Areas of Parallelograms and Triangles

While you may not see the similarities between
parallelograms and
triangles initially, we will come to see that they are actually quite
related when it comes to area.
But, just how similar can they be if one is a three-sided polygon, and the other
is a specific type of quadrilateral? Let’s begin drawing some connections between
parallelograms and triangles by finding out how to measure the areas of parallelograms
first.

Areas of Parallelograms

Recall that parallelograms are special a type of
quadrilateral whose opposite sides do not intersect. Parallelograms come
in the form of
rectangles, rhombuses, and squares. When trying to determine the area of
a parallelogram, it will be necessary to identify two main components: the base
and the height of the parallelogram. The base of a parallelogram can be on
any side of the figure. The height of a parallelogram is the perpendicular distance
between any two parallel bases. Let’s look at these different parts in the figure
below.

We could have also chosen the left and right sides of the parallelogram as our bases.
In this case, the height would run horizontally.

The area of a parallelogram is given by the product of the base and the height.
That is, areas of parallelograms can be expressed as

where A represents the area, b is the base, and h
is the height.

Let’s practice using this formula by doing the exercise below.

Exercise 1

Find the area of parallelogram ABCD.

Answer:

First, we want to choose a side as the base of our parallelogram. In this case,
we can choose the side labeled 12 inches as our base.

The height of our parallelogram is a segment perpendicular to the base we have chosen.
In the diagram, we see that there is a right angle at the intersection of DC
and the dotted, blue line with length of 8 inches. Now that
we have the measures of our base and height, we can plug these values directly into
our area formula. So, we have

The area of parallelogram ABCD is 96 square inches.
That means we could fit 96 one inch by one inch squares perfectly
into the parallelogram.

Let’s try one more example to make sure we understand how to use the area formula
for parallelograms.

Exercise 2

If the area of parallelogram EFGH is 112 square meters, what must the
value of x be?

Answer:

We cannot simply “plug and chug” for this exercise as we did in the first example.
In this exercise, we are given the area of the parallelogram and must work backwards.

We are given that the height of the parallelogram is x meters. This
is the variable we will solve for. We are also given that the base of our parallelogram,
HG, is split up into two smaller lengths: one that is 11
meters long and another that is 5 meters long. Let’s
combine these to find out what our base is:

So, we’ve found that our base, b, is 16 meters long.
Let’s plug what we know into our area formula and solve for x.

Thus, the height of parallelogram EFGH is 7 meters.
(Notice that we did not say the height was 7 meters squared because
we are not talking about area; we are talking about the height.)

Now, let’s examine the area properties of triangles in order to establish the connections
they have with parallelograms.

Areas of Triangles

Similar to parallelograms, the main components of a triangle’s area are the base
and height of the triangle. The base of a triangle can be any side. The height of
a triangle is the length of the triangle’s altitude drawn to the chosen base.
The altitude of a triangle is the line a through a vertex which is perpendicular
to the opposite side of the triangle. Let’s examine these parts in the triangle
below.

The area of a triangle is one-half the product of the base and its corresponding
height. Thus, the formula is

Let’s practice using this formula by doing the following exercises.

Exercise 3

Find the area of ?KLM.

Answer:

The base of the triangle, which is segment ML, has a length of 24
centimeters. Therefore, we will be able to plug in 24 for
b when we want to use our area formula. The length of the altitude,
or the height, is also given. We see that the height here is 7 centimeters,
so we can plug 7 in for h. We have both components of
our formula, so we can plug these numbers in to find the area.

So, the area of triangle KLM is 84 square centimeters.
Let’s try one more example.

Exercise 4

Find the area of ?NPQ.

Answer:

Like we noticed in the previous exercise, finding the area of the triangle is as
simple as finding the lengths of the base and height. However, for this exercise,
we will have to be careful in order to not mistake the values of our base and height.

Let’s look for our base first. The segment NQ is our base with a length
of 10 feet. However, we see that there is a dotted line extending
from Q in the direction of point R. This line is not
part of our triangle. The only purpose it serves it to help us find the height of
our triangle, as we will see.

In this case, our height (and altitude) lies on the outside of our triangle. Recall
that, by definition, an altitude is the segment perpendicular to a base of a triangle
that passes through the opposite point. Since NQ was chosen as our
base, we needed to extend that segment out to point R in order for
it to be possible to have a line that is perpendicular to NQ that
passes through P. We see that the height of ?NPQ is
6 feet. Now, we are ready to plug the values of our base and height
into the area formula for triangles.

So, the area of ?NPQ is 30 square feet.

Now, that we are familiar with the area formulas of parallelograms and triangles,
let’s figure out what the relationship between them is.

Relationship Between Parallelograms and Triangles

Looking strictly at their formulas, we notice that the formula for a triangle has
an extra factor of one-half, whereas the area formula for a parallelogram is just
the base multiplied by the height. Let’s figure out how this extra factor of one-half
comes into play.

Triangles are the most basic type of polygon that make up all other types of polygons.
In other words, we can fit a certain number of congruent triangles into any kind
of polygon. More specifically, we learned that (n-2) triangles can
fit into an n-gon (see
polygons). Thus, when we are working with parallelograms, which are a type
of quadrilateral, we know that we can fit (4-2)=2 congruent triangles
into the parallelogram.

It is possible to split up parallelogram ABCD into two congruent triangles: ?ABD
and ?CDB.

Therefore, if we have a parallelogram of the same base and height as a triangle,
we know we can put two of these triangles together to create the parallelogram.
Let’s take a look at this by using the diagrams below.