## Squares and Square Roots Class 8 Extra Questions Maths Chapter 6

**Extra Questions for Class 8 Maths Chapter 6 Squares and Square Roots**

### Squares and Square Roots Class 8 Extra Questions Very Short Answer Type

Question 1.

Find the perfect square numbers between 40 and 50.

Solution:

Perfect square numbers between 40 and 50 = 49.

Question 2.

Which of the following 24^{2}, 49^{2}, 77^{2}, 131^{2} or 189^{2} end with digit 1?

Solution:

Only 49^{2}, 131^{2} and 189^{2} end with digit 1.

Question 3.

Find the value of each of the following without calculating squares.

(i) 27^{2} – 26^{2}

(ii) 118^{2} – 117^{2}

Solution:

(i) 27^{2} – 26^{2} = 27 + 26 = 53

(ii) 118^{2} – 117^{2} = 118 + 117 = 235

Question 4.

Write each of the following numbers as difference of the square of two consecutive natural numbers.

(i) 49

(ii) 75

(iii) 125

Solution:

(i) 49 = 2 × 24 + 1

49 = 25^{2} – 24^{2}

(ii) 75 = 2 × 37 + 1

75 = 38^{2} – 37^{2}

(iii) 125 = 2 × 62 + 1

125 = 63^{2} – 62^{2}

Question 5.

Write down the following as sum of odd numbers.

(i) 7^{2}

(ii) 9^{2}

Solution:

(i) 7^{2} = Sum of first 7 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) 9^{2} = Sum of first 9 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

Question 6.

Express the following as the sum of two consecutive integers.

(i) 15^{2}

(ii) 19^{2}

Solution:

Question 7.

Find the product of the following:

(i) 23 × 25

(ii) 41 × 43

Solution:

(i) 23 × 25 = (24 – 1) (24 + 1) = 24^{2} – 1 = 576 – 1 = 575

(ii) 41 × 43 = (42 – 1) (42 + 1) = 42^{2} – 1 = 1764 – 1 = 1763

Question 8.

Find the squares of:

(i)

(ii)

Solution:

Question 9.

Check whether (6, 8, 10) is a Pythagorean triplet.

Solution:

2m, m^{2} – 1 and m^{2} + 1 represent the Pythagorean triplet.

Let 2m = 6 ⇒ m = 3

m^{2} – 1 = (3)^{2} – 1 = 9 – 1 = 8

and m^{2} + 1 = (3)^{2} + 1 = 9 + 1 = 10

Hence (6, 8, 10) is a Pythagorean triplet.

Alternative Method:

(6)^{2} + (8)^{2} = 36 + 64 = 100 = (10)^{2}

⇒ (6, 8, 10) is a Pythagorean triplet.

Question 10.

Using property, find the value of the following:

(i) 19^{2} – 18^{2}

(ii) 23^{2} – 22^{2}

Solution:

(i) 19^{2} – 18^{2} = 19 + 18 = 37

(ii) 23^{2} – 22^{2} = 23 + 22 = 45

### Squares and Square Roots Class 8 Extra Questions Short Answer Type

Question 11.

Using the prime factorisation method, find which of the following numbers are not perfect squares.

(i) 768

(ii) 1296

Solution:

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, 3 is not in pair.

768 is not a perfect square.

1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3

Here, there is no number left to make a pair.

1296 is a perfect square.

Question 12.

Which of the following triplets are Pythagorean?

(i) (14, 48, 50)

(ii) (18, 79, 82)

Solution:

We know that 2m, m^{2} – 1 and m^{2} + 1 make Pythagorean triplets.

(i) For (14, 48, 50),

Put 2m =14 ⇒ m = 7

m^{2} – 1 = (7)^{2} – 1 = 49 – 1 = 48

m^{2} + 1 = (7)^{2} + 1 = 49 + 1 = 50

Hence (14, 48, 50) is a Pythagorean triplet.

(ii) For (18, 79, 82)

Put 2m = 18 ⇒ m = 9

m^{2} – 1 = (9)^{2} – 1 = 81 – 1 = 80

m^{2 }+ 1 = (9)^{2} + 1 = 81 + 1 = 82

Hence (18, 79, 82) is not a Pythagorean triplet.

Question 13.

Find the square root of the following using successive subtraction of odd numbers starting from 1.

(i) 169

(ii) 81

(iii) 225

Solution:

(i) 169 – 1 = 168, 168 – 3 = 165, 165 – 5 = 160, 160 – 7 = 153, 153 – 9 = 144, 144 – 11 = 133, 133 – 13 = 120, 120 – 15 = 105, 105 – 17 = 88, 88 – 19 = 69,

69 – 21 = 48, 48 – 23 = 25, 25 – 25 = 0

We have subtracted odd numbers 13 times to get 0.

√169 = 13

(ii) 81 – 1 = 80, 80 – 3 = 77, 77 – 5 = 72, 72 – 7 = 65, 65 – 9 = 56, 56 – 11 = 45, 45 – 13 = 32, 32 – 15 = 17, 17 – 17 = 0

We have subtracted 9 times to get 0.

√81 = 9

(iii) 225 – 1 = 224, 224 – 3 = 221, 221 – 5 = 216, 216 – 7 = 209, 209 – 9 = 200, 200 – 11 = 189, 189 – 13 = 176, 176 – 15 = 161, 161 – 17 = 144, 144 – 19 = 125,

125 – 21 = 104, 104 – 23 = 81, 81 – 25 = 56, 56 – 27 = 29, 29 – 29 = 0

We have subtracted 15 times to get 0.

√225 = 15

Question 14.

Find the square rootofthe following using prime factorisation

(i) 441

(ii) 2025

(iii) 7056

(iv) 4096

Solution:

(i) 441 = 3 × 3 × 7 × 7

√441 = 3 × 7 = 21

(ii) 2025 = 3 × 3 × 3 × 3 × 5 × 5

√2025 = 3 × 3 × 5 = 45

(iii) 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

√7056 = 2 × 2 × 3 × 7 = 84

(iv) 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

Question 15.

Find the least square number which is divisible by each of the number 4, 8 and 12.

Solution:

LCM of 4, 8, 12 is the least number divisible by each of them.

LCM of 4, 8 and 12 = 24

24 = 2 × 2 × 2 × 3

To make it perfect square multiply 24 by the product of unpaired numbers, i.e., 2 × 3 = 6

Required number = 24 × 6 = 144

Question 16.

Find the square roots of the following decimal numbers

(i) 1056.25

(ii) 10020.01

Solution:

Question 17.

What is the least number that must be subtracted from 3793 so as to get a perfect square? Also, find the square root of the number so obtained.

Solution:

First, we find the square root of 3793 by division method.

Here, we get a remainder 72

61^{2} < 3793

Required perfect square number = 3793 – 72 = 3721 and √3721 = 61

Question 18.

Fill in the blanks:

(а) The perfect square number between 60 and 70 is …………

(b) The square root of 361 ends with digit …………..

(c) The sum of first n odd numbers is …………

(d) The number of digits in the square root of 4096 is ………..

(e) If (-3)^{2} = 9, then the square root of 9 is ……….

(f) Number of digits in the square root of 1002001 is …………

(g) Square root of is ………..

(h) The value of √(63 × 28) = …………

Solution:

(a) 64

(b) 9

(c) n^{2}

(d) 2

(e) ±3

(f) 4

(g)

(h) 42

Question 19.

Simplify: √900 + √0.09 + √0.000009

Solution:

We know that √(ab) = √a × √b

√900 = √(9 × 100) = √9 × √100 = 3 × 10 = 30

√0.09 = √(0.3 × 0.3) = 0.3

√0.000009 = √(0.003 × 0.003) = 0.003

√900 + √0.09 + √0.000009 = 30 + 0.3 + 0.003 = 30.303

### Squares and Square Roots Class 8 Extra Questions Higher Order Thinking Skills (HOTS)

Question 20.

Find the value of x if

Solution:

Question 21.

Simplify:

Solution:

Question 22.

A ladder 10 m long rests against a vertical wall. If the foot of the ladder is 6 m away from the wall and the ladder just reaches the top of the wall, how high is the wall? (NCERT Exemplar)

Solution:

Let AC be the ladder.

Therefore, AC = 10 m

Let BC be the distance between the foot of the ladder and the wall.

Therefore, BC = 6 m

∆ABC forms a right-angled triangle, right angled at B.

By Pythagoras theorem,

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

10^{2 }= AB^{2} + 6^{2}

or AB^{2} = 10^{2} – 6^{2} = 100 – 36 = 64

or AB = √64 = 8m

Hence, the wall is 8 m high.

Question 23.

Find the length of a diagonal of a rectangle with dimensions 20 m by 15 m. (NCERT Exemplar)

Solution:

Using Pythagoras theorem, we have Length of diagonal of the rectangle = units

Hence, the length of the diagonal is 25 m.

Question 24.

The area of a rectangular field whose length is twice its breadth is 2450 m2. Find the perimeter of the field.

Solution:

Let the breadth of the field be x metres. The length of the field 2x metres.

Therefore, area of the rectangular field = length × breadth = (2x)(x) = (2x^{2}) m^{2}

Given that area is 2450 m^{2}.

Therefore, 2x^{2} = 2450

⇒ x^{2} = 1225

⇒ x = √1225 or x = 35 m

Hence, breadth = 35 m

and length = 35 × 2 = 70 m

Perimeter of the field = 2 (l + b ) = 2(70 + 35) m = 2 × 105 m = 210 m.