### Video Transcript

In this video, we’ll learn how to
identify triangles that have the same area when their bases are equal in length and
their vertices opposite to these bases are on a parallel line to them. Let’s begin by recalling the
formula that helps us to calculate the area of a triangle.

Given a triangle whose base
measures 𝑏 units and whose perpendicular height measures ℎ units, its area is a
half 𝑏ℎ or 𝑏ℎ divided by two. The area units are square units,
such as square centimeters and square meters. Be aware though, the height
dimension must be perpendicular to the base. In any triangle, we might be given
a superfluous measurement, for example, this measurement 𝑙 units. At first glance, it looks like it
might be the height. But it isn’t perpendicular to the
base of the triangle. So this measurement is
irrelevant. So given this information, how can
we deduce where the two triangles have an equal area? Let’s look at an example that will
help us.

Which of the following has the same
area as triangle 𝑁𝑀𝐾? Is it (A) triangle 𝐻𝑁𝑍? (B) triangle 𝐶𝑁𝐻. Is it (C) 𝑍𝑂𝑋𝐻? Is it (D) 𝐻𝑁𝐾𝐶? Or (E) triangle 𝐶𝑍𝐻.

Let’s simply begin by highlighting
the shape we’re looking at. Triangle 𝑁𝑀𝐾 is highlighted
here. We’re looking to find a shape with
an equal area. So let’s recall how to find the
area of a triangle. For a triangle whose base is 𝑏
units and whose perpendicular height is ℎ units, its area is a half base times
height. We also have different kinds of
shapes here, but we’ll consider them in a moment.

For now, we’re simply going to
begin by finding another shape which has the same area as our triangle. We recall, of course, that the base
doesn’t necessarily need to be at the bottom of the triangle. It very much depends on its
orientation. Here, let’s say the base is equal
to the length of the line segment 𝑀𝑁. The height is then the
perpendicular distance between the line segment 𝑀𝑁 and the line that has point 𝐾
on it.

The lines passing through 𝑀𝑁𝐻𝑋
and 𝐾𝐶𝑍𝑂𝐷 are parallel. This means the perpendicular
distance between our two lines will always be ℎ units. It doesn’t matter where on the line
we’re looking. And this is really useful because
we now know that the perpendicular height of any of our triangles will be ℎ.

There are two triangles we’re going
to look at as well as a number of other shapes. Let’s look at triangle 𝐶𝑍𝐻 and
triangle 𝑋𝐷𝑂. Both of these triangles are drawn
by connecting two points on one parallel line to a third point or vertex on the
second parallel line. And this means their heights, as we
saw, are all the same; they’re ℎ units.

In fact, we can go further than
this. There are dashes between 𝑀𝑁,
𝐶𝑍, and 𝑂𝐷. These dashes indicate to us that
each of these lines are congruent. They’re equal in length. And so we can say that the base of
all three triangles must be 𝑏 units.

Remember, we defined the line
segment 𝑀𝑁 to be equal to 𝑏 units. And we’ve just seen that 𝐶𝑍 and
𝑂𝐷 are the same length. The area of each of our triangles
must, therefore, be a half 𝑏ℎ. And so we’ve shown that these three
triangles are all equal in area. But what about the other shapes we
have?

We’re specifically interested in
two quadrilaterals, 𝑍𝑂𝑋𝐻 and 𝐻𝑁𝐾𝐶. 𝑍𝑂𝑋𝐻 is here, and 𝐻𝑁𝐾𝐶 is
this one. We have two quadrilaterals which
have one pair of parallel sides each. And so they are trapezoids. And we recall that the area of a
trapezoid is half the sum of the parallel sides times the perpendicular height
between them.

For the area of triangle 𝑁𝑀𝐾 to
be equal to the area of either of these trapezoids, we need to be able to prove that
the base 𝑏 is equal to the sum of the parallel sides in either of our
trapezoids. There is nothing to indicate that
this is the case. It might be, but we can’t be
sure. And so all we can be sure about is
that the area of our triangle 𝑁𝑀𝐾 is equal to the area of the other two triangles
we considered. From our list, we see that that’s
(E), triangle 𝐶𝑍𝐻.

There were two other triangles on
our list. These were 𝐻𝑁𝑍 and 𝐶𝑁𝐻. Now, in each of these triangles, we
do have the same perpendicular height. But we don’t know anything about
the base of these triangles. So we can’t deduce whether their
areas are the same as triangle 𝑁𝑀𝐾. The answer is (E).

Now, in fact, we can generalize
what we’ve seen here. These concepts rely on working with
pairs of parallel sides. The first rule is that if two
triangles share the same base and their vertices lie on a straight line parallel to
the base, then the two triangles are equal in area. Let’s visualize this by considering
two parallel lines. Our two triangles are going to
share the same base.

Now, that didn’t actually happen in
our previous example. But we saw that the bases were of
equal length and lay on those parallel lines. The base of our two triangles is
the line segment 𝐴𝐵. We have vertex 𝐶, which lies on
the straight line parallel to our base, and vertex 𝐷, which also lies on this same
straight line, meaning that the area of triangle 𝐴𝐵𝐶 has to be equal to the area
of triangle 𝐴𝐵𝐷.

Now, be careful. This doesn’t mean the triangles are
congruent. What it does mean is that their
bases and perpendicular heights are equal in length, and thus their areas are. We can also say that if two
triangles share the same vertex and have bases of equal length along the same line,
then the heights of the triangles must be equal. And the areas of the two triangles
must also be equal.

To demonstrate, we might start with
a vertex 𝐴. This is the shared vertex. We could add vertices 𝐵 and 𝐶 on
this line. We can also add vertices 𝐷 and 𝐸
on this line. As long as line segments 𝐵𝐶 and
𝐷𝐸 are equal in length, then we can say that the area of triangle 𝐴𝐵𝐶 is equal
to the area of triangle 𝐴𝐷𝐸. Now that we have a formal
definition for the equality of the areas of two triangles, let’s have a look at an
example that uses it.

Which of the following has the same
area as triangle 𝐷𝐸𝐵? Is it (A) triangle 𝐹𝐵𝐶? (B) Triangle 𝐸𝐷𝐶, (C) triangle
𝐸𝐹𝐶. Is it (D) 𝐴𝐷𝐹𝐸? Or (E) 𝐷𝐵𝐶𝐸.

Let’s begin by highlighting
triangle 𝐷𝐸𝐵 in our diagram. Triangle 𝐷𝐸𝐵 is this one. We need to find the shape which has
the same area. And so we remember that if two
triangles share the same base and their vertices lie on a straight line parallel to
that base, then the triangles are equal in area. We do indeed have a pair of
parallel lines in our diagram. Line 𝐷𝐸 is parallel to line
𝐵𝐶. And so we could take the line 𝐷𝐸
as being the base of our triangle.

The vertex of this triangle does
indeed lie on the other parallel line; it’s 𝐵. There’s another vertex that lies on
this line; it’s 𝐶. And we can then join point 𝐶 to
the line segment 𝐷𝐸 to create a second triangle. These triangles share the same
base. They share the base 𝐷𝐸. And their vertices 𝐵 and 𝐶 lie on
a straight line parallel to that base. And this, therefore, means that the
area of triangle 𝐷𝐸𝐵 must be equal to the area of triangle 𝐸𝐷𝐶. The correct answer in this example
then is triangle 𝐸𝐷𝐶.

Let’s consider another example.

Which triangle has the same area as
triangle 𝐿𝐵𝐶?

We’re going to begin by
highlighting triangle 𝐿𝐵𝐶 on our diagram and recalling what we know about the
equality of the areas of two triangles. Triangle 𝐿𝐵𝐶 is this one. The rule we’re going to use is that
if two triangles share the same vertex and have bases of equal length on the same
line, then their areas must be equal.

Let’s take the vertex 𝐿. The vertex 𝐿 is a vertex for
several triangles in this diagram. There are specifically three
triangles which share vertex 𝐿 and have bases on the same line. One of these is the triangle we
highlighted. One is triangle 𝐿𝐶𝐷, and the
other is triangle 𝐿𝐷𝑋. We’ve shown that these three
triangles share the same vertex and they have bases on the same line. We need to find the bases which are
of equal length. And actually, that’s quite
clear. We can see that these little dashes
on line segments 𝐵𝐶 and 𝐷𝑋 indicate to us that they’re of the same length. And so what we’ve shown is that
triangle 𝐿𝐵𝐶 and 𝐿𝐷𝑋 have the same length bases and the same perpendicular
heights. And so their areas must absolutely
be the same. The triangle that has the same area
as triangle 𝐿𝐵𝐶 is triangle 𝐿𝐷𝑋.

In our next example, we’ll look at
how we can use the converse of these definitions.

If the areas of triangle 𝐿𝑁𝐴 and
triangle 𝑌𝐴𝐺 are the same, which of the following must be true? Is it (A) 𝑌𝐿 equals 𝑁𝐺? (B) Line segment 𝑌𝐿 is parallel
to line segment 𝑁𝐺. Is it (C) 𝐴𝑁 equals 𝐴𝐺? (D) 𝑌𝐺 equals 𝑁𝐿. Or (E) line segment 𝑌𝐺 is
parallel to line segment 𝑁𝐿.

Now, be a little bit careful. Each of these lines is drawn on a
two-dimensional plane. This is not a 3D shape. It does look a little bit like a
square-based pyramid, but it’s certainly not. We’re told that the areas of
triangle 𝐿𝑁𝐴 and triangle 𝑌𝐴𝐺 are the same. Let’s highlight these in pink and
yellow, respectively. This is triangle 𝐿𝑁𝐴, and we
have triangle 𝑌𝐴𝐺 here.

And so let’s recall what we know
about the equality of the areas of two triangles. We know that the areas of two
triangles will be equal if these triangles share the same base and their vertices
lie on a straight line parallel to the base, or if the triangles share the same
vertex and have bases of equal length along the same line. And so what we’re going to do to be
able to use one of these definitions is split our triangles up.

If we split triangle 𝐿𝑁𝐴 and
𝑌𝐴𝐺 as shown by adding the line segments 𝐷𝑁 and 𝐺𝐸, respectively, then we can
begin by considering a shared vertex 𝐴. This corresponds to the second
definition. These triangles share the same
vertex. And so we’ll be able to say that
the area of triangle 𝐴𝐺𝐸 will be equal to the area of triangle 𝐴𝐷𝑁 if their
bases are equal length and they lie along the same line. Well, in fact, the dashes on line
segment 𝐺𝐸 and 𝐷𝑁 indicate that they’re the same length. And we can quite clearly see from
the picture that they do lie along the same line. This must mean that the area of
triangle 𝐴𝐺𝐸 is equal to the area of triangle 𝐴𝐷𝑁.

Well, since the areas of our larger
triangles are the same, this in turn means that the area of triangle 𝐺𝐸𝑌 must be
equal to the area of triangle 𝐷𝑁𝐿. These triangles don’t have a shared
vertex. But they do have bases of equal
length, which lie on the same line. That’s 𝐺𝐸 and 𝐷𝑁. And so we can use our first
definition.

Now, it doesn’t really matter that
𝐺𝐸 and 𝐷𝑁 are different line segments. The simple fact that they lie on
the same line and they’re equal in length is good enough as saying that they share
the same base. And so we can say that for triangle
𝐺𝐸𝑌 to have an equal area to triangle 𝐷𝑁𝐿, the vertices 𝑌 and 𝐿 — that’s
these two — must lie on a line parallel to the base. And so the lines passing through
𝐺𝐸𝐷𝑁 and 𝑌𝐿 must be parallel. If we go to our options, we see
that corresponds to option (D). The line segment 𝑌𝐿 must be
parallel to the line segment 𝑁𝐺. The answer is (B).

In our very final example, we’ll
look at applying these definitions to help us actually calculate an area.

Given that the area of 𝐷𝑀𝑌𝐶 is
equal to 68 square centimeters and 𝐵𝑋 is equal to 𝐶𝑌, find the area of
𝐴𝑀𝑋𝐵.

We’re given information about the
area of a quadrilateral in our shape. So let’s begin by highlighting that
quadrilateral. It’s this one. And we’re being asked to find the
area of 𝐴𝑀𝑋𝐵, which is this one. And so let’s begin by looking at
our shape. It’s a quadrilateral with one pair
of parallel sides. So that tells us it’s a
trapezium. We might also notice that the line
segments 𝐴𝐶 and 𝐵𝐷 are the diagonals of this trapezium or trapezoid.

Now, what happens is that when we
divide our trapezoid into four triangles by using its diagonals, the area of the two
triangles opposite one another whose bases are not the parallel sides are equal. In other words, the area of
triangle 𝐷𝑀𝐶 must be equal to the area of triangle 𝐴𝑀𝐵. But that’s not good enough. We still have a little bit of our
shape 𝐷𝑀𝑌𝐶 left over. And so we recall that if two
triangles share the same vertex and then have bases of equal lengths on the same
line, then their areas must be equal.

If we consider vertex 𝑀 in the
middle of our triangle, we can then see that there are two triangles coming off of
vertex 𝑀 who have bases of equal length on the same line. Those bases are made up of line
segments 𝐶𝑌 and 𝑋𝐵. And so this means that the area of
triangle 𝐶𝑀𝑌 has to be equal to the area of 𝑋𝑀𝐵. And we can now consider the fact
that the quadrilateral 𝐷𝑀𝑌𝐶 is made up of the triangles 𝐷𝑀𝐶 and 𝐶𝑀𝑌. The area of 𝐷𝑀𝑌𝐶 is 68 square
centimeters. So the sum of the area of triangle
𝐷𝑀𝐶 and 𝐶𝑀𝑌 is also 68 square centimeters.

If we replace the area of triangle
𝐷𝑀𝐶 with the area of triangle 𝐴𝑀𝐵, because we said that they’re the same, and
similarly triangle 𝐶𝑀𝑌 with triangle 𝑋𝑀𝐵, we see that the sum of the areas of
triangle 𝐴𝑀𝐵 and 𝑋𝑀𝐵 must also be equal to 68 square centimeters. But we also said that 𝐴𝑀𝑋𝐵, the
quadrilateral we highlighted earlier, is made up of these two triangles. And so that the area of 𝐴𝑀𝑋𝐵 is
the same as the area of 𝐷𝑀𝑌𝐶. It’s 68 square centimeters.

In this video, we used the formula
for area of a triangle to generalize a definition for the equality of the areas of
two triangles. We said that if two triangles have
the same base and their vertices lie on a straight line parallel to the base, then
their areas are equal. We also saw that if two triangles
share the same vertex and have bases of equal length along the same line, then the
heights of the triangles must be equal and the areas of the two triangles must also
be equal. We also saw how we can reverse
these definitions to use the fact that the area of two triangles are equal to deduce
properties about parallel lines and equal lengths.