NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise  6.1

Question 1. What will be the unit digit of the squares of the following numbers?

1. 81
2. 272
3. 799
4. 3853
5. 1234
6. 20387
7. 52698
8. 99880
9. 12796
10. 55555

Solution:

1. Unit digit of 81² = 1
2. Unit digit of 272² = 4
3. Unit digit of 799² = 1
4. Unit digit of 3853² = 9
5. Unit digit of 1234² = 6
6. Unit digit of 26387² = 9
7. Unit digit of 52698² = 4
8. Unit digit of 99880² = 0
9. Unit digit of 12796² = 6
10. Unit digit of 55555² = 5

Question 2. The following numbers are not perfect squares. Give reason.

(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050

Solution:

(i) 1057 ends with 7 at unit place. So it is not a perfect square number.
(ii) 23453 ends with 3 at unit place. So it is not a perfect square number.
(iii) 7928 ends with 8 at unit place. So it is not a perfect square number.
(iv) 222222 ends with 2 at unit place. So it is not a perfect square number.
(v) 64000 ends with 3 zeros. So it cannot a perfect square number.
(vi) 89722 ends with 2 at unit place. So it is not a perfect square number.
(vii) 22000 ends with 3 zeros. So it can not be a perfect square number.
(viii) 505050 ends with 1 zero. So it is not a perfect square number.

Question 3. The squares of which of the following would be odd numbers?

(i) 431
(ii) 2826
(iii) 7779
(iv) 82004

Solution:

(i) 431² is an odd number.
(ii) 2826² is an even number.
(iii) 7779² is an odd number.
(iv) 82004² is an even number.

Question 4. Observe the following pattern and find the missing digits.

11² = 121
101² = 10201
1001² = 1002001
100001² = 1…2…1
10000001² = …………………….

Solution:

According to the above pattern, we have
100001² = 10000200001
10000001² = 100000020000001

Question 5. Observe the following pattern and supply the missing numbers.

11² = 121
101² = 10201
10101² = 102030201
1010101² = ……….
……….² = 10203040504030201

Solution:

According to the above pattern, we have
1010101² = 1020304030201
101010101² = 10203040504030201

Question 6. Using the given pattern, find the missing numbers.

1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + ….² = 21²
5² + ….² + 30² = 31²
6² + 7² + …..² = ……²

Solution:

According to the given pattern, we have

4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²

Question 7. Without adding, find the sum.

(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 + 21 + 23

Solution:

We know that the sum of n odd numbers = n²
(i) 1 + 3 + 5 + 7 + 9 = (5)² = 25 [∵ n = 5]
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)² = 100 [∵ n = 10]
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = (12)² = 144 [∵ n = 12]

Question 8.

(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.

Solution:

(i) Express 49 as the sum of 7 odd numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (n = 7)

(ii) Express 121 as the sum of 11 odd numbers.
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (n = 11)

Question 9. How many numbers lie between squares of the following numbers?

(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100.

Solution:

(i) 12 and 13
We know that numbers between n² and (n + 1)² = 2n
Numbers between 12² and 13² = (2n) = 2 × 12 = 24

(ii) 25 and 26
Numbers between 25² and 26² = 2 × 25 = 50 (∵ n = 25)

(iii) 99 and 100
Numbers between 99² and 100² = 2 × 99 = 198 (∵ n = 99)

Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.2

Question 1. Find the square of the following numbers.

(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46

Solution:

(i) 32 = 30 + 2
(32)² = (30 + 2)²
= 30(30 + 2) + 2(30 + 2)
= 30² + 30 × 2 + 2 × 30 + 2²
= 900 + 60 + 60 + 4
= 1024
Thus (32)² = 1024

(ii) 35 = (30 + 5)
(35)² = (30 + 5)²
= 30(30 + 5) + 5(30 + 5)
= (30)² + 30 × 5 + 5 × 30 + (5)²
= 900 + 150 + 150 + 25
= 1225
Thus (35)² = 1225

(iii) 86 = (80 + 6)
86² = (80 + 6)²
= 80(80 + 6) + 6(80 + 6)
= (80)² + 80 × 6 + 6 × 80 + (6)²
= 6400 + 480 + 480 + 36
= 7396
Thus (86)² = 7396

(iv) 93 = (90+ 3)
93² = (90 + 3)²
= 90 (90 + 3) + 3(90 + 3)
= (90)² + 90 × 3 + 3 × 90 + (3)²
= 8100 + 270 + 270 + 9
= 8649
Thus (93)² = 8649

(v) 71 = (70 + 1)
71² = (70 + 1)²
= 70 (70 + 1) + 1(70 + 1)
= (70)² + 70 × 1 + 1 × 70 + (1)²
= 4900 + 70 + 70 + 1
= 5041
Thus (71)² = 5041

(vi) 46 = (40+ 6)
46² = (40 + 6)²
= 40 (40 + 6) + 6(40 + 6)
= (40)² + 40 × 6 + 6 × 40 + (6)²
= 1600 + 240 + 240 + 36
= 2116
Thus (46)² = 2116

Question 2. Write a Pythagorean triplet whose one member is.

(i) 6
(ii) 14
(iii) 16
(iv) 18

Solution:

(i) Let m² – 1 = 6
[Triplets are in the form 2m, m² – 1, m² + 1]
m² = 6 + 1 = 7
So, the value of m will not be an integer.
Now, let us try for m² + 1 = 6
⇒ m² = 6 – 1 = 5
Also, the value of m will not be an integer.
Now we let 2m = 6 ⇒ m = 3 which is an integer.
Other members are:
m² – 1 = 3² – 1 = 8 and m² + 1 = 3² + 1 = 10
Hence, the required triplets are 6, 8 and 10

(ii) Let m² – 1 = 14 ⇒ m² = 1 + 14 = 15
The value of m will not be an integer.
Now take 2m = 14 ⇒ m = 7 which is an integer.
The member of triplets are 2m = 2 × 7 = 14
m² – 1 = (7)² – 1 = 49 – 1 = 48
and m² + 1 = (7)² + 1 = 49 + 1 = 50
i.e., (14, 48, 50)

(iii) Let 2m = 16 m = 8
The required triplets are 2m = 2 × 8 = 16
m² – 1 = (8)² – 1 = 64 – 1 = 63
m² + 1 = (8)² + 1 = 64 + 1 = 65
i.e., (16, 63, 65)

(iv) Let 2m = 18 ⇒ m = 9
Required triplets are:
2m = 2 × 9 = 18
m² – 1 = (9)² – 1 = 81 – 1 = 80
and m² + 1 = (9)² + 1 = 81 + 1 = 82
i.e., (18, 80, 82)

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise  6.3

Question 1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025

Solution:

(i) One’s digit in the square root of 9801 maybe 1 or 9.
(ii) One’s digit in the square root of 99856 maybe 4 or 6.
(iii) One’s digit in the square root of 998001 maybe 1 or 9.
(iv) One’s digit in the square root of 657666025 can be 5.

Question 2. Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153
(ii) 257
(iii) 408
(iv) 441

Solution:

We know that the numbers ending with 2, 3, 7 or 8 are not perfect squares.
(i) 153 is not a perfect square number. (ending with 3)
(ii) 257 is not a perfect square number. (ending with 7)
(iii) 408 is not a perfect square number. (ending with 8)
(iv) 441 is a perfect square number.

Question 3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Solution:

Using the method of repeated subtraction of consecutive odd numbers, we have
(i) 100 – 1 = 99, 99 – 3 = 96, 96 – 5 = 91, 91 – 7 = 84, 84 – 9 = 75, 75 – 11 = 64, 64 – 13 = 51, 51 – 15 = 36, 36 – 17 = 19, 19 – 19 = 0
(Ten times repetition)
Thus √100 = 10

(ii) 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
(Thirteen times repetition)
Thus √169 = 13

Question 4. Find the square roots of the following numbers by the prime factorisation Method.

(i) 729
(ii) 400
(iii) 1764
(iv) 4096
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100

Solution:

(i) We have 729
The Prime factors of 729
729 = 3 × 3 × 3 × 3 × 3 × 3 = 3² × 3² × 3²
√729 = 3 × 3 × 3 = 27

(ii) We have 400
The Prime factors of 400
400 = 2 × 2 × 2 × 2 × 5 × 5 = 2² × 2² × 5²
√400 = 2 × 2 × 5 = 20

(iii) 1764
1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2² × 3² × 7²
√1764 = 2 × 3 × 7 = 42

(iv) 4096
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2² × 2² × 2² × 2² × 2² × 2²
√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

(v) Prime factorisation of 7744 is 7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
= 2² × 2² × 2² × 11²
√7744 = 2 × 2 × 2 × 11 = 88

(vi) Prime factorisation of 9604
9604 = 2 × 2 × 7 × 7 × 7 × 7 = 2² × 7² × 7²
√9604 = 2 × 7 × 7 = 98

(vii) Prime factorisation of 5929 is 5929 = 7 × 7 × 11 × 11 = 7² × 11²
√5929 = 7 × 11 = 77

(viii) Prime factorisation of 9216

9216 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 3 × 3
= 2² × 2² × 2² × 2² × 2² × 3²
√9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96

(ix) Prime factorisation of 529 is
529 = 23 × 23 = 23²
√529 = 23

(x) Prime factorisation of 8100
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = 2² × 3² × 3² × 5²
√8100 = 2 × 3 × 3 × 5 = 90

Question 5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained.

(i) 252
(ii) 180
(iii) 1008
(iv) 2028
(v) 1458
(vi) 768

Solution:

(i) Prime factorisation of 252 is
252 = 2 × 2 × 3 × 3 × 7
Here, the prime factorisation is not in pair. 7 has no pair.
Thus, 7 is the smallest whole number by which the given number is multiplied to get a perfect square number.
The new square number is 252 × 7 = 1764
Square root of 1764 is
√1764 = 2 × 3 × 7 = 42

(ii) Primp factorisation of 180 is
180 = 2 × 2 × 3 × 3 × 5
Here, 5 has no pair.
New square number = 180 × 5 = 900
The square root of 900 is
√900 = 2 × 3 × 5 = 30
Thus, 5 is the smallest whole number by which the given number is multiplied to get a square number.

(iii) Prime factorisation of 1008 is
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
Here, 7 has no pair.
New square number = 1008 × 7 = 7056
Thus, 7 is the required number.
Square root of 7056 is
√7056 = 2 × 2 × 3 × 7 = 84

(iv) Prime factorisation of 2028 is
2028 = 2 × 2 × 3 × 13 × 13
Here, 3 is not in pair.
3 is the required smallest whole number.
New square number = 2028 × 3 = 6084
Square root of 6084 is
√6084 = 2 × 13 × 3 = 78

(v) Prime factorisation of 1458 is
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, 2 is not in pair.
2 is the required smallest whole number.
New square number = 1458 × 2 = 2916
Square root of 1458 is
√2916 = 3 × 3 × 3 × 2 = 54

(vi) Prime factorisation of 768 is
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, 3 is not in pair.
Thus, 3 is the required whole number.
New square number = 768 × 3 = 2304
Square root of 2304 is
√2304 = 2 × 2 × 2 × 2 × 3 = 48

Question 6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained.

(i) 252
(ii) 2925
(iii) 396
(iv) 2645
(v) 2800
(vi) 1620

Solution:

(i) Prime factorisation of 252 is
252 = 2 × 2 × 3 × 3 × 7
Here 7 has no pair.
7 is the smallest whole number by which 252 is divided to get a square number.
New square number = 252 ÷ 7 = 36
Thus, √36 = 6

(ii) Prime factorisation of 2925 is
2925 = 3 × 3 × 5 × 5 × 13
Here, 13 has no pair.
13 is the smallest whole number by which 2925 is divided to get a square number.
New square number = 2925 ÷ 13 = 225
Thus √225 = 15

(iii) Prime factorisation of 396 is
396 = 2 × 2 × 3 × 3 × 11
Here 11 is not in pair.
11 is the required smallest whole number by which 396 is divided to get a square number.
New square number = 396 ÷ 11 = 36
Thus √36 = 6

(iv) Prime factorisation of 2645 is
2645 = 5 × 23 × 23
Here, 5 is not in pair.
5 is the required smallest whole number.
By which 2645 is multiplied to get a square number
New square number = 2645 ÷ 5 = 529
Thus, √529 = 23

(v) Prime factorisation of 2800 is
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
Here, 7 is not in pair.
7 is the required smallest number.
By which 2800 is multiplied to get a square number.
New square number = 2800 ÷ 7 = 400
Thus √400 = 20

(vi) Prime factorisation of 1620 is
1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
Here, 5 is not in pair.
5 is the required smallest prime number.
By which 1620 is multiplied to get a square number = 1620 ÷ 5 = 324
Thus √324 = 18

Question 7. The students of class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Solution:

Total amount of money donated = ₹ 2401
Total number of students in the class = √2401
= 72×72−−−−−−√
= 7×7×7×7−−−−−−−−−−√
= 7 × 7
= 49

Question 8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.

Solution:

Total number of rows = Total number of plants in each row = √2025
= 3×3×3×3×5×5−−−−−−−−−−−−−−−−√
= 32×32×52−−−−−−−−−−√
= 3 × 3 × 5
= 45
Thus the number of rows and plants = 45

Question 9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution:

LCM of 4, 9, 10 = 180
The least number divisible by 4, 9 and 10 = 180
Now prime factorisation of 180 is
180 = 2 × 2 × 3 × 3 × 5
Here, 5 has no pair.
The required smallest square number = 180 × 5 = 900

Question 10. Find the smallest number that is divisible by each of the numbers 8, 15 and 20.

Solution:

The smallest number divisible by 8, 15 and 20 is equal to their LCM.
LCM = 2 × 2 × 2 × 3 × 5 = 120
Here, 2, 3 and 5 have no pair.
The required smallest square number = 120 × 2 × 3 × 5 = 120 × 30 = 3600

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NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise  6.4

Additional Question and Answer on Squares and Square Roots

1. What is square root?

The square root of any number is the number whose square is equal to the original number. For example, if we have a number x, then its square root is the number y such that x^2 y^2. If we want to find the square root of a number, we simply take the square root of both sides of the equation. So, if we wanted to find the square root for 12, we would do (12)^(1/2) 6.

2. How does square roots work?

Square roots are useful in many different situations. One of the most popular uses is to solve equations. If we have two numbers, say x and y, and we know their product, z xy, then we can use the following formula to find the square root:

z^(1/2)x+y

This means that if we know the sum of the square roots of two numbers, we can calculate the individual square roots of those numbers.

3. Why should I learn about squares and square roots?

Squares and square roots are often used in math problems. In fact, they are the basis for many mathematical operations. You might even encounter them in everyday life. For instance, if you were trying to figure out how much money you need to buy something, you could multiply the price times the quantity. Then you could divide the result by the cost per unit to get the total amount of money you need.

4. How do I remember square roots?

You can easily memorize the rules for finding square roots. Just think of the word “square” and the number 1. When you add the first letter of each word together, you get the number 2. Now, just keep adding the letters of the words together until you reach the number you are looking for.

5. Where do I find square roots?

There are several websites where you can find square roots online. However, some sites require you to download software to access the information. If you don’t have time to download software, you can always look at the answers to math questions on Khan Academy.