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If the sum of all the sides of a right angled triangle is 30cm. Then what will be the length of the hypotenuse? Total completely calculation to be needed..

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closed as off-topic by M. Winter, mfl, Ethan Bolker, Shailesh, Namaste Jul 26 '17 at 14:27

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    $\begingroup$ It's impossible to say without more information. You could have a $\{5,12,13\}$ triangle or a $\{7.5,10,12.5\}$ (a $\{3,4,5\}$ scaled up by $2.5$). $\endgroup$ – T. Linnell Jul 26 '17 at 12:07
  • $\begingroup$ Assume $c$ is the hypotenuse and $a$ one of the cathetus. Then $30=c+a+\sqrt{c^2-a^2}$. From there we get that $(30-c-a)^2=900+c^2+a^2-60c-60a+2ac=c^2-a^2$. Therefore $c=\frac{900-60a+2a^2}{60-2a}$. So, for each $0<a<30/2$ the formula before gives us the length of a possible hypotenuse. $\endgroup$ – Nina Simone Jul 26 '17 at 12:10
  • $\begingroup$ as @T. Linell points out really all we can say is the hypotenuse has to be less than half the sum. $\endgroup$ – user451844 Jul 26 '17 at 12:11
  • $\begingroup$ If there were some other condition, it would be possible to calculate the actual value of the hypotenuse using the formula @NinaSimone derived. This condition could be the length of one of the cathetus, or it could be one of the non-right angles, or even the area. (I messed up and hit enter too early!!!) $\endgroup$ – T. Linnell Jul 26 '17 at 12:13
  • $\begingroup$ If $c$ is the hypotenuse and $A$ is one of the angles $<90^\circ\implies0<A<90^\circ$ we have $$c(1+\cos A+\sin A)=30$$ $\endgroup$ – lab bhattacharjee Jul 26 '17 at 12:14
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If we say that $a,b,c$ are the lengths of the sides of the triangle, with $c$ being the hypotenuse.

We know that \begin{align}a+b+c&=30\\ a^2+b^2&=c^2\end{align}

We can also say that \begin{align}0<a<30\\ 0<b<30\\ 0<c<30\end{align}

If we put all of this into WolframAlpha and solve for $c$, we can see that $$c=\frac{b^2-30b+450}{30-b}$$ where $0<b<15$ and $a=\dfrac{30(b-15)}{b-30}$

Therefore, there are infinitely many solutions to this problem without specifying more constraints on $a$ and $b$.

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If you mean to solve it in natural numbers, we obtain.

$$d^2(2mn+m^2+n^2+m^2-n^2)=30,$$ where $m>n$, $gcd(m,n)=1$ and $mn$ is even.

Thus, $d^2m(m+n)=15$, which says that $d=1$, $m+n=5$ and $m=3$.

Id est, $n=2$ and length of hypotenuse is $m^2+n^2=13$.

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