Class 8 - Mathematics
Squares and Square Roots - Exercise 6.1
Top Block 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
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555
Answer :
(i) The number 81 contains its unit’s place digit 1. So, square of 1 is 1.
Hence, unit’s digit of square of 81 is 1.
(ii) The number 272 contains its unit’s place digit 2. So, square of 2 is 4.
Hence, unit’s digit of square of 272 is 4.
(iii) The number 799 contains its unit’s place digit 9. So, square of 9 is 81.
Hence, unit’s digit of square of 799 is 1.
(iv) The number 3853 contains its unit’s place digit 3. So, square of 3 is 9.
Hence, unit’s digit of square of 3853 is 9.
(v) The number 1234 contains its unit’s place digit 4. So, square of 4 is 16.
Hence, unit’s digit of square of 1234 is 6.
(vi) The number 26387 contains its unit’s place digit 7. So, square of 7 is 49.
Hence, unit’s digit of square of 26387 is 9.
(vii) The number 52698 contains its unit’s place digit 8. So, square of 8 is 64.
Hence, unit’s digit of square of 52698 is 4.
(viii) The number 99880 contains its unit’s place digit 0. So, square of 0 is 0.
Hence, unit’s digit of square of 99880 is 0.
(ix) The number 12796 contains its unit’s place digit 6. So, square of 6 is 36.
Hence, unit’s digit of square of 12796 is 6.
(x) The number 55555 contains its unit’s place digit 5. So, square of 5 is 25.
Hence, unit’s digit of square of 55555 is 5.
(i) The number 81 contains its unit’s place digit 1. So, square of 1 is 1.
Hence, unit’s digit of square of 81 is 1.
(ii) The number 272 contains its unit’s place digit 2. So, square of 2 is 4.
Hence, unit’s digit of square of 272 is 4.
(iii) The number 799 contains its unit’s place digit 9. So, square of 9 is 81.
Hence, unit’s digit of square of 799 is 1.
(iv) The number 3853 contains its unit’s place digit 3. So, square of 3 is 9.
Hence, unit’s digit of square of 3853 is 9.
(v) The number 1234 contains its unit’s place digit 4. So, square of 4 is 16.
Hence, unit’s digit of square of 1234 is 6.
(vi) The number 26387 contains its unit’s place digit 7. So, square of 7 is 49.
Hence, unit’s digit of square of 26387 is 9.
(vii) The number 52698 contains its unit’s place digit 8. So, square of 8 is 64.
Hence, unit’s digit of square of 52698 is 4.
(viii) The number 99880 contains its unit’s place digit 0. So, square of 0 is 0.
Hence, unit’s digit of square of 99880 is 0.
(ix) The number 12796 contains its unit’s place digit 6. So, square of 6 is 36.
Hence, unit’s digit of square of 12796 is 6.
(x) The number 55555 contains its unit’s place digit 5. So, square of 5 is 25.
Hence, unit’s digit of square of 55555 is 5.
Question:2. The following numbers are obviously not perfect squares. Give reasons.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050
Answer :
(i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0.
Therefore 1057 is not a perfect square because its unit’s place digit is 7.
(ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 23453 is not a perfect square because its unit’s place digit is 3.
(iii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 7928 is not a perfect square because its unit’s place digit is 8.
(iv) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222222 is not a perfect square because its unit’s place digit is 2.
(v) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 64000 is not a perfect square because its unit’s place digit is single 0.
(vi) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 89722 is not a perfect square because its unit’s place digit is 2.
(vii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222000 is not a perfect square because its unit’s place digit is triple 0.
(viii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 505050 is not a perfect square because its unit’s place digit is 0.
(i) Since, perfect square numbers contain their unit’s place digit 1, 4, 5, 6, 9 and even numbers of 0.
Therefore 1057 is not a perfect square because its unit’s place digit is 7.
(ii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 23453 is not a perfect square because its unit’s place digit is 3.
(iii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 7928 is not a perfect square because its unit’s place digit is 8.
(iv) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222222 is not a perfect square because its unit’s place digit is 2.
(v) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 64000 is not a perfect square because its unit’s place digit is single 0.
(vi) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 89722 is not a perfect square because its unit’s place digit is 2.
(vii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 222000 is not a perfect square because its unit’s place digit is triple 0.
(viii) Since, perfect square numbers contain their unit’s place digit 0, 1, 4, 5, 6, 9 and even number of 0. Therefore 505050 is not a perfect square because its unit’s place digit is 0.
Question:3. The squares of which of the following would be odd number:
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Answer :
(i) 431 – Unit’s digit of given number is 1 and square of 1 is 1. Therefore, square of 431 would be an odd number.
(ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36. Therefore, square of 2826 would not be an odd number.
(iii) 7779 – Unit’s digit of given number is 9 and square of 9 is 81. Therefore, square of 7779 would be an odd number.
(iv) 82004 – Unit’s digit of given number is 4 and square of 4 is 16. Therefore, square of 82004 would not be an odd number.
(i) 431 – Unit’s digit of given number is 1 and square of 1 is 1. Therefore, square of 431 would be an odd number.
(ii) 2826 – Unit’s digit of given number is 6 and square of 6 is 36. Therefore, square of 2826 would not be an odd number.
(iii) 7779 – Unit’s digit of given number is 9 and square of 9 is 81. Therefore, square of 7779 would be an odd number.
(iv) 82004 – Unit’s digit of given number is 4 and square of 4 is 16. Therefore, square of 82004 would not be an odd number.
Question:4. Observe the following pattern and find the missing digits:
11² = 121
101² = 10201
1001² = 1002001
10001² = 1…….2…….1
10000001² = 1……………………
11² = 121
101² = 10201
1001² = 1002001
10001² = 1…….2…….1
10000001² = 1……………………
Answer :
11² = 121
101² = 10201
1001² = 1002001
10001² = 10000200001
10000001² = 100000020000001
11² = 121
101² = 10201
1001² = 1002001
10001² = 10000200001
10000001² = 100000020000001
Question:5. Observe the following pattern and supply the missing numbers:
11² = 121
101² = 10201
10101² = 102030201
1010101² = ………………………
………..² = 10203040504030201
11² = 121
101² = 10201
10101² = 102030201
1010101² = ………………………
………..² = 10203040504030201
Answer :
11² = 121
101² = 10201
10101² = 102030201
1010101² = 1020304030201
101010101² = 10203040504030201
11² = 121
101² = 10201
10101² = 102030201
1010101² = 1020304030201
101010101² = 10203040504030201
Question:6. Using the given pattern, find the missing numbers:
1² + 2² + 3² = 6²
2² + 3² + 6² = 6²
3² + 4² + 12² = 13²
4² + 5² + __² = 21²
5² + __² + 30² = 31²
6² + __² + __² = 43²
1² + 2² + 3² = 6²
2² + 3² + 6² = 6²
3² + 4² + 12² = 13²
4² + 5² + __² = 21²
5² + __² + 30² = 31²
6² + __² + __² = 43²
Answer :
1² + 2² + 3² = 6²
2² + 3² + 6² = 6²
3² + 4² + 12² = 13²
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²
1² + 2² + 3² = 6²
2² + 3² + 6² = 6²
3² + 4² + 12² = 13²
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²
Mddle block 1
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
(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 :
(i) Here, there are five odd numbers. Therefore square of 5 is 25.
∴ 1 + 3 + 5 + 7 + 9 = 5² = 25
(ii) Here, there are ten odd numbers. Therefore square of 10 is 100.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10² = 100
(iii) Here, there are twelve odd numbers. Therefore square of 12 is 144.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12² = 144
(i) Here, there are five odd numbers. Therefore square of 5 is 25.
∴ 1 + 3 + 5 + 7 + 9 = 5² = 25
(ii) Here, there are ten odd numbers. Therefore square of 10 is 100.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10² = 100
(iii) Here, there are twelve odd numbers. Therefore square of 12 is 144.
∴ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12² = 144
Question:8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer :
(i) 49 is the square of 7. Therefore it is the sum of 7 odd numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 is the square of 11. Therefore it is the sum of 11 odd numbers
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
(i) 49 is the square of 7. Therefore it is the sum of 7 odd numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 is the square of 11. Therefore it is the sum of 11 odd numbers
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Question:9. How many numbers lie between squares of the following numbers:
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100
Answer :
(i) Since, non-perfect square numbers between 𝓃² and (𝓃+1)² are Here, 𝓃 = 12
Therefore, non-perfect square numbers between 12 and 13 = 2𝓃 = 2 x 12 = 24
(ii) Since, non-perfect square numbers between 𝓃² and (𝓃+1)² are Here, 𝓃 = 25 Therefore, non-perfect square numbers between 25 and 26 = 2𝓃 = 2 x 25 = 50
(iii) Since, non-perfect square numbers between 𝓃² and (𝓃+1)² are Here, 𝓃 = 99 Therefore, non-perfect square numbers between 99 and 100 = 2𝓃 = 2 x 99 = 198
(i) Since, non-perfect square numbers between 𝓃² and (𝓃+1)² are Here, 𝓃 = 12
Therefore, non-perfect square numbers between 12 and 13 = 2𝓃 = 2 x 12 = 24
(ii) Since, non-perfect square numbers between 𝓃² and (𝓃+1)² are Here, 𝓃 = 25 Therefore, non-perfect square numbers between 25 and 26 = 2𝓃 = 2 x 25 = 50
(iii) Since, non-perfect square numbers between 𝓃² and (𝓃+1)² are Here, 𝓃 = 99 Therefore, non-perfect square numbers between 99 and 100 = 2𝓃 = 2 x 99 = 198