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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 $x$?

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Do you know pythagoras theorem? – user22705 Feb 1 '12 at 16:48
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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... – Steven Stadnicki Feb 1 '12 at 16:49
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How about adding the homework tag? Plus, not every right angled triangle is 30-60-90. – Inquest Feb 1 '12 at 17:02

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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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So by taking the positive solution $x=1$, we get the three sides of the triangle as $x=1$, $x+2 = 3$ and $\sqrt{10}$. If we use the negative solution, the sides are $x=-3$, $x+2=-1$, and $\sqrt{10}$, which in effect gives the same lengths (in different order) for the sides if we take length to mean magnitude (i.e., ignore the negative signs). T.is is as it should be since squaring in the Pythagorean Theorem gets rid of the negative signs, if any, associated with the length, – Dilip Sarwate Feb 1 '12 at 17:17
@Dilip Sarwate Nice catch. I was thinking of the problem in purely geometric terms and was interpreting "length" as nonnegative. – David Mitra Feb 1 '12 at 17:25

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