Let $\overline{AB}$ be a line segment of length $1$; extend $\overline{AB}$ past $B$ to a point $C$ such that $|BC| = 2010$. Let $O$ be the midpoint of $\overline{AC}$; construct a circle with center $O$ and radius $|AO| = \frac{2011}{2}$. Let $\ell$ be a line which contains point $B$ and is perpendicular to $\overline{AB}$, and let $\ell$ intersect circle $O$ at $D$. Then $AC$ is the diameter of circle $O$ and $\triangle ACD$ is a right triangle with $DB \perp AC$ so $|DB|^{2} = |AB|\cdot |BC| \implies |DB| = \sqrt{2010}$. [asy][asy] defaultpen(fontsize(10)); real r = 6; pair A=(0,0), B=(1,0), C=(r+1,0), D=(1,sqrt(r)), O=(r+1)/2; draw(arc(O,abs(A-O),0,180)); draw(B--A--D--C--B--D); label("$A$",A,(-1,-1));label("$B$",B,(0,-1));label("$C$",C,(1,-1));label("$O$",O,(0,-1));label("$D$",D,(-1,1)); dot(A^^B^^C^^D^^O); [/asy][/asy](figure not to scale)

minsoens wrote: so $|DB|^{2} = |AB|\cdot |BC| $. Why? What theorem is that?

If you do not know the name of the theorem, you may see that $\triangle ADB \sim \triangle ACD$ ,and get that, in a right-angled triangle the square of the altitude equals the product of the segments in which the hypotenuse is divided by this altitude. Best regards, sunken rock

More generally you can construct the length $\sqrt{N} \forall N \in Z^+$ given a line segment of length $1$. Just construct a right-angled triangle of base and height $1$ so hypotenuse $\sqrt{2}$ then from this construct a right-angled triangle of sides $1,\sqrt{2},\sqrt{3}$ and then repeat the process of constructing the length $\sqrt{n+1}$ form the lengths $1$ and $\sqrt{n}$. The key thing in the construction is being able to construct a line perpendicular to a given line (say $AB$) at a given point (say $B$). This is done by drawing the ray $AB$ and a circle centre $B$ through $A$, this will meet the ray again at $C$ then just consruct the perpendicular bisector of $AC$ which is a line perpendicular to $AC$ (so to $AB$) and intersects with it at the midpoint of $AC$ (which is $B$).