# Lesson Video: Applications of Triangle Congruence Mathematics • 8th Grade

In this video, we will learn how to use the triangle congruence criteria SSS, SAS, ASA, and RHS to find unknown angles or sides in geometry problems.

13:34

### Video Transcript

In this video, we’re going to look at congruent triangles. We’re going to use the SSS, the SAS, and the ASA rules to find congruence. We’ll then look at how we can use this congruency to find missing angles or sides in congruent triangles.

Let’s begin by reminding ourselves of these congruency rules. The first congruency rule is the SSS rule, which stands for showing that we have three pairs of corresponding sides congruent. So, if we take these two triangles, we would start by establishing if we have a pair of congruent sides, then a second pair of congruent sides, and then finally a third pair of congruent sides. So, demonstrating that there are three pairs of congruent sides in two triangles would fulfill the SSS rule and prove that triangles are congruent.

The important thing in any of these congruency relationships is that it doesn’t matter if the triangles are rotated or flipped. They would still be congruent. They’re still the same shape and the same size no matter what orientation they’re in.

The second rule is the SAS rule. This time the A stands for the included angle between two sides. So, in our two triangles, we would demonstrate that there are two corresponding sides congruent and the angle in between them is congruent. That would show that these two triangles are congruent.

The ASA rule is the angle-side-angle rule. And this time, the side is included between two angles. So, we’d show a pair of sides congruent, like this, and two pairs of congruent angles, remembering that the side is in between or included between these two angles.

The AAS rule is similar in that it’s two pairs of corresponding angles congruent and a pair of sides congruent. However, when we’re using the AAS rule, the side does not have to be included between the two angles.

The final congruency rule is the RHS rule, which applies in right triangles. The R stands for right angle, and the H stands for hypotenuse, which is the longest side in a right triangle. To show congruence using this rule, we’d need to show that both triangles have a right angle, the hypotenuse on both triangles is congruent. And that out of the other two sides on the triangle, there’s a pair of congruent sides.

Now that we have the five rules listed, we’re going to use the first three rules in particular to solve problems. Knowing that we have congruent triangles can help us to find any missing sides or angles. Let’s have a look at our first question.

Are two triangles congruent if both triangles have the same side lengths?

Let’s start by reminding ourselves that congruent means the same shape and size. All the corresponding pairs of sides in these triangles would be the same length, and all the corresponding pairs of angles would be the same size. Sometimes, when we’re answering a question like this, it can be helpful to draw out a few triangles to investigate.

Let’s draw out this triangle which has lengths of four, five, and six units. We could draw another triangle that also has the same lengths of four, five, and six. We could even draw another one that looks like this. So, even though these triangles are all in different orientations, are they still congruent? And the answer is yes. We couldn’t draw a differently shaped triangle that has these side lengths of four, five, and six.

We can also apply the congruency rule SSS. This stands for three pairs of corresponding sides congruent. So, we can see that we have got a set of sides here in these three triangles which are all of length four. We have another corresponding set of lengths five units and a third set of corresponding length of six units.

In this example, we use the lengths of four, five, and six units. But this works for any size of triangles. If we can show that there are three pairs of corresponding side lengths or that the triangles in other words here have the same side lengths, then we would say that these are congruent. And so, our answer to the question would be yes.

In the next question, we’re told that there’s a congruency. And we’ll need to work out a missing angle.

Given that triangle 𝐴𝐵𝐶 is congruent to triangle 𝑋𝑌𝑍, find the measure of angle 𝐵.

Because we’re told that these two triangles are congruent, that means that they will have pairs of corresponding angles congruent. What we need to do here is to work out exactly which angles are congruent. Sometimes, it can be very easy to tell from a diagram and other times not quite so easy. But we can use the order of the letters to help us.

As we’re told that triangle 𝐴𝐵𝐶 is congruent to triangle 𝑋𝑌𝑍, that means that the angle at 𝐴 is corresponding with the angle at 𝑋. So, both the angle at 𝐴 and the angle at 𝑋 will be 54 degrees. In the same way, this angle 𝐵 in triangle 𝐴𝐵𝐶 is corresponding to the angle 𝑌 in triangle 𝑋𝑌𝑍. And both of these will be 52 degrees. As we were asked to find the measure of angle 𝐵, then our answer would be that this is 52 degrees.

For completeness, we can note that the angle at 𝐶 would be corresponding to the angle at 𝑍. And these would both be 74 degrees. But as we’re asked for 𝐵, that’s 52 degrees.

In the next question, before finding a missing length, we’ll need to prove that the triangles are congruent.

In the figure, 𝐵𝐷 meets 𝐴𝐸 at 𝐶, which is also the midpoint of 𝐵𝐷. Find the length of 𝐶𝐸.

The missing length that we need to find out is the length 𝐶𝐸. It may not be immediately obvious how to find this length of 𝐶𝐸. However, these triangles look very close to being the same shape and the same size, in other words, congruent. Let’s see if we have enough information about any sides or angles to prove congruence.

In the question, we’re told that 𝐵𝐷 meets 𝐴𝐸 at this point 𝐶. 𝐶 is the midpoint of 𝐵𝐷. As it’s the midpoint, then the length 𝐵𝐶 of 27 will also be the same for the length 𝐶𝐷. Then, if we consider our triangles 𝐸𝐷𝐶 and 𝐴𝐵𝐶, we could write that the length 𝐷𝐶 is equal or congruent to 𝐵𝐶 as they’re both of the length 27.

Let’s have a look at some angles next. We have this right angle at angle 𝐴𝐵𝐶, but will there be a right angle in triangle 𝐸𝐷𝐶? Well, yes, there will be. But let’s think about why. We have this line 𝐷𝐵, which is perpendicular to the line 𝐴𝐵, and so creating a right angle. But there’s also a parallel line to 𝐴𝐵. And that’s the line 𝐸𝐷. And that’s why we’ll also have a right angle here at 𝐸𝐷𝐶.

Now that we’ve shown that we have a pair of sides congruent and a pair of angles congruent, what else can we see from the diagram? We have this length 𝐴𝐵, which is 36. But we can’t say for sure that there’s any other length that’s also the same length on triangle 𝐸𝐷𝐶.

But let’s have a think about this angle at 𝐵𝐶𝐴. There would, in fact, be a congruent angle. Angle 𝐷𝐶𝐸 would be equal to the angle 𝐵𝐶𝐴 because these are vertically opposite angles.

If we look at what we’ve shown on these two triangles, we’ve got a pair of corresponding sides, a pair of congruent angles, and another pair of congruent angles. The side here is included between the two angles, so we could use the ASA rule to say that triangle 𝐸𝐷𝐶 is congruent to triangle 𝐴𝐵𝐶.

It’s important to remember that even though we have a right triangle, the RHS congruency rule would not be applicable here. To use the RHS rule, we need to show that there’s a right angle, hypotenuse, and side congruent. As we don’t know the hypotenuse length here, then we can’t use this rule. But let’s go ahead and see if we can work out the length of 𝐶𝐸.

The length that corresponds to 𝐶𝐸 in triangle 𝐴𝐵𝐶 will be this length of 𝐴𝐶. We’re not told what this length of 𝐴𝐶 is, but there’s a way that we can work it out. The Pythagorean theorem tells us that the square on the hypotenuse is equal to the sum of the squares on the other two sides. If we take a look at the triangle 𝐴𝐵𝐶, we don’t know the hypotenuse. We want to find that out. So, let’s define this length as 𝑥.

With the other two sides of 27 and 36, we can fill these into the Pythagorean theorem to give us 𝑥 squared equals 27 squared plus 36 squared. Evaluating our squares, we’ll have 𝑥 squared equals 729 plus 1296. And adding these, we have 𝑥 squared equals 2025. Taking the square root of both sides of our equation will give us that 𝑥 equals 45. Now that we have found this length of 𝐶𝐴 is 45, we can say that the corresponding length of 𝐶𝐴 on triangle 𝐸𝐷𝐶 will also be 45. And that’s our answer for the question to find the length of 𝐶𝐸.

In the final question, we’ll see how we can use congruency to help us find the area of a triangle.

The two triangles in the given figure are congruent. Work out the area of triangle 𝐴𝐵𝐶.

We’re told here that the two triangles are congruent. That means that pairs of corresponding angles will be equal and pairs of corresponding sides will be equal. We’ll need to use this fact to help us work out the area of triangle 𝐴𝐵𝐶.

We can recall that to find the area of a triangle, we multiply half times the base times the perpendicular height. When we look at triangle 𝐴𝐵𝐶, we can see that we don’t know the base length of this triangle, which is why we’ll need to use the fact that this is congruent with triangle 𝐷𝐸𝐹 to help us work out the length of 𝐵𝐶.

In this question, we weren’t given a congruency relationship, so we’ll need to establish which sides correspond to which sides. Let’s start with the hypotenuse, the longest side on triangle 𝐴𝐵𝐶. This will correspond with the longest side or hypotenuse on our other triangle. So, 𝐴𝐶 and 𝐷𝐹 will be congruent.

On triangle 𝐴𝐵𝐶, if we go from the hypotenuse down to the right angle along the line 𝐴𝐵, this corresponds to the same journey or path from the hypotenuse down to the right angle on triangle 𝐷𝐸𝐹. So, 𝐴𝐵 and 𝐷𝐸 will be the same length of 5.1. The final pair of sides 𝐵𝐶 and 𝐸𝐹 will also be congruent, and they’ll be of length 4.1.

We now have enough information to work out the area of triangle 𝐴𝐵𝐶. Filling in the values for our base length of 4.1 and the perpendicular height of 5.1, we’ll have a half times 4.1 times 5.1. We can work out 4.1 times 5.1 by calculating 41 times 51. As our values had a total of two decimal digits, then our answer will also have two decimal digits. Half of 20.91 will give us 10.455. The units here would be square units. This is our answer for the area of triangle 𝐴𝐵𝐶. Note that if we worked out the area of triangle 𝐷𝐸𝐹 instead, we would’ve got the same answer as both of these triangles are congruent.

We can now summarize what we’ve learned in this video. We saw that triangles can still be congruent even if they’re reflected or rotated; they don’t have to be in the same orientation. We reminded ourselves of the congruency rules: SSS, SAS, ASA, AAS, and RHS. And finally, when we’re proving triangles are congruent, we may need to use other angle rules, for example, remembering that vertically opposite angles are equal.