Image: SqrtGeom

Description: Straightedge and compass construction of the length x=a{\displaystyle x={\sqrt {a}}}, given the length a{\displaystyle a} and the unit length. The proof relies on Thales's theorem (blue circle) to obtain a right triangle with hypothenuse length a+1{\displaystyle a+1}, and on the Pythagorean theorem to set up equations about its height and side lengths, viz. a2+x2=a2+x22{\displaystyle a^{2}+x^{2}={\sqrt {a^{2}+x^{2}}}^{2}} (from the left triangle; trivially satisfied), 12+x2=x2+12{\displaystyle 1^{2}+x^{2}={\sqrt {x^{2}+1}}^{2}} (from the right triangle; trivially satisfied), and a2+x22+x2+12=(a+1)2{\displaystyle {\sqrt {a^{2}+x^{2}}}^{2}+{\sqrt {x^{2}+1}}^{2}=(a+1)^{2}} (from the big triangle). Their unique positive solution is x=a{\displaystyle x={\sqrt {a}}}.
Title: SqrtGeom
Credit: Own work, inspired from File:TomoyukiMogi(SquareRoot).gif.
Author: Jochen Burghardt
Usage Terms: Creative Commons Attribution-Share Alike 4.0
License: CC BY-SA 4.0
License Link: https://creativecommons.org/licenses/by-sa/4.0
Attribution Required?: Yes