1458 can be factorised as follows.
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, prime factor 2 does not have its pair. If 2 gets a pair, then the number will become a perfect square. Therefore, 1458 has to be multiplied with 2 to obtain a perfect square.
Therefore, 1458 × 2 = 2916 is a perfect square.
1458 × 2 = 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
∴ `sqrt(2916)` = 2 x 3 x 3 x 3 = 54
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Ex 6.3, 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. (v) 1458 Prime factorizing 1458 By prime factorization, 1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3 Since 2 does not occur in pair we multiply by 2 to make it a pair So, our number becomes 1458 × 2 = 2 × 3 × 3 × 3 × 3 × 3 × 3 × 2 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 Square root of 2916 ∴ √2916 = 2 × 3 × 3 × 3 = 6 × 9 = 54 ∴ The smallest whole number to be multiplied = 2 and square root of new number = 54
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Solution:
We have to find the smallest whole number by which the number should be multiplied so as to get a perfect square number
To get a perfect square, each factor of the given number must be paired.
(i) 252
Hence, prime factor 7 does not have its pair. If 7 gets a pair, then the number becomes a perfect square. Therefore, 252 has to be multiplied by 7 to get a perfect square.
So, perfect square is 252 × 7 = 1764
1764 = 2 × 2 × 3 × 3 × 7 × 7
Thus, √1764 = 2 × 3 × 7 = 42
(ii) 180
Hence, prime factor 5 does not have its pair. If 5 gets a pair, then the number becomes a perfect square. Therefore, 180 has to be multiplied by 5 to get a perfect square.
So, perfect square is 180 × 5 = 900
900 = 2 × 2 × 3 × 3 × 5 × 5
Thus, √900 = 2 × 3 × 5 = 30
(iii) 1008
Hence, prime factor 7 does not have its pair. If 7 gets a pair, then the number becomes a perfect square. Therefore, 1008 has to be multiplied by 7 to get a perfect square.
So, perfect square is 1008 × 7 = 7056
7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
Thus, √7056 = 2 × 2 × 3 × 7 = 84
(iv) 2028
Hence, prime factor 3 does not have its pair. If 3 gets a pair, then the number becomes a perfect square. Therefore, 2028 has to be multiplied by 3 to get a perfect square.
So, perfect square is 2028 × 3 = 6084
6084 = 2 × 2 × 13 × 13 × 3 × 3
Thus, √6084 = 2 × 13 × 3 = 78
(v) 1458
Hence, prime factor 2 does not have its pair. If 2 gets a pair, then the number becomes a perfect square. Therefore, 1458 has to be multiplied by 2 to get a perfect square.
So, perfect square is 1458 × 2 = 2916
2916 = 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2
Thus, √2916 = 3 × 3 × 3 × 2 = 54
(vi) 768
Hence, prime factor 3 does not have its pair. If 3 gets a pair, then the number becomes a perfect square. Therefore, 768 has to be multiplied by 3 to get a perfect square.
So, perfect square is 768 × 3 = 2304
2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
Thus, √2304 = 2 × 2 × 2 × 2 × 3 = 48
☛ Check: NCERT Solutions for Class 8 Maths Chapter 6
Video Solution:
NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.3 Question 5
Summary:
For each of the following numbers, (i) 252 (ii) 180 (iii) 1008 (iv) 2028 (v) 1458 (vi) 768 the smallest whole number by which it should be multiplied so as to get a perfect square number and the square root of the square number so obtained are as follows: (i) 7; √1764 = 42 (ii) 5; √900 = 30 (iii) 7; √7056 = 84 (iv) 3; √6084 = 78 (v) 2; √2916 = 54 and (vi) 3; √2304 = 48
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Find the smallest number by which 1458 must be divided so that the quotient is a perfect square. Find the square root of the quotient.
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