Associated Topics || Dr. Math Home || Search Dr. Math

### Area of Trapezoid Given Only the Side Lengths

```Date: 12/15/2008 at 11:08:48
From: Jeromy
Subject: sides are 1, 2, 3, and 4. find the area of the trapezoid

How would you find the area of a trapezoid without the height, but
when all sides are given and they don't tell you which sides are
parallel?

Ex. given trapezoid has side lengths of 1,2,3, and 4.  Find the area.

How would you find the area without the height?

```

```
Date: 12/15/2008 at 11:51:54
From: Doctor Peterson
Subject: Re: sides are 1, 2, 3, and 4. find the area of the trapezoid

Hi, Jeromy.

Interesting question!

You can find the area without knowing the height ahead of time, but
you have to know which sides are parallel.  Different pairs of
parallel sides may yield different areas.

To find the area given only the sides AND which are parallel, you can
draw a parallelogram within the trapezoid:

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

You will now know all three sides of the triangle, from which you can
find its height.

Let's try doing that with the obvious first choice of 2 and 4 for the
parallel sides:

2
+---------+
/         /  \
1/        1/     \3
/   2     /   2    \
+---------+-----------+
4

Looking closely at the triangle, we see that this is impossible!  The
Triangle Inequality Theorem says that the sum of any two sides must be
greater than the other side, but 1+2 = 3.  So the 2 and 4 sides can't
be parallel.

Can we come up with a rule to figure out which pairs of parallel
sides are possible?  Let's use variables this time:

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

In order for the triangle to be possible (assuming, as shown, that
b>a), we must have (by the Triangle Inequality Theorem)

c + d > b - a
c + (b-a) > d, so d-c < b-a
d + (b-a) > c, so c-d < b-a

The last two facts can be combined, and the assumption that b>a
eliminated, by writing

|d-c| < |b-a|

The first inequality can be written similarly as

|b-a| < c+d

So we're looking for a pair of sides whose difference is between the
sum and the difference of the other two sides:

|d-c| < |b-a| < c+d

There are 6 ways to choose a pair of parallel sides; let's see which
fit this requirement:

bases   sides
------- -------
a   b   c   d  d-c b-a c+d
--- --- --- --- --- --- ---
1   2   3   4   1 = 1 < 7  no
1   3   2   4   2 = 2 < 6  no
1   4   2   3   1 < 3 < 5  yes
2   3   1   4   3 > 1 < 5  no
2   4   1   3   2 = 2 < 4  no
3   4   1   2   1 = 1 < 3  no

So there's only one way to place the sides that will work.  (That may
not always be true!)

1
+----+
/    /  \
2/   2/     \3
/ 1  /   3    \
+----+-----------+
4

Now we need to find the height:

+
/: \
2/ :h  \3
/  :     \
+---+-------+
3

We can use the Pythagorean theorem to find the two parts of the base
in terms of h, and then solve for h:

sqrt(2^2 - h^2) + sqrt(3^2 - h^2) = 3

Try solving this; you'll get a fairly nice radical expression after
some pretty hard work.  Then you can use this in the formula for the
area of a trapezoid to find the area you want.

An alternative way to get the area is to find the area of the
triangle using Heron's formula (see our FAQ on Formulas) and then add
on the area of the parallelogram.

If you have any further questions, feel free to write back.

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

```Date: 8/3/2013 at 4:30:25
From: Denis
Subject: sides are 1, 2, 3, and 4. find the area of the trapezoid
As Denis wrote to Dr. Math
On 08/03/2013 at 13:25:04 (Eastern Time),

There is an easier way to finish solving the problem than either way
you suggested:

1
+----+
/|    | \
2/ |    |h  \3
/  |    |     \
+---+----+-------+
3-x   1     x
. . . . 4 . . . .

Using the left triangle:  h^2 = 2^2 - (3 - x)^2
Using the right triangle: h^2 = 3^2 - x^2

Combine these equations, you can solve for x, then find h and the area.

```

```Date: 12/15/2008 at 17:35:58
From: Doctor Peterson
Subject: sides are 1, 2, 3, and 4. find the area of the trapezoid

Hi, Denis,

Yes, that is much easier than my radical equation. It's a good example
of coming at a problem from a different direction -- in this case,
starting with two variables rather than one so that you have more
freedom in how to solve it, then seeing that eliminating h (resulting
in a polynomial) is easier than eliminating x (resulting in radicals).
My equation actually amounts to doing the latter, because I
essentially turned your equations into

3 - x = sqrt(2^2 - h^2)
x = sqrt(3^2 - h^2)

Then I added them together. (My picture contains the same two triangles
as yours.)

I also notice that my alternative idea can be improved, for this
particular case, because my triangle happens to be isosceles:

+
1/  \
+ .  \3
1/    h  .\
+-----------+
3

We can first find the area of the triangle, by taking base 2 and
finding the height to that base, as shown above, by the Pythagorean
theorem.

+
/: \
2/ :h  \3
/  :     \
+---+-------+
3

Then, taking 3 as the base as shown here, we can use the known area to
find the height h, and use that to find the area of the parallelogram.

Thanks for your contribution!

- Doctor Peterson, The Math Forum

```
Associated Topics:
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

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/