NCERT Solution for Squares and Square Roots

(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

∴ The unit’s digit of (81)^{2} will be 1.

(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

The unit’s digit of (55555)^{2} will be 5.

(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).

∴89722 is not a perfect square.

(viii) 222000

Since, the number of zeros is odd.

∴222000 is not a perfect square.

(viii) 505050

The unit’s digit is odd zero.

∴505050 can not be a perfect square.

(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.]

11^{2} =

121

101^{2} =

10201

1001^{2} =

1002001

100001^{2} =

1.............2 ............. 1

10000001^{2} =

............

Sol: Observing the above pattern, we have

(i) (100001)^{2} - 10000200001

(ii) (10000001)^{2} = 100000020000001

11^{2} = 121

101^{2} = 10201

10101^{2} = 102030201

1010101^{2} =

..............^{2} = 10203040504030201

Sol: Observing the above, we have

(i) (1010101)^{2} = 1020304030201

(ii) 10203040504030201 = (101010101)^{2}

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}

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) 4^{2} + 5^{2} + 20^{2} = 21^{2}

(ii) 5^{2} + 2^{2} + 30^{2} = 31^{2}

(iii) 6^{2} + 7^{2} + 42^{2} = 43^{2}

(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 = 5^{2}

= 25

(ii) The sum of first 10 odd numbers = 10^{2}

= 100

(iii) The sum of first 12 odd numbers = 12^{2}

= 144

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

Sol: (i) 49 = 7^{2} = Sum of first 7 odd numbers

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

(ii) 121 = 11^{2} = Sum of first 11 odd numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100

Sol: Since between n^{2} and (n + 1)^{2}, there are 2n non-square numbers.

∴ (i) Between 12^{2} and 13^{2}, there are 2 × 12, i.e. 24 numbers

(ii) Between 25^{2} and 26^{2}, there are 2 × 25, i.e. 50 numbers

(iii) Between 99^{2} and 100^{2}, there are 2 × 99, i.e. 198 numbers

(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

Sol: (i) (32)^{2} = (30 + 2)^{2}

= 30^{2} + 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

35^{2} = 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 2n =6 ∴n=3

Now, n^{2} – 1 = 3^{2} –1 = 8

and n^{2} + 1 = 3^{2} + 1 = 10

Thus, the required Pythagorean triplet is 6, 8, 10.

(ii) Let 2n = 14 ∴n = 7

Now, n^{2} – 1 = 7^{2} – 1 = 48

and n^{2} + 1 = 7^{2} + 1 = 50

Thus, the required Pythagorean triplet is 14, 48, 50.

(iii) Let 2n = 16 /n = 8

Now, n^{2} – 1 = 8^{2} – 1

= 64 – 1 = 63

and n^{2} + 1 = 82 + 1

=64 + 1 = 65

∴ The required Pythagorean triplet is 16, 63, 65.

(iv) Let 2n = 18 n = 9

Now, n^{2} – 1 = 9^{2} – 1

=81 – 1 = 80

and n^{2} + 1 = 92 + 1

= 81 + 1 = 82

∴ The required Pythagorean triple is 18, 80, 82.

(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]

(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.

(i) 729

(ii) 400

(iii) 1764

(iv) 4096

(v) 7744

(vi) 9604

(vii) 5929

(viii) 9216

(ix) 529

(x) 8100

(i) 252

(ii) 180

(iii) 1008

(iv) 2028

(v) 1458

(vi) 768

Sol: (i) We have

252 = 2 × 2 × 3 × 3 × 7

∵ The prime factor 7 has no pair.

∴ [252] × 7= [2 × 2 × 3 × 3 × 7] × 7

or 1764 = 2 × 2 × 3 × 3 × 7 × 7

Thus, the required smallest whole number = 7

Also, the square root of 1764 is 42.

(ii) We have

Thus, the required smallest whole number = 5

Also, the square root of 900 is 30.

(iii) We have

Thus, the required smallest whole number = 7

Also the square root of 7056 is 84.

(iv) We have

(v) We have

(vi) We have

(i) 252

(ii) 2925

(iii) 396

(iv) 2645

(v) 2800

(vi) 1620

Sol: (i) 252

We have

∵ The prime factor 7 is unpaired, so the given number should be divided by 7.

(ii) 2925

We have

∵ The prime factor 13 is unpaired, so the given number should be divided by 13.

(iii) 396

We have

(iv) 2645

We have

∵ The prime factor 5 is unpaired.

∴ Dividing the given number by 5, we have

or 529 = 23 × 23

Thus, 529 is a perfect square, and 529 = 23.

(iv) 2800

We have

(v) 1620

We have

1620 = 2 × 2 × 3 × 3 × 3 × 3 ×5

∵ The prime factor 5 is unpaired.

∴ Dividing the given number by 5, we have

Sol: Let the number of students = x

Each student donated Rs x.

Total amount donated by the class = Rs x × x = Rs x^{2}

Thus, x^{2} = 2401

Soln: Let the number of rows = x

∴ Number of plants is a row = x

So, the number of plants to be planted = x × x = x^{2}

Thus, the required number of rows = 45

Also, number of plants in a row = 45.

Sol: 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

∵ All the prime factors of 900 are paired.

∴ 900 is a perfect square.

Thus, the required number = 900.

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.

(i) 2304

(ii) 4489

(iii) 3481

(iv) 529

(v) 3249

(vi) 1369

(vii) 5776

(viii) 7921

(ix) 576

(x) 1024

(xi) 3136

(xii) 900

(i) 64

(ii) 144

(iii) 4489

(iv) 27225

(v) 390625

Sol: If ‘n’ stands for number of digits in the given number, then

(i) 2.56

(ii) 7.29

(iii) 51.84

(iv) 42.25

(v) 31.36

Here, number of decimal places, are already even.

∴ We mark off the periods and find the square root.

Here, number of decimal places are already even.

Therefore, we mark off the periods and find the square root.

(i) 402

(ii) 1989

(iii) 3250

(iv) 825

(v) 4000

Sol: (i) On proceeding to find the square root of 402, we have

Since, we get a remainder 2

The required least number to be subtracted from 402 is 2.

(ii) Since, we get a remainder of 53

The least number to be subtracted from the given number - 53 84

(iii) Since, we get a remainder 1.

The smallest number to be subtracted from the given number = 1

(iv) Since, we get a remainder 41.

The required smallest number tcI be subtracted from the given number = 41

(v) Since, we get a remainder 31,

The required smallest number to he subtracted from om the given number = 31

(i) 525

(ii) 1750

(iii) 252

(iv) 1825

(v) 6412

Sol: (i) Since, we get a remainder 41.

and next square number is 23.

∴ The required number to be added = 232 – 525

= 529 – 525 = 4

Now, 524 + 4 = 529, and = 23.

(ii) Since, we get a remainder 69.

and next square number is 422.

∴ The required number to be added = 422 – 1750

= 1764 – 1750 = 14

Now, 1750 + 14 = 1764, and

(ii) Since, we get a remainder 27.

(iv) Since, we get a remainder, 61.

Next square number = 43

The required number to be added = (43)^{2} – 1825

= 1849 – 1825 = 24

Now, 1825 + 24 = 1849, and

(v) Since, we get a remainder 12.

Sol: Let the side of the square = x metre

Area = side × side

(a) If AB = 6 cm, BC = 8 cm, find AC.

(b) If AC = 13 cm, BC = 5 cm, find AB. Solution:

Sol:

I. In a right triangle, the longest side is called the hypotenuse.

II. (Hypotenuse)^{2} = [Sum of the squares of other two sides]

(a) ∠B = 90°

Hypotenuse = AC

AC^{2} = AB^{2} + BC^{2}

= 8^{2} + 6^{2}

= 64 + 36 = 100

AC = 10

Thus, AC = 10 cm

(b) Here B = 90°

∴ Hypotenuse = AC

AC^{2} = AB^{2} + BC^{2}

or 13^{2} = AB^{2} = 5^{2}

= 169 – 25 = 144

or AB = 12

Thus, AB = 12 cm

Sol: Since, the number of plants in a row and the number of columns are the same.

∴ Their product must be a square number.

The gardener has 1000 plants.

1000 is not a perfect square, and (31)^{2} < 1000

(There is a remainder of 39).

Obviously the next square number = 32

Number of plants required to be added = (32)^{2} – 1000

= 1024 – 1000 = 24

Sol: Since, the number of rows and the number of columns are same.

Total number (i.e. their product) must be a square number, we have

Since, we get a remainder of 16

500 > (22)^{2} or 500 – 16 = (22)^{2}

Thus, the required number of children to be left out = 16