Class 8 NCERT Solutions - Chapter 6 Squares and Square Roots - Exercise 6.1
Last Updated : 30 Aug, 2024
Chapter 6 of the Class 8 NCERT Mathematics textbook, titled "Squares and Square Roots," introduces students to the concepts of squares and square roots. This chapter covers the properties of square numbers, methods for finding squares and square roots, and their applications. Exercise 6.1 focuses on helping students understand how to identify and calculate the squares of numbers, as well as recognizing patterns in square numbers.
NCERT Solutions for Class 8 - Chapter 6 Squares and Square Roots - Exercise 6.1
This section provides detailed solutions for Exercise 6.1 from Chapter 6 of the Class 8 NCERT Mathematics textbook. The exercise includes problems that require students to find the squares of numbers, identify perfect squares, and understand the properties of square numbers. These solutions are designed to strengthen students' foundational knowledge of squares and square roots, which is essential for solving more complex mathematical problems.
Problem 1: What will be the unit digit of the squares of the following numbers?
Solution:
To find the unit digit of the square of any number, find the square of the unit digit of that number, and
its unit digit will be the unit digit of the square of the actual number.
eg:- if a number is 2007, the square of the unit digit of 2007 is 49 (7 square). and the unit digit of 49 is 9. Therefore, the unit digit of the square of 2007 is 9.
(i) 81
Ans: 1
Reason: If a number has 1 or 9 in the units place, then its square ends in 1.
(or) Unit digit of 81 is 1. and 12 = 1, therefore unit digit of square of 81 is 1.
(ii) 272
Ans: 4
Reason: Unit digit of 272 is 2 and 22 = 4. Therefore unit digit of the square of 272 is 4.
(iii) 799
Ans: 1
Reason: If a number has 1 or 9 in the units place, then its square ends in 1. (or) Unit digit of 799 is 9 and 92 = 81, and unit digit of 81 is 1. Therefore unit digit of square of 799 is 1.
(iv) 3853
Ans: 9
Reason: Unit digit of 3853 is 3 and 32 = 9, therefore unit digit of square of 3853 is 9.
(v) 1234
Ans: 6
Reason: Unit digit of 1234 is 4 and 42 = 16, and the unit digit of 16 is 6. Therefore the unit digit of the square of 1234 is 6.
(vi) 26387
Ans: 9
Reason: Unit digit of 26387 is 7 and 72 = 49, and the unit digit of 49 is 9. Therefore the unit digit of the square of 26387 is 9.
(vii) 52698
Ans: 4
Reason: Unit digit of 52698 is 8 and 82 = 64, and the unit digit of 64 is 4. Therefore the unit digit of the square of 52698 is 4.
(viii) 99880
Ans: 0
Reason: If a number has 0 in its unit place then square of its number will also have 0 in its units place (since 02 = 0).
(ix) 12796
Ans: 6
Reason: Unit digit of 12796 is 6 and 62 = 36, and the unit digit of 36 is 6. Therefore the unit digit of the square of 12796 is 6.
(x) 55555
Ans: 5
Reason: The unit digit of 55555 is 5 and 52 = 25, and the unit digit of 25 is 5. Therefore the unit digit of the square of 55555 is 5.
Problem 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:
- If a number ends with 2 (or) 3 (or) 7 (or) 8 at units place then we can tell it is not a perfect square. From the above rule, we can tell 1057, 23453, 7928, 222222, 89722 are not perfect squares.
- For a number to be a perfect square it should have an even number of zeros at the end. From rule 2 we can tell 64000, 222000, 505050 are not perfect squares as they are having an odd number of zeros at the end.
Problem 3: The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Ans: 431 and 7779
Reason: The square of an odd number will be an odd number, and the square of an even number will be an even number.
Problem 4: Observe the following pattern and find the missing digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 ......... 2 ......... 1
100000012 = ...........................
Ans:
1000012 = 10000200001
100000012 = 100000020000001
Problem 5: Observe the following pattern and supply the missing numbers.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = ...........................
............2 = 10203040504030201
Ans:
10101012 = 1020304030201
1010101012 = 10203040504030201
Problem 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:
To find pattern third number is related to first and second number. How?
If we multiply the first and second numbers we will get the third number.
The fourth number is related to the third number. How?
Forth number = third number + 1
42 + 52 + __2 = 212
Ans: 4 * 5 = 20
52 + __2 + 302 = 312
Ans: 5 * x = 30
=> x = (30 / 5) = 6
62 + 72 + x2 = y2
Ans: x = 6 * 7 = 42
y = x + 1 = 43
Problem 7: Without adding, find the sum.
Solution:
The Sum of first n odd natural numbers is n2
(i) 1 + 3 + 5 + 7 + 9
Ans: 25
Sum of first 5 odd natural numbers is 52, and 52 = 25.
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
Ans: 100
Sum of first 10 odd natural numbers is 102, and 102 = 100.
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Ans: 144
Sum of first 12 odd natural numbers is 122, and 122 = 144.
Problem 8
(i) Express 49 as the sum of 7 odd numbers.
Ans: 1 + 3 + 5 + 7 + 9 + 11 + 13
Explanation: 49 = 72 therefore the sum of the first 7 odd natural numbers is 49.
(ii) Express 121 as the sum of 11 odd numbers
Ans: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Explanation: 121 = 112 therefore sum of first 11 odd natural numbers is 121.
Problem 9: How many numbers lie between squares of the following numbers?
In between n2 and (n + 1)2, there will be 2n natural numbers
(i) 12 and 13
Ans: 24 (as n is 12, 2n will be 24)
(ii) 25 and 26
Ans: 50 (as n is 25, 2n will be 50)
(iii) 99 and 100
Ans: 198 (as n is 99, 2n will be 198)
Summary
Chapter 6 of the Class 8 NCERT Mathematics textbook, "Squares and Square Roots," introduces students to the concepts of square numbers and their properties. Exercise 6.1 focuses on identifying and calculating squares, recognizing perfect squares, and understanding the patterns in square numbers. This exercise helps students build a strong foundation in working with squares and square roots, essential for more advanced mathematical concepts.
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