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Topic: Equilateral triangle areas
Replies: 6   Last Post: Apr 25, 2006 11:15 PM

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Eric Bainville

Posts: 10
Registered: 12/6/04
Re: Equilateral triangle areas
Posted: Apr 25, 2006 12:50 PM
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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 ?




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