Find the maximum perimeter of a right angled triangle with hypotenuse 1 This is a question from the grade 7 Math Competition. I can solve it by considering one of the angles, say $\theta$, besides the right angle. We have the perimeter given by $1+\cos\theta+\sin\theta = 1+\sqrt{2}\cos(\theta-\pi/4 )$ and the rest is easy. However, I believe there should be an elementary but elegant way to solve it. Any inspiration?
 A: I don't know if this is more elementary, but we can replace the trigonometric functions using the Pythagorean Theorem.
If one of the legs is $x$, then the remaining leg is $\sqrt{1-x^2}$. The perimeter is then
$$
1+x+\sqrt{1-x^2}\tag1
$$
Note that
$$
\begin{align}
\left(x+\sqrt{1-x^2}\right)^2
&=1+\sqrt{4x^2-4x^4}\tag{2a}\\
&=1+\sqrt{1-\left(2x^2-1\right)^2}\tag{2b}
\end{align}
$$
$\text{(2b)}$ is no more than $2$ since $\sqrt{1-\left(2x^2-1\right)^2}\le1$. $\text{(2b)}$ equals $2$ when $x=\frac1{\sqrt2}$. This means that
$$
x+\sqrt{1-x^2}\le\sqrt2\tag3
$$
and $(3)$ equals $\sqrt2$ when $x=\frac1{\sqrt2}$. Therefore,
$$
1+x+\sqrt{1-x^2}\le1+\sqrt2\tag4
$$
and $(4)$ equals $1+\sqrt2$ when $x=\frac1{\sqrt2}$.
A: Assume the sides of the right triangle are $a,b,1$. We have $a^2+b^2=1$ and we should find the maximum value of $a+b+1$.Hence we need to maximize  $a+b$ or $(a+b)^2$ which is equal to $a^2+b^2+2ab=1+4S$ (where $S$ is area of right triangle).

Now from the image, can you see when the area of the right triangle is maximum?
