Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Area of an Unspecified Triangle

Date: 6/1/96 at 20:24:19
From: Anonymous
Subject: Area of an unspecified triangle

A square ABCD, with side "y", and an equilateral triangle DEF, with 
the same side "y", are in contact (at point "d") and their bases CD 
and DE are a continuous line segment. A line segment is drawn from 
point "b" of the square to point "e" of the triangle (it is the 
hypotenuse of the right-angled triangle BCE formed by the sides BC of 
the square and CE of the square plus the triangle DEF). There is a 
"space" between the side DA of the square and the side FD of the 
equilateral triangle. Another triangle is formed between the 
hypotenuse already drawn and this space (one of the angles of this new 
triangle is 30 degrees, the difference between 90, the supplementary 
angle of the square, and 60, the internal angle of the equilateral 

The question is: what is the area of this last triangle?

          b      a
           xxxxxx  f
           x    x  x
           x    x x x <sides measuring "y" 
          c      d    e

Date: 6/1/96 at 20:56:22
From: Doctor Pete
Subject: Re: Area of an unspecified triangle

Draw the figure  as follows:

   B __________ A     F
    |-._       |     /\ 
    |   -._    |    /  \
    |       -._|H  /    \
    |          |-./_     \
    |          | /G  --._ \
   C           D            E

So you want the area of triangle DGH.  As you pointed out, angle GDH 
is 30 degrees.  But if CD = DE, then 2BC = CE, so BE intersects AD at 
the midpoint H.  So DH = y/2.  

Also note that angle GHD = EBC.  But tan(EBC) = 2/1 = 2.

  H |-._
    |x   -._
    |        -._
  K |-----------/ G
    |    h     /
    |         /
    |        /
    |       /
    |      /        tan(x) = 2
    |     /
    |    /
    |   /
    | /
  D |/
So we want h.  Since GKD is a 30-60-90 right triangle, DK = sqrt(3)*h.  
So HK = y/2 - DK = y/2 - sqrt(3)*h.  But tan(x) = h/HK = 2, so

       h/(y/2-sqrt(3)*h) = 2
   ==> h = y - 2*sqrt(3)*h
   ==> (1+2*sqrt(3))h = y
   ==> h = y/(1+2*sqrt(3)) = (2*sqrt(3)-1)*y/11.

So the area of triangle GHD = 1/2 * y/2 * h = (2*sqrt(3)-1)*y^2/44.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum