Provide detailed instructions how to construct a right triangle with edge lengths in geometric progression.
(In reply to
re: A Start- I concur by Steven Lord)
There is a method to "calculate" the squareroot using compass and straight edge, this method is spelled out here
https://www.geogebra.org/m/edtecfcv
As pointed out by Steven, we need to make a right triangle with legs of lengths 1 and sqrt((1+sqrt(5))/2)
1) start with a unit length line, call this line L1
2) replicate this line 5 times to get a line of length 5, call this L2
3) use the above method to get a line of length sqrt(5), call this L3
4) add a line of length L1 to L3 to get a line of length 1+sqrt(5), call this L4
5) bisect line L4 to get a line of length (1+sqrt(5))/2, call this L5
6) construct the square root of L5 to get a line of length sqrt((1+sqrt(5))/2), call this L6
7) now to make the triangle. Draw a line of unit length, then construct a perpendicular line at one of the endpoints of length equal to L6.
8) connect the other end point to the opposite side of the perpendicular made in step 7.
You now have a right triangle whose side lengths are in geometric progression.
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Posted by Daniel
on 2018-09-06 06:35:24 |
re(2): A Start- I concur
