Some of the examples where we can observe similar pairs:

**1**. $\sqrt{9}$

= $\sqrt{3\times3}$ [here square root of 9 is 3×3 that means 3 is in pair]

= $\sqrt{3^2}$ [here square roots of 3^{2} ]

= 3

**2**. $\sqrt{12}$

= $\sqrt{2\times2\times3}$ [here 2×2 is in pair so 2 will go outside the square root]

= 2 $\sqrt{3}$ [here 2 is outside the square root and 3 is inside the square root]

**3**. $\sqrt{18}$

= $\sqrt{2\times3\times3}$ [here 3×3 is in pair so 3 will go outside the square root]

= 3 $\sqrt{2}$ [here 3 is outside the square root and 2 is inside the square root]

**4**. $\sqrt{49}$

= $\sqrt{7\times7}$ [here 7×7 is in pair so 7 will go outside the square root]

= 7

**5**. $\sqrt{324}$

= $\sqrt{2\times2\times3\times3\times3\times3}$ [here we can observe that 2×2×3×3×3×3 ,that means 2, 3, 3 numbers are in pairs so one number from each pair will go outside the square root]

= 2×3×3

= 18

**6**. $\sqrt{720}$

= $\sqrt{2\times2\times2\times2\times3\times3\times5}$ [here similarly 2×2×2×2×3×3×5 and we can observe that 2, 2, 3 are in pairs so one number from each pair will go outside the square root and we can also see that 5 is not in pairs so 5 will be inside the square root]

= 2×2×3 $\sqrt{5}$

= 12 $\sqrt{5}$

**7**. Find the square root of the perfect square 5184.

5184 = 2×2×2×2×2×2×3×3×3×3

[First we find the prime factors of the given perfect square. Since the number is a perfect square, therefore, we get pairs of similar prime factors].

$\sqrt{5184}$ =2×2×2×3×3 [here choose one factor from each pair and multiply together].

= 72 [the result will be the square root of the given number].

**8**. Find the value of $\sqrt{\frac{25}{64}}$.

$\sqrt{\frac{25}{64}}$ = $\frac{\sqrt{25}}{\sqrt{64}}$ [here both are equal].

= $\frac{\sqrt{5\times5}}{\sqrt{8\times8}}$ [here first we find the prime factors of the numerator and the denominator. Since the number is a perfect square, therefore, we get pairs of similar prime factors].

= $\frac{5}{8}$ [here one factor is chosen from each pair]

**9**. $\sqrt{18}$

= $\sqrt{2\times3\times3}$ [here 3×3 is in pair so 3 will go outside the square root]

= 3 $\sqrt{2}$ [here 3 is outside the square root and 2 is inside the square root]

**10**. Find the smallest number by which 252 must be multiplied so that the product becomes a perfect square.

252 = 2×2×3×3×7

We see that, in prime factorization of 252, there exists a number 7 which is unpaired.

Hence , to make a pair of 7, we will have to multiply 252 by 7

Hence, 7 is the required number.

So, 252 × 7 = 1764 now you can notice that 2×2×3×3×7×7 [is a perfect square] = 2×3×7 = 42.