# Construction of $\sqrt{ab}$ user ruler and compass [duplicate]

Q. Given two line segments of length a and b. Draw a line segment of length $$\sqrt{ab}$$ using a ruler and compass.

I didn't get any idea how to approach to the solution.

## marked as duplicate by Jaap Scherphuis, Lord Shark the Unknown, Trần Thúc Minh Trí, Javi, ThéophileApr 3 at 16:34

Let the base of the large triangle be $$a+b$$, and the height $$h$$. By similarity of the small triangles,

$$\frac ha=\frac bh$$ so that $$h=\sqrt{ab}.$$

Draw a line of length $$a+b$$. Construct the perpendicular line in the point they joint. The semicircle over $$a+b$$ cuts that perpendicular. Now the distance between that point and the joining point is $$\sqrt{ab}$$ due to Euclid.

• If you don't mind , Can you show it analytically please?? – user579689 Apr 2 at 11:16
• See en.wikipedia.org/wiki/Geometric_mean_theorem, please. – Michael Hoppe Apr 2 at 11:29
• The word "diameter" is maybe missing in "The semi-circle over" – Jean Marie Apr 2 at 13:53
• Being not a native speaker of the English language: What is ambiguous in the phrase "semicircle over a line segment"? – Michael Hoppe Apr 2 at 13:56
• Nothing ambiguous... Don't bother with my remark. – Jean Marie Apr 2 at 13:58

Join segments of length a and length b together on the same line. Call where they join Point J. Call their Midpoint M. Construct a circle centered at M through either end point of a+b. Construct a line perpendicular to a+b through point J. Call where it intersects the circle point Q. The length of QJ is $$\sqrt{ab}$$ as proven elsewhere. These above constructions follow from Euclid's postulates I, III, SAS, and ASA. So the constructions should be valid in neutral geometry.

A related approach. Construct a rectangle having one sidelength a and one side length b. Find a square having the same area as the starting rectangle. The side length of thes square will be $$\sqrt{ab}$$.