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

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

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Answer 1 · 2016-08-06T20:00:23

Not a right triangle.

Explanation:

This makes use of $converse of Pythagoras' theorem$ $color(blue)"converse of Pythagoras' theorem"$

If this is a right triangle then.

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

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

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

Since $1 \neq \frac{241}{400} this is not a right triangle$ $1≠241/400" this is not a right triangle"$
$--------------------------------------------------------$ $color(blue)"--------------------------------------------------------"$

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 : $\frac{3}{4} + \frac{1}{5} = \frac{19}{20} \to 1 > \frac{19}{20} not valid$ $3/4+1/5=19/20to1>19/20" not valid"$
$--------------------------------------------------------------$ $color(blue)"--------------------------------------------------------------"$

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

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

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