Rationalizing denominator of $\frac{18}{\sqrt{162}}$. Cannot match textbook solution

Question

I am given this expression and asked to simplify by rationalizing the denominator:

$$\frac{18}{\sqrt{162}}$$

The solution is provided:

$\sqrt{2}$

I arrived at:

$$\frac{\sqrt{162}}{9}$$

Here is my thought process to arrive at this incorrect answer:

$\frac{18}{\sqrt{162}}$

= $\frac{18}{\sqrt{162}}$ * $\frac{\sqrt{162}}{\sqrt{162}}$

= $\frac{18\sqrt{162}}{162}$

= $\frac{\sqrt{162}}{9}$

How can I arrive at $\sqrt{2}$ ?

Answer 1

$\frac{\sqrt{162}}{9} = \frac{\sqrt{2 \cdot 9^2}}{9} = \frac{9\sqrt{2}}{9} = \sqrt{2}$.

Answer 2

$$\require{cancel}\frac{18}{\sqrt{162}}=\frac{2\cdot3^2}{\sqrt{2\cdot3^4}}=\frac{2\cdot\cancel{3^2}}{\sqrt{2}\cdot\cancel{3^2}}=\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\cancel2\sqrt{2}}{\cancel{2}}=\sqrt{2}$$

Answer 3

Your thought process is good. But just continue with factorizing $162=2*81=2*3^4$.

So $\sqrt {162}=\sqrt {2*3^4}=\sqrt {2}\sqrt {3^4}=\sqrt 2*3^2=9\sqrt 2$ and from there.... it's just mechanics.

Answer 4

$ \sqrt{162} $ needs to be simplified further, as $ 162 $ is the product containing a perfect square (i.e. $ 81 $). Thus $$ \sqrt{162} = \sqrt{2 \cdot 81} = \sqrt{2}\sqrt{81} = 9\sqrt{2}$$

and hence $$ \frac{\sqrt{162}}{9} = \frac{9\sqrt{2}}{9} = \sqrt{2} $$

Answer 5

Alternatively if $a = \frac{18}{\sqrt{162}}$, $a^2 = \frac{18^2}{\sqrt{162}^2} = \frac{324}{162} = 2$. Since $a$ is a positive number, $a = \sqrt{2}$.