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### 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

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

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/
```
Associated Topics:
High School Triangles and Other Polygons

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