# What is the length of the hypotenuse when the sum of all sides in a right triangle is 30cm? [closed]

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..

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – M. Winter, mfl, Ethan Bolker, Shailesh, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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 $2.5$). – T. Linnell Jul 26 '17 at 12:07
• 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. – Nina Simone Jul 26 '17 at 12:10
• as @T. Linell points out really all we can say is the hypotenuse has to be less than half the sum. – user451844 Jul 26 '17 at 12:11
• 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!!!) – T. Linnell Jul 26 '17 at 12:13
• If $c$ is the hypotenuse and $A$ is one of the angles $<90^\circ\implies0<A<90^\circ$ we have $$c(1+\cos A+\sin A)=30$$ – lab bhattacharjee Jul 26 '17 at 12:14

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 $b$.

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 $m^2+n^2=13$.