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(i) 252

(ii) 2925

(iii) 396

(iv) 2645

(v) 2800

(vi) 1620

Answer :

(i) We have prime factors of 252 as:

\(\qquad \begin{array}{c|lcr}
2 & 252\\
\hline
2 & 126\\
\hline
3 & 63\\
\hline
3 & 21\\
\hline
7 & 7\\
\hline
& 1\\
\end{array}
\)

\(252 = 2 \times 2 \times 3 \times 3 \times 7\)

Here, the prime factors are not in pair as 7 has no pair.
Thus, the smallest whole number by which the given number is to be divided to get a perfect square number is 7.

The new square number is \(252 ÷ 7 = 36\)

∴Square root of 36 is \(\sqrt{36} = 6\)

(ii) We have prime factors of 2925 as:

\(\qquad \begin{array}{c|lcr}
3 & 2925\\
\hline
3 & 975\\
\hline
5 & 325\\
\hline
5 & 65\\
\hline
13 & 13\\
\hline
& 1\\
\end{array}
\)

\(2925 = 3 \times 3 \times 5 \times 5 \times 13\)

Here, 13 has no pair.So, the smallest whole number by which 2925 is divided to get a square number is 13.

New square number =\( 2925 ÷ 13 = 225\)

∴ \(\sqrt{225} = 15\)

(iii)We have prime factors of 396 as:

\(\qquad \begin{array}{c|lcr}
2 & 396\\
\hline
2 & 198\\
\hline
3 & 99\\
\hline
3 & 33\\
\hline
11 & 11\\
\hline
& 1\\
\end{array}
\)

\(396 = 2 \times 2 \times 3 \times 3 \times 11\)

Here 11 is not in pair.So,the required smallest whole number by which 396 is divided to get a square number is 11.

New square number = \(396 ÷ 11 = 36\)

∴ \(\sqrt{36} = 6\)

(iv)We have prime factors of 2645 as:

\(\qquad \begin{array}{c|lcr}
5 & 2645\\
\hline
23 & 529\\
\hline
23 & 23\\
\hline
& 1\\
\end{array}
\)

\(2645 = 5 \times 23 \times 23\)

Here, 5 is not in pair.So, 5 is the required smallest whole number by which 2645 is divided to get a square number

New square number = \(2645 ÷ 5 = 529\)

∴ \(\sqrt{529} = 23\)

(v) We have prime factors of 2800 as:

\(\qquad \begin{array}{c|lcr}
2 & 2800\\
\hline
2 & 1400\\
\hline
2 & 700\\
\hline
2 & 350\\
\hline
5 & 175\\
\hline
5 & 35\\
\hline
7 & 7\\
\hline
& 1\\
\end{array}
\)

\(2800 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 7\)

Here, 7 is not in pair.So, 7 is the required smallest number by which 2800 is divided to get a square number.

New square number =\( 2800 ÷ 7 = 400\)

∴ \(\sqrt{400} = 20\)

(vi)We have prime factors of 1620 as:

\(\qquad \begin{array}{c|lcr}
2 & 1620\\
\hline
2 & 810\\
\hline
3 & 405\\
\hline
3 & 135\\
\hline
3 & 45\\
\hline
3 & 15\\
\hline
5 & 5\\
\hline
& 1\\
\end{array}
\)

\(1620 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5\)

Here, 5 is not in pair.So, 5 is the required smallest prime number by which 1620 is divided to get a square number = \(1620 ÷ 5 = 324\)

∴ \(\sqrt{324} = 18\)

- 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
- 2.Without doing any calculation, find the numbers which are surely not perfect squares. (i) 153 (ii) 257 (iii) 408 (iv) 441
- 3.Find the square roots of 100 and 169 by the method of repeated subtraction.
- 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
- 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
- 7.The students of class VIII of a school donated ₹ 2401 in all, for the 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.
- 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.
- 9.Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
- 10.Find the smallest number that is divisible by each of the numbers 8, 15 and 20.

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