
Re: Equilateral triangle areas
Posted:
Apr 25, 2006 12:50 PM


The triangle DBE has constant area, meaning that BD*BE is constant, with a special position when D is on A: BD = 10cm, implying BE = 2cm to get 1/5 of the total area, so BE = 20 / BD. Taking D free on segment AB, we can construct E on BC, then get the parallel to DE at distance DE of it, and its intersection G with AC (so the distance between G and line DE is distance DE). F is then constructed as intersection between DE and a perpendicular to DE through G.
Doing this, we obtain all points depending only on D, and can get an approximation of the result using Cabri Geometry measures and calculator. I got DE = 4.98715... cm
 Eric
At 20:12 21/04/2006, ontadian@hotmail.com wrote: >ABC is an equilateral triangle with 10 cm. side lengths. D is on the >side between A and B. E is on the side between B and C. G is on the >base side between A and C. F is a point on a line DE and FG is a >line perpendicular to DE. Area DBE is equal to 1/5 of the total >area. Area ADFG = area GFEC. DE = FG. What is the lengths DE,FG ?

