# Squares And Square Roots

## Exercise 5.1

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) 81

The unit digit in the square of 81 is 1.

(ii) 272

The unit digit in the square of 272 is 4.

(iii) 799

The unit digit in the square of 799 is 1.

(iv) 3853

The unit digit in the square of 3853 is 9.

(v) 1234

The unit digit in the square of 1234 is 6. (vi) 26387

The unit digit in the square of 26387 is 9.

(vii) 52698

The unit digit in the square of 52698 is 4.

(viii) 99880

The unit digit in the square of 99880 is 0.

(ix) 12796

The unit digit in the square of 12796 is 6.

(x) 55555

The unit digit in the square of 55555 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:

(i) 1057

The digit in the unit’s place is 7.

Therefore, it is not a perfect square.

(ii) 23453

The digit in the unit’s place is 3.

Therefore, it is not a perfect square.

(iii) 7928

The digit in the unit’s place is 8.

Therefore, it is not a perfect square.

(iv) 222222

The digit in the unit’s place is 2.

Therefore, it is not a perfect square.

(v) 64000

The number contains three 0’s at the end.

Therefore, it is not a perfect square.

(vi) 89722

The digit in the unit’s place is 2.

Therefore, it is not a perfect square.

(vii) 222000

The number contains three 0’s at the end.

Therefore, it is not a perfect square.

(viii) 505050

The number contains three 0’s at the end.

Therefore, it is not a perfect square.

3). The squares of which of the following would be odd numbers?

(i) 431 (ii) 2826

(iii) 7779 (iv) 82004

Solution:

(i) 431

Square of 431 is odd.

(ii) 2826

Square of 2826 is even.

(iii) 7779

Square of 7779 is odd.

(iv) 82004

Square of 82004 is even.

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

11^{2} = 121

101^{2} = 10201

1001^{2} = 1002001

100001^{2} = 1 ……… 2 ……… 1

10000001^{2} = ………………………

Solution:

11^{2} = 121

101^{2} = 10201

1001^{2} = 1002001

100001^{2} = 10000200001

10000001^{2} =1__000000__2__000000__1

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

11^{2} = 1 2 1

101^{2} = 1 0 2 0 1

10101^{2} = 102030201

1010101^{2} = ………………………

…………^{2} = 10203040504030201

Solution:

11^{2} = 1 2 1

101^{2} = 1 0 2 0 1

10101^{2} = 102030201

1010101^{2} = __1020304030201__

__ 101010101__^{2} = 10203040504030201

6). Using the given pattern, find the missing numbers.

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} + _^{2} = 21^{2}

5^{2} + _^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + _^{2} = __^{2}

Solution:

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} +__20__^{2} = 21^{2}

5^{2} + __6__^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + __42__^{2} = __43__^{2}

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

We know that sum of *n* odd numbers = *n*^{2}

1 + 3 + 5 + 7 + 9 = 5^{2} = 25 (*n* = 5)

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19

We know that sum of *n* odd numbers = *n*^{2}

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 = 10^{2} (*n* = 10)

= 100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

We know that sum of *n* odd numbers = *n*^{2}

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 + 21 + 23 = 12^{2}

= 144 (*n* = 12)

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.

*n* = 11

49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Express 121 as the sum of 11 odd numbers.

*n* = 11

121 = 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:

(i) 12 and 13

We know that between the squares of *n* and (*n+1*) there are 2*n* non-perfect square numbers.

2*n* = 2 X 12 = 24

24 numbers lie between squares of 12 and 13.

(ii) 25 and 26

We know that between the squares of *n* and (*n+1*) there are 2*n* non-perfect square numbers.

2*n* = 2 X 25 = 50

50 numbers lie between squares of 25 and 26.

(iii) 99 and 100

We know that between the squares of *n* and (*n+1*) there are 2*n* non-perfect square numbers.

2*n* = 2 X 99 = 198

198 numbers lie between squares of 99 and 100.

Click here for the solutions of Std 8 Maths

1). Rational Numbers

2). Linear Equations in One Variable

3). Understanding Quadrilaterals

4). Data Handling

5). Squares and Square Roots