Area of an Unusual Hexagon

Date: 12/05/96 at 20:31:47
From: James Fletcher
Subject: Trig problem?

Dear Dr. Math

Please take a look at the following problem.  I think it involves 
trigonometry, which I haven't learned yet - but I'm not certain. 

A certain person owned the three square fields shown in the figure 
containing 18, 20, and 26 acres.  In order to fence the property, the 
person bought the four intervening triangular fields.  What is the 
whole area of the estate?

In a nutshell, the figure looks like a hexagon with three squares of 
18, 20, and 26 acres joined in the middle by a triangle formed by one 
of each their sides and three more triangular fields on the outskirts 
consisting of one unknown side and respective sides of sqrt18 and 
sqrt20, sqrt20 and sqrt26, and sqrt26 and sqrt18.

       

Thank you,
James Fletcher

Date: 12/11/96 at 21:14:44
From: Doctor Rob
Subject: Re: Trig problem?

James,

You certainly could use trigonometry to compute the area of the whole
estate.  A useful formula would be the Law of Cosines.  You can do  
the problem without this, however.

1. I would convert all the areas to square feet, so that I could deal
   with the lengths of the sides of the square fields in feet.

2. You already know the areas of the square fields.

3. You can find the area of the central triangle immediately using
   Hero's (or Heron's) formula, once you know the lengths of the three
   sides.  A = Sqrt[s*(s-a)*(s-b)*(s-c)], where s = (a+b+c)/2, and 
   a, b and c are the three sides.

4. To find the areas of the three outer triangles, you need only find
   the lengths of the outer sides.  To do this consider the following
   diagram:

              A
             /|`
            / |\ `
           /  | |  `
        c /  h| \    ` r
         /    |  |b    `
        /     |  \       `
       /  a-x | x |    a   `
       ----------------------
      B       Q   C          P

I got this by considering the central triangle ABC with sides a, b, c
and the outer triangle which touches it at vertex C.  This outer 
triangle has two known sides, a and b, and an unknown side r.  I 
rotated the outer triangle counterclockwise 90 degrees until the two 
sides of length b coincided.  This rotated triangle is ACP, and the 
angle BCP is a straight angle because the two angles BCA and ACP must 
add up to 180 degrees, since they, plus two right angles at C in the 
original diagram, add up to 360 degrees.  Then I dropped a 
perpendicular from A to line BP of length h hitting BP at Q.  I let 
the length of QC be x, so the length of BQ is a-x.

Now in this diagram, we can use the Pythagorean theorem on the 
following three triangles:

ABQ:  c^2 = h^2 + (a-x)^2
ACQ:  b^2 = h^2 + x^2
APQ:  r^2 = h^2 + (a+x)^2

Now add the first equation to the third (to eliminate the terms linear 
in x) and subtract twice the second (to eliminate the x^2 term):

c^2 - 2*b^2 + r^2 = 2*a^2

Solving for r^2, we get:

r^2 = 2*a^2 + 2*b^2 - c^2

Since we know a, b, and c, we can easily find:

r = Sqrt[2*a^2 + 2*b^2 - c^2]

The same technique works for finding the outer sides of the other two
triangles.  They are:

  p = Sqrt[2*b^2 + 2*c^2 - a^2]
and
  q = Sqrt[2*a^2 + 2*c^2 - b^2]

Now Hero's formula will get you the answer you want.  It may involve  
unpleasant square roots, however.

I hope this helps.  If not, write back, and we'll give it another 
shot!

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    

Date: 06/14/2003 at 02:51:04
From: David
Subject: Area of an unusual hexagon

Hi,

Note that Dr. Rob's triangle ABP is ABC (the centre triangle) plus 
ACP (an outer triangle). They have the same base length and height, 
hence the same area, so we don't need to do the calculations. 

The same applies to all 3 outer triangles. So the answer is 4 times 
the area of the centre triangle plus the squares. 

Cute, isn't it?

- David