## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.2

Question 1.

Write five numbers which you can decide by looking at their one’s digit that they are not square numbers.

Solution:

We know that a number which ends with the digits 2,3, 7 or 8

at its unit places, is not a perfect square.

For example:

372, 563, 111, 978, 1282 are not square numbers.

Question 2.

What will be the unit digit of the squares of the following numbers?

(i) 951

(ii) 502

(iii) 329

(iv) 643

(v) 5124

(vi) 7625

(vii) 68327

(viii) 95628

(ix) 99880

(x) 12796

Solution:

The unit digit of the square of the following numbers will be

(i) 951: Its square will have unit digit = 1

(ii) 502: Its square will have unit digit = 4

(iii) 329: Its square will have unit digit = 1

(iv) 643: Its square will have unit digit = 9

(v) 5124: Its square will have unit digit = 6

(vi) 7625: Its square will have unit digit = 5

(vii) 68327: Its square will have unit digit = 9

(viii) 95628: Its square will have unit digit = 4

(ix) 99880: Its square will have unit digit = 0

(x) 12796: Its square will have unit digit = 6

Question 3.

The following numbers are obviously not perfect. Give reason.

(i) 567

(ii) 2453

(iii) 5298

(iv) 46292

(v) 74000

Solution:

We know that if the square of a number does not have

2, 3, 7, 8 or 0 (in an odd number) as its unit digit.

So, the squares 567, 2453, 5208, 46292 and 74000 can’t be the

perfect squares as they have 7, 2, 8, 2 digits at the unit place.

Question 4.

The square of which of the following numbers would be an odd number or an even number? Why?

(i) 573

(ii) 4096

(iii) 8267

(iv) 37916

Solution:

We know that the square of an odd number is odd and

a square of an even number is even. Therefore:

(i) 573, (iii) 8262 are odd numbers.

So, their squares will also be an odd number and

(ii) 4096 and (iv) 37916 are even numbers.

So, their square will be also even-numbered.

Question 5.

How many natural numbers lie between square of the following numbers?

(i) 12 and 13

(ii) 90 and 91

Solution:

(i) Numbers of natural number between

the squares of 12 and 13

= (13^{2} – 12^{2}) – 1 = (13 + 12 – 1)

= 25 – 1 = 24

(ii) Between 90 and 91

= (91^{2} – 90^{2}) – 1 = (91 + 90 – 1)

= 181 – 1 = 180

Question 6.

Without adding, find the sum.

(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

Solution:

Find the sum of:

(i) 1+3 + 5 + 7 + 9 + 11 + 13 + 15 = n^{2}

Here n = 8

Sum = (8)^{2} = 64

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = n^{2}

Sum = 15^{2} (Here n = 15)

Sum = 225

Question 7.

(i) Express 64 as the sum of 8 odd numbers.

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

Solution:

(i) 64 as the sum of odd number

= (8)^{2} = n^{2}

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 (Here n = 8)

(ii) 121 = (11)^{2}

= 1 + 3 + 5 + 7 + 9 +11 + 13 + 15 +17 + 19 + 21 = n^{2} (Here n = 11)

Question 8.

Express the following as the sum of two consecutive integers:

(i) 19^{2}

(ii) 33^{2}

(iii) 47^{2}

Solution:

Question 9.

Find the squares of the following numbers without actual multiplication:

(i) 31

(ii) 42

(iii) 86

(iv) 94

Solution:

Using (a + b)^{2} – a^{2} + 2ab + b^{2}

(i) (31)^{2} = (30 + 1)^{2}

= (30)^{2} + 2 × 30 + 1 + (1)^{2}

= 900 + 60 + 1 = 961

(ii) (42)^{2} = (40 + 2)^{2}

= (40)^{2} + 2 × 40 × 2 + (2)^{2}

= 1600 + 160 + 4 = 1764

(iii) (86)^{2} = (80 + 6)^{2}

= (80)^{2} + 2 × 80 × 6 + (6)^{2}

= 6400 + 960 + 36 = 7396

(iv) (94)^{2} = (90 + 4)^{2}

= (90)^{2} + 2 × 90 × 4 + (4)^{2}

= 8100 + 720 + 16 = 8836

Question 10.

Find the squares of the following numbers containing 5 in unit’s place:

(i) 45

(ii) 305

(iii) 525

Solution:

(i) (45)^{2} = (n5)^{2}

= (n)(n + 1) hundred + (5)^{2}

= 4 × 5 hundred + 25

= 2000 + 25 = 2025

(ii) (305)^{2} = (30 × 31) hundred + 25

= 93000 + 25 = 93025

(iii) (525)^{2} = (52 × 53) hundred + 25

= 275600 + 25 = 275625

Question 11.

Write a Pythagorean triplet whose one number is

(i) 8

(ii) 15

(iii) 63

(iv) 80

Solution:

Pythagoras triplet whose one number is

(i) 8

Let n = 8, then the triplet will be

2n, n^{2} – 1, n^{2} + 1

If 2n = 8, then n = \(\frac{8}{2}\) = 4

∴ n^{2} – 1 = 4^{2} – 1 = 16 – 1 = 15

and n^{2} + 1 = 4^{2} + 1 = 16 + 1 = 17

∴ Triplet is 8, 15, 17

(ii) 15

Let 2n = 15, then n = \(\frac{n}{2}\) which is not possible

or n^{2} – 1 = 15 ⇒ n^{2} = 15 + 1 = 16 = (4)^{2}

∴ n = 4

Now, 2n = 2 × 4 = 8

n^{2} – 1 = 15

n^{2} + 1 = 4^{2} + 1 = 16 + 1 = 17

∴ Triplet is 8, 15, 17

(iii) 63

Let n^{2} – 1 = 63 ⇒ n^{2} = 63 + 1 = 64 = (8)^{2}

∴ n = 8

Now, 2n = 2 × 8 = 16

n^{2} – 1 = 63

n^{2} + 1 = 8^{2} + 1 = 64 + 1 = 65

∴ Triplet is 16, 63, 65

(iv) 80

Let 2n = 80 ⇒ n = \(\frac{80}{2}\) =40

∴ n^{2} – 1 = 40^{2} – 1 = 1600 – 1 = 1599

and n^{2} + 1 = 40^{2} + 1 = 1600 + 1 = 1601

∴ Triplet is 80, 1599, 1601

Question 12.

Observe the following pattern and find the missing digits:

21^{2} = 441

201^{2} = 40401

2001^{2} = 4004001

20001^{2} = 4 – – – 4 – – – 1

200001^{2} = ————–

Solution:

21^{2} = 441

201^{2} = 40401

2001^{2} = 4004001

Similarly, 20001^{2} = 400040001

200001^{2} = 40000400001

Question 13.

Observe the following pattern and find the missing digits:

9^{2} = 81

99^{2} = 9801

999^{2} = 998001

9999^{2} = 99980001

99999^{2} = 9——–8———01

999999^{2} = 9——–0———1

Solution:

9^{2} = 81

99^{2} = 9801

999^{2} = 998001

9999^{2} = 99980001

Similarly, 99999^{2} = 9999800001

and 999999^{2} = 9999998000001

Question 14.

Observe the following pattern and find the missing digits:

7^{2} = 49

67^{2} = 4489

667^{2} = 444889

6667^{2} = 44448889

66667^{2} = 4 ———–8 ————– 9

666667^{2} = 4———–8————8 –

Solution:

7^{2} = 49

67^{2} = 4489

667^{2} = 444889

6667^{2} = 44448889

Similarly, 66667^{2} = 4444488889

666667^{2} = 444444888889