### Video Transcript

In this video, weβll 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. Letβs begin by recalling the
formula that helps us to calculate the area of a triangle.

Given a triangle whose base
measures π units and whose perpendicular height measures β units, its area is a
half πβ or πβ 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 π units. At first glance, it looks like it
might be the height. But it isnβt 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? Letβs look at an example that will
help us.

Which of the following has the same
area as triangle πππΎ? Is it (A) triangle π»ππ? (B) triangle πΆππ». Is it (C) ππππ»? Is it (D) π»ππΎπΆ? Or (E) triangle πΆππ».

Letβs simply begin by highlighting
the shape weβre looking at. Triangle πππΎ is highlighted
here. Weβre looking to find a shape with
an equal area. So letβs recall how to find the
area of a triangle. For a triangle whose base is π
units and whose perpendicular height is β units, its area is a half base times
height. We also have different kinds of
shapes here, but weβll consider them in a moment.

For now, weβre simply going to
begin by finding another shape which has the same area as our triangle. We recall, of course, that the base
doesnβt necessarily need to be at the bottom of the triangle. It very much depends on its
orientation. Here, letβs say the base is equal
to the length of the line segment ππ. The height is then the
perpendicular distance between the line segment ππ and the line that has point πΎ
on it.

The lines passing through πππ»π
and πΎπΆπππ· are parallel. This means the perpendicular
distance between our two lines will always be β units. It doesnβt matter where on the line
weβre looking. And this is really useful because
we now know that the perpendicular height of any of our triangles will be β.

There are two triangles weβre going
to look at as well as a number of other shapes. Letβs look at triangle πΆππ» and
triangle ππ·π. 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; theyβre β units.

In fact, we can go further than
this. There are dashes between ππ,
πΆπ, and ππ·. These dashes indicate to us that
each of these lines are congruent. Theyβre equal in length. And so we can say that the base of
all three triangles must be π units.

Remember, we defined the line
segment ππ to be equal to π units. And weβve just seen that πΆπ and
ππ· are the same length. The area of each of our triangles
must, therefore, be a half πβ. And so weβve shown that these three
triangles are all equal in area. But what about the other shapes we
have?

Weβre specifically interested in
two quadrilaterals, ππππ» and π»ππΎπΆ. ππππ» is here, and π»ππΎπΆ 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 πππΎ to
be equal to the area of either of these trapezoids, we need to be able to prove that
the base π 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 canβt be
sure. And so all we can be sure about is
that the area of our triangle πππΎ is equal to the area of the other two triangles
we considered. From our list, we see that thatβs
(E), triangle πΆππ».

There were two other triangles on
our list. These were π»ππ and πΆππ». Now, in each of these triangles, we
do have the same perpendicular height. But we donβt know anything about
the base of these triangles. So we canβt deduce whether their
areas are the same as triangle πππΎ. The answer is (E).

Now, in fact, we can generalize
what weβve 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. Letβs visualize this by considering
two parallel lines. Our two triangles are going to
share the same base.

Now, that didnβt 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 π΄π΅. We have vertex πΆ, which lies on
the straight line parallel to our base, and vertex π·, which also lies on this same
straight line, meaning that the area of triangle π΄π΅πΆ has to be equal to the area
of triangle π΄π΅π·.

Now, be careful. This doesnβt 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 π΄. This is the shared vertex. We could add vertices π΅ and πΆ on
this line. We can also add vertices π· and πΈ
on this line. As long as line segments π΅πΆ and
π·πΈ are equal in length, then we can say that the area of triangle π΄π΅πΆ is equal
to the area of triangle π΄π·πΈ. Now that we have a formal
definition for the equality of the areas of two triangles, letβs have a look at an
example that uses it.

Which of the following has the same
area as triangle π·πΈπ΅? Is it (A) triangle πΉπ΅πΆ? (B) Triangle πΈπ·πΆ, (C) triangle
πΈπΉπΆ. Is it (D) π΄π·πΉπΈ? Or (E) π·π΅πΆπΈ.

Letβs begin by highlighting
triangle π·πΈπ΅ in our diagram. Triangle π·πΈπ΅ 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 π·πΈ is parallel to line
π΅πΆ. And so we could take the line π·πΈ
as being the base of our triangle.

The vertex of this triangle does
indeed lie on the other parallel line; itβs π΅. Thereβs another vertex that lies on
this line; itβs πΆ. And we can then join point πΆ to
the line segment π·πΈ to create a second triangle. These triangles share the same
base. They share the base π·πΈ. And their vertices π΅ and πΆ lie on
a straight line parallel to that base. And this, therefore, means that the
area of triangle π·πΈπ΅ must be equal to the area of triangle πΈπ·πΆ. The correct answer in this example
then is triangle πΈπ·πΆ.

Letβs consider another example.

Which triangle has the same area as
triangle πΏπ΅πΆ?

Weβre going to begin by
highlighting triangle πΏπ΅πΆ on our diagram and recalling what we know about the
equality of the areas of two triangles. Triangle πΏπ΅πΆ is this one. The rule weβre 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.

Letβs take the vertex πΏ. The vertex πΏ is a vertex for
several triangles in this diagram. There are specifically three
triangles which share vertex πΏ and have bases on the same line. One of these is the triangle we
highlighted. One is triangle πΏπΆπ·, and the
other is triangle πΏπ·π. Weβve 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, thatβs quite
clear. We can see that these little dashes
on line segments π΅πΆ and π·π indicate to us that theyβre of the same length. And so what weβve shown is that
triangle πΏπ΅πΆ and πΏπ·π 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 πΏπ΅πΆ is triangle πΏπ·π.

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

If the areas of triangle πΏππ΄ and
triangle ππ΄πΊ are the same, which of the following must be true? Is it (A) ππΏ equals ππΊ? (B) Line segment ππΏ is parallel
to line segment ππΊ. Is it (C) π΄π equals π΄πΊ? (D) ππΊ equals ππΏ. Or (E) line segment ππΊ is
parallel to line segment ππΏ.

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 itβs certainly not. Weβre told that the areas of
triangle πΏππ΄ and triangle ππ΄πΊ are the same. Letβs highlight these in pink and
yellow, respectively. This is triangle πΏππ΄, and we
have triangle ππ΄πΊ here.

And so letβs 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 weβre going to do to be
able to use one of these definitions is split our triangles up.

If we split triangle πΏππ΄ and
ππ΄πΊ as shown by adding the line segments π·π and πΊπΈ, respectively, then we can
begin by considering a shared vertex π΄. This corresponds to the second
definition. These triangles share the same
vertex. And so weβll be able to say that
the area of triangle π΄πΊπΈ will be equal to the area of triangle π΄π·π if their
bases are equal length and they lie along the same line. Well, in fact, the dashes on line
segment πΊπΈ and π·π indicate that theyβre 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 π΄πΊπΈ is equal to the area of triangle π΄π·π.

Well, since the areas of our larger
triangles are the same, this in turn means that the area of triangle πΊπΈπ must be
equal to the area of triangle π·ππΏ. These triangles donβt have a shared
vertex. But they do have bases of equal
length, which lie on the same line. Thatβs πΊπΈ and π·π. And so we can use our first
definition.

Now, it doesnβt really matter that
πΊπΈ and π·π are different line segments. The simple fact that they lie on
the same line and theyβre equal in length is good enough as saying that they share
the same base. And so we can say that for triangle
πΊπΈπ to have an equal area to triangle π·ππΏ, the vertices π and πΏ β thatβs
these two β must lie on a line parallel to the base. And so the lines passing through
πΊπΈπ·π and ππΏ must be parallel. If we go to our options, we see
that corresponds to option (D). The line segment ππΏ must be
parallel to the line segment ππΊ. The answer is (B).

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

Given that the area of π·πππΆ is
equal to 68 square centimeters and π΅π is equal to πΆπ, find the area of
π΄πππ΅.

Weβre given information about the
area of a quadrilateral in our shape. So letβs begin by highlighting that
quadrilateral. Itβs this one. And weβre being asked to find the
area of π΄πππ΅, which is this one. And so letβs begin by looking at
our shape. Itβs a quadrilateral with one pair
of parallel sides. So that tells us itβs a
trapezium. We might also notice that the line
segments π΄πΆ and π΅π· 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 π·ππΆ must be equal to the area of triangle π΄ππ΅. But thatβs not good enough. We still have a little bit of our
shape π·πππΆ 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 π in the
middle of our triangle, we can then see that there are two triangles coming off of
vertex π who have bases of equal length on the same line. Those bases are made up of line
segments πΆπ and ππ΅. And so this means that the area of
triangle πΆππ has to be equal to the area of πππ΅. And we can now consider the fact
that the quadrilateral π·πππΆ is made up of the triangles π·ππΆ and πΆππ. The area of π·πππΆ is 68 square
centimeters. So the sum of the area of triangle
π·ππΆ and πΆππ is also 68 square centimeters.

If we replace the area of triangle
π·ππΆ with the area of triangle π΄ππ΅, because we said that theyβre the same, and
similarly triangle πΆππ with triangle πππ΅, we see that the sum of the areas of
triangle π΄ππ΅ and πππ΅ must also be equal to 68 square centimeters. But we also said that π΄πππ΅, the
quadrilateral we highlighted earlier, is made up of these two triangles. And so that the area of π΄πππ΅ is
the same as the area of π·πππΆ. Itβs 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.