# Is a triangle with sides of 3/4, 1, 1/5 a right triangle?

Aug 6, 2016

Not a right triangle.

#### Explanation:

This makes use of $\textcolor{b l u e}{\text{converse of Pythagoras' theorem}}$

If this is a right triangle then.

${1}^{2} = {\left(\frac{3}{4}\right)}^{2} + {\left(\frac{1}{5}\right)}^{2} \text{ would have to be true}$

left side $= {1}^{2} = 1$

and right side $= {\left(\frac{3}{4}\right)}^{2} + {\left(\frac{1}{5}\right)}^{2} = \frac{9}{16} + \frac{1}{25} = \frac{241}{400}$

Since 1≠241/400" this is not a right triangle"
$\textcolor{b l u e}{\text{--------------------------------------------------------}}$

Additional Note Oversight on my part.

This is not a valid triangle. In any triangle, any side must always be less than the sum of the other 2 sides.

consider side of 1 : $\frac{3}{4} + \frac{1}{5} = \frac{19}{20} \to 1 > \frac{19}{20} \text{ not valid}$
$\textcolor{b l u e}{\text{--------------------------------------------------------------}}$

However: the method I have described above would hold for a valid triangle.