RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square Roots are available here. The subject experts at BYJUâ€™S outline the concepts in a clear and precise manner based on the IQ level of students. Our solution module utilizes numerous shortcut tips and practical examples to explain all the exercise questions in a simple and easily understandable language. If you wish to obtain an excellent score, solving RD Sharma Class 8 Solutions is a must.
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EXERCISE 3.1 PAGE NO: 3.4
1. Which of the following numbers are perfect squares?
(i) 484
(ii) 625
(iii) 576
(iv) 941
(v) 961
(vi) 2500
Solution:
(i) 484
First find the prime factors for 484
484 = 2Ã—2Ã—11Ã—11
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (11Ã—11)
By observation, none of the prime factors are left out.
âˆ´ 484 is a perfect square.
(ii) 625
First find the prime factors for 625
625 = 5Ã—5Ã—5Ã—5
By grouping the prime factors in equal pairs we get,
= (5Ã—5) Ã— (5Ã—5)
By observation, none of the prime factors are left out.
âˆ´ 625 is a perfect square.
(iii) 576
First find the prime factors for 576
576 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3)
By observation, none of the prime factors are left out.
âˆ´ 576 is a perfect square.
(iv) 941
First find the prime factors for 941
941 = 941 Ã— 1
We know that 941 itself is a prime factor.
âˆ´ 941 is not a perfect square.
(v) 961
First find the prime factors for 961
961 = 31Ã—31
By grouping the prime factors in equal pairs we get,
= (31Ã—31)
By observation, none of the prime factors are left out.
âˆ´ 961 is a perfect square.
(vi) 2500
First find the prime factors for 2500
2500 = 2Ã—2Ã—5Ã—5Ã—5Ã—5
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (5Ã—5) Ã— (5Ã—5)
By observation, none of the prime factors are left out.
âˆ´ 2500 is a perfect square.
2. Show that each of the following numbers is a perfect square. Also find the number whose square is the given number in each case:
(i) 1156
(ii) 2025
(iii) 14641
(iv) 4761
Solution:
(i) 1156
First find the prime factors for 1156
1156 = 2Ã—2Ã—17Ã—17
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (17Ã—17)
By observation, none of the prime factors are left out.
âˆ´ 1156 is a perfect square.
To find the square of the given number
1156 = (2Ã—17) Ã— (2Ã—17)
= 34 Ã— 34
= (34)^{2}
âˆ´ 1156 is a square of 34.
(ii) 2025
First find the prime factors for 2025
2025 = 3Ã—3Ã—3Ã—3Ã—5Ã—5
By grouping the prime factors in equal pairs we get,
= (3Ã—3) Ã— (3Ã—3) Ã— (5Ã—5)
By observation, none of the prime factors are left out.
âˆ´ 2025 is a perfect square.
To find the square of the given number
2025 = (3Ã—3Ã—5) Ã— (3Ã—3Ã—5)
= 45 Ã— 45
= (45)^{2}
âˆ´ 2025 is a square of 45.
(iii) 14641
First find the prime factors for 14641
14641 = 11Ã—11Ã—11Ã—11
By grouping the prime factors in equal pairs we get,
= (11Ã—11) Ã— (11Ã—11)
By observation, none of the prime factors are left out.
âˆ´ 14641 is a perfect square.
To find the square of the given number
14641 = (11Ã—11) Ã— (11Ã—11)
= 121 Ã— 121
= (121)^{2}
âˆ´ 14641 is a square of 121.
(iv) 4761
First find the prime factors for 4761
4761 = 3Ã—3Ã—23Ã—23
By grouping the prime factors in equal pairs we get,
= (3Ã—3) Ã— (23Ã—23)
By observation, none of the prime factors are left out.
âˆ´ 4761 is a perfect square.
To find the square of the given number
4761 = (3Ã—23) Ã— (3Ã—23)
= 69 Ã— 69
= (69)^{2}
âˆ´ 4761 is a square of 69.
3. Find the smallest number by which the given number must be multiplied so that the product is a perfect square:
(i) 23805
(ii) 12150
(iii) 7688
Solution:
(i) 23805
First find the prime factors for 23805
23805 = 3Ã—3Ã—23Ã—23Ã—5
By grouping the prime factors in equal pairs we get,
= (3Ã—3) Ã— (23Ã—23) Ã— 5
By observation, prime factor 5 is left out.
So, multiply by 5 we get,
23805 Ã— 5 = (3Ã—3) Ã— (23Ã—23) Ã— (5Ã—5)
= (3Ã—5Ã—23) Ã— (3Ã—5Ã—23)
= 345 Ã— 345
= (345)^{2}
âˆ´ Product is the square of 345.
(ii) 12150
First find the prime factors for 12150
12150 = 2Ã—3Ã—3Ã—3Ã—3Ã—3Ã—5Ã—5
By grouping the prime factors in equal pairs we get,
= 2Ã—3 Ã— (3Ã—3) Ã— (3Ã—3) Ã— (5Ã—5)
By observation, prime factor 2 and 3 are left out.
So, multiply by 2Ã—3 = 6 we get,
12150 Ã— 6 = 2Ã—3 Ã— (3Ã—3) Ã— (3Ã—3) Ã— (5Ã—5) Ã— 2 Ã— 3
= (2Ã—3Ã—3Ã—3Ã—5) Ã— (2Ã—3Ã—3Ã—3Ã—5)
= 270 Ã— 270
= (270)^{2}
âˆ´ Product is the square of 270.
(iii) 7688
First find the prime factors for 7688
7688 = 2Ã—2Ã—31Ã—31Ã—2
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (31Ã—31) Ã— 2
By observation, prime factor 2 is left out.
So, multiply by 2 we get,
7688 Ã— 2 = (2Ã—2) Ã— (31Ã—31)Ã— (2Ã—2)
= (2Ã—31Ã—2) Ã— (2Ã—31Ã—2)
= 124 Ã— 124
= (124)^{2}
âˆ´ Product is the square of 124.
4. Find the smallest number by which the given number must be divided so that the resulting number is a perfect square:
(i) 14283
(ii) 1800
(iii) 2904
Solution:
(i) 14283
First find the prime factors for 14283
14283 = 3Ã—3Ã—3Ã—23Ã—23
By grouping the prime factors in equal pairs we get,
= (3Ã—3) Ã— (23Ã—23) Ã— 3
By observation, prime factor 3 is left out.
So, divide by 3 to eliminate 3 we get,
14283/3 = (3Ã—3) Ã— (23Ã—23)
= (3Ã—23) Ã— (3Ã—23)
= 69 Ã— 69
= (69)^{2}
âˆ´ Resultant is the square of 69.
(ii) 1800
First find the prime factors for 1800
1800 = 2Ã—2Ã—5Ã—5Ã—3Ã—3Ã—2
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (5Ã—5) Ã— (3Ã—3) Ã— 2
By observation, prime factor 2 is left out.
So, divide by 2 to eliminate 2 we get,
1800/2 = (2Ã—2) Ã— (5Ã—5) Ã— (3Ã—3)
= (2Ã—5Ã—3) Ã— (2Ã—5Ã—3)
= 30 Ã— 30
= (30)^{2}
âˆ´ Resultant is the square of 30.
(iii) 2904
First find the prime factors for 2904
2904 = 2Ã—2Ã—11Ã—11Ã—2Ã—3
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (11Ã—11) Ã— 2 Ã— 3
By observation, prime factor 2 and 3 are left out.
So, divide by 6 to eliminate 2 and 3 we get,
2904/6 = (2Ã—2) Ã— (11Ã—11)
= (2Ã—11) Ã— (2Ã—11)
= 22 Ã— 22
= (22)^{2}
âˆ´ Resultant is the square of 22.
5. Which of the following numbers are perfect squares?
11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121
Solution:
11 it is a prime number by itself.
So it is not a perfect square.
12 is not a perfect square.
16= (4)^{2}
16 is a perfect square.
32 is not a perfect square.
36= (6)^{2}
36 is a perfect square.
50 is not a perfect square.
64= (8)^{2}
64 is a perfect square.
79 it is a prime number.
So it is not a perfect square.
81= (9)^{2}
81 is a perfect square.
111 it is a prime number.
So it is not a perfect square.
121= (11)^{2}
121 is a perfect square.
6. Using prime factorization method, find which of the following numbers are perfect squares?
189, 225, 2048, 343, 441, 2961, 11025, 3549
Solution:
189 prime factors are
189 = 3^{2}Ã—3Ã—7
Since it does not have equal pair of factors 189 is not a perfect square.
225 prime factors are
225 = (5Ã—5) Ã— (3Ã—3)
Since 225 has equal pair of factors. âˆ´ It is a perfect square.
2048 prime factors are
2048 = (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— 2
Since it does not have equal pair of factors 2048 is not a perfect square.
343 prime factors are
343 = (7Ã—7) Ã— 7
Since it does not have equal pair of factors 2048 is not a perfect square.
441 prime factors are
441 = (7Ã—7) Ã— (3Ã—3)
Since 441 has equal pair of factors. âˆ´ It is a perfect square.
2961 prime factors are
2961 = (3Ã—3) Ã— (3Ã—3) Ã— (3Ã—3) Ã— (2Ã—2)
Since 2961 has equal pair of factors. âˆ´ It is a perfect square.
11025 prime factors are
11025 = (3Ã—3) Ã— (5Ã—5) Ã— (7Ã—7)
Since 11025 has equal pair of factors. âˆ´ It is a perfect square.
3549 prime factors are
3549 = (13Ã—13) Ã— 3 Ã— 7
Since it does not have equal pair of factors 3549 is not a perfect square.
7. By what number should each of the following numbers by multiplied to get a perfect square in each case? Also find the number whose square is the new number.
(i)Â 8820
(ii) 3675
(iii) 605
(iv) 2880
(v) 4056
(vi) 3468
(vii) 7776
Solution:
(i) 8820
First find the prime factors for 8820
8820 = 2Ã—2Ã—3Ã—3Ã—7Ã—7Ã—5
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (3Ã—3) Ã— (7Ã—7) Ã— 5
By observation, prime factor 5 is left out.
So, multiply by 5 we get,
8820 Ã— 5 = (2Ã—2) Ã— (3Ã—3) Ã— (7Ã—7) Ã— (5Ã—5)
= (2Ã—3Ã—7Ã—5) Ã— (2Ã—3Ã—7Ã—5)
= 210 Ã— 210
= (210)^{2}
âˆ´ Product is the square of 210.
(ii) 3675
First find the prime factors for 3675
3675 = 5Ã—5Ã—7Ã—7Ã—3
By grouping the prime factors in equal pairs we get,
= (5Ã—5) Ã— (7Ã—7) Ã— 3
By observation, prime factor 3 is left out.
So, multiply by 3 we get,
3675 Ã— 3 = (5Ã—5) Ã— (7Ã—7) Ã— (3Ã—3)
= (5Ã—7Ã—3) Ã— (5Ã—7Ã—3)
= 105 Ã— 105
= (105)^{2}
âˆ´ Product is the square of 105.
(iii) 605
First find the prime factors for 605
605 = 5Ã—11Ã—11
By grouping the prime factors in equal pairs we get,
= (11Ã—11) Ã— 5
By observation, prime factor 5 is left out.
So, multiply by 5 we get,
605 Ã— 5 = (11Ã—11) Ã— (5Ã—5)
= (11Ã—5) Ã— (11Ã—5)
= 55 Ã— 55
= (55)^{2}
âˆ´ Product is the square of 55.
(iv) 2880
First find the prime factors for 2880
2880 = 5Ã—3Ã—3Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2
By grouping the prime factors in equal pairs we get,
= (3Ã—3) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— 5
By observation, prime factor 5 is left out.
So, multiply by 5 we get,
2880 Ã— 5 = (3Ã—3) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (5Ã—5)
= (3Ã—2Ã—2Ã—2Ã—5) Ã— (3Ã—2Ã—2Ã—2Ã—5)
= 120 Ã— 120
= (120)^{2}
âˆ´ Product is the square of 120.
(v) 4056
First find the prime factors for 4056
4056 = 2Ã—2Ã—13Ã—13Ã—2Ã—3
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (13Ã—13) Ã— 2 Ã— 3
By observation, prime factors 2 and 3 are left out.
So, multiply by 6 we get,
4056 Ã— 6 = (2Ã—2) Ã— (13Ã—13) Ã— (2Ã—2) Ã— (3Ã—3)
= (2Ã—13Ã—2Ã—3) Ã— (2Ã—13Ã—2Ã—3)
= 156 Ã— 156
= (156)^{2}
âˆ´ Product is the square of 156.
(vi) 3468
First find the prime factors for 3468
3468 = 2Ã—2Ã—17Ã—17Ã—3
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (17Ã—17) Ã— 3
By observation, prime factor 3 is left out.
So, multiply by 3 we get,
3468 Ã— 3 = (2Ã—2) Ã— (17Ã—17) Ã— (3Ã—3)
= (2Ã—17Ã—3) Ã— (2Ã—17Ã—3)
= 102 Ã— 102
= (102)^{2}
âˆ´ Product is the square of 102.
(vii) 7776
First find the prime factors for 7776
7776 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—2Ã—3
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3) Ã— (3Ã—3) Ã— 2 Ã— 3
By observation, prime factors 2 and 3 are left out.
So, multiply by 6 we get,
7776 Ã— 6 = (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3) Ã— (3Ã—3) Ã— (2Ã—2) Ã— (3Ã—3)
= (2Ã—2Ã—3Ã—3Ã—2Ã—3) Ã— (2Ã—2Ã—3Ã—3Ã—2Ã—3)
= 216 Ã— 216
= (216)^{2}
âˆ´ Product is the square of 216.
8. By What numbers should each of the following be divided to get a perfect square in each case? Also, find the number whose square is the new number.
(i) 16562
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575
Solution:
(i) 16562
First find the prime factors for 16562
16562 = 7Ã—7Ã—13Ã—13Ã—2
By grouping the prime factors in equal pairs we get,
= (7Ã—7) Ã— (13Ã—13) Ã— 2
By observation, prime factor 2 is left out.
So, divide by 2 to eliminate 2 we get,
16562/2 = (7Ã—7) Ã— (13Ã—13)
= (7Ã—13) Ã— (7Ã—13)
= 91 Ã— 91
= (91)^{2}
âˆ´ Resultant is the square of 91.
(ii) 3698
First find the prime factors for 3698
3698 = 2Ã—43Ã—43
By grouping the prime factors in equal pairs we get,
= (43Ã—43) Ã— 2
By observation, prime factor 2 is left out.
So, divide by 2 to eliminate 2 we get,
3698/2 = (43Ã—43)
= (43)^{2}
âˆ´ Resultant is the square of 43.
(iii) 5103
First find the prime factors for 5103
5103 = 3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—7
By grouping the prime factors in equal pairs we get,
= (3Ã—3) Ã— (3Ã—3) Ã— (3Ã—3) Ã— 7
By observation, prime factor 7 is left out.
So, divide by 7 to eliminate 7 we get,
5103/7 = (3Ã—3) Ã— (3Ã—3) Ã— (3Ã—3)
= (3Ã—3Ã—3) Ã— (3Ã—3Ã—3)
= 27 Ã— 27
= (27)^{2}
âˆ´ Resultant is the square of 27.
(iv) 3174
First find the prime factors for 3174
3174 = 2Ã—3Ã—23Ã—23
By grouping the prime factors in equal pairs we get,
= (23Ã—23) Ã— 2 Ã— 3
By observation, prime factor 2 and 3 are left out.
So, divide by 6 to eliminate 2 and 3 we get,
3174/6 = (23Ã—23)
= (23)^{2}
âˆ´ Resultant is the square of 23.
(v) 1575
First find the prime factors for 1575
1575 = 3Ã—3Ã—5Ã—5Ã—7
By grouping the prime factors in equal pairs we get,
= (3Ã—3) Ã— (5Ã—5) Ã— 7
By observation, prime factor 7 is left out.
So, divide by 7 to eliminate 7 we get,
1575/7 = (3Ã—3) Ã— (5Ã—5)
= (3Ã—5) Ã— (3Ã—5)
= 15 Ã— 15
= (15)^{2}
âˆ´ Resultant is the square of 15.
9. Find the greatest number of two digits which is a perfect square.
Solution:
We know that the two digit greatest number is 99
âˆ´ Greatest two digit perfect square number is 99-18 = 81
10. Find the least number of three digits which is perfect square.
Solution:
We know that the three digit greatest number is 100
To find the square root of 100
âˆ´ the least number of three digits which is a perfect square is 100 itself.
11. Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.
Solution:
First find the prime factors for 4851
4851 = 3Ã—3Ã—7Ã—7Ã—11
By grouping the prime factors in equal pairs we get,
= (3Ã—3) Ã— (7Ã—7) Ã— 11
âˆ´ The smallest number by which 4851 must be multiplied so that the product becomes a perfect square is 11.
12. Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.
Solution:
First find the prime factors for 28812
28812 = 2Ã—2Ã—3Ã—7Ã—7Ã—7Ã—7
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— 3 Ã— (7Ã—7) Ã— (7Ã—7)
âˆ´ The smallest number by which 28812 must be divided so that the quotient becomes a perfect square is 3.
13. Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number.
Solution:
First find the prime factors for 1152
1152 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3) Ã— 2
âˆ´ The smallest number by which 1152 must be divided so that the quotient becomes a perfect square is 2.
The number after division, 1152/2 = 576
prime factors for 576 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3
By grouping the prime factors in equal pairs we get,
= (2Ã—2) Ã— (2Ã—2) Ã— (2Ã—2) Ã— (3Ã—3)
= 2^{6} Ã— 3^{2}
= 24^{2}
âˆ´ The resulting number is the square of 24.
EXERCISE 3.2 PAGE NO: 3.18
1. The following numbers are not perfect squares. Give reason.
(i) 1547
(ii) 45743
(iii)8948
(iv) 333333
Solution:
The numbers ending with 2, 3, 7 or 8 is not a perfect square.
So, (i) 1547
(ii) 45743
(iii) 8948
(iv) 333333
Are not perfect squares.
2. Show that the following numbers are not, perfect squares:
(i) 9327
(ii) 4058
(iii)22453
(iv) 743522
Solution:
The numbers ending with 2, 3, 7 or 8 is not a perfect square.
So, (i) 9327
(ii) 4058
(iii) 22453
(iv) 743522
Are not perfect squares.
3. The square of which of the following numbers would be an old number?
(i) 731
(ii) 3456
(iii)5559
(iv) 42008
Solution:
We know that square of an even number is even number.
Square of an odd number is odd number.
(i) 731
Since 731 is an odd number, the square of the given number is also odd.
(ii) 3456
Since 3456 is an even number, the square of the given number is also even.
(iii) 5559
Since 5559 is an odd number, the square of the given number is also odd.
(iv) 42008
Since 42008 is an even number, the square of the given number is also even.
4. What will be the unitâ€™s digit of the squares of the following numbers?
(i) 52
(ii) 977
(iii) 4583
(iv) 78367
(v) 52698
(vi) 99880
(vii) 12796
(viii) 55555
(ix) 53924
Solution:
(i) 52
Unit digit of (52)^{2} = (2^{2}) = 4
(ii) 977
Unit digit of (977)^{2} = (7^{2}) = 49 = 9
(iii) 4583
Unit digit of (4583)^{2} = (3^{2}) = 9
(iv) 78367
Unit digit of (78367)^{2} = (7^{2}) = 49 = 9
(v) 52698
Unit digit of (52698)^{2} = (8^{2}) = 64 = 4
(vi) 99880
Unit digit of (99880)^{2} = (0^{2}) = 0
(vii) 12796
Unit digit of (12796)^{2} = (6^{2}) = 36 = 6
(viii) 55555
Unit digit of (55555)^{2} = (5^{2}) = 25 = 5
(ix) 53924
Unit digit of (53924)^{2} = (4^{2}) = 16 = 6
5. Observe the following pattern
1+3 = 2^{2}
1+3+5 = 3^{2}
1+3+5+7 = 4^{2}
And write the value of 1+3+5+7+9+â€¦â€¦â€¦ up to n terms.
Solution:
We know that the pattern given is the square of the given number on the right hand side is equal to the sum of the given numbers on the left hand side.
âˆ´ The value of 1+3+5+7+9+â€¦â€¦â€¦ up to n terms = n^{2} (as there are only n terms).
6. Observe the following pattern
2^{2} -1^{2} = 2 + 1
3^{2} â€“ 2^{2} = 3 + 2
4^{2} â€“ 3^{2} = 4 + 3
5^{2} â€“ 4^{2} = 5 + 4
And find the value of
(i) 100^{2} -99^{2}
(ii)111^{2} – 109^{2}
(iii) 99^{2} â€“ 96^{2}
Solution:
(i) 100^{2} -99^{2}
100 + 99 = 199
(ii) 111^{2} â€“ 109^{2}
(111^{2} â€“ 110^{2}) + (110^{2} â€“ 109^{2})
(111 + 110) + (100 + 109)
440
(iii) 99^{2} â€“ 96^{2}
(99^{2} â€“ 98^{2}) + (98^{2} â€“ 97^{2}) + (97^{2} â€“ 96^{2})
(99 + 98) + (98 + 97) + (97 + 96)
585
7. Which of the following triplets are Pythagorean?
(i) (8, 15, 17)
(ii) (18, 80, 82)
(iii) (14, 48, 51)
(iv) (10, 24, 26)
(v) (16, 63, 65)
(vi) (12, 35, 38)
Solution:
(i) (8, 15, 17)
LHS = 8^{2} + 15^{2}
= 289
RHS = 17^{2}
= 289
LHS = RHS
âˆ´ The given triplet is a Pythagorean.
(ii) (18, 80, 82)
LHS = 18^{2} + 80^{2}
= 6724
RHS = 82^{2}
= 6724
LHS = RHS
âˆ´ The given triplet is a Pythagorean.
(iii) (14, 48, 51)
LHS = 14^{2} + 48^{2}
= 2500
RHS = 51^{2}
= 2601
LHS â‰ RHS
âˆ´ The given triplet is not a Pythagorean.
(iv) (10, 24, 26)
LHS = 10^{2} + 24^{2}
= 676
RHS = 26^{2}
= 676
LHS = RHS
âˆ´ The given triplet is a Pythagorean.
(v) (16, 63, 65)
LHS = 16^{2} + 63^{2}
= 4225
RHS = 65^{2}
= 4225
LHS = RHS
âˆ´ The given triplet is a Pythagorean.
(vi) (12, 35, 38)
LHS = 12^{2} + 35^{2}
= 1369
RHS = 38^{2}
= 1444
LHS â‰ RHS
âˆ´ The given triplet is not a Pythagorean.
8. Observe the following pattern
(1Ã—2) + (2Ã—3) = (2Ã—3Ã—4)/3
(1Ã—2) + (2Ã—3) + (3Ã—4) = (3Ã—4Ã—5)/3
(1Ã—2) + (2Ã—3) + (3Ã—4) + (4Ã—5) = (4Ã—5Ã—6)/3
And find the value of
(1Ã—2) + (2Ã—3) + (3Ã—4) + (4Ã—5) + (5Ã—6)
Solution:
(1Ã—2) + (2Ã—3) + (3Ã—4) + (4Ã—5) + (5Ã—6) = (5Ã—6Ã—7)/3 = 70
9. Observe the following pattern
1 = 1/2 (1Ã—(1+1))
1+2 = 1/2 (2Ã—(2+1))
1+2+3 = 1/2 (3Ã—(3+1))
1+2+3+4 = 1/2 (4Ã—(4+1))
And find the values of each of the following:
(i) 1+2+3+4+5+â€¦+50
(ii) 31+32+â€¦.+50
Solution:
We know that R.H.S =Â 1/2Â [No. of terms in L.H.S Ã— (No. of terms + 1)] (if only when L.H.S starts with 1)
(i) 1+2+3+4+5+â€¦+50 = 1/2 (5Ã—(5+1))
25 Ã— 51 = 1275
(ii) 31+32+â€¦.+50 = (1+2+3+4+5+â€¦+50) â€“ (1+2+3+â€¦+30)
1275 â€“ 1/2 (30Ã—(30+1))
1275 â€“ 465
810
10. Observe the following pattern
1^{2} = 1/6 (1Ã—(1+1)Ã—(2Ã—1+1))
1^{2}+2^{2} = 1/6 (2Ã—(2+1)Ã—(2Ã—2+1)))
1^{2}+2^{2}+3^{2} = 1/6 (3Ã—(3+1)Ã—(2Ã—3+1)))
1^{2}+2^{2}+3^{2}+4^{2} = 1/6 (4Ã—(4+1)Ã—(2Ã—4+1)))
And find the values of each of the following:
(i) 1^{2}+2^{2}+3^{2}+4^{2}+â€¦+10^{2}
(ii) 5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}+11^{2}+12^{2}
Solution:
RHS = 1/6Â [(No. of terms in L.H.S) Ã— (No. of terms + 1) Ã— (2 Ã— No. of terms + 1)]
(i) 1^{2}+2^{2}+3^{2}+4^{2}+â€¦+10^{2} = 1/6 (10Ã—(10+1)Ã—(2Ã—10+1))
= 1/6 (2310)
= 385
(ii) 5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}+11^{2}+12^{2} = 1^{2}+2^{2}+3^{2}+â€¦+12^{2} â€“ (1^{2}+2^{2}+3^{2}+4^{2})
1/6 (12Ã—(12+1)Ã—(2Ã—12+1)) – 1/6 (4Ã—(4+1)Ã—(2Ã—4+1))
650-30
620
11. Which of the following numbers are squares of even numbers?
121, 225, 256, 324, 1296, 6561, 5476, 4489, 373758
Solution:
We know that only even numbers be the squares of even numbers.
So, 256, 324, 1296, 5476, 373758 are even numbers, since 373758 is not a perfect square
âˆ´ 256, 324, 1296, 5476 are squares of even numbers.
12. By just examining the units digits, can you tell which of the following cannot be whole squares?
(i) 1026
(ii) 1028
(iii)1024
(iv) 1022
(v) 1023
(vi) 1027
Solution:
We know that numbers ending with 2, 3, 7, 8 cannot be a perfect square.
âˆ´ 1028, 1022, 1023, and 1027 cannot be whole squares.
13. Which of the numbers for which you cannot decide whether they are squares.
Solution:
We know that the natural numbers such as 0, 1, 4, 5, 6 or 9 cannot be decided surely whether they are squares or not.
14. Write five numbers which you cannot decide whether they are square just by looking at the unitâ€™s digit.
Solution:
We know that any natural number ending with 0, 1, 4, 5, 6 or 9 can be or cannot be a square number.
Here are the five examples which you cannot decide whether they are square or not just by looking at the units place:
(i) 2061
The unit digit is 1. So, it may or may not be a square number
(ii) 1069
The unit digit is 9. So, it may or may not be a square number
(iii) 1234
The unit digit is 4. So, it may or may not be a square number
(iv) 56790
The unit digit is 0. So, it may or may not be a square number
(v) 76555
The unit digit is 5. So, it may or may not be a square number
15. Write true (T) or false (F) for the following statements.
(i) The number of digits in a square number is even.
(ii) The square of a prime number is prime.
(iii) The sum of two square numbers is a square number.
(iv) The difference of two square numbers is a square number.
(v) The product of two square numbers is a square number.
(vi) No square number is negative.
(vii) There is no square number between 50 and 60.
(viii) There are fourteen square number up to 200.
Solution:
(i) False, because 169 is a square number with odd digit.
(ii) False, because square of 3(which is prime) is 9(which is not prime).
(iii) False, because sum of 2^{2}Â and 3^{2}Â is 13 which is not square number.
(iv) False, because difference of 3^{2}Â and 2^{2}Â is 5, which is not square number.
(v) True, because the square of 2^{2}Â and 3^{2}Â is 36 which is square of 6
(vi) True, because (-2)^{2} is 4, which is not negative.
(vii) True, because as there is no square number between them.
(viii) True, because the fourteen numbers up to 200 are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196.
EXERCISE 3.3 PAGE NO: 3.32
1. Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication:
(i) 25
(ii) 37
(iii) 54
(iv) 71
(v) 96
Solution:
(i) 25
So here, a = 2 and b = 5
Column I | Column II | Column III |
a^{2}
4 +2 6 |
2ab
20 +2 22 |
b^{2}
25 |
6 | 2 | 5 |
âˆ´ 25^{2}Â = 625
Where, it can be expressed as
25^{2}Â = 25Ã— 25 = 625
(ii) 37
So here, a = 3 and b = 7
Column I | Column II | Column III |
a^{2}
9 +4 13 |
2ab
42 +4 46 |
b^{2}
49 |
13 | 6 | 9 |
âˆ´ 37^{2}Â = 1369
Where, it can be expressed as
25^{2}Â = 37Ã— 37 = 1369
(iii) 54
So here, a = 5 and b = 4
Column I | Column II | Column III |
a^{2}
25 +4 29 |
2ab
40 +1 41 |
b^{2}
16 |
29 | 1 | 6 |
âˆ´ 54^{2}Â = 2916
Where, it can be expressed as
54^{2}Â = 54 Ã— 54 = 2916
(iv) 71
So here, a = 7 and b = 1
Column I | Column II | Column III |
a^{2}
49 +1 50 |
2ab
14 +0 14 |
b^{2}
01 |
50 | 4 | 1 |
âˆ´ 71^{2}Â = 5041
Where, it can be expressed as
71^{2}Â = 71 Ã— 71 = 5041
(v) 96
So here, a = 9 and b = 6
Column I | Column II | Column III |
a^{2}
81 +11 92 |
2ab
108 +3 111 |
b^{2}
36 |
92 | 1 | 6 |
âˆ´ 96^{2}Â = 9216
Where, it can be expressed as
96^{2}Â = 96 Ã— 96 = 9216
2. Find the squares of the following numbers using diagonal method:
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171
Solution:
(i) 98
Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.
Step 2: Draw square and divide it into n^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.
Step 3: Draw the diagonals of each sub-square.
Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.
Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.
Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.
Step 7: Obtain the required square by writing the digits from the left-most side.
âˆ´ 98^{2}Â = 9604
(ii) 273
Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.
Step 2: Draw square and divide it into n^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.
Step 3: Draw the diagonals of each sub-square.
Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.
Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.
Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.
Step 7: Obtain the required square by writing the digits from the left-most side.
âˆ´ 273^{2}Â = 74529
(iii) 348
Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.
Step 2: Draw square and divide it into n^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.
Step 3: Draw the diagonals of each sub-square.
Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.
Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.
Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.
Step 7: Obtain the required square by writing the digits from the left-most side.
âˆ´ 348^{2}Â = 121104
(iv) 295
Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.
Step 2: Draw square and divide it into n^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.
Step 3: Draw the diagonals of each sub-square.
Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.
Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.
Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.
Step 7: Obtain the required square by writing the digits from the left-most side.
âˆ´ 295^{2}Â = 87025
(v) 171
Step 1: Obtain the number and count the number of digits in it. Let there be n digits in the number to be squared.
Step 2: Draw square and divide it into n^{2}Â sub-squares of the same size by drawing (n – 1) horizontal and (n – 1) vertical lines.
Step 3: Draw the diagonals of each sub-square.
Step 4: Write the digits of the number to be squared along left vertical side sand top horizontal side of the squares.
Step 5: Multiply each digit on the left of the square with each digit on top of the column one-by-one. Write the units digit of the product below the diagonal and tens digit above the diagonal of the corresponding sub-square.
Step 6: Starting below the lowest diagonal sum the digits along the diagonals so obtained. Write the units digit of the sum and take carry, the tens digit (if any) to the diagonal above.
Step 7: Obtain the required square by writing the digits from the left-most side.
âˆ´ 171^{2}Â = 29241
3. Find the squares of the following numbers:
(i) 127
(ii) 503
(iii) 450
(iv) 862
(v) 265
Solution:
(i) 127
127^{2} = 127 Ã— 127 = 16129
(ii) 503
503^{2} = 503 Ã— 503 = 253009
(iii) 450
450^{2} = 450 Ã— 450 = 203401
(iv) 862
862^{2} = 862 Ã— 862 = 743044
(v) 265
265^{2} = 265 Ã— 265 = 70225
4. Find the squares of the following numbers:
(i) 425
(ii) 575
(iii)405
(iv) 205
(v) 95
(vi) 745
(vii) 512
(viii) 995
Solution:
(i)425
425^{2} = 425 Ã— 425 = 180625
(ii) 575
575^{2} = 575 Ã— 575 = 330625
(iii)405
405^{2} = 405 Ã— 405 = 164025
(iv) 205
205^{2} = 205 Ã— 205 = 42025
(v) 95
95^{2} = 95 Ã— 95 = 9025
(vi) 745
745^{2} = 745 Ã— 745 = 555025
(vii) 512
512^{2} = 512 Ã— 512 = 262144
(viii) 995
995^{2} = 995 Ã— 995 = 990025
5. Find the squares of the following numbers using the identityÂ (a+b)^{ 2}= a^{2}+2ab+b^{2}:
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605
Solution:
(i) 405
We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}
405 = (400+5)^{2}
= (400)^{2}Â + 5^{2}Â + 2 (400) (5)
= 160000 + 25 + 4000
= 164025
(ii) 510
We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}
510 = (500+10)^{2}
= (500)^{2}Â + 10^{2}Â + 2 (500) (10)
= 250000 + 100 + 10000
= 260100
(iii) 1001
We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}
1001 = (1000+1)^{2}
= (1000)^{2}Â + 1^{2}Â + 2 (1000) (1)
= 1000000 + 1 + 2000
= 1002001
(iv) 209
We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}
209 = (200+9)^{2}
= (200)^{2}Â + 9^{2}Â + 2 (200) (9)
= 40000 + 81 + 3600
= 43681
(v) 605
We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}
605 = (600+5)^{2}
= (600)^{2}Â + 5^{2}Â + 2 (600) (5)
= 360000 + 25 + 6000
= 366025
6. Find the squares of the following numbers using the identityÂ (a-b)^{ 2}= a^{2}-2ab+b^{2}
(i) 395
(ii) 995
(iii)495
(iv) 498
(v) 99
(vi) 999
(vii)599
Solution:
(i) 395
We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}
395 = (400-5)^{2}
= (400)^{2}Â + 5^{2}Â – 2 (400) (5)
= 160000 + 25 â€“ 4000
= 156025
(ii) 995
We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}
995 = (1000-5)^{2}
= (1000)^{2}Â + 5^{2}Â – 2 (1000) (5)
= 1000000 + 25 â€“ 10000
= 990025
(iii) 495
We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}
495 = (500-5)^{2}
= (500)^{2}Â + 5^{2}Â – 2 (500) (5)
= 250000 + 25 â€“ 5000
= 245025
(iv) 498
We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}
498 = (500-2)^{2}
= (500)^{2}Â + 2^{2}Â – 2 (500) (2)
= 250000 + 4 â€“ 2000
= 248004
(v) 99
We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}
99 = (100-1)^{2}
= (100)^{2}Â + 1^{2}Â – 2 (100) (1)
= 10000 + 1 â€“ 200
= 9801
(vi) 999
We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}
999 = (1000-1)^{2}
= (1000)^{2}Â + 1^{2}Â – 2 (1000) (1)
= 1000000 + 1 â€“ 2000
= 998001
(vii) 599
We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}
599 = (600-1)^{2}
= (600)^{2}Â + 1^{2}Â – 2 (600) (1)
= 360000 + 1 â€“ 1200
= 358801
7. Find the squares of the following numbers by visual method:
(i) 52
(ii) 95
(iii) 505
(iv) 702
(v) 99
Solution:
(i) 52
We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}
52 = (50+2)^{2}
= (50)^{2}Â + 2^{2}Â + 2 (50) (2)
= 2500 + 4 + 200
= 2704
(ii) 95
We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}
95 = (100-5)^{2}
= (100)^{2}Â + 5^{2}Â – 2 (100) (5)
= 10000 + 25 – 1000
= 9025
(iii) 505
We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}
505 = (500+5)^{2}
= (500)^{2}Â + 5^{2}Â + 2 (500) (5)
= 250000 + 25 + 5000
= 255025
(iv) 702
We know, (a+b)^{ 2}= a^{2}+2ab+b^{2}
702 = (700+2)^{2}
= (700)^{2}Â + 2^{2}Â + 2 (700) (2)
= 490000 + 4 + 2800
= 492804
(v) 99
We know, (a-b)^{ 2}= a^{2}-2ab+b^{2}
99 = (100-1)^{2}
= (100)^{2}Â + 1^{2}Â – 2 (100) (1)
= 10000 + 1 – 200
= 9801
EXERCISE 3.4 PAGE NO: 3.38
1.Write the possible unitâ€™s digits of the square root of the following numbers. Which of these numbers are odd square roots?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i)Â 9801
We know that unit digit of 9801 is 1
Unit digit of square root = 1 or 9
Since the number is odd, square root is also odd
(ii)Â 99856
We know that unit digit of 99856 = 6
Unit digit of square root = 4 or 6
Since the number is even, square root is also even
(iii)Â 998001
We know that unit digit of 998001 = 1
Unit digit of square root = 1 or 9
Since the number is odd, square root is also odd
(iv)Â 657666025
We know that unit digit of 657666025 = 5
Unit digit of square root = 5
Since the number is odd, square root is also odd
2. Find the square root of each of the following by prime factorization.
(i) 441 (ii) 196
(iii) 529 (iv) 1764
(v) 1156 (vi) 4096
(vii) 7056 (viii) 8281
(ix) 11664 (x) 47089
(xi) 24336 (xii) 190969
(xiii) 586756 (xiv) 27225
(xv) 3013696
Solution:
(i) 441
Firstly letâ€™s find the prime factors for
441 = 3Ã—3Ã—7Ã—7
= 3^{2}Ã—7^{2}
âˆš441 = 3Ã—7
= 21
(ii) 196
Firstly letâ€™s find the prime factors for
196 = 2Ã—2Ã—7Ã—7
= 2^{2}Ã—7^{2}
âˆš196 = 2Ã—7
= 14
(iii) 529
Firstly letâ€™s find the prime factors for
529 = 23Ã—23
= 23^{2}
âˆš529 = 23
(iv) 1764
Firstly letâ€™s find the prime factors for
1764 = 2Ã—2Ã—3Ã—3Ã—7Ã—7
= 2^{2}Ã—3^{2}Ã—7^{2}
âˆš1764 = 2Ã—3Ã—7
= 42
(v) 1156
Firstly letâ€™s find the prime factors for
1156 = 2Ã—2Ã—17Ã—17
= 2^{2}Ã—17^{2}
âˆš1156 = 2Ã—17
= 34
(vi) 4096
Firstly letâ€™s find the prime factors for
4096 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2
= 2^{12}
âˆš4096 = 2^{6}
= 64
(vii) 7056
Firstly letâ€™s find the prime factors for
7056 = 2Ã—2Ã—2Ã—2Ã—21Ã—21
= 2^{2}Ã—2^{2}Ã—21^{2}
âˆš7056 = 2Ã—2Ã—21
= 84
(viii) 8281
Firstly letâ€™s find the prime factors for
8281 = 91Ã—91
= 91^{2}
âˆš8281 = 91
(ix) 11664
Firstly letâ€™s find the prime factors for
11664 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3
= 2^{2}Ã—2^{2}Ã—3^{2}Ã—3^{2}Ã—3^{2}
âˆš11664 = 2Ã—2Ã—3Ã—3Ã—3
= 108
(x) 47089
Firstly letâ€™s find the prime factors for
47089 = 217Ã—217
= 217^{2}
âˆš47089 = 217
(xi) 24336
Firstly letâ€™s find the prime factors for
24336 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—13Ã—13
= 2^{2}Ã—2^{2}Ã—3^{2}Ã—13^{2}
âˆš24336 = 2Ã—2Ã—3Ã—13
= 156
(xii) 190969
Firstly letâ€™s find the prime factors for
190969 = 23Ã—23Ã—19Ã—19
= 23^{2}Ã—19^{2}
âˆš190969 = 23Ã—19
= 437
(xiii) 586756
Firstly letâ€™s find the prime factors for
586756 = 2Ã—2Ã—383Ã—383
= 2^{2}Ã—383^{2}
âˆš586756 = 2Ã—383
= 766
(xiv) 27225
Firstly letâ€™s find the prime factors for
27225 = 5Ã—5Ã—3Ã—3Ã—11Ã—11
= 5^{2}Ã—3^{2}Ã—11^{2}
âˆš27225 = 5Ã—3Ã—11
= 165
(xv) 3013696
Firstly letâ€™s find the prime factors for
3013696 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—217Ã—217
= 2^{6}Ã—217^{2}
âˆš3013696 = 2^{3}Ã—217
= 1736
3.Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.
Solution:
Firstly letâ€™s find the prime factors for
180 = (2 Ã— 2) Ã— (3 Ã— 3) Ã— 5
=2^{2}Â Ã— 3^{2}Â Ã— 5
To make the unpaired 5 into paired, multiply the number with 5
180 Ã— 5 = 2^{2}Â Ã— 3^{2}Â Ã— 5^{2}
âˆ´ Square root of âˆš (180 Ã— 5) = 2Â Ã— 3Â Ã— 5
= 30
4. Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Firstly letâ€™s find the prime factors for
147 = (7 Ã— 7) Ã— 3
=7^{2}Â Ã— 3
To make the unpaired 3 into paired, multiply the number with 3
147 Ã— 3 = 7^{2}Â Ã— 3^{2}
âˆ´ Square root of âˆš (147 Ã— 3) = 7Â Ã— 3
= 21
5. Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.
Solution:
Firstly letâ€™s find the prime factors for
3645 = (3 Ã— 3) Ã— (3 Ã— 3) Ã— (3 Ã— 3) Ã— 5
=3^{2}Â Ã— 3^{2} Ã— 3^{2}Â Ã— 5
To make the unpaired 5 into paired, the number 3645 has to be divided by 5
3645 Ã· 5 = 3^{2} Ã— 3^{2} Ã— 3^{2}
âˆ´ Square root of âˆš (3645 Ã· 5) = 3Â Ã— 3 Ã— 3
= 27
6. Find the smallest number by which 1152 must be divided so that it becomes a square. Also, find the square root of the number so obtained.
Solution:
Firstly letâ€™s find the prime factors for
1152 = (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— 2 Ã— (3 Ã— 3)
=2^{2}Â Ã— 2^{2} Ã— 2^{2} Ã— 3^{2} Ã— 2
To make the unpaired 2 into paired, the number 1152 has to be divided by 2
1152 Ã· 2 = 2^{2}Â Ã— 2^{2 }Ã— 2^{2} Ã— 3^{2}
âˆ´ Square root of âˆš (1152 Ã· 2) = 2Â Ã— 2 Ã— 2 Ã— 3
= 24
7. The product of two numbers is 1296. If one number is 16 times the other, find the numbers.
Solution:
Let us consider two numbers a and b
So we know that one of the number, a =16b
a Ã— b = 1296
16b Ã— b = 1296
16b^{2} = 1296
b^{2} = 1296/16 = 81
b = 9
a = 16b
= 16(9)
= 144
âˆ´ a =144 and b =9
8. A welfare association collected Rs 202500 as donation from the residents. If each paid as many rupees as there were residents, find the number of residents.
Solution:
Let us consider total residents as a
So, each paid Rs. a
Total collection = a (a) = a^{2}
We know that the total Collection = 202500
a =Â âˆš 202500
a = âˆša(2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 5 Ã— 5 Ã— 5 Ã— 5)
= 2 Ã— 3 Ã— 3 Ã— 5 Ã— 5a
= 450
âˆ´ Total residents = 450
9. A society collected Rs 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute?
Solution:
Let us consider there were few members, each attributed a paise
a (a), i.e. total cost collected = 9216 paise
a^{2}Â = 9216
a =Â âˆš9216
= 2 Ã— 2 Ã— 2 Ã— 12
= 96
âˆ´ There were 96 members in the society and each contributed 96 paise
10. A society collected Rs 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school?
Solution:
Let us consider number of school students as a
each student contributed a paise
Total money obtained = a^{2}paise
= 2304 paise
a =Â âˆš2304
a = âˆš2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3
a = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3
a = 48
âˆ´ There were 48 students in the school
11. The area of a square field is 5184 m^{2}. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.
Solution:
Let us consider the side of square field as a
a^{2}Â = 5184 m^{2}
a =Â âˆš5184m
a = 2 Ã— 2 Ã—2 Ã— 9
= 72 m
Perimeter of square = 4a
= 4(72)
= 288 m
Perimeter of rectangle = 2 (l + b) = perimeter of the square field
= 288 m
2 (2b + b) = 288
b = 48 and l = 96
Area of rectangle = 96 Ã— 48 m^{2}
= 4608 m^{2}
12. Find the least square number, exactly divisible by each one of the numbers: (i) 6, 9, 15 and 20 (ii) 8, 12, 15 and 20
Solution:
(i) 6, 9, 15 and 20
Firstly take L.C.M for 6, 9, 15, 20 which is 180
So the prime factors of 180 = 2^{2}Â Ã— 3^{2}Â Ã— 5
To make it a perfect square, we have to multiply the number with 5
180 Ã— 5 = 2^{2}Â Ã— 3^{2}Â Ã— 5^{2}
âˆ´ 900 is the least square number divisible by 6, 9, 15 and 20
(ii) 8, 2, 15 and 20
Firstly take L.C.M for 8, 2, 15, 20 which is 360
So the prime factors of 360 = 2^{2}Â Ã— 3^{2}Â Ã—2 Ã— 5
To make it a perfect square, we have to multiply the number with 2 Ã— 5 = 10
360 Ã— 10 = 2^{2}Â Ã— 3^{2}Â Ã— 5^{2}Â Ã— 2^{2}
âˆ´ 3600 is the least square number divisible by 8, 12, 15 and 20
13. Find the square roots of 121 and 169 by the method of repeated subtraction.
Solution:
Let us find the square roots of 121 and 169 by the method of repeated subtraction
121 â€“ 1 = 120
120 â€“ 3 = 117
117 â€“ 5 = 112
112 â€“ 7 = 105
105 â€“ 9 = 96
96 â€“ 11 = 85
85 â€“ 13 = 72
72 â€“ 15 = 57
57 â€“ 17 = 40
40 â€“ 19 = 21
21 â€“ 21 = 0
Clearly, we have performed operation 11 times
âˆ´ âˆš121Â = 11
169 â€“ 1 = 168
168 â€“ 3 = 165
165 â€“ 5 = 160
160 â€“ 7 = 153
153 â€“ 9 = 144
144 â€“ 11 = 133
133 â€“ 13 = 120
120 â€“ 15 = 105
105 â€“ 17 = 88
88 â€“ 19 = 69
69 â€“ 21 = 48
48 â€“ 23 = 25
25 â€“ 25 = 0
Clearly, we have performed subtraction 13 times
âˆ´ âˆš169Â = 13
14. Write the prime factorization of the following numbers and hence find their square roots.
(i) 7744
(ii) 9604
(iii) 5929
(iv) 7056
Solution:
(i)Â 7744
Prime factors of 7744 is
7744 = 2^{2}Â Ã— 2^{2}Â Ã— 2^{2}Â Ã— 11^{2}
âˆ´ The square root of 7744 is
âˆš7744Â = 2 Ã— 2 Ã— 2 Ã— 11
= 88
(ii)Â 9604
Prime factors of 9604 is
9604 = 2^{2}Â Ã— 7^{2}Â Ã— 7^{2}
âˆ´ The square root of 9604 is
âˆš9604Â = 2 Ã— 7 Ã— 7
= 98
(iii)Â 5929
Prime factors of 5929 is
5929 = 11^{2}Â Ã— 7^{2}
âˆ´ The square root of 5929 is
âˆš5929Â = 11 Ã— 7
= 77
(iv)Â 7056
Prime factors of 7056 is
7056 = 2^{2}Â Ã— 2^{2}Â Ã— 7^{2}Â Ã— 3^{2}
âˆ´ The square root of 7056 is
âˆš7056Â = 2 Ã— 2 Ã— 7 Ã— 3
= 84
15. The students of class VIII of a school donated Rs 2401 for PMâ€™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:
Let us consider number of students as a
Each student denoted a rupee
So, total amount collected is a Ã— a rupees = 2401
a^{2}Â = 2401
a =Â âˆš2401
a = 49
âˆ´ There are 49 students in the class.
16. A PT teacher wants to arrange maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.
Solution:
Let us consider number of rows as a
No. of columns = a
Total number of students who sat in the field = a^{2}
Total students a^{2}Â + 71 = 6000
a^{2}Â = 5929
a =Â âˆš5929
a = 77
âˆ´ total number of rows are 77.
EXERCISE 3.5 PAGE NO: 3.43
1.Find the square root of each of the following by long division method:
(i) 12544 (ii) 97344
(iii) 286225 (iv) 390625
(v) 363609 (vi) 974169
(vii) 120409 (viii) 1471369
(ix) 291600 (x) 9653449
(xi) 1745041 (xii) 4008004
(xiii) 20657025 (xiv) 152547201
(xv) 20421361 (xvi)62504836
(xvii) 82264900 (xviii) 3226694416
(xix) 6407522209 (xx) 3915380329
Solution:
(i) 12544
By using long division method
âˆ´ the square root of 12544
âˆš12544 = 112
(ii) 97344
By using long division method
âˆ´ the square root of 97344
âˆš97344 = 312
(iii) 286225
By using long division method
âˆ´ the square root of 286225
âˆš286225 = 535
(iv) 390625
By using long division method
âˆ´ the square root of 390625
âˆš390625 = 625
(v) 363609
By using long division method
âˆ´ the square root of 363609
âˆš36369 = 603
(vi) 974169
By using long division method
âˆ´ the square root of 974169
âˆš974169 = 987
(vii) 120409
By using long division method
âˆ´ the square root of 120409
âˆš120409 = 347
(viii) 1471369
By using long division method
âˆ´ the square root of 1471369
âˆš1471369 = 1213
(ix) 291600
By using long division method
âˆ´ the square root of 291600
âˆš291600 = 540
(x) 9653449
By using long division method
âˆ´ the square root of 9653449
âˆš9653449 = 3107
(xi) 1745041
By using long division method
âˆ´ the square root of 1745041
âˆš1745041 = 1321
(xii) 4008004
By using long division method
âˆ´ the square root of 4008004
âˆš4008004 = 2002
(xiii) 20657025
By using long division method
âˆ´ the square root of 20657025
âˆš20657025 = 4545
(xiv) 152547201
By using long division method
âˆ´ the square root of 152547201
âˆš152547201 = 12351
(xv) 20421361
By using long division method
âˆ´ the square root of 20421361
âˆš20421361 = 4519
(xvi) 62504836
By using long division method
âˆ´ the square root of 62504836
âˆš62504836 = 7906
(xvii) 82264900
By using long division method
âˆ´ the square root of 82264900
âˆš82264900 = 9070
(xviii) 3226694416
By using long division method
âˆ´ the square root of 3226694416
âˆš3226694416 = 56804
(xix) 6407522209
By using long division method
âˆ´ the square root of 6407522209
âˆš6407522209 = 80047
(xx) 3915380329
By using long division method
âˆ´ the square root of 3915380329
âˆš3915380329 = 62573
2. Find the least number which must be subtracted from the following numbers to make them a perfect square:
(i) 2361
(ii) 194491
(iii) 26535
(iv) 161605
(v) 4401624
Solution:
(i) 2361
By using long division method
âˆ´ 57 has to be subtracted from 2361 to get a perfect square.
(ii) 194491
By using long division method
âˆ´ 10 has to be subtracted from 194491 to get a perfect square.
(iii) 26535
By using long division method
âˆ´ 291 has to be subtracted from 26535 to get a perfect square.
(iv) 161605
By using long division method
âˆ´ 1 has to be subtracted from 161605 to get a perfect square.
(v) 4401624
By using long division method
âˆ´ 20 has to be subtracted from 4401624 to get a perfect square.
3. Find the least number which must be added to the following numbers to make them a perfect square:
(i) 5607
(ii)4931
(iii) 4515600
(iv) 37460
(v) 506900
Solution:
(i) 5607
By using long division method
The remainder is 131
Since, (74)^{2}Â < 5607
We take, the next perfect square number i.e., (75)^{2}
(75)^{2}Â = 5625 > 5607
So, the number to be added = 5625 â€“ 5607 = 18
(ii) 4931
By using long division method
The remainder is 31
Since, (70)^{2}Â < 4931
We take, the next perfect square number i.e., (71)^{2}
(71)^{2}Â = 5041 > 4931
So, the number to be added = 5041 â€“ 4931 = 110
(iii) 4515600
By using long division method
The remainder is 4224
Since, (2124)^{2}Â < 4515600
We take, the next perfect square number i.e., (2125)^{2}
(2125)^{2}Â = 4515625 > 4515600
So, the number to be added = 4515625 â€“ 4515600 = 25
(iv) 37460
By using long division method
The remainder is 211
Since, (193)^{2}Â < 37460
We take, the next perfect square number i.e., (194)^{2}
(194)^{2}Â = 37636 > 37460
So, the number to be added = 37636 â€“ 37460 = 176
(v) 506900
By using long division method
The remainder is 1379
Since, (711)^{2}Â < 506900
We take, the next perfect square number i.e., (712)^{2}
(712)^{2}Â = 506944 > 506900
So, the number to be added = 506944 â€“ 506900 = 44
4. Find the greatest number of 5 digits which is a perfect square.
Solution:
We know that the greatest 5 digit number is 99999
By using long division method
The remainder is 143
So, the greatest 5 digit perfect square number is:
99999 â€“ 143 = 99856
âˆ´ 99856 is the required greatest 5 digit perfect square number.
5. Find the least number of 4 digits which is a perfect square.
Solution:
We know that the least 4 digit number is 1000
By using long division method
The remainder is 39
Since, (31)^{2}Â < 1000
We take, the next perfect square number i.e., (32)^{2}
(32)^{2}Â = 1024 > 1000
âˆ´ 1024 is the required least number 4 digit number which is a perfect square.
6. Find the least number of six digits which is a perfect square.
Solution:
We know that the least 6 digit number is 100000
By using long division method
The remainder is 144
Since, (316)^{2}Â < 100000
We take, the next perfect square number i.e., (317)^{2}
(317)^{2}Â = 100489 > 100000
âˆ´ 100489 is the required least number 6 digit number which is a perfect square.
7. Find the greatest number of 4 digits which is a perfect square.
Solution:
We know that the greatest 4 digit number is 9999
By using long division method
The remainder is 198
So, the greatest 4 digit perfect square number is:
9999 â€“ 198 = 9801
âˆ´ 9801 is the required greatest 4 digit perfect square number.
8. A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8160, find the number of soldiers in each row
Solution:
We know that the total number of soldiers = 8160
Number of soldiers left out = 60
Number of soldiers arranged in rows to form a perfect square = 8160 â€“ 60 = 8100
âˆ´ number of soldiers in each row =Â âˆš8100
=Â âˆš (9Ã—9Ã—10Ã—10)
= 9Ã—10
= 90
9. The area of a square field is 60025m^{2}. A man cycles along its boundary at 18 Km/hr. In how much time will he return at the starting point?
Solution:
We know that the area of square field = 60025 m^{2}
Speed of cyclist = 18 km/h
= 18 Ã—Â (1000/60Ã—60)
= 5 m/s^{2}
Area = 60025 m^{2}
Side^{2}Â = 60025
Side =Â âˆš60025
= 245
We know, Total length of boundary = 4 Ã— Side
= 4 Ã— 245
= 980 m
âˆ´ Time taken to return to the starting point =Â 980/5
= 196 seconds
= 3 minutes 16 seconds
10. The cost of levelling and turning a square lawn at Rs 2.50 per m^{2}Â is Rs13322.50 Find the cost of fencing it at Rs 5 per metre.
Solution:
We know that the cost of levelling and turning a square lawn = 2.50 per m^{2}
Total cost of levelling and turning = Rs. 13322.50
Total area of square lawn =Â 13322.50/2.50
= 5329 m^{2}
Side^{2}Â = 5329
Side of square lawn =Â âˆš5329
= 73 m
So, total length of lawn = 4 Ã— 73
= 292 m
âˆ´ Cost of fencing the lawn at Rs 5 per metre = 292 Ã— 5
= Rs. 1460
11. Find the greatest number of three digits which is a perfect square.
Solution:
We know that the greatest 3 digit number is 999
By using long division method
The remainder is 38
So, the greatest 3 digit perfect square number is:
999 â€“ 38 = 961
âˆ´ 961 is the required greatest 3 digit perfect square number.
12. Find the smallest number which must be added to 2300 so that it becomes a perfect square.
Solution:
By using long division method letâ€™s find the square root of 2300
The remainder is 91
Since, (47)^{2}Â < 2300
We take, the next perfect square number i.e., (48)^{2}
(48)^{2}Â = 2304 > 2300
âˆ´ The smallest number required to be added to 2300 to get a perfect square is
2304 â€“ 2300 = 4
EXERCISE 3.6 PAGE NO: 3.48
1. Find the square root of:
(i) 441/961
(ii) 324/841
(iii) 4 29/29
(iv) 2 14/25
(v) 2 137/196
(vi) 23 26/121
(vii) 25 544/729
(viii) 75 46/49
(ix) 3 942/2209
(x) 3 334/3025
(xi) 21 2797/3364
(xii) 38 11/25
(xiii) 23 394/729
(xiv) 21 51/169
(xv) 10 151/225
Solution:
(i) 441/961
The square root of
âˆš441/961 = 21/31
(ii) 324/841
The square root of
âˆš324/841= 18/29
(iii) 4 29/29
The square root of
âˆš(4 29/29) = âˆš(225/49) = 15/7
(iv) 2 14/25
The square root of
âˆš(2 14/25) = âˆš(64/25) = 8/5
(v) 2 137/196
The square root of
âˆš2 137/196 = âˆš (529/196) = 23/14
(vi) 23 26/121
The square root of
âˆš(23 26/121) = âˆš(2809/121) = 53/11
(vii) 25 544/729
The square root of
âˆš(25 544/729) = âˆš(18769/729) = 137/27
(viii) 75 46/49
The square root of
âˆš(75 46/49) = âˆš(3721/49) = 61/7
(ix) 3 942/2209
The square root of
âˆš(3 942/2209) = âˆš(7569/2209) = 87/47
(x) 3 334/3025
The square root of
âˆš(3 334/3025) = âˆš(9409/3025) = 97/55
(xi) 21 2797/3364
The square root of
âˆš(21 2797/3364) = âˆš(73441/3364) = 271/58
(xii) 38 11/25
The square root of
âˆš(38 11/25) = âˆš(961/25) = 31/5
(xiii) 23 394/729
The square root of
âˆš(23 394/729) = âˆš(17161/729) = 131/27 = 4 23/27
(xiv) 21 51/169
The square root of
âˆš(21 51/169) = âˆš(3600/169) = 60/13 = 4 8/13
(xv) 10 151/225
The square root of
âˆš(10 151/225) = âˆš(2401/225) = 49/15 = 3 4/15
2. Find the value of:
(i) âˆš80/âˆš405
(ii) âˆš441/âˆš625
(iii) âˆš1587/âˆš1728
(iv) âˆš72 Ã—âˆš338
(v) âˆš45 Ã— âˆš20
Solution:
(i) âˆš80/âˆš405
âˆš80/âˆš405 = âˆš16/âˆš81 = 4/9
(ii) âˆš441/âˆš625
âˆš441/âˆš625 = 21/25
(iii) âˆš1587/âˆš1728
âˆš1587/âˆš1728 = âˆš529/âˆš576 = 23/24
(iv) âˆš72 Ã—âˆš338
âˆš72 Ã—âˆš338 = âˆš(2Ã—2Ã—2Ã—3Ã—3) Ã—âˆš(2Ã—13Ã—13)
By using the formula âˆša Ã— âˆšb = âˆš(aÃ—b)
= âˆš(2Ã—2Ã—2Ã—3Ã—3Ã—2Ã—13Ã—13)
= 2^{2} Ã— 3 Ã— 13
= 156
(v) âˆš45 Ã— âˆš20
âˆš45 Ã— âˆš20 = âˆš(5Ã—3Ã—3) Ã— âˆš(5Ã—2Ã—2)
By using the formula âˆša Ã— âˆšb = âˆš(aÃ—b)
= âˆš(5Ã—3Ã—3Ã—5Ã—2Ã—2)
= 5 Ã— 3 Ã— 2
= 30
3. The area of a square field is 80 244/729Â square metres. Find the length of each side of the field.
Solution:
We know that the given area = 80 244/729Â m^{2}
=Â 58564/729Â m^{2}
If L is length of each side
L^{2}Â =Â 58564/729
L =Â âˆš (58564/729) = âˆš58564/âˆš729
=Â 242/27
= 8 26/27
âˆ´ Length isÂ 8 26/27
4. The area of a square field isÂ 30 1/4m^{2}. Calculate the length of the side of the square.
Solution:
We know that the given area = 30 1/4Â m^{2}
=Â 121/4Â m^{2}
If L is length of each side then,
L^{2}Â =Â 121/4
L = âˆš(121/4)Â Â =Â âˆš121/âˆš4
=Â 11/2
âˆ´ Length isÂ 11/2
5. Find the length of a side of a square playground whose area is equal to the area of a rectangular field of dimensions 72m and 338 m.
Solution:
By using the formula
Area of rectangular field = l Ã— b
= 72 Ã— 338 m^{2}
= 24336 m^{2}
Area of square, L^{2}Â = 24336 m^{2}
L =Â âˆš24336
= 156 m
âˆ´ Length of side of square playground is 156 m.
EXERCISE 3.7 PAGE NO: 3.52
Find the square root of the following numbers in decimal form:
1. 84.8241
Solution:
By using long division method
âˆ´ the square root of 84.8241
âˆš84.8241 = 9.21
2. 0.7225
Solution:
By using long division method
âˆ´ the square root of 0.7225
âˆš0.7225 = 0.85
3. 0.813604
Solution:
By using long division method
âˆ´ the square root of 0.813604
âˆš0.813604 = 0.902
4. 0.00002025
Solution:
By using long division method
âˆ´ the square root of 0.00002025
âˆš0.00002025 = 0.0045
5. 150.0625
Solution:
By using long division method
âˆ´ the square root of 150.0625
âˆš150.0625 = 12.25
6. 225.6004
Solution:
By using long division method
âˆ´ the square root of 225.6004
âˆš225.6004 = 15.02
7. 3600.720036
Solution:
By using long division method
âˆ´ the square root of 3600.720036
âˆš3600.720036 = 60.006
8. 236.144689
Solution:
By using long division method
âˆ´ the square root of 236.144689
âˆš236.144689 = 15.367
9. 0.00059049
Solution:
By using long division method
âˆ´ the square root of 0.00059049
âˆš0.00059049 = 0.0243
10. 176.252176
Solution:
By using long division method
âˆ´ the square root of 176.252176
âˆš176.252176 = 13.276
11. 9998.0001
Solution:
By using long division method
âˆ´ the square root of 9998.0001
âˆš9998.0001 = 99.99
12. 0.00038809
Solution:
By using long division method
âˆ´ the square root of 0.00038809
âˆš0.00038809 = 0.0197
13. What is that fraction which when multiplied by itself gives 227.798649?
Solution:
Let us consider a number a
Where, a = âˆš227.798649
= 15.093
By using long division method let us verify
âˆ´ 15.093 is the fraction which when multiplied by itself gives 227.798649.
14. The area of a square playground is 256.6404 square meter. Find the length of one side of the playground.
Solution:
We know that the given area of a square playground = 256.6404
i.e., L^{2} = 256.6404 m^{2}
L = âˆš256.6404
= 16.02m
By using long division method let us verify
âˆ´ length of one side of the playground is 16.02m.
15. What is the fraction which when multiplied by itself gives 0.00053361?
Solution:
Let us consider a number a
Where, a = âˆš0.00053361
= 0.0231
By using long division method let us verify
âˆ´ 0.0231 is the fraction which when multiplied by itself gives 0.00053361.
16. Simplify:
(i) (âˆš59.29 – âˆš5.29)/ (âˆš59.29 + âˆš5.29)
(ii) (âˆš0.2304 + âˆš0.1764)/ (âˆš0.2304 – âˆš0.1764)
Solution:
(i) (âˆš59.29 – âˆš5.29)/ (âˆš59.29 + âˆš5.29)
Firstly let us find the square root âˆš59.29 and âˆš5.29
âˆš59.29 = âˆš5929/ âˆš100
= 77/10
= 7.7
âˆš5.29 = âˆš5.29/ âˆš100
= 23/10
= 2.3
So, (7.7 â€“ 2.3)/ (7.7 + 2.3)
= 54/10
= 0.54
(ii) (âˆš0.2304 + âˆš0.1764)/ (âˆš0.2304 – âˆš0.1764)
Firstly let us find the square root âˆš0.2304 and âˆš0.1764
âˆš0.2304 = âˆš2304/ âˆš10000
= 48/100
= 0.48
âˆš0.1764 = âˆš1764/ âˆš10000
= 42/100
= 0.42
So, (0.48 + 0.42)/ (0.48 â€“ 0.42)
= 0.9/0.06
= 15
17. Evaluate âˆš50625Â and hence find the value of âˆš506.25 + âˆš5.0625
Solution:
By using long division method let us find the âˆš50625Â
So now, âˆš506.25 = âˆš50625/ âˆš100
= 225/10
= 22.5
âˆš5.0625 = âˆš50625/ âˆš10000
= 225/100
= 2.25
So equating in the above equation we get,
âˆš506.25 + âˆš5.0625 = 22.5 + 2.25
= 24.75
18. Find the value of âˆš103.0225Â and hence find the value of
(i) âˆš10302.25
(ii)Â âˆš1.030225
Solution:
By using long division method let us find the
âˆš103.0225 = âˆš(1030225/10000) = âˆš1030225/âˆš10000
So now, (i)âˆš10302.25 = âˆš(1030225/ 100)
= 1015/ 10
= 101.5
(ii)âˆš1.030225 = âˆš1030225/ âˆš1000000
= 1015/1000
= 1.015
EXERCISE 3.8 PAGE NO: 3.56
1. Find the square root of each of the following correct to three places of decimal.
(i) 5 (ii) 7
(iii) 17 (iv) 20
(v) 66 (vi) 427
(vii) 1.7 (viii) 23.1
(ix) 2.5 (x) 237.615
(xi) 15.3215 (xii) 0.9
(xiii) 0.1 (xiv) 0.016
(xv) 0.00064 (xvi) 0.019
(xvii)Â 7/8Â (xviii)Â 5/12
(xix)Â 2 1/2Â (xx)Â 287 5/8
Solution:
(i) 5
By using long division method
âˆ´ the square root of 5 is 2.236
(ii) 7
By using long division method
âˆ´ the square root of 7 is 2.646
(iii) 17
By using long division method
âˆ´ the square root of 17 is 4.123
(iv) 20
By using long division method
âˆ´ the square root of 20 is 4.472
(v) 66
By using long division method
âˆ´ the square root of 66 is 8.124
(vi) 427
By using long division method
âˆ´ the square root of 427 is 20.664
(vii) 1.7
By using long division method
âˆ´ the square root of 1.7 is 1.304
(viii) 23.1
By using long division method
âˆ´ the square root of 23.1 is 4.806
(ix) 2.5
By using long division method
âˆ´ the square root of 2.5 is 1.581
(x) 237.615
By using long division method
âˆ´ the square root of 237.615 is 15.415
(xi) 15.3215
By using long division method
âˆ´ the square root of 15.3215 is 3.914
(xii) 0.9
By using long division method
âˆ´ the square root of 0.9 is 0.949
(xiii) 0.1
By using long division method
âˆ´ the square root of 0.1 is 0.316
(xiv) 0.016
By using long division method
âˆ´ the square root of 0.016 is 0.126
(xv) 0.00064
By using long division method
âˆ´ the square root of 0.00064 is 0.025
(xvi) 0.019
By using long division method
âˆ´ the square root of 0.019 is 0.138
(xvii)Â 7/8
By using long division method
âˆ´ the square root of 7/8 is 0.935
(xviii)Â 5/12
By using long division method
âˆ´ the square root of 5/12 is 0.645
(xix)Â 2 1/2Â
By using long division method
âˆ´ the square root of 5/2 is 1.581
(xx)Â 287 5/8
By using long division method
âˆ´ the square root of 2301/8 is 16.960
2. Find the square root of 12.0068 correct to four decimal places.
Solution:
By using long division method
âˆ´ the square root of 12.0068 is 3.4651
3. Find the square root of 11 correct to five decimal places.
Solution:
By using long division method
âˆ´ the square root of 11 is 3.31662
4. Give that: âˆš2 = 1.414, âˆš3 = 1.732, âˆš5 = 2.236 and âˆš7 = 2.646, evaluate each of the following:
(i)Â âˆš (144/7)
(ii)Â âˆš (2500/3)
Solution:
(i)Â âˆš (144/7)
Now let us simplify the given equation
âˆš (144/7) = âˆš (12Ã—12)/ âˆš7
= 12/ 2.646
= 4.535
(ii)Â âˆš (2500/3)
Now let us simplify the given equation
âˆš (2500/3) = âˆš (5Ã—5Ã—10Ã—10) /âˆš3
= 5Ã—10/ 1.732
= 50/1.732
= 28.867
5. Given that âˆš2 = 1.414, âˆš3 = 1.732, âˆš5 = 2.236 and âˆš7 = 2.646Â find the square roots of the following:
(i)Â 196/75
(ii)Â 400/63
(iii)Â 150/7
(iv)Â 256/5
(v)Â 27/50
Solution:
(i)Â 196/75
Let us find the square root for196/75
âˆš(196/75) = âˆš(196)/ âˆš (75)
= âˆš(14Ã—14)/ âˆš (5Ã—5Ã—3)
= 14/ (5âˆš3)
= 14/ (5Ã—1.732)
= 14/8.66
= 1.617
(ii)Â 400/63
Let us find the square root for400/63
âˆš(400/63) = âˆš(400)/ âˆš (63)
= âˆš(20Ã—20)/ âˆš (3Ã—3Ã—7)
= 20/ (3âˆš7)
= 20/ (3Ã—2.646)
= 20/7.938
= 2.520
(iii)Â 150/7
Let us find the square root for150/7
âˆš(150/7) = âˆš(150)/ âˆš (7)
= âˆš(3Ã—5Ã—5Ã—2)/ âˆš (7)
= (5âˆš3Ã—âˆš2)/ (âˆš7)
= 5Ã—1.732Ã—1.414/ (2.646)
= 12.245/2.646
= 4.628
(iv)Â 256/5
Let us find the square root for256/5
âˆš(256/5) = âˆš(256)/ âˆš (5)
= âˆš(16Ã—16)/ âˆš (5)
= 16/ (âˆš5)
= 16/2.236
= 7.155
(v)Â 27/50
Let us find the square root for27/50
âˆš(27/50) = âˆš(27)/ âˆš (50)
= âˆš(3Ã—3Ã—3)/ âˆš (5Ã—5Ã—2)
= (3âˆš3)/ (5âˆš2)
= (3Ã—1.732)/ (5Ã—1.414)
= 5.196/7.07
= 0.735
EXERCISE 3.9 PAGE NO: 3.61
Square Root Table |
|||||||||
Number |
Square Root(âˆš) |
Number |
Square Root(âˆš) |
Number |
Square Root(âˆš) |
Number |
Square Root(âˆš) |
Number |
Square Root(âˆš) |
1 |
1 |
21 |
4.583 |
41 |
6.403 |
61 |
7.81 |
81 |
9 |
2 |
1.414 |
22 |
4.69 |
42 |
6.481 |
62 |
7.874 |
82 |
9.055 |
3 |
1.732 |
23 |
4.796 |
43 |
6.557 |
63 |
7.937 |
83 |
9.11 |
4 |
2 |
24 |
4.899 |
44 |
6.633 |
64 |
8 |
84 |
9.165 |
5 |
2.236 |
25 |
5 |
45 |
6.708 |
65 |
8.062 |
85 |
9.22 |
6 |
2.449 |
26 |
5.099 |
46 |
6.782 |
66 |
8.124 |
86 |
9.274 |
7 |
2.646 |
27 |
5.196 |
47 |
6.856 |
67 |
8.185 |
87 |
9.327 |
8 |
2.828 |
28 |
5.292 |
48 |
6.928 |
68 |
8.246 |
88 |
9.381 |
9 |
3 |
29 |
5.385 |
49 |
7 |
69 |
8.307 |
89 |
9.434 |
10 |
3.162 |
30 |
5.477 |
50 |
7.071 |
70 |
8.367 |
90 |
9.487 |
11 |
3.317 |
31 |
5.568 |
51 |
7.141 |
71 |
8.426 |
91 |
9.539 |
12 |
3.464 |
32 |
5.657 |
52 |
7.211 |
72 |
8.485 |
92 |
9.592 |
13 |
3.606 |
33 |
5.745 |
53 |
7.28 |
73 |
8.544 |
93 |
9.644 |
14 |
3.742 |
34 |
5.831 |
54 |
7.348 |
74 |
8.602 |
94 |
9.695 |
15 |
3.873 |
35 |
5.916 |
55 |
7.416 |
75 |
8.66 |
95 |
9.747 |
16 |
4 |
36 |
6 |
56 |
7.483 |
76 |
8.718 |
96 |
9.798 |
17 |
4.123 |
37 |
6.083 |
57 |
7.55 |
77 |
8.775 |
97 |
9.849 |
18 |
4.243 |
28 |
6.164 |
58 |
7.616 |
78 |
8.832 |
98 |
9.899 |
19 |
4.359 |
29 |
6.245 |
59 |
7.681 |
79 |
8.888 |
99 |
9.95 |
20 |
4.472 |
40 |
6.325 |
60 |
7.746 |
80 |
8.944 |
100 |
10 |
Using square root table, find the square roots of the following:
1. 7
Solution:
From square root table we know,
Square root of 7 is:
âˆš7 = 2.645
âˆ´ The square root of 7 is 2.645
2. 15
Solution:
We know that,
15 = 3 x 5
So, âˆš15 = âˆš3 x âˆš5
From square root table we know,
Square root of 3 and 5 are:
âˆš3 = 1.732 and âˆš5 = 2.236
â‡’ âˆš15 = 1.732 x 2.236 = 3.873
âˆ´ The square root of 15 is 3.873
3. 74
Solution:
We know that,
74 = 2 x 37
So, âˆš74 = âˆš2 x âˆš37
From square root table we know,
Square root of 2 and 37 are:
âˆš2 = 1.414 and âˆš37 = 6.083
â‡’ âˆš74 = 1.414 x 6.083 = 8.602
âˆ´ The square root of 74 is 8.602
4. 82
Solution:
We know that,
82 = 2 x 41
So, âˆš82 = âˆš2 x âˆš41
From square root table we know,
Square root of 2 and 41 are:
âˆš2 = 1.414 and âˆš41 = 6.403
â‡’ âˆš82 = 1.414 x 6.403 = 9.055
âˆ´ The square root of 82 is 9.055
5. 198
Solution:
We know that,
198 = 2 x 9 x 11
So, âˆš198 = âˆš2 x âˆš9 x âˆš11
From square root table we know,
Square root of 2, 9 and 11 are:
âˆš2 = 1.414, âˆš9 = 3 and âˆš11 = 3.317
â‡’ âˆš198 = 1.414 x 3 x 3.317 = 14.071
âˆ´ The square root of 198 is 14.071
6. 540
Solution:
We know that,
540 = 6 x 9 x 10
So, âˆš540 = âˆš6 x âˆš9 x âˆš10
From square root table we know,
Square root of 6, 9 and 10 are:
âˆš6 = 2.449, âˆš9 = 3 and âˆš10 = 3.162
â‡’ âˆš540 = 2.449 x 3 x 3.162 = 23.24
âˆ´ The square root of 540 is 23.24
7. 8700
Solution:
We know that,
8700 = 87 x 100
So, âˆš8700 = âˆš87 x âˆš100
From square root table we know,
Square root of 87 and 100 are:
âˆš8700 = 9.327 and âˆš100 = 10
â‡’ âˆš8700 = 9.327 x 10 = 93.27
âˆ´ The square root of 8700 is 93.27
8. 3509
Solution:
We know that,
3509 = 121 x 29
So, âˆš3509 = âˆš121 x âˆš29
From square root table we know,
Square root of 121 and 29 are:
âˆš121 = 11 and âˆš29 = 5.385
â‡’ âˆš3509 = 11 x 10 = 5.385
âˆ´ The square root of 3509 is 59.235
9. 6929
Solution:
We know that,
6929 = 169 x 41
So, âˆš6929 = âˆš169 x âˆš41
From square root table we know,
Square root of 169 and 41 are:
âˆš169 = 13 and âˆš41 = 6.403
â‡’ âˆš6929 = 13 x 6.403 = 83.239
âˆ´ The square root of 6929 is 83.239
10. 25725
Solution:
We know that,
25725 = 3 x 7 x 25 x 49
So, âˆš25725 = âˆš3 x âˆš7 x âˆš25 x âˆš49
From square root table we know,
Square root of 3, 7, 25 and 49 are:
âˆš3 = 1.732, âˆš7 = 2.646, âˆš25 = 5 and âˆš49 = 7
â‡’ âˆš25725 = 1.732 x 2.646 x 5 x 7 = 160.41
âˆ´ The square root of 25725 is 160.41
11. 1312.
Solution:
We know that,
1312 = 2 x 16 x 41
So, âˆš1312 = âˆš2 x âˆš16 x âˆš41
From square root table we know,
Square root of 2, 16 and 41 are:
âˆš2 = 1.414, âˆš16 = 4 and âˆš41 = 6.403
â‡’ âˆš1312 = 1.414 x 4 x 6.403 = 36.22
âˆ´ The square root of 1312 is 36.22
12. 4192
Solution:
We know that,
4192 = 2 x 16 x 131
So, âˆš4192 = âˆš2 x âˆš16 x âˆš131
From square root table we know,
Square root of 2 and16 are:
âˆš2 = 1.414 and âˆš16 = 4
The square root of 131 is not listed in the table
Thus, letâ€™s apply long division to find it
So, square root of 131 is 11.445
Now,
â‡’ âˆš4192 = 1.414 x 4 x 11.445 = 64.75
âˆ´ The square root of 4192 is 64.75
13. 4955
Solution:
We know that,
4955 = 5 x 991
So, âˆš4955 = âˆš5 x âˆš991
From square root table we know,
Square root of 5 is:
âˆš5 = 2.236
The square root of 991 is not listed in the table
Thus, letâ€™s apply long division to find it
So, square root of 991 is 31.480
Now,
â‡’ âˆš4955 = 2.236 x 31.480 = 70.39
âˆ´ The square root of 4955 is 70.39
14. 99/144
Solution:
We know that,
99/144 = (9 x 11) / (12 x 12)
So, âˆš(99/144) = âˆš[(9 x 11) x (12 x 12)]
= 3/12 x âˆš11
From square root table we know,
Square root of 11 is:
âˆš11 = 3.317
â‡’ âˆš(99/144) = 3/12 x 3.317 = 3.317/4 = 0.829
âˆ´ The square root of 99/144 is 0.829
15. 57/169
Solution:
We know that,
57/169 = (3 x 19) / (13 x 13)
So, âˆš(57/169) = âˆš[(3 x 19) x (13 x 13)]
= âˆš3 x âˆš19 x 1/13
From square root table we know,
Square root of 3 and 19 is:
âˆš3 = 1.732 and âˆš19 = 4.359
â‡’ âˆš(57/169) = 1.732 x 4.359 x 1/13 = 0.581
âˆ´ The square root of 57/169 is 0.581
16. 101/169
Solution:
We know that,
101/169 = 101 / (13 x 13)
So, âˆš(101/169) = âˆš[101 / (13 x 13)]
= âˆš101/13
From square root table we donâ€™t have the square root of 101
Thus, we have to manipulate the number such that we get the square root of a number less than 100
âˆš101 = âˆš(1.01 x 100)
= âˆš1.01 x 10
Now, we have to find the square of 1.01
We know that,
âˆš1 = 1 and âˆš2 = 1.414 (From the square root table)
Their difference = 1.414 â€“ 1 = 0.414
Hence, for a difference of 1 (2 – 1), the difference in the value of the square root is 0.414
So,
For the difference of 0.01, the difference in the value of the square roots will be
0.01 x 1.414 = 0.00414
âˆ´âˆš1.01 = 1 + 0.00414 = 1.00414
Then, âˆš101 = 1.00414 x 10 = 10.0414
â‡’ âˆš(101/169) = âˆš101/13 = 10.0414/13
âˆ´ The square root of 101/169 is 0.773
17. 13.21
Solution:
We need to find âˆš13.21
From square root table we know,
Square root of 13 and 14 are:
âˆš13 = 3.606 and âˆš14 = 3.742
Their difference = 3.742 â€“ 3.606 = 0.136
Hence, for a difference of 1 (14 – 13), the difference in the value of the square root is 0.136
So,
For the difference of 0.21, the difference in the value of the square roots will be
0.136 x 0.21 = 0.0286
â‡’ âˆš13.21 = 3.606 + 0.0286 = 3.635
âˆ´ The square root of 13.21 is 3.635
18. 21.97
Solution:
We need to find âˆš21.97
From square root table we know,
Square root of 21 and 22 are:
âˆš21 = 4.583 and âˆš22 = 4.690
Their difference = 4.690 â€“ 4.583 = 0.107
Hence, for a difference of 1 (23 – 22), the difference in the value of the square root is 0.107
So,
For the difference of 0.97, the difference in the value of the square roots will be
0.107 x 0.97 = 0.104
â‡’ âˆš21.97 = 4.583 + 0.104 = 4.687
âˆ´ The square root of 21.97 is 4.687
19. 110
Solution:
We know that,
110 = 11 x 10
So, âˆš110 = âˆš11 x âˆš10
From square root table we know,
Square root of 11 and 10 are:
âˆš11 = 3.317 and âˆš10 = 3.162
â‡’ âˆš110 = 3.317 x 3.162 = 10.488
âˆ´ The square root of 110 is 10.488
20. 1110
Solution:
We know that,
1110 = 37 x 30
So, âˆš1110 = âˆš37 x âˆš30
From square root table we know,
Square root of 37 and 30 are:
âˆš37 = 6.083 and âˆš30 = 5.477
â‡’ âˆš1110 = 6.083 x 5.477 = 33.317
âˆ´ The square root of 1110 is 33.317
21. 11.11
Solution:
We need to find âˆš11.11
From square root table we know,
Square root of 11 and 12 are:
âˆš11 = 3.317 and âˆš12 = 3.464
Their difference = 3.464 â€“ 3.317 = 0.147
Hence, for a difference of 1 (12 – 11), the difference in the value of the square root is 0.147
So,
For the difference of 0.11, the difference in the value of the square roots will be
0.11 x 0.147 = 0.01617
â‡’ âˆš11.11 = 3.317 + 0.0162 = 3.333
âˆ´ The square root of 11.11 is 3.333
22. The area of a square field is 325m^{2}. Find the approximate length of one side of the field.
Solution:
We know that the given area of the field = 325 m^{2}
To find the approximate length of the side of the field we will have to calculate the square root of 325
âˆš325 = âˆš25 x âˆš13
From the square root table, we know
âˆš25 = 5 and âˆš13 = 3.606
â‡’ âˆš325 = 5 x 3.606 = 18.030
âˆ´ The approximate length of one side of the field is 18.030 m
23. Find the length of a side of a square, whose area is equal to the area of a rectangle with sides 240 m and 70 m.
Solution:
We know that from the question,
Area of square = Area of rectangle
Side^{2} = 240 Ã— 70
Side = âˆš(240 Ã— 70)
= âˆš(10 Ã— 10 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 7)
= 20âˆš42
Now, from the square root table, we know âˆš42 = 6.481
= 20 Ã— 6.48
= 129.60 m
âˆ´ The length of side of the square is 129.60 m
RD Sharma solutions For Class 8 Maths Chapter 3 – Squares and Square Roots
Chapter 3, Squares and Square Roots contains nine exercises. RD Sharma Solutions are given here which include answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.
- Perfect Square or Square numbers.
- Properties and patterns of some Square numbers.
- Product of two consecutive odd or consecutive even numbers.
- Column method for squaring two-digit numbers.
- Diagonal method for squaring a number.
- Definition of square roots.
- The square root of a perfect square by the prime factorisation method.
- The square root of a perfect square by the long division method.
- Square roots of a number in decimal form.
- The square root of fractions.
Chapter Brief of RD Sharma Solutions for Class 8 Maths Chapter 3 – Squares and Square roots
The RD Sharma Solutions for Class 8 Maths Chapter 3 â€“ Squares and Square Roots deal with the properties of perfect squares, which in turn help to solve problems easily and quickly. By learning these concepts thoroughly, we can find squares and square roots of the given number without using a calculator.