How do you find the lengths of the sides of a right triangle given the legs x and x and the hypotenuse 8?

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Answer 1 · 2016-03-07T12:16:08

$x = 4 \sqrt{2}$ $x=4sqrt2$

The Pythagorean Theorem gives the relations between the legs $a, b$ $a,b$ and hypotenuse $c$ $c$ of a right triangle:

$a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$

Here, since the legs are both $x$ $x$ , $a = x$ $a=x$ and $b = x$ $b=x$ , and since the hypotenuse is $8$ $8$ , $c = 8$ $c=8$ .

$x^{2} + x^{2} = 8^{2}$ $x^2+x^2=8^2$

Combine the $x^{2}$ $x^2$ terms. Recall that $8^{2} = 64$ $8^2=64$ .

$2 x^{2} = 64$ $2x^2=64$

Divide both sides by $2$ $2$ .

$x^{2} = 32$ $x^2=32$

Take the square root of both sides. (Only take the positive root--a negative side length wouldn't make any sense.)

$x = \sqrt{32}$ $x=sqrt32$

Simplify the term in the square root.

$x = \sqrt{16} \sqrt{2}$ $x=sqrt16sqrt2$

$x = 4 \sqrt{2}$ $x=4sqrt2$