# Question #95e73

Mar 9, 2017

$\frac{\sqrt{2}}{2}$

#### Explanation:

$\sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{1}{2}\right)}^{2}}$

First find out what ${\left(\frac{1}{2}\right)}^{2}$ is:

${\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}$

So, we now know that the expression is now $\sqrt{\frac{2}{4}}$ which is the same as $\sqrt{\frac{1}{2}}$.

Then, we rationalize the denominator as we cannot have a root as
the denominator:

$\sqrt{1} = 1$, so the expression is now $\frac{1}{\sqrt{2}}$

Multiply both the numerator and the denominator by $\sqrt{2}$

Denominator $= \sqrt{2} \cdot \sqrt{2} = 2$

Numerator $= 1 \cdot \sqrt{2} = \sqrt{2}$

Hence,

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

Mar 9, 2017

The expression is equivalent to $\frac{\sqrt{2}}{2}$.

#### Explanation:

The square of $\frac{1}{2}$ is $\frac{1}{4}$ because $\left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = \frac{1}{2 \cdot 2} = \frac{1}{4}$.

Therefore,

$\sqrt{\frac{1}{4} + \frac{1}{4}}$

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

$\sqrt{\frac{1}{2}}$

We can separate the radicals.

$\frac{\sqrt{1}}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}$

I would recommend you rationalize the denominators.

$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

$\frac{\sqrt{2}}{2}$

Hopefully this helps!