Squares and Square Roots ICSE Class-8th Concise Selina

Squares and Square Roots ICSE Class-8th Concise Selina Chapter-3. We provide step by step Solutions of Exercise / lesson-3 Squares and Square Roots ICSE Class-8th Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-3 A , Exe-3 B and Exe-3 C,  to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8.

Squares and Square Roots ICSE Class-8th Concise Selina Chapter-3

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–: Select Topics :–

 

Exercise-3 A ,

Exercise-3 B

Exercise-3 C

 

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Exercise – 3 A Squares and Square Roots ICSE Class-8th Concise Selina

Question 1.

Find the square of :
(i) 59
(ii) 63
(iii) 15

Answer 1:-

(i) Square of 59=59×59=3481

(ii) Square of 6.3=6.3×6.3=39.69

(iii) Square of 15=15×15=225

 

Question 2 :-

By splitting into prime factors, find the square root of :
(i) 11025
(ii) 396900
(iii) 194481

Answer 2 :-

(i)

= 5 x 7 x 3 = 105

Question 3 :-

(i) Find the smallest number by which 2592 be multiplied so that the product is a perfect square.
(ii) Find the smallest number by which 12748 be multiplied so that the product is a perfect square?

Answer 3 :-

 

Question 4 :-

Find the smallest number by which 10368 be divided, so that the result is a perfect square. Also, find the square root of the resulting numbers.

Answer 4 :-

10368 =

On grouping the prime factors of 10368 as shown; one factors i.e. 2 is left which cannot be paired with equal factor.

Question 5 :-

Find the square root of :
(i) 0.1764
(ii) 
(iii) 0.0169

Answer 5 :-

(i)

= 0.42

(ii)

(iii) 0.0169

Question 6 :-

Evaluate :
(i)

(ii)

(iii)

(iv)
Answer 6:-

(i)

= 0.8
(ii) 

17	28900
17	1700
10	100
	10

(iii)

(iv) 

Question 7 :-
Evaluate:
(i) 
(ii)
(iii)

(iv)

(v)

Answer 7:-

(i)

(ii)

(iii)

(iv)

(v)

Question 8 :-

A man, after a tour, finds that he had spent every day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all ₹ 1,296
Answer 8:-

Let the number of days he had spent = x
Number of rupees spent in each day = x
Total money spent = x x x = x² = 1,296 (given)

∴ x= √1296

⇒ x = √4×4×9×9

x = 4 × 9

⇒ x = 36

Hence required number of days = 36

 

Question 9 :-

Out of 745 students, maximum are to be arranged in the school field for a P.T. display, such that the number of rows is equal to the number of columns. Find the number of rows if 16 students were left out after the arrangement.

Answer 9 :-

Question 10 :-

13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Find two more such pairs.

Answer :-

(13⁾² = 169 and (31)² = 961
Similarly, two such number can be 12 and 21
∴ (12)² = 144 and (21)² =441
and 102, 201
(102)² = 102 x 102 =  10404
and (201)² = 201 x 201 = 40401

 

Question 11 :-

Find the smallest perfect square divisible by 3, 4, 5 and 6.

Answer 11 :-

L.C.M. of 3, 4, 5, 6 = 2 x 2 x 3 x 5 = 60

in which 3 and 5 are not in pairs L.C.M. = 2 x 3 x 2 x 5 = 60

We should multiple it by 3 x 5 i.e. by 15

Required perfect square = 60 x 15 = 900

Question 12 :-

If √784 = 28, find the value of:
(i) √7.84 + √78400
(ii) √0.0784 + √0.000784

Answer 12 :-

(i)

(ii)

 

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 Exercise- 3 B Squares and Square Roots ICSE Class-8th Concise Selina

Question 1 :-

Find the square root of:
(i) 4761
(ii) 7744
(iii) 15129
(iv) 0.2916
(v) 0.001225
(vi) 0.023104
(vii) 27.3529
Answer 1 :-

(i) 4761

Required square root = 69

(ii) 7744

Required square root = 88

(iii) 15129

Required Square root = 123

(iv) 0.2916

Required square root = 0.54

(v) 0.001225

Required square root = 0.035

(vi) 0.023104

Required square root = 0.152

(vii) 27.3529

Required square root = 5.23

Question 2 :-

Find the square root of:

(i) 4.2025
(ii) 531.7636
(iii) 0.007225
Answer 2 :-

(i) 4.2025

Required square root = 2.05

(ii) 531.7636

Required square root = 23.06

(iii) 0.007225

Required square root = 0.085

Question 3 :-

Find the square root of:
(i) 245 correct to two places of decimal.
(ii) 496 correct to three places of decimal.
(iii) 82.6 correct to two places of decimal.
(iv) 0.065 correct to three places of decimal.
(v) 5.2005 correct to two places of decimal.
(vi) 0.602 correct to two places of decimal
Answer 3 :-

(i) 245

Required square root = 15.65 up to two places of decimal.

(ii) 496

Required square root = 22.2708 = 22.271 up to three places of decimals.

(iii) 82.6

Required square root = 9.088 = 9.09 up to two places of decimal.

(iv) 0.065

Required square root = 0.255 up to three places of decimal.

(v) 5.2005

Required square root = 2.28 up to two places of decimal.

(vi) 0.602

Required square root = 0.78 up to two places of decimals.

Question 4 :-

Find the square root of each of the following correct to two decimal places:
(i) 3  4/5
(ii) 6  7/8
Answer 4 :-

(i) 3  4/5

= 3.80

Required. square root = 1.949 = 1.95 up to two places of decimal.

(ii) 6  7/8

Question 5 :-

For each of the following, find the least number that must be subtracted so that the resulting number is a perfect square.
(i) 796
(ii) 1886
(iii) 23497
Answer 5 :-
(i) 796

Taking square root of 796, we find that 12 has been left

∴ Least number to be subtracted = 12

(ii) 1886

Taking square root of 1886, we find that 37 has been left

∴ Least number to be subtracted = 37

(iii) 23497

Taking square root of 23497, we find that 88 has been left

∴ Least number to be subtracted = 88

Question 6 :-

For each of the following, find the least number that must be added so that the resulting number is a perfect square.
(i) 511
(ii) 7172
(iii) 55078
Answer 6 :-

(i) 511

Taking square root of 511, we find that 27 has been left We see that 511 is greater than (22)² 

On adding the required number to 511, we get (23)² i.e., 529
So, the required number = 529 – 511 = 18

(ii) 7172

Taking square root of 7172, we find that 116 has been left
We see that 7172 is greater than (84)²

∴ On adding the required number to 7172, we get (85)² i.e., 7225
Required number = 7225 – 7172 = 53

(iii) 55078

Taking square root of 55078, we find that 322 has been left
We see that 55078 is greater than (234)²
On adding the required number to 55078, we get (235)² i.e., 55225
Required number = 55225 – 55078 = 147

Question 7 :-

Find the square root of 7 correct to two decimal places; then use it to find the value of  correct to three significant digits.
Answer 7 :-

√7 = 2.645 = 2.65

Question 8 :-

Find the value of √5 correct to 2 decimal places; then use it to find the square root of  correct to 2 significant digits.
Answer 8 :-

√5 = 2.236 = 2.24

 

Question 9 :-

Find the square root of:
(i) ¹⁷⁶⁴⁄2809
(ii) ⁵⁰⁷⁄4107
(iii)
(iv) 0.01 + √0.0064

Answer 9 :-

(i) ¹⁷⁶⁴⁄2809

Question 10 :-

Find the square root of 7.832 correct to :
(i) 2 decimal places
(ii) 2 significant digits.

Answer 10 :-

(i)

Square root of 7.832

√7.832= 2.80 up to two decimal places

(ii)

Square root of 7.832

= 2.8 up to two significant digits

 

Question 11 :-

Find the least number which must be subtracted from 1205 so that the resulting number is a perfect square.

Answer 11 :-

Clearly, if 49 is subtracted from 1205, the number will be a perfect square.

 

∴ 1205 – 49 = 1156 and √1156 = 34

 

Question 12 :-

Find the least number which must be added to 1205 so that the resulting number is a perfect square.
Answer 12 :-

Clearly, 1205 is greater than 34²

∴ On adding the required number to 1205, we shall be getting 35² i.e., 1225
∴ The required number = 1225 – 1225 = 20

 

Question 13 :-

Find the least number which must be subtracted from 2037 so that the resulting number is a perfect square.
Answer 13 :-

Find the least number which must be added to 5483 so that the resulting number is a perfect square.
Answer 14 :-

Clearly, 5483 is greater than 74²

∴ On adding the required number to 5483, we shall be getting 75² i.e. 5625.
Hence, the required number = 5625 – 5483
= 142

 

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Exercise – 3 C Concise Selina solutions of Squares and Square Roots for ICSE Class-8th 

Question 1 :-

Seeing the value of the digit at unit’s place, state which of the following can be square of a number :
(i) 3051
(ii) 2332
(iii) 5684
(iv) 6908
(v) 50699

Answer 1 :-

We know that the ending digit (the digit at units place) of the square of a number is 0, 1, 4, 5, 6, or 9
So, the following numbers can be squares: 3051, 5684, and 50699 i.e., (i), (iii), and (v)

 

Question 2 :-

Squares of which of the following numbers will have 1 (one) at their unit’s place :
(i) 57
(ii) 81
(iii) 139
(iv) 73
(v) 64

Answer 2 :-

The square of the following numbers will have 1 at their units place as (1)² = 1, (9)² = 81
81 and 139 i.e., (i) and (iii)

 

Question 3 :-

Which of the following numbers will not have 1 (one) at their unit’s place :
(i) 322
(ii) 572
(iii) 692
(iv) 3212
(v) 2652
Answer 3 :-

The square of the following numbers will not have 1 at their units place: as only (1)² = 1, (9)² = 81 have 1 at then units place
322, 572, 2652 i.e., (i), (ii) and (v)

 

Question 4 :-

Square of which of the following numbers will not have 6 at their unit’s place :
(i) 35
(ii) 23
(iii) 64
(iv) 76
(v) 98

Answer 4 :-

 

Question 5 :-

Which of the following numbers will have 6 at their unit’s place :
(i) 262
(ii) 492
(iii) 342
(iv) 432
(v) 2442

Answer 5 :-

 

Question 6 :-

If a number ends with 3 zeroes, how many zeroes will its square have?
Answer 6 :-

We know that if a number ends with n zeros, then its square will have 2n zeroes at their ends
A number ends with 3 zeroes, then its square will have 3 x 2 = 6 zeroes

 

Question 7 :-

If the square of a number ends with 10 zeroes, how many zeroes will the number have?
Answer 7 :-

We know that if a number ends with n zeros Then its square will have 2n zeroes Conversely, if square of a number have 2n zeros at their ends then the number will have n zeroes
The square of a number ends 10 zeroes, then the number will have 10/2 = 5 zeroes

 

Question 8 :-

Is it possible for the square of a number to end with 5 zeroes ? Give reason.
Answer 8 :-

 

Question 9 :-

Give reason to show that none of the numbers, given below, is a perfect square.
(i) 2162
(ii) 6843
(iii) 9637
(iv) 6598
Answer 9 :-

 

Question 10 :-

State, whether the square of the following numbers is even or odd?
(i) 23
(ii) 54
(ii) 76
(iv) 75
Answer 10 :-

(i) 23 – odd

(ii) 54 – even

(ii) 76 – odd

(iv) 75 – even

 

Question 11 :-

Give reason to show that none of the numbers 640, 81000 and 3600000 is a perfect square.
Answer 11 :-

 

Question 12 :-

Evaluate:
(i) 37² – 36²
(ii) 85² – 84²
(iii) 101² – 100²

Answer 12 :-

Question 13 :-

Without doing the actual addition, find the sum of:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(ii) 1 + 3 + 5 + 7 + 9 + ……………… + 39 + 41
(iii) 1 + 3 + 5 + 7 + 9 + ………………… + 51 + 53
Answer 13 :-

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

= Sum of first 12 odd natural numbers = 122 = 144

(ii) 1 + 3 + 5 + 7 + 9 + ……………… + 39 + 41

 

Question 14 :-

Write three sets of Pythagorean triplets such that each set has numbers less than 30.
Answer 14 :-

The three sets of Pythagorean triplets such that each set has numbers less than 30 are 3, 4 and 5; 6, 8 and 10; 5, 12 and 13
Proof:
In 3, 4 and 5
3² + 4² = 5²
⇒ 9 + 16 = 25
⇒ 25 = 25
In 6, 8 and 10
6² + 8² = 10²
⇒ 36 + 64 = 100
⇒ 100 = 100
In 5, 12, and 13
5² + 12² = 13²
⇒ 25 + 144 = 169
⇒ 169 = 169

 

 

— End of Squares and Square Roots Solutions :–

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