How do you determine if the lengths $\sqrt{65}, 6 \sqrt{2}, \sqrt{97}$ $sqrt65, 6sqrt2, sqrt97$ form a right triangle?

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Answer 1 · 2017-02-03T15:01:24

$137 \neq 97$ $137 !=97$

This is not a right-angled triangle.

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

Test to see whether $a^{2} + b^{2} = c^{2}$ $color(red)(a^2 +b^2) = color(blue)(c^2)$ is true.

$color(red)((sqrt65)^2 + (6sqrt2)^2)color(white)(..........)color(blue) ((sqrt97)^2)$

$color(red)(=65+36xx2)color(white)(................)color(blue) (=97)$

$color(red)(=65+72)$

$color(red)(=137)$

How do you determine if the lengths sqrt65, 6sqrt2, sqrt97√65,6√2,√97sqrt65, 6sqrt2, sqrt97 form a right triangle?