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Chapter 3

Squares and Square Roots — Exercise 3.2

Class - 8 ML Aggarwal Understanding ICSE Mathematics

Exercise 3.2

Question 1

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

Answer

According to the property of square numbers, a number having 2, 3, 7 or 8 at its unit place is never a square number.

The numbers 142, 273, 347, 878 and 1273 have 2, 3, 7, 8 and 3 at their unit place respectively.

Hence, 142, 273, 347, 878 and 1273 are five numbers which are not square numbers.

Question 2(i)

What will be the unit digit of the square of the following number?

951

Answer

The digit at unit's place of 951 is 1 and 1² = 1.

Hence, the unit digit of the square of 951 is 1.

Question 2(ii)

What will be the unit digit of the square of the following number?

502

Answer

The digit at unit's place of 502 is 2 and 2² = 4.

Hence, the unit digit of the square of 502 is 4.

Question 2(iii)

What will be the unit digit of the square of the following number?

329

Answer

The digit at unit's place of 329 is 9 and 9² = 81.

Hence, the unit digit of the square of 329 is 1.

Question 2(iv)

What will be the unit digit of the square of the following number?

643

Answer

The digit at unit's place of 643 is 3 and 3² = 9.

Hence, the unit digit of the square of 643 is 9.

Question 2(v)

What will be the unit digit of the square of the following number?

5124

Answer

The digit at unit's place of 5124 is 4 and 4² = 16.

Hence, the unit digit of the square of 5124 is 6.

Question 2(vi)

What will be the unit digit of the square of the following number?

7625

Answer

The digit at unit's place of 7625 is 5 and 5² = 25.

Hence, the unit digit of the square of 7625 is 5.

Question 2(vii)

What will be the unit digit of the square of the following number?

68327

Answer

The digit at unit's place of 68327 is 7 and 7² = 49.

Hence, the unit digit of the square of 68327 is 9.

Question 2(viii)

What will be the unit digit of the square of the following number?

95628

Answer

The digit at unit's place of 95628 is 8 and 8² = 64.

Hence, the unit digit of the square of 95628 is 4.

Question 2(ix)

What will be the unit digit of the square of the following number?

99880

Answer

The digit at unit's place of 99880 is 0 and 0² = 0.

Hence, the unit digit of the square of 99880 is 0.

Question 2(x)

What will be the unit digit of the square of the following number?

12796

Answer

The digit at unit's place of 12796 is 6 and 6² = 36.

Hence, the unit digit of the square of 12796 is 6.

Question 3

The following numbers are obviously not perfect squares. Give reason.

(i) 567 (ii) 2453 (iii) 5298 (iv) 46292 (v) 74000

Answer

According to the property of square numbers, a number having 2, 3, 7 or 8 at its unit place is never a square number and a number ending in an odd number of zeros is never a perfect square.

(i) Hence, 567 is not a perfect square as its unit digit is 7.

(ii) Hence, 2453 is not a perfect square as its unit digit is 3.

(iii) Hence, 5298 is not a perfect square as its unit digit is 8.

(iv) Hence, 46292 is not a perfect square as its unit digit is 2.

(v) The number 74000 ends in three zeros.

Hence, 74000 is not a perfect square as it has an odd number of zeros at the end.

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

Answer

According to the property of square numbers, squares of even natural numbers are even and squares of odd natural numbers are odd.

(i) Hence, (573)² is an odd number because 573 is an odd number.

(ii) Hence, (4096)² is an even number because 4096 is an even number.

(iii) Hence, (8267)² is an odd number because 8267 is an odd number.

(iv) Hence, (37916)² is an even number because 37916 is an even number.

Question 5(i)

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

12 and 13

Answer

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

Here, n = 12

Natural numbers between 12² and 13² = 2 × 12 = 24

Hence, there are 24 natural numbers between 12² and 13².

Question 5(ii)

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

90 and 91

Answer

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

Here, n = 90

Natural numbers between 90² and 91² = 2 × 90 = 180

Hence, there are 180 natural numbers between 90² and 91².

Question 6(i)

Without adding, find the sum.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

Answer

As we know that, the sum of first n odd natural numbers = n².

Here, the given numbers are the first 8 odd natural numbers.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8² = 64

Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8² = 64.

Question 6(ii)

Without adding, find the sum.

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

Answer

As we know that, the sum of first n odd natural numbers = n².

Here, the given numbers are the first 15 odd natural numbers.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = 15² = 225

Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = 15² = 225.

Question 7(i)

Express 64 as the sum of 8 odd numbers.

Answer

As we know that, the sum of first n odd natural numbers = n².

64 = 8²

So, 64 is the sum of the first 8 odd natural numbers.

Hence, 64 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15.

Question 7(ii)

Express 121 as the sum of 11 odd numbers.

Answer

As we know that, the sum of first n odd natural numbers = n².

121 = 11²

So, 121 is the sum of the first 11 odd natural numbers.

Hence, 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.

Question 8(i)

Express the following as the sum of two consecutive integers:

19²

Answer

We know that the square of an odd number n can be expressed as the sum of the two consecutive integers $\dfrac{n^2 - 1}{2}$ and $\dfrac{n^2 + 1}{2}$ .

Since 19² = 361

First number $= \dfrac{19^2 - 1}{2} = \dfrac{361 - 1}{2}$

$= \dfrac{360}{2} = 180$

Second number $= \dfrac{19^2 + 1}{2} = \dfrac{361 + 1}{2}$

$= \dfrac{362}{2} = 181$

Hence, 19² = 180 + 181.

Question 8(ii)

Express the following as the sum of two consecutive integers:

33²

Answer

We know that the square of an odd number n can be expressed as the sum of the two consecutive integers $\dfrac{n^2 - 1}{2}$ and $\dfrac{n^2 + 1}{2}$ .

Since 33² = 1089

First number $= \dfrac{33^2 - 1}{2} = \dfrac{1089 - 1}{2}$

$= \dfrac{1088}{2} = 544$

Second number $= \dfrac{33^2 + 1}{2} = \dfrac{1089 + 1}{2}$

$= \dfrac{1090}{2} = 545$

Hence, 33² = 544 + 545.

Question 8(iii)

Express the following as the sum of two consecutive integers:

47²

Answer

We know that the square of an odd number n can be expressed as the sum of the two consecutive integers $\dfrac{n^2 - 1}{2}$ and $\dfrac{n^2 + 1}{2}$ .

Since 47² = 2209

First number $= \dfrac{47^2 - 1}{2} = \dfrac{2209 - 1}{2}$

$= \dfrac{2208}{2} = 1104$

Second number $= \dfrac{47^2 + 1}{2} = \dfrac{2209 + 1}{2}$

$= \dfrac{2210}{2} = 1105$

Hence, 47² = 1104 + 1105.

Question 9(i)

Find the square of the following number containing 5 in unit's place:

45

Answer

We know that (a5)² = a(a + 1) hundreds + 25.

Comparing 45 with a5, we get a = 4.

(45)² = 4(4 + 1) hundreds + 25 = (4 × 5) hundreds + 25 = 20 hundreds + 25 = 2000 + 25 = 2025

Hence, (45)² = 2025.

Question 9(ii)

Find the square of the following number containing 5 in unit's place:

305

Answer

We know that (a5)² = a(a + 1) hundreds + 25.

Comparing 305 with a5, we get a = 30.

(305)² = 30(30 + 1) hundreds + 25 = (30 × 31) hundreds + 25 = 930 hundreds + 25 = 93000 + 25 = 93025

Hence, (305)² = 93025.

Question 9(iii)

Find the square of the following number containing 5 in unit's place:

525

Answer

We know that (a5)² = a(a + 1) hundreds + 25.

Comparing 525 with a5, we get a = 52.

(525)² = 52(52 + 1) hundreds + 25 = (52 × 53) hundreds + 25 = 2756 hundreds + 25 = 275600 + 25 = 275625

Hence, (525)² = 275625.

Question 10(i)

Write a Pythagorean triplet whose one number is 8.

Answer

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let 2m = 8

⇒ m = 4

∴ m² - 1 = 16 - 1 = 15 and m² + 1 = 16 + 1 = 17

So, (8, 15, 17) is a Pythagorean triplet.

Again, let m² + 1 = 8

⇒ m² = 7, so m is not a natural number.

Let m² - 1 = 8

⇒ m² = 9

⇒ m = 3

∴ 2m = 2 × 3 = 6 and m² + 1 = 9 + 1 = 10

So, (6, 8, 10) is also a Pythagorean triplet.

Hence, (8, 15, 17) or (6, 8, 10) is a Pythagorean triplet whose one number is 8.

Question 10(ii)

Write a Pythagorean triplet whose one number is 15.

Answer

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let 2m = 15

⇒ m = 7.5, which is not a natural number.

Let m² + 1 = 15

⇒ m² = 14, so m is not a natural number.

Let m² - 1 = 15

⇒ m² = 16

⇒ m = 4

∴ 2m = 2 × 4 = 8 and m² + 1 = 16 + 1 = 17

Hence, (8, 15, 17) is a Pythagorean triplet whose one number is 15.

Question 10(iii)

Write a Pythagorean triplet whose one number is 63.

Answer

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let 2m = 63

⇒ m = 31.5, which is not a natural number.

Let m² + 1 = 63

⇒ m² = 62, so m is not a natural number.

Let m² - 1 = 63

⇒ m² = 64

⇒ m = 8

∴ 2m = 2 × 8 = 16 and m² + 1 = 64 + 1 = 65

Hence, (16, 63, 65) is a Pythagorean triplet whose one number is 63.

Question 10(iv)

Write a Pythagorean triplet whose one number is 80.

Answer

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let m² - 1 = 80

⇒ m² = 81

⇒ m = 9

∴ 2m = 2 × 9 = 18 and m² + 1 = 81 + 1 = 82

So, (18, 80, 82) is a Pythagorean triplet.

Again, let 2m = 80

⇒ m = 40

∴ m² - 1 = 1600 - 1 = 1599 and m² + 1 = 1600 + 1 = 1601

So, (80, 1599, 1601) is also a Pythagorean triplet.

Hence, (18, 80, 82) or (80, 1599, 1601) is a Pythagorean triplet whose one number is 80.