Can the sides 10, 40, 42 be a right triangle?

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Answer 1 · 2015-07-01T08:28:17

The following numbers do not form a Pythagorean triplet.

Let the three sides be denoted as: $a, b, c$ $a,b,c$ ,in which $a > b, c$ $color(blue)(a)> color(green)(b,c$ .

If the sides form a Pythagorean triplet they should satisfy the Pythagoras theorem:

$a^{2} = b^{2} + c^{2}$ $color(blue)(a^2) = color(green)(b^2 +c^2$

Here $a = 42, b = 40, c = 10$ $color(blue)(a=42, color(green)(b = 40, c=10$

As per theorem:

$a^{2}$ $color(blue)(a^2$ should be equal to $b^{2} + c^{2}$ $color(green)( b^2 +c^2$

But the sum doesn't follow the Pythagoras theorem, as:

$42^{2} \neq 40^{2} + 10^{2}$ $42^2 != 40^2 + 10^2$
$1764 \neq 1600 + 100$ $1764 != 1600+ 100$
$1764 \neq 1700$ $1764!=1700$

So, the following numbers do not form a Pythagorean triplet.