# Construction of square root of 3

I came up with this construction for $$\sqrt{3}$$, and want to know if it's valid.

Start with a unit segment $$AB$$. At points $$A$$ and $$B$$, draw two circles with radius 1 like so:

Then, mark the points where these circles intersect as $$C$$ and $$D$$. Draw line $$CD$$ between the points. The length of line $$CD$$ should be precisely $$\sqrt{3}$$. I think this is so because by the Pythagorean theorem, the height of a triangle drawn between points $$A$$, $$C$$, and $$B$$ should be $$\frac{\sqrt{3}}{2}$$. Since the length of segment $$CD$$ should be twice that, and thus $$\sqrt{3}$$. Is my thinking correct here?

• Yes, this is correct. Aug 6, 2022 at 2:25
• – lhf
Aug 6, 2022 at 16:46

Note that AB is crossed by CD exactly in the middle and vice versa. Assuming that E is the midpoint of A and B, therefore midpoint of C and D, the triangle ACE is rectangle, AC is unit segment, AE is half-unit segment ($$AE=AB/2$$) and $$CE=CD/2$$.
$$AC^2=AE^2+CE^2 \implies 1^2=(1/2)^2+(CD/2)^2 \implies CD=\sqrt{3}$$.