Area, Perimeter, and Congruency of Quadrilaterals

Date: 06/23/2003 at 21:32:14
From: Martha
Subject: Area, Perimeter, and congruency of quadrilaterals

I'm trying to find out how to prove or disprove if it is always true 
that if two quadrilaterals have the same area and the same perimeter, 
they will always be congruent.

I know how to figure it out for rectangles.  If I have a fixed 
perimeter, I can make a table to show that as one side increases, the 
area increases until the side becomes 1/4 of the perimeter (and your 
rectangle becomes a square) - so if a rectangle has the same perimeter 
AND area, it has to be congruent. I don't know how to go about it for 
other quadrilaterals, though.

In order to find the perimeters of parallelograms and trapezoids with 
slanty sides, I'm going to be using the Pythagorean theorem a lot and 
therefore dealing with irrational numbers. It's a lot more complicated 
than comparing the perimeter and areas of rectangles to see if they 
are always congruent. I'm not sure where to start on this part. (I'm 
an 8th grade teacher, by the way.)

Date: 06/23/2003 at 23:23:13
From: Doctor Peterson
Subject: Re: Area, Perimeter, and congruency of quadrilaterals

Hi, Martha.

Interesting question!

I'm inclined to doubt that this is true for quadrilaterals in general; 
it sounds as if it will be barely true for rectangles, and since 
quadrilaterals have more wiggle, they will probably slip out of our 
grasp. So let's try making a proof for rectangles, and then a 
counterexample for quadrilaterals in general.

For rectangles with sides a by b and c by d, if the areas and 
perimeters are both equal, we have

  ab = cd
  a+b = c+d

Take a and b as given, and we have two equations in two unknowns, from 
which we should be able to solve for c and d. (That's the basic 
observation that leads me to expect this to be true.) In fact,

  ab = c(a+b-c)
  c^2 - (a+b)c + ab = 0
  (c - a)(c - b) = 0

and c is equal to either a or b. (That's a neat bit of work: it's not 
often that you can factor an equation so neatly, but in this case I 
expected this answer so it was easy to see. I almost made a fool of 
myself by using the quadratic formula!)

This is what we wanted: c and d are either a and b, or b and a, 
respectively.

Now how about a counterexample for two arbitrary quadrilaterals? I'll 
try making one a rectangle and the other an isosceles trapezoid, to 
keep it simple:

                            d
  +-------------+       +-------+
  |             |      /|        \
  |             |b    / |h        \
  |             |    /  |          \
  +-------------+   +---+-----------+
         a                  c

Our equations now are

  ab = (c+d)h/2
  2(a+b) = c+d + 2 sqrt(h^2+(c-d)^2/4)

The mere fact that we now have two equations with three unknowns (c, 
d, and h), is the "wiggle" I mentioned: there are more variables even 
under the constraint I've given myself in specifying what kind of 
quads we have!

This will take a lot more work to solve. My claim is that I should be 
able to specify the value of one variable (say, h) and find at least 
one pair of c and d that will make both equations true. If it turned 
out that this would only work if h=b and c=d=a, then I would be 
proven wrong. I'd better make sure my guess is right.

We can first plug the value for c+d from the first equation into the 
second:

  2(a+b) = 2ab/h + 2 sqrt(h^2+(c-d)^2/4)

Let's multiply this by h:

  2(a+b)h - 2ab = h sqrt(4h^2 + (c-d)^2)

Square both sides and do a little more work and

  4[(a+b)h - ab]^2 = h^2(4h^2 + (c-d)^2)

  4[(a+b)h - ab]^2/h^2 = 4h^2 + (c-d)^2

  c-d = 2 sqrt([(a+b)h - ab]^2/h^2 - h^2)

Combining this with our equation for c+d, we can solve for each 
variable; for example,

  c = [(c+d)+(c-d)]/2
    = ab/h + sqrt([(a+b)h - ab]^2/h^2 - h^2)

As a check, if h=b(making the trapezoid as high as the rectangle), we 
get

  c = a + sqrt(b^2 - b^2) = a

as I would expect.
---------------------

As a follow-up, I wanted to find a nice specific counterexample, which 
would have whole numbers for all the sides. I realized that if I took 
c, d, and h as the givens, I could choose them so they were all whole 
and then solve for a and b; and if I took h and c-d instead as given, 
I could then adjust d so that a and b would be whole. I took h=4 and 
c-d=6 so that I had a 3-4-5 right triangle at each end of the 
trapezoid; then the area of the trapezoid is 4d+12 and the perimeter 
is 2d+16. Solving for a and b, I found that d=3 made a good choice.

That gives these two quadrilaterals that have the same area (24) and 
perimeter (22):

                      3
                    +---+
  +---------+      /|    \
  |         |3   5/ |4    \
  |         |    /  |      \
  +---------+   +---+-------+
       8              9

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 06/25/2003 at 18:50:11
From: Martha
Subject: Area, Perimeter, and congruency of quadrilaterals

Dear Dr. Peterson,

Thanks so much for your reply - it was very helpful and thanks also 
for the counterexample - I had found the same one by doing some 
guessing and testing!

One more thing I discovered to help me out by reading another Dr. Math 
archive is that if you draw and cut out a parallelogram, cut it along 
the diagonal, reflect one half of it, and put it back together along 
the diagonal, you have another quadrilateral of equal area and 
perimeter. Depending upon which diagonal you use, your new 
quadrilateral may be a kite or a chevron.  Another pretty quick 
counterexample.

Thanks again!

Date: 06/25/2003 at 22:41:39
From: Doctor Peterson
Subject: Re: Area, Perimeter, and congruency of quadrilaterals

Hi, Martha.

Yes, that's a nice simple counterexample. And I misread it the first 
time, giving me another:

     +--+-----+      +--+--+
    /   |    /      /   |   \
   /    |   /      /    |    \
  +-----+--+      +-----+-----+

There you have a parallelogram and a trapezoid with the same area and 
perimeter.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/