IV OBM

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 $\frac{BC}{AB}=\frac{AB}{BD}⟺BC=\frac{A{B}^2}{BD}=A{B}^2=a^2+1$. 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.