What is the length of the hypotenuse of a right triangle if the two other sides are of lengths 7 and 1?

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Answer 1 · 2015-12-21T01:23:19

The length of the hypotenuse is $5 \sqrt{2}$ $5sqrt 2$ .

Use the Pythagorean theorem, $c^{2} = a^{2} + b^{2}$ $c^2=a^2+b^2$ , where $c$ $c$ is the hypotenuse and $a and b$ $a and b$ are the other two sides.

Substitute $7$ $7$ and $1$ $1$ into the equation for $a and b$ $a and b$ .

$c^{2} = 7^{2} + 1^{2}$ $c^2=7^2+1^2$

Simplify.

$c^{2} = 49 + 1$ $c^2=49+1$

Simplify.

$c^{2} = 50$ $c^2=50$

Take the square root of both sides.

$c = \sqrt{50}$ $c=sqrt50$

Factor $\sqrt{50}$ $sqrt 50$ .

$c = \sqrt{2 \times 5 \times 5}$ $c=sqrt(2xx5xx5)$

Simplify.

$c = \sqrt{2 \times 5^{2}}$ $c=sqrt (2xx5^2)$

Apply the square root rule $\sqrt{a^{2}} = a$ $sqrt(a^2)=a$ .

$c = 5 \sqrt{2}$ $c=5sqrt2$