While keeping one side constant in a right angle triangle, when we change the other side, which would increase more, the side or the hypotenuse?

Question

While keeping one side constant in a right angle triangle, when we change the other side, which would increase more, the side or the hypotenuse?

Let the sides be X,Y and Hypotenuse be P. Keeping the length of X same, if we increase Y to (Y+Z), hypotenuse becomes (P+Q). Z and Q are changes in length of the side of length Y and hypotenuse.

Question is, which is greater, Z or Q?

Answer 1

We have the equations \begin{align*}(P+Q)^{2}&=X^{2}+(Y+Z)^{2},\\ P^{2}&=X^{2}+Y^{2},\end{align*} by applying the Pythagorean theorem to the original and new triangles. This gives $$2PQ+Q^{2}=2YZ+Z^{2}$$ when we expand the squares in the first equation and subtract the second equation. Now we can solve for $Q$ in terms of $Z,$ since this is a quadratic equation, and we get $$Q=-P\pm\sqrt{P^{2}+Z(Z+2Y)}.$$ Since $Q>0,$ the only choice that makes sense in the equation above is $-P+\sqrt{P^{2}+Z(Z+2Y)},$ and thus the question is whether or not this is $<Z.$ This is equivalent to $\sqrt{P^{2}+Z(Z+2Y)}<Z+P,$ and squaring both sides (note that $x\mapsto x^{2}$ is an increasing function on the positive real numbers), we get the proposed inequality $$P^{2}+Z(Z+2Y)<(Z+P)^{2},$$ which again is equivalent to $Q<Z.$ Expanding the square and subtracting $Z^{2}+P^{2}$ from both sides, we get the equivalent inequality $2YZ<2ZP$, or $$2Z(Y-P)<0.$$ But this last inequality is true, since the hypotenuse $P$ is always greater than either $X$ or $Y,$ and $Z>0.$ This gives the final conclusion: $$Q\text{ is always less than }Z.$$

Answer 2

For any $a>0$ we have that $f(x)=\sqrt{x^2+a^2}$ is a convex function on $\mathbb{R}^+$.
In particular $f'(x)$ is increasing from $0$ to $1$ on $\mathbb{R}^+$ and for any $x,h>0$

$$ f(x+h)-f(x) \stackrel{\text{Lagrange}}{=} h\cdot f'(\xi) < h.\qquad \left[\xi\in(x,x+h)\right] $$ In particular, by increasing the length of $x$ by $h$ the increment of the length of the hypotenuse is always less than $h$.