Lesson Video: Equality of the Areas of Two Triangles Mathematics • 6th Grade

In this video, we will learn how to identify triangles that have the same area when their bases are equal in length and the vertices opposite to these bases are on a parallel line to them.

16:18

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.

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