ML AGGARWAL CLASS 8 CHAPTER 3 SQUARES AND ROOTS Exercise 3.1

 Exercise 3.1

Question 1

(i) 729

Sol :

Prime fractorisation

$\begin{array}{r|l}3&729\\ \hline 3& 243 \\ \hline 3&81\\ \hline 3& 27\\ \hline 3& 9 \\ \hline 3&3\\ \hline&1\end{array}$

⇒729=3×3×3×3×3×3

∴$729=27^{2}$

because 729 can be expressed as product of pairs of equal prime factors. 


(ii) 5488

Sol:

$\begin{array}{r|l}2& 5488\\ \hline 2& 2744 \\ \hline 2&1372\\ \hline 2& 686\\ \hline 7& 343 \\ \hline 7&49\\ \hline 7&7\\ \hline&1\end{array}$


⇒5488= 2×2×2×2×7×7×7

Since 7 left unpaired 5488 in not a perfect square

 

(iii)1024 

Sol:

 $\begin{array}{r|l}2&1024\\ \hline 2& 512 \\ \hline 2&256\\ \hline 2& 128\\ \hline 2& 64 \\ \hline 2&32\\ \hline2&16 \\ \hline 2&8\\ \hline 2&4\\ \hline 2&2\\ \hline&1\end{array}$

 ⇒1024= 2x2x2x2x2x2x2x2x2x2x2 

since 1024 is expressed as the product of pairs of equal prime number so it is a perfect square 

(iv) 243

Sol:

$\begin{array}{r|l}3&243\\ \hline 3& 81 \\ \hline 3&27\\ \hline 3& 9\\ \hline 3& 3 \\ \hline&1\end{array}$

$243=3 \times 3 \times 3 \times 3 \times 3$

As ' 3 ' let unpaired, So 243 is not a square (perfect)


Question 2 


(i) 1296

 $\begin{array}{r|l}2&1246\\ \hline 2& 648 \\ \hline 2&324\\ \hline 2& 162\\ \hline 3& 81 \\ \hline 3&27\\ \hline3&9 \\ \hline 3&3\\ \hline&1\end{array}$

⇒1296=2x2x2x2x3x3x3x3

since 1296 is expressed as the product of pairs of equal prime numbers so it is a perfect square 

1296= $2^{2} \times 2^{2} \times 3^{2} \times 3^{2}$

1296= $(2 \times 2 \times 3 \times 3)^{2}=36^{2}$

∴ 1296 is square of 36 

 

(ii) 1784

Sol:

$\begin{array}{r|l}2&1784\\ \hline 2& 892 \\ \hline 2&446\\ \hline 223& 223\\ \hline&1\end{array}$

1784= $=2 \times 2 \times 2 \times 223$

As 1784  can not be expressed as product of pairs of

 equal prime factors, so is not a perfect square 


(iii) 3025 

Sol:

$\begin{array}{r|l}5&3025\\ \hline 5& 605 \\ \hline 11&121\\ \hline&1\end{array}$

$3025=5 \times 5 \times 11 \times 11$

Since 3025 can be expressed as the product of pairs of

equal prime factors.

$3025=(5 \times 11)^{2}=55^{2}$

Hence, 55 is a number whose square is 3025


(iv) 3969

Sol: 

$\begin{array}{r|l}3&3969\\ \hline 3& 1323 \\ \hline 3&441\\ \hline 3& 147\\ \hline 3& 49 \\ \hline 7&7\\ \hline&1\end{array}$

$3969=3 \times 3 \times 3 \times 3 \times 7 \times 7$

3969 Can be expressed as product of pairs of

equal prime numbers.

$3969=3^{4} \times 3^{2} \times 7^{2}$

$3969=(3 \times 3 \times 7)^{2}$

$3969=63^{2}$

Hence, 63 is the number whose square is 3969


Question 3 

1008

Sol:
$\begin{array}{r|l}2& 1008\\ \hline 2& 504 \\ \hline 2&252\\ \hline 2& 126\\ \hline 7& 63 \\ \hline 3&9\\  \hline 3&3\\ \hline&1\end{array}$

$1008=2 \times 2 \times 2 \times 2 \times 7 \times 3 \times 3$

Since' 7 ' is left unpaired, so to make 1008 a
Perfect square it should be mutiplied by 7

Question 4 

5808

Sol:
⇒ $\begin{array}{r|l}2& 5808\\ \hline 2& 2904 \\ \hline 2&1452\\ \hline 2& 726\\ \hline 3& 363 \\ \hline 11&121\\  \hline 11&11\\ \hline&1\end{array}$

$5808=2 \times 2 \times 2 \times 2 \times 3 \times 11 \times 11$

Since ' 3' let unpaired. To make 5808 a perfect square
it should be divided by '3'.

so divide 5808 by '3'

$\frac{5808}{3}=\frac{2 \times 2 \times 2 \times 2 \times 3 \times 11 \times 11}{3}$

$1936=(2 \times 2 \times 11)^{2}=(44)^{2}$

so 44 is a number whose square is 1936

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