Brazilian Mathematical Olympiad

We show first how to get the square and the square root. Take $ABC$ with $\angle A = 90^\circ$, altitude $AD$ of length $a$ and $BD = 1$. Then by similar triangles $\frac{BC}{AB} = \frac{AB}{BD} \iff BC = \frac{AB^2}{BD} = AB^2 = a^2 + 1$. Hence $CD = a^2$.

Conversely, we can take $BD = 1$, $CD = a$ and then construct $A$ (as the intersection of the perpendicular at $D$ and the circle with diameter $BC$) to obtain $AD$ of length $\sqrt{a}$.

So now given $a, b$ construct $a^2, b^2$. Then take a right-angled triangle with those lengths as its shorter sides and the hypotenuse is $\sqrt{a^4 + b^4}$. Finally, take the square root to get the required length.