Question

4r/√2 is equal to 2r√2...how?

Answer 1

We know that $(\sqrt{2})^{2} = 2$

We also know that $2^{2} = 4$

So, 4 can be written as,

$4 = 2^{2} = (\sqrt{2}.\sqrt{2} ) ^{2}$  

          $= (\sqrt{2}.\sqrt{2}.)(\sqrt{2}.\sqrt{2})$

Now coming to your question,

$\frac{4r}{\sqrt{2}}$

From above observations, this can be written as

$= \frac{\sqrt{2}.\sqrt{2}.\sqrt{2}.\sqrt{2} . r}{\sqrt{2}}$

Now one $\sqrt{2}$ in the numerator and the one in the denominator cancels out.

Hence we get,

$\sqrt{2}.\sqrt{2}.\sqrt{2}.r$

Now, see it this way:

$(\sqrt{2}.\sqrt{2}) \sqrt{2}r$

= $(\sqrt{2})^{2}$$\sqrt{2}.r$

= $2\sqrt{2}r$

= $2r\sqrt{2}$

Hope that is clear!