check divisibility by 4 of the number formed by the last two digits)
10. 99 numbers (Difference between squares of consecutive numbers)
11. 1 (Look at the unit digit of the square)
12. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 (Use consecutive odd numbers)
13. 740 (Sum of an arithmetic progression)
14. 122, 142 (Pythagorean triples)
15. Yes, it’s the square of 111111111 (Observe the pattern of digits)