Given a unit segment, construct a segment for the number $4 + \sqrt3$.
Given a unit segment, construct a segment for the number $7 + \sqrt{4 + \sqrt3}$.
I get how to use a unit segment and create a segment $\sqrt2$ and then use another unit segment at a right angle and get a segment $\sqrt3$. Then do I just tack on $4$ unit lengths to that line to construct a long segment that is $4 + \sqrt3$? And then it seems the next part builds on the last, but I’m not sure how.
From the segment $4+\sqrt 3$ build a rectangle having height $1$ so its area is $4+\sqrt 3$
Now we build an equivalent square, so its side will be $\sqrt{4+\sqrt 3}$
With center in the midpoint $M$ of the segment $AB$ build a circle and the perpendicular to $AB$ in the point $H$. The intersection $C$ is the vertex of a triangle were
$CH^2=AH\cdot HB\to CH =\sqrt{4+\sqrt 3}$