The perimeter in a right triangle is known and also one side. Am I right thinking that the hypotenuse and the other side can only have one set of lengths?
Using Geogebra I was able to approximate the lengths but I wonder if someone can help me find a formula.
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Let $c$ be the length of the hypothenuse and let $a$ and $b$ be the lengths of the legs. Assume that we know the value of $b$ and the perimeter $P$. Then the perimeter is given by $$ P = a+b+c, $$ and by Pythagoras, we know that $$ c^2 = a^2+b^2, $$ so $$ P = a+b+\sqrt{a^2+b^2} \quad \Leftrightarrow \quad P-a-b = \sqrt{a^2+b^2}. $$ Squaring both sides, we obtain $$ P^2+a^2+b^2-2Pa-2Pb+2ab = a^2+b^2 \quad \Leftrightarrow \quad a = \frac{2Pb-P^2}{2b-2P} = \frac{P}{2}\frac{2b-P}{b-P}. $$ You can find $c$ by using Pythagoras.
You are right. Let $b$ be the known side, being $a$ the other side and $c$ the hypotenuse. Let $p$ be the perimeter. Then $a$ and $c$ are thee solution of the system$$\left\{\begin{array}{l}a+c=p-b\\c^2-a^2=b^2.\end{array}\right.$$This solution is$$a=\frac{p(p-2b)}{2(p-b)}\text{ and }c=\frac{2b^2-2bp+p^2}{2(p-b)}.$$