NCERT Solutions for Class 8 Maths Chapter 5 – Squares and Square Roots Ex 5.1
NCERT Solutions for Class 8 Maths Chapter 5 – Squares and Square Roots Ex 5.1 are given here in simple, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. When you multiply a number by itself, you get its square. For example, the square of 6 is 6 × 6 = 36, written as 62. A number like 36 that is the square of a whole number is called a perfect square or a square number. Exercise 5.1 teaches the special properties of these square numbers.
This exercise has 9 questions. Most of them can be answered just by knowing the properties of squares, without long working. You will learn which numbers can never be perfect squares, how to tell if a square is odd or even, how many numbers lie between two squares, and a neat pattern for adding odd numbers. Every question below is solved in short steps, with the final answer in bold. Students can use this page for homework and revision, parents can use it to check the work at home, and teachers can use it in class. A free printable PDF of these solutions is also available for offline study.
NCERT Solutions for Class 8 Maths Chapter 5 – Squares and Square Roots Ex 5.1
Squares and Square Roots
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Properties of Square Numbers
- A perfect square never ends in 2, 3, 7 or 8. So any number ending in these digits is not a perfect square.
- A perfect square never ends in an odd number of zeros.
- The square of an even number is even, and the square of an odd number is odd.
- Between the squares of two numbers n and (n + 1), there are exactly 2n numbers that are not perfect squares.
- The sum of the first n odd numbers is n2. For example, 1 + 3 + 5 = 9 = 32.
- Pythagorean triple: for any number m greater than 1, (2m, m2 − 1, m2 + 1) are the three sides of a right-angled triangle.
Exercise 5.1 – Questions and Answers
Q1. 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
Answer:
The unit digit of a square depends only on the unit digit of the number.
(i) 81 – The number ends with 1. Its square will end with 1 × 1 = 1. So, the unit digit of the square of 81 will be 1.
(ii) 272 – The number ends with 2. Its square will end with 2 × 2 = 4. So, the unit digit of the square of 272 will be 4.
(iii) 799 – The number ends with 9. Its square will end with 9 × 9 = 81. So, the unit digit of the square of 799 will be 1.
(iv) 3853 – The number ends with 3. Its square will end with 3 × 3 = 9. So, the unit digit of the square of 3853 will be 9.
(v) 1234 – The number ends with 4. Its square will end with 4 × 4 = 16. So, the unit digit of the square of 1234 will be 6.
(vi) 26387 – The number ends with 7. Its square will end with 7 × 7 = 49. So, the unit digit of the square of 26387 will be 9.
(vii) 52698 – The number ends with 8. Its square will end with 8 × 8 = 64. So, the unit digit of the square of 52698 will be 4.
(viii) 99880 – The number ends with 0. Its square will end with 0 × 0 = 0. So, the unit digit of the square of 99880 will be 0.
(ix) 12796 – The number ends with 6. Its square will end with 6 × 6 = 36. So, the unit digit of the square of 12796 will be 6.
(x) 55555 – The number ends with 5. Its square will end with 5 × 5 = 25. So, the unit digit of the square of 55555 will be 5.
Final answer: (i) 1 (ii) 4 (iii) 1 (iv) 9 (v) 6 (vi) 9 (vii) 4 (viii) 0 (ix) 6 (x) 5
Q2. 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
Answer:
A perfect square always ends with one of the digits 0, 1, 4, 5, 6 or 9 at its unit's place. Also, a perfect square ends with an even number of zeroes.
(i) 1057 – The unit digit is 7. Therefore, it is not a perfect square.
(ii) 23453 – The unit digit is 3. Therefore, it is not a perfect square.
(iii) 7928 – The unit digit is 8. Therefore, it is not a perfect square.
(iv) 222222 – The unit digit is 2. Therefore, it is not a perfect square.
(v) 64000 – A perfect square should have an even number of zeroes at the end, but 64000 has three zeroes at the end, which is an odd number. Therefore, it is not a perfect square.
(vi) 89722 – The unit digit is 2. Therefore, it is not a perfect square.
(vii) 222000 – A perfect square should have an even number of zeroes at the end, but 222000 has three zeroes at the end, which is an odd number. Therefore, it is not a perfect square.
(viii) 505050 – A perfect square should have an even number of zeroes at the end, but 505050 has one zero at the end, which is an odd number. Therefore, it is not a perfect square.
Final answer: None of the given numbers is a perfect square.
Q3. The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Answer:
Here, 431 and 7779 are odd numbers, and 2826 and 82004 are even numbers.
We know that the square of an odd number is odd and the square of an even number is even.
Thus, the squares of 431 and 7779 will be odd numbers.
Final answer: (i) 431 and (iii) 7779
Q4. Observe the following pattern and find the missing digits.
11² = 121
101² = 10201
1001² = 1002001
100001² = 1 ......... 2 ......... 1
10000001² = ...........................
Answer:
We can observe that the squares of the given numbers have the same number of zeroes before and after the digit 2 as there were in the original number.
In 100001 there are four zeroes between the 1s, so the square has four zeroes before 2 and four zeroes after 2.
In 10000001 there are six zeroes between the 1s, so the square has six zeroes before 2 and six zeroes after 2.
Therefore,
100001² = 10000200001
10000001² = 100000020000001
Final answer: 100001² = 10000200001 and 10000001² = 100000020000001
Q5. Observe the following pattern and supply the missing numbers.
11² = 1 2 1
101² = 1 0 2 0 1
10101² = 102030201
1010101² = ...........................
............² = 10203040504030201
Answer:
We can observe that the squares of the given numbers have the same number of zeroes after every digit as there were in the original number. The digits in the square rise 1, 2, 3, ... up to the number of 1s in the original number and then come back down.
Therefore,
1010101² = 1020304030201
101010101² = 10203040504030201
Hence, the missing numbers are:
1010101² = 1020304030201 and 101010101² = 10203040504030201
Q6. 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
Answer:
Clearly, the third number is the product of the first two numbers, and the fourth number can be obtained by adding 1 to the third number.
For the fourth line: 4 × 5 = 20, and 20 + 1 = 21.
For the fifth line: 30 ÷ 5 = 6, and 30 + 1 = 31.
For the sixth line: 6 × 7 = 42, and 42 + 1 = 43.
Thus, the missing numbers in the pattern will be as follows.
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²
Check: 6² + 7² + 42² = 36 + 49 + 1764 = 1849 = 43². ✔
Final answer: 20, 6, 42 and 43
Q7. 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
Answer:
We know that the sum of the first n odd natural numbers is n².
(i) Here, there are 5 odd numbers. Therefore, 1 + 3 + 5 + 7 + 9 = (5)² = 25.
(ii) Here, there are 10 odd numbers. Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)² = 100.
(iii) Here, there are 12 odd numbers. Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = (12)² = 144.
Final answer: (i) 25 (ii) 100 (iii) 144
Q8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer:
(i) We know that the sum of the first n odd natural numbers is n². Therefore, 49 is the sum of the first 7 odd natural numbers.
49 = (7)²
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) We know that the sum of the first n odd natural numbers is n². Therefore, 121 is the sum of the first 11 odd natural numbers.
121 = (11)²
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Final answer: 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 and 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Q9. How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
Answer:
We know that there will be 2n numbers between the squares of the numbers n and (n + 1).
(i) Between 12² and 13², there will be 2 × 12 = 24 numbers.
(ii) Between 25² and 26², there will be 2 × 25 = 50 numbers.
(iii) Between 99² and 100², there will be 2 × 99 = 198 numbers.
Check: 13² − 12² − 1 = 169 − 144 − 1 = 24. ✔
Final answer: (i) 24 (ii) 50 (iii) 198