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.
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Lines have infinite length. You mean segments (or line segments). – Steven Alexis Gregory Apr 02 '19 at 10:56
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Take a look at https://math.stackexchange.com/q/708 – Jean Marie Apr 02 '19 at 13:56
3 Answers
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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 $$h=\sqrt{ab}.$$
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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.
Michael Hoppe
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See https://en.wikipedia.org/wiki/Geometric_mean_theorem, please. – Michael Hoppe Apr 02 '19 at 11:29
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Being not a native speaker of the English language: What is ambiguous in the phrase "semicircle over a line segment"? – Michael Hoppe Apr 02 '19 at 13:56
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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 $\sqrt{ab}$.
TurlocTheRed
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