How can we divide a segment of an arbitrary length in the ratio $1:\sqrt a$, where $a$ is not a perfect square?

Question

The question in the title is asking in general. The question given:

Divide a segment of length 8cm in the ratio $1:\sqrt{12}$

How can it be done?

Answer 1

The trick is that $\sqrt{12}=2\sqrt3$ and it is easy to get a segment length $\sqrt3$ using equilateral triangles.

Draw equilateral triangles $ABC,DBC$ side 2 (with $A,D$ on opposite sides of $BC$). Then $AD=2\sqrt3$. Take $E$ on the line $AD$ with $D$ between $A$ and $E$ so that $DE=1$.

Now if $AF$ has length 8 (or any other length), draw a line through $D$ parallel to $EF$ intersecting $AF$ at $X$. Evidently $AX/XF=AD/DE=2\sqrt3$.