Class 8th mathematics Squares and Square Roots chapter 6 exercise 6.1 solution

 Here you can read Class 8th MATHEMATICS "SQUARE AND SQUARE ROOTS" chapter 6 Exercise 6.1 Solution Based on NCERT and CBSE book. After "SQUARE AND SQUARE ROOTS" chapter 6 you can get links to Class 8th and other classes SCIENCE, MATHEMATICS and OTHER NCERT book Notes, Solutions, Practice Papers, etc.

Scroll down for NCERT Book Classes MATHEMATICS Book & important study material.

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.1 solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.2 solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.3 solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.4 solution 

SQUARE AND SQUARE ROOTS (Chapter-6) SOLUTION

EXERCISE 6.1 

Question 1. What will be the unit digit of the squares of the following numbers? 
(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555 

Solution- According to the chapter we read that when we find a square of a number than the unit digit of a number is multiply by the unit digit of same number,
the unit or one's digit of a square number 

which numbers have 1 or 9 unit digit, square number have 1 unit digit.
which numbers have 2 or 8 unit digit, square number have 4 unit digit.
which numbers have 3 or 7 unit digit, square number have 9 unit digit.
which numbers have 4 or 6 unit digit, square number have 6 unit digit.
which numbers have 5 unit digit, square number have 5 unit digit.  

which numbers have 0 unit digit, square number have 0 unit digit. 

 

(i) In 81, The unit digit is 1.

Then, 1 X 1 = 1

Hence, The unit digit of the square of the 81 is 1.

(ii) In 272, The unit digit is 2.

Then, 2 X 2 = 4

Hence, The unit digit of the square of the 272 is 4.

(iii) In 799, The unit digit is 9.

Then, 9 X 9 = 81

Hence, The unit digit of the square of the 799 is 1.

(iv) In 3853 The unit digit is 3.

Then, 3 X 3 = 9

Hence, The unit digit of the square of the 3853 is 9.

(v) In 1234 The unit digit is 4.

Then, 4 X 4 = 16

Hence, The unit digit of the square of the 1234 is 6.

(vi) In 26387 The unit digit is 7.

Then, 7 X 7 = 49

Hence, The unit digit of the square of the 26387 is 9.

(vii) In 52698 The unit digit is 8.

Then, 8 X 8 = 64

Hence, The unit digit of the square of the 52698 is 4

(viii) In 99880 The unit digit is 0.

Then, 0 X 0 = 0

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

(ix) In 12796 The unit digit is 6.

Then, 6 X 6 = 36

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

(x) In 55555 The unit digit is 5.

Then, 5 X 5 = 25

Hence, The unit digit of the square of the 55555 is 5.

Question 2. The following numbers are obviously not perfect squares. Give reason. 
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050 

Solution- All the square numbers end with 0, 1, 4, 5, 6 or 9 at units place. None of these end with 2, 3, 7 or 8 at unit’s place. So we Can say that if a number ends in 0, 1, 4, 5, 6 or 9 at units place, then it must be a square number. And if a number ends in 2, 3, 7 or 8 at units place, then it must be not a square number.
All the square numbers can only have even number of zeros at the end.

(i) 1057 end with 7 at units place.

(ii) 23453 end with 3 at units place.

(iii) 7928 end with 8 at units place.

(iv) 222222 end with 2 at units place.

(v) 64000 have odd number of zeros at the end.

(vi) 89722 end with 2 at units place.

(vii) 222000 have odd number of zeros at the end.

(viii) 505050 have odd number of zeros at the end.

Question 3. The squares of which of the following would be odd numbers? 
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004 

Solution- As we know, when we multiply odd numbers with odd numbers product is always odd number, and when we multiply even numbers with even numbers product is always even number.

(i) 431, The unit digit of 431 is 1. 1 is a odd number. After square we get odd number.

(ii) 2826, The unit digit of 2826 is 6. 6 is a even number. After square we get even number.

(iii) 7779, The unit digit of 7779 is 9. 9 is a odd number. After square we get odd number. 

(iv) 82004, The unit digit of 82004 is 4. 4 is a even number. After square we get 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 observe above pattern we find the missing numbers.

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 observe above pattern we find the missing numbers.

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- According to chapter, We can also say that if a natural number can be expressed as a sum of successive odd natural numbers starting with 1, then it is a perfect square.

The sum of first n odd natural numbers is n2 .

So,

(i) 1 + 3 + 5 + 7 + 9, 

Here n = 5,
Then n2 = 52 = 25 

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 

Here n = 10,
Then n2 = 102 = 100 

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

Here n = 12,
Then n2 = 122 = 144 

Question 8. (i) Express 49 as the sum of 7 odd numbers. 
    (ii) Express 121 as the sum of 11 odd numbers.

Solution-  According to chapter, We can also say that if a natural number can be expressed as a sum of successive odd natural numbers starting with 1, then it is a perfect square.

n2  is the sum of first n odd natural numbers.

So,

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

Here  n2 = 49, and n = 7

Then, 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49

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

Here  n2 = 49, and n = 7

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

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- Numbers lie between n2 and (n + 1)2 there are 2n numbers which is 1 less than the difference of two squares.

(i) 12 and 13 

Here n = 12 and (n+1) = 13

132 - 122 = 169 - 144 = 25

25 - 1 = 24

or 

2n = 2(12) = 24

Numbers lie between squares of the 12 and 13 is 24. 

(ii) 25 and 26 

Here n = 25 and (n+1) = 26

262 - 252 = 676 - 625 = 51

51 - 1 = 50

or 

2n = 2(25) = 50

Numbers lie between squares of the 25 and 26 is 50.

(iii) 99 and 100 

Here n = 99 and (n+1) = 100

1002 - 992 = 10000 - 9801 = 199

199 - 1 = 198

or 

2n = 2(99) = 198

Numbers lie between squares of the 99 and 100 is 198.

       THANKS FOR READING    

please follow and comment about the content which you Read above.

To read more chapter of 8th Class and other classes NCERT book based Notes and solution, Please click below link as per Your Choice.

 

From this students study the NCERT SOLUTION to get knowledge of the type of questions asked from the chapter, "SQUARE AND SQUARE ROOTS". This notes and solution of class 8th is very helpful to understand the Science subject and chapter 6 in a Better manner. We provided the chapter wise link below of class 8th MATHEMATICS book. These NCERT SOLUTIOS are Based on the latest CBSE and RBSE syllabus. All the study material was prepared to help you understand the topic easy and better way. If you like our resources, please share the post. NCERT SOLUTION for Class 8th MATHEMATICS Chapter 6 "SQUARE AND SQUARE ROOTS" is an outstanding study Materials which is needed for the students of CBSE and RBSE of Class 8th. The NCERT Solutions for Class 8th Science Chapter 6 has good weightage. Thorough knowledge and good practice will help you score full marks on the questions asked from this chapter.

Learn more about "SQUARE AND SQUARE ROOTS" by exploring CBSE/RBSE based NCERT Notes for Class 8th MATHEMATICS Chapter 6. These CBSE/RBSE based NCERT notes are comprehensive and detailed yet concise enough to glance through for exam preparations.

 

NCERT Solutions For Class 8th MATHEMATICS Chapter 6 SQUARE AND SQUARE ROOTS

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.1 solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.2 solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.3 solution

Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.4 solution 

You can Read the class 7th NCERT SCIENCE notes based on CBSE and RBSE pattern.

NUTRITION IN PLANTS Class 7th SCIENCE chapter 1 notes

NUTRITION IN ANIMALS Class 7th SCIENCE chapter 2 notes

FIBRE TO FABRIC Class 7th SCIENCE chapter 3 notes

HEAT Class 7th SCIENCE chapter 4 notes

ACIDS, BASES AND SALTS Class 7th SCIENCE chapter 5 notes

PHYSICAL AND CHEMICAL CHANGES Class 7th SCIENCE chapter 6 notes

WEATHER, CLIMATE AND ADAPTATIONS OF ANIMALS TO CLIMATE Class 7th SCIENCE chapter 7 notes

WINDS, STROMS AND CYCLONES Class 7th SCIENCE chapter 8 notes

Class 8th Study Material  Solution based on CBSE and RBSE.

coming soon

For Class 8th Science subject you can get Revision Notes, Important Solutions. 

coming soon

Here students can get classes & chapters wise Class 8th and other classes NCERT books notes based on CBSE and RBSE pattern, 

coming soon