# How do you determine if the lengths sqrt65, 6sqrt2, sqrt97 form a right triangle?

Feb 3, 2017

$137 \ne 97$

This is not a right-angled triangle.

#### Explanation:

The theorem of Pythagoras only holds true in a right-angled triangle.

Test to see whether $\textcolor{red}{{a}^{2} + {b}^{2}} = \textcolor{b l u e}{{c}^{2}}$ is true.

$\textcolor{red}{{\left(\sqrt{65}\right)}^{2} + {\left(6 \sqrt{2}\right)}^{2}} \textcolor{w h i t e}{\ldots \ldots \ldots .} \textcolor{b l u e}{{\left(\sqrt{97}\right)}^{2}}$

$\textcolor{red}{= 65 + 36 \times 2} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .} \textcolor{b l u e}{= 97}$

$\textcolor{red}{= 65 + 72}$

$\textcolor{red}{= 137}$