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A right triangle has a hypotenuse of $\sqrt{10}$, one of the legs is $x+2$, and the shortest leg is $x$. How do I find $x$?

Thanks.

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    Note that while this is a right triangle, it isn't necessarily a 30-60-90 triangle - in fact, in this case it turns out not to be. In any case, the Pythagorean Theorem should be all you need to solve this problem...2012-02-01

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The sum of the squares of lengths of the non-hypotenuse sides is the square of the length of the hypotenuse$^\dagger$: $ (x)^2+(x+2)^2=(\sqrt {10})^2. $ Whence $ x^2+(x^2+4x+4)=10. $ Putting the above in standard form and solving for $x$: $ 2x^2+4x-6=0\iff x^2+2x-3=0\iff(x-1)(x+3)=0\iff x=1, x=-3. $ We take the positive solution: $x=1$.


$^\dagger$ Please do not simply remember the Pythagorean Theorem as "$a^2+b^2=c^2$"; evilly minded instructors will try to trick you by labeling the hypotenuse $b$ and one of the legs $c$.

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    @Dilip Sarwate Nice catch. I was thinking of the problem in purely geometric terms and was interpreting "length" as nonnegative.2012-02-01