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Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.1 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.2 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.3 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.4 solution
SQUARE AND SQUARE ROOTS (Chapter-6) Exercise 6.3 SOLUTION
EXERCISE 6.3
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
Solution- According to the chapter we know that when we find the unit or one's digit of a square number than we get-
which numbers have 1 or 9 unit digit, square number have 1 unit digit.
which numbers have 2 or 8 unit digit, square number have 4 unit digit.
which numbers have 3 or 7 unit digit, square number have 9 unit digit.
which numbers have 4 or 6 unit digit, square number have 6 unit digit.
which numbers have 5 unit digit, square number have 5 unit digit.
which numbers have 0 unit digit, square number have 0 unit digit.If process becomes opposite then-
If square numbers have 1 unit digit, square root will be 1 or 9 unit digit.
If square numbers have 4 unit digit, square root will be 2 or 8 unit digit.
If square numbers have 9 unit digit, square root will be 3 or 7 unit digit.
If square numbers have 6 unit digit, square root will be 4 or 6 unit digit.
If square numbers have 5 unit digit, square root will be 5 unit digit.
If square numbers have 0 unit digit, square root will be 0 unit digit.(i) 9801 have 1 unit digit, square root will be 1 or 9 unit digit.
(ii) 99856 have 6 unit digit, square root will be 4 or 6 unit digit.
(iii) 998001 have 1 unit digit, square root will be 1 or 9 unit digit.
(iv) 657666025 have 5 unit digit, square root will be 5 unit digit.
2. Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153 (ii) 257 (iii) 408 (iv) 441
Solution- All the square numbers end with 0, 1, 4, 5, 6 or 9 at units place. None of these end with 2, 3, 7 or 8 at unit’s place. So we Can say that if a number ends in 0, 1, 4, 5, 6 or 9 at units place, then it must be a square number. And if a number ends in 2, 3, 7 or 8 at units place, then it must be not a square number.
(i) 153 is not a perfect square because 153 end with 3.
(ii) 257 is not a perfect square because 257 end with 7.
(iii) 408 is not a perfect square because 408 end with 8.
(iv) 441 is a perfect square because 441 end with 1.
3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution- The sum of the first n odd natural numbers is n 2 ? That is, every square number can be expressed as a sum of successive odd natural numbers starting from 1.
√100
(i) 100 – 1 = 99 (ii) 99 – 3 = 96 (iii) 96 – 5 = 91 (iv) 91 – 7 = 84 (v) 84 – 9 = 75 (vi) 75 – 11 = 64 (vii) 64 – 13 = 51 (viii) 51 – 15 = 36 (ix) 36 – 17 = 19 (x) 19 – 19 = 0
From 100 we have subtracted successive odd numbers starting from 1 and obtained 0 at 10th step. Therefore √100= 10
√169
(i) 169 – 1 = 168 (ii) 168 – 3 = 165 (iii) 165 – 5 = 160 (iv) 160 – 7 = 153 (v) 153 – 9 = 144 (vi) 144 – 11 = 133 (vii) 133 – 13 = 120 (viii) 120 – 15 = 105 (ix) 105 – 17 = 88 (x) 88 – 19 = 69 (xi) 69 – 21= 48 (xii) 48 – 23 = 25 (xiii) 25 – 25 = 0
From 169 we have subtracted successive odd numbers starting from 1 and obtained 0 at 13th step. Therefore √169 = 13
4. Find the square roots of the following numbers by the Prime Factorisation Method.
(i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744 (vi) 9604 (vii) 5929 (viii) 9216 (ix) 529 (x) 8100
Solution-
(i) 729
Prime factorisation of 729 is
729 = 3 × 3 × 3 × 3 × 3 × 3
By pairing the prime factors we get,
729 = 3 × 3 × 3 × 3 × 3 × 3 = (3 × 3 × 3)²
Therefore, √729 = 3 × 3 × 3 = 27
(ii) 400
Prime factorisation of 400 is
400 = 2 × 2 × 2 × 2 × 5 × 5
By pairing the prime factors we get,
400 = 2 × 2 × 2 × 2 × 5 × 5 = (2 × 2 × 5)²
Therefore, √400 = 2 × 2 × 5 = 20
(iii) 1764
Prime factorisation of 1764 is1764 = 2 × 2 × 3 × 3 × 7 × 7
By pairing the prime factors we get,
1764 = 2 × 2 × 3 × 3 × 7 × 7 = (2 × 3 × 7)²
Therefore, √1764 = 2 × 3 × 7 = 42
(iv) 4096Prime factorisation of 4096 is
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
By pairing the prime factors we get,
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2 × 2 × 2× 2 × 2)²
Therefore, √4096 = 2 × 2 × 2 × 2 × 2 × 2= 64
(v) 7744Prime factorisation of 7744 is
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
By pairing the prime factors we get,
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11 = (2 × 2 × 2 × 11)²
Therefore, √7744 = 2 × 2 × 2 × 11 = 88
(vi) 9604Prime factorisation of 9604 is
9604 = 2 × 2 × 7 × 7 × 7 × 7
By pairing the prime factors we get,
9604 = 2 × 2 × 7 × 7 × 7 × 7 = (2 × 7 × 7)²
Therefore, √9604 = 2 × 7 × 7 = 98
(vii) 5929Prime factorisation of 5929 is
5929 = 7 × 7 × 11 × 11
By pairing the prime factors we get,
5929 = 7 × 7 × 11 × 11 = (7 × 11)²
Therefore, √5929 = 7 × 11 = 77
(viii) 9216Prime factorisation of 9216 is
9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
By pairing the prime factors we get,
9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = (2 × 2 × 2 × 2 × 2 × 3)²
Therefore, √9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96
(ix) 529
Prime factorisation of 529 is
529 = 23 × 23
By pairing the prime factors we get,
529 = 23 × 23 = (23)²
Therefore, √529 = 23
(x) 8100Prime factorisation of 8100 is
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
By pairing the prime factors we get,
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = (2 × 3 × 3 × 5)²
Therefore, √8100 = 2 × 3 × 3 × 5 = 90
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
Solution-
(i) We have 252 = 2 × 2 × 3 × 3 × 7
As the prime factor 7 has no pair, 252 is not a perfect square.
If 7 gets a pair then the number will become perfect square.
So, we multiply 252 by 7 to get,
252 × 7 = 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 252 × 7 = 1764 is a perfect square.
Thus the required smallest multiple of 252 is 1764 which is a perfect square.
And, √1764 = 2 × 3 × 7 = 42
(ii) 180 We have 180 = 2 × 2 × 3 × 3 × 5
As the prime factor 5 has no pair, 180 is not a perfect square.
If 5 gets a pair then the number will become perfect square.
So, we multiply 180 by 5 to get,
180 × 5 = 2 × 2 × 3 × 3 × 5 × 5
Now each prime factor is in a pair. Therefore, 180 × 5 = 900 is a perfect square.
Thus the required smallest multiple of 180 is 900 which is a perfect square.
And, √900 = 2 × 3 × 5 = 30
(iii) 1008 We have 1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
As the prime factor 7 has no pair, 1008 is not a perfect square.
If 7 gets a pair then the number will become perfect square.
So, we multiply 1008 by 7 to get,
1008 × 7 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 1008 × 7 = 7056 is a perfect square.
Thus the required smallest multiple of 1008 is 7056 which is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(iv) We have 2028 = 2 × 2 × 3 × 13 × 13
As the prime factor 3 has no pair, 2028 is not a perfect square.
If 3 gets a pair then the number will become perfect square.
So, we multiply 2028 by 3 to get,
2028 × 3 = 2 × 2 × 3 × 3 × 13 × 13
Now each prime factor is in a pair. Therefore, 2028 × 3 = 6084 is a perfect square.
Thus the required smallest multiple of 2028 is 6084 which is a perfect square.
And, √6084 = 2 × 3 × 13 = 78
(v) We have 1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
As the prime factor 2 has no pair, 1458 is not a perfect square.
If 2 gets a pair then the number will become perfect square.
So, we multiply 1458 by 2 to get,
1458 × 2 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Now each prime factor is in a pair. Therefore, 1458 × 2 = 2916 is a perfect square.
Thus the required smallest multiple of 1458 is 2916 which is a perfect square.
And, √2916 = 2 × 3 × 3 × 3 = 54
(vi) 768 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If 3 gets a pair then the number will become perfect square.
So, we multiply 2352 by 3 to get,
2352 × 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 × 3 = 7056 is a perfect square.
Thus the required smallest multiple of 2352 is 7056 which is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
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
Solution-
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(ii) 2925 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(iii) 396 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(iv) 2645 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(v) 2800 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
(vi) 1620 We have 2352 = 2 × 2 × 2 × 2 × 3 × 7 × 7
As the prime factor 3 has no pair, 2352 is not a perfect square.
If we Remove 3 then the number will become perfect square.
So, we divide 2352 by 3 to get,
2352 ÷ 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Now each prime factor is in a pair. Therefore, 2352 ÷ 3 = 7056 is a perfect square.
And, √7056 = 2 × 2 × 3 × 7 = 84
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.
Solution- According to the question,
The students donated Total `2401 rupees
The number of students in the class = x
Each student donated as many rupees =x
Then,
x² = 2401
x = √2401 = 49
Hence, The number of students in the class is 49.
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.
Solution-
Total number of plants are to be planted in a garden = 2025
The number of rows = x
The number of rows and the number of plants in each row = x
Then,
x² = 2025
x = √2025= 45
Hence, The number of rows and the number of plants in each row is 45.
9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution-
First find the smallest common multiple and then find the square number needed.
The least number divisible by each one of 4, 9 and 10 is their LCM.
The LCM of 4, 9 and 10 is 2 × 2 × 3 × 3 × 5 = 180.
Prime factorisation of 180 is 180 = 2 × 2 × 3 × 3 × 5.
We see that prime factors 5 are not in pairs.
Therefore 180 is not a perfect square. In order to get a perfect square, each factor of 180 must be paired.
So we need to make pairs of 5.
Therefore, 180 should be multiplied by 5.
Hence, the required square number is 180 × 5 = 900.
10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Solution-
First find the smallest common multiple and then find the square number needed.
The least number divisible by each one of 8, 15 and 20 is their LCM.
The LCM of 8, 15 and 20 is 2 × 2 × 2 × 3 × 5 = 120.
Prime factorisation of 120 is 120 = 2 × 2 × 2 × 3 × 5.
We see that prime factors 2, 3 and 5 are not in pairs.
Therefore 120 is not a perfect square. In order to get a perfect square, each factor of 120 must be paired.
So we need to make pairs of 2, 3 and 5.
Therefore, 120 should be multiplied by 2 × 3 × 5, i.e., 30.
Hence, the required square number is 120 × 30 = 3600.
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