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
triangle).

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"
xxxxxxxxxxx
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
|     /
|    /
|   /
|30/
| /
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
Math Forum Home || Math Library || Quick Reference || Math Forum Search