Class 8 - Mathematics
Squares and Square Roots - Exercise 6.3
Top Block 1
Question: 1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers:
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Answer :
Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are:
(i) 1 (ii) 6 (iii) 1 (iv) 5
Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are:
(i) 1 (ii) 6 (iii) 1 (iv) 5
Question: 2. Without doing any calculation, find the numbers which are surely not perfect squares:
(i) 153
(ii) 257
(iii) 408
(iv) 441
(i) 153
(ii) 257
(iii) 408
(iv) 441
Answer :
Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.
(i) But given number 153 has its unit digit 3. So it is not a perfect square number.
(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.
(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.
(iv) Given number 441 has its unit digit 1. So it would be a perfect square number
Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.
(i) But given number 153 has its unit digit 3. So it is not a perfect square number.
(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.
(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.
(iv) Given number 441 has its unit digit 1. So it would be a perfect square number
Question: 3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Answer :
By successive subtracting odd natural numbers from 100,
100 – 1 = 99
99 – 3 = 96
96 – 5 = 91
91 – 7 = 84
84 – 9 = 75
75 – 11 = 64
64 – 13 = 51
51 – 15 = 36
36 – 17 = 19
19 – 19 = 0
This successive subtraction is completed in 10 steps.
Therefore √100 = 10
By successive subtracting odd natural numbers from 169,
169 – 1 = 168
168 – 3 = 165
165 – 5 = 160
160 – 7 = 153
153 – 9 = 144
144 – 11 = 133
133 – 13 = 120
120 – 15 = 105
105 – 17 = 88
88 – 19 = 69
69 – 21 = 48
48 – 23 = 25
25 – 25 = 0
This successive subtraction is completed in 13 steps.
Therefore √169 = 13
By successive subtracting odd natural numbers from 100,
100 – 1 = 99
99 – 3 = 96
96 – 5 = 91
91 – 7 = 84
84 – 9 = 75
75 – 11 = 64
64 – 13 = 51
51 – 15 = 36
36 – 17 = 19
19 – 19 = 0
This successive subtraction is completed in 10 steps.
Therefore √100 = 10
By successive subtracting odd natural numbers from 169,
169 – 1 = 168
168 – 3 = 165
165 – 5 = 160
160 – 7 = 153
153 – 9 = 144
144 – 11 = 133
133 – 13 = 120
120 – 15 = 105
105 – 17 = 88
88 – 19 = 69
69 – 21 = 48
48 – 23 = 25
25 – 25 = 0
This successive subtraction is completed in 13 steps.
Therefore √169 = 13
Question: 4. Find the square roots of the following numbers by the Prime Factorization method:
(i) 729
(ii) 400
(iii) 1764
(iv) 4096
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100
(i) 729
(ii) 400
(iii) 1764
(iv) 4096
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100
Answer :
(i) 729
√729 = √3 x 3 x 3 x 3 x 3 x 3
= 3 x 3 x 3
= 27
(i) 729
√729 = √3 x 3 x 3 x 3 x 3 x 3
= 3 x 3 x 3
= 27
(ii) 400
√400 = √2 x 2 x 2 x 2 x 5 x 5
= 2 x 2 x 5
= 20
√400 = √2 x 2 x 2 x 2 x 5 x 5
= 2 x 2 x 5
= 20
Mddle block 1
(iii) 1764
√1764 = √2 x 2 x 3 x 3 x 7 x 7
= 2 x 3 x 7
= 42
√1764 = √2 x 2 x 3 x 3 x 7 x 7
= 2 x 3 x 7
= 42
(iv) 4096
√4096 = √2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 2 x 2 x 2 x 2 x 2 x 2
= 64
√4096 = √2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= 2 x 2 x 2 x 2 x 2 x 2
= 64
(v) 7744
√7744 = √2 x 2 x 2 x 2 x 2 x 2 x 11 x 11
= 2 x 2 x 2 x 11
= 88
√7744 = √2 x 2 x 2 x 2 x 2 x 2 x 11 x 11
= 2 x 2 x 2 x 11
= 88
(vi) 9604
√9604 = √2 x 2 x 7 x 7 x 7 x 7
= 2 x 7 x 7
= 98
√9604 = √2 x 2 x 7 x 7 x 7 x 7
= 2 x 7 x 7
= 98
(vii) 5929
√5929 = √7 x 7 x 11 x 11
= 7 x 11
= 77
√5929 = √7 x 7 x 11 x 11
= 7 x 11
= 77
(viii) 9216
√9216 = √2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
= 2 x 2 x 2 x 2 x 2 x 3
= 96
√9216 = √2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
= 2 x 2 x 2 x 2 x 2 x 3
= 96
(ix) 529
√529 = √23 x 23
= 23
√529 = √23 x 23
= 23
(x) 8100
√8100 = √2 x 2 x 3 x 3 x 3 x 3 x 5 x 5
= 2 x 3 x 3 x 5
= 90
√8100 = √2 x 2 x 3 x 3 x 3 x 3 x 5 x 5
= 2 x 3 x 3 x 5
= 90
Question: 5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained:
(i) 252 (ii) 180
(iii) 1008 (iv) 2028
(v) 1458 (vi) 768
(i) 252 (ii) 180
(iii) 1008 (iv) 2028
(v) 1458 (vi) 768
Answer :
(i) 252 = 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square.
∴ 252 x 7 = 1764
And (i)√1764 = 2 x 3 x 7 = 42
(i) 252 = 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square.
∴ 252 x 7 = 1764
And (i)√1764 = 2 x 3 x 7 = 42
(ii) 180 = 2 x 2 x 3 x 3 x 5
Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.
∴ 180 x 5 = 900
And √900 = 2 x 3 x 5 = 30
Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.
∴ 180 x 5 = 900
And √900 = 2 x 3 x 5 = 30
(iii) 1008 = 2 x 2 x 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 1008 must be multiplied by 7 to make it a perfect square.
∴1008 x 7 = 7056
And √7056 = 2 x 2 x 3 x 7 = 84
Here, prime factor 7 has no pair. Therefore 1008 must be multiplied by 7 to make it a perfect square.
∴1008 x 7 = 7056
And √7056 = 2 x 2 x 3 x 7 = 84
(iv) 2028 = 2 x 2 x 3 x 13 x 13
Here, prime factor 3 has no pair. Therefore 2028 must be multiplied by 3 to make it a perfect square.
∴ 2028 x 3 = 6084
And √6084 = 2 x 2 x 3 x 3 x 13 x 13
= 78
Here, prime factor 3 has no pair. Therefore 2028 must be multiplied by 3 to make it a perfect square.
∴ 2028 x 3 = 6084
And √6084 = 2 x 2 x 3 x 3 x 13 x 13
= 78
(v) 1458 = 2 x 3 x 3 x 3 x 3 x 3 x 3
Here, prime factor 2 has no pair. Therefore 1458 must be multiplied by 2 to make it a perfect square.
∴ 1458 x 2 = 2916
And √2916 = 2 x 3 x 3 x 3 = 54
Here, prime factor 2 has no pair. Therefore 1458 must be multiplied by 2 to make it a perfect square.
∴ 1458 x 2 = 2916
And √2916 = 2 x 3 x 3 x 3 = 54
(vi) 768 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3
Here, prime factor 3 has no pair. Therefore 768 must be multiplied by 3 to make it a perfect square.
∴ 768 x 3 = 2304
And √2304 = 2 x 2 x 2 x 2 x 3
= 48
Here, prime factor 3 has no pair. Therefore 768 must be multiplied by 3 to make it a perfect square.
∴ 768 x 3 = 2304
And √2304 = 2 x 2 x 2 x 2 x 3
= 48
Question: 6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained:
(i) 252
(ii) 2925
(iii) 396
(iv) 2645
(v) 2800
(vi) 1620
(i) 252
(ii) 2925
(iii) 396
(iv) 2645
(v) 2800
(vi) 1620
Answer :
(i) 252 = 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 252 must be divided by 7 to make it a perfect square.
∴ 252 ÷ 7 = 36
And √36 = 2 x 3 = 6
(i) 252 = 2 x 2 x 3 x 3 x 7
Here, prime factor 7 has no pair. Therefore 252 must be divided by 7 to make it a perfect square.
∴ 252 ÷ 7 = 36
And √36 = 2 x 3 = 6
(ii) 2925 = 3 x 3 x 5 x 5 x 13
Here, prime factor 13 has no pair. Therefore 2925 must be divided by 13 to make it a perfect square.
∴ 2925 ÷ 13 = 225
And √225 = 3 x 5 = 15
∴ 2925 ÷ 13 = 225
And √225 = 3 x 5 = 15
(iii) 396 = 2 x 2 x 3 x 3 x 11
Here, prime factor 11 has no pair. Therefore 396 must be divided by 11 to make it a perfect square.
∴ 396 ÷ 11 = 36
And √36 = 2 x 3 = 6
Here, prime factor 11 has no pair. Therefore 396 must be divided by 11 to make it a perfect square.
∴ 396 ÷ 11 = 36
And √36 = 2 x 3 = 6
(iv) 2645 = 5 x 23 x 23
Here, prime factor 5 has no pair. Therefore 2645 must be divided by 5 to make it a perfect square.
∴ 2645 ÷ 5 = 529
And √529 = 23
Here, prime factor 5 has no pair. Therefore 2645 must be divided by 5 to make it a perfect square.
∴ 2645 ÷ 5 = 529
And √529 = 23
(v) 2800 = 2 x 2 x 2 x 2 x 5 x 5 x 7
Here, prime factor 7 has no pair. Therefore 2800 must be divided by 7 to make it a perfect square.
∴ 2800 ÷ 7 = 400
And √400 = 2 x 2 x 5 = 20
Here, prime factor 7 has no pair. Therefore 2800 must be divided by 7 to make it a perfect square.
∴ 2800 ÷ 7 = 400
And √400 = 2 x 2 x 5 = 20
(vi) 1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5
Here, prime factor 5 has no pair. Therefore 1620 must be divided by 5 to make it a perfect square.
∴ 1620 ÷ 5 = 324
And √324 = 2 x 3 x 3 = 18
Here, prime factor 5 has no pair. Therefore 1620 must be divided by 5 to make it a perfect square.
∴ 1620 ÷ 5 = 324
And √324 = 2 x 3 x 3 = 18
Question: 7. The students of Class VIII of a school donated ` 2401 in all, for Prime Minister’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.
Answer :
Here, Donated money = `2401
Let the number of students be 𝓍
Therefore donated money = 𝓍 x 𝓍
According to question,
𝓍² = 2401
⇒ 𝓍 = √2401 = √7 x 7 x 7 x 7
𝓍 = 7 x 7 = 49
Hence, the number of students is 49.
Here, Donated money = `2401
Let the number of students be 𝓍
Therefore donated money = 𝓍 x 𝓍
According to question,
𝓍² = 2401
⇒ 𝓍 = √2401 = √7 x 7 x 7 x 7
𝓍 = 7 x 7 = 49
Hence, the number of students is 49.
Question: 8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Answer :
Here, Number of plants = 2025
Let the number of rows of planted plants be 𝓍
And each row contains number of plants = 𝓍
According to question,
𝓍² = 2025
𝓍 = √2025 = √3 x 3 x 3 x 3 x 5 x 5
𝓍 = 3 x 3 x 5 = 45
Hence, each row contains 45 plants.
Here, Number of plants = 2025
Let the number of rows of planted plants be 𝓍
And each row contains number of plants = 𝓍
According to question,
𝓍² = 2025
𝓍 = √2025 = √3 x 3 x 3 x 3 x 5 x 5
𝓍 = 3 x 3 x 5 = 45
Hence, each row contains 45 plants.
Question: 9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Answer :
L.C.M. of 4, 9 and 10 is 180.
Prime factors of 180 = 2 x 2 x 3 x 3 x 5
Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.
∴ 180 x 5 = 900
Hence, the smallest square number which is divisible by 4, 9 and 10 is 900.
L.C.M. of 4, 9 and 10 is 180.
Prime factors of 180 = 2 x 2 x 3 x 3 x 5
Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.
∴ 180 x 5 = 900
Hence, the smallest square number which is divisible by 4, 9 and 10 is 900.
Question: 10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Answer :
L.C.M. of 8, 15 and 20 is 120.
Prime factors of 120 = 2 x 2 x 2 x 3 x 5
Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be multiplied by
2 x 3 x 5 to make it a perfect square.
∴ 120 x 2 x 3 x 5 = 3600
Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600.
L.C.M. of 8, 15 and 20 is 120.
Prime factors of 120 = 2 x 2 x 2 x 3 x 5
Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be multiplied by
2 x 3 x 5 to make it a perfect square.
∴ 120 x 2 x 3 x 5 = 3600
Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600.
