The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Math Topics » geometry.puzzles

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Eric Bainville

Posts: 10
Registered: 12/6/04
Re: Equilateral triangle areas
Posted: Apr 25, 2006 12:50 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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
of the result using Cabri Geometry measures and calculator. I got DE
= 4.98715... cm

-- Eric

At 20:12 21/04/2006, 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 ?

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2017. All Rights Reserved.