Similar Triangles - Maths GCSE Revision
If the similarity ratio of two similar figures is a: b, then. 1. the ratio of their perimeters is also a: b. 2. the ratio of their areas is a2: b2. SWBAT demonstrate the proportional relationship between the corresponding sides and angles of right triangles. • SWBAT solve problems using similar. COMMON CORE STATE STANDARDS. G-SRT Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric.
So we can write, triangle ACE is going to be similar to triangle-- and we want to get the letters in the right order. So where the blue angle is here, the blue angle there is vertex B. Then we go to the wide angle, C, and then we go to the unlabeled angle right over there, BCD.
So we did that first one. Now let's do this one right over here. This is kind of similar, but it looks, just superficially looking at it, that YZ is definitely not parallel to ST.
So we won't be able to do this corresponding angle argument, especially because they didn't even label it as parallel. And so you don't want to look at things just by the way they look. You definitely want to say, what am I given, and what am I not given? If these weren't labeled parallel, we wouldn't be able to make the statement, even if they looked parallel.
One thing we do have is that we have this angle right here that's common to the inner triangle and to the outer triangle, and they've given us a bunch of sides. So maybe we can use SAS for similarity, meaning if we can show the ratio of the sides on either side of this angle, if they have the same ratio from the smaller triangle to the larger triangle, then we can show similarity. So let's go, and we have to go on either side of this angle right over here. Let's look at the shorter side on either side of this angle.
So the shorter side is two, and let's look at the shorter side on either side of the angle for the larger triangle. Well, then the shorter side is on the right-hand side, and that's going to be XT.
So what we want to compare is the ratio between-- let me write it this way. We want to see, is XY over XT equal to the ratio of the longer side? Or if we're looking relative to this angle, the longer of the two, not necessarily the longest of the triangle, although it looks like that as well.
Is that equal to the ratio of XZ over the longer of the two sides-- when you're looking at this angle right here, on either side of that angle, for the larger triangle-- over XS? And it's a little confusing, because we've kind of flipped which side, but I'm just thinking about the shorter side on either side of this angle in between, and then the longer side on either side of this angle.
Tenth grade Lesson Error Analysis: Finding Sides of Similar Triangles
So these are the shorter sides for the smaller triangle and the larger triangle. These are the longer sides for the smaller triangle and the larger triangle. And we see XY. XZ is 3, and XS is 6.
Determining similar triangles
So the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle, for both triangles, the ratio is the same. So by SAS we know that the two triangles are congruent. But we have to be careful on how we state the triangles.
We want to make sure we get the corresponding sides. And I'm running out of space here. Let me write it right above here.
We can write that triangle XYZ is similar to triangle-- so we started up at X, which is the vertex at the angle, and we went to the shorter side first. So now we want to start at X and go to the shorter side on the large triangle. So you go to XTS.
Now, let's look at this right over here. So in our larger triangle, we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it. And if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything.
And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity. So there's no similarity going on here.
Determining similar triangles (video) | Khan Academy
There are some shared angles. This guy-- they both share that angle, the larger triangle and the smaller triangle. So there could be a statement of similarity we could make if we knew that this definitely was a right angle.
Then we could make some interesting statements about similarity, but right now, we can't really do anything as is. Let's try this one out, this pair right over here. So these are the first ones that we have actually separated out the triangles. So they've given us the three sides of both triangles.
- Division of a right triangle into similar triangles using an altitude.
- Introduction to triangle similarity
- Similar Triangles
So let's just figure out if the ratios between corresponding sides are a constant. Triangle similarity Video transcript In this first problem over here, we're asked to find out the length of this segment, segment CE. And we have these two parallel lines. AB is parallel to DE. And then, we have these two essentially transversals that form these two triangles.
So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles. So they are going to be congruent. So we have this transversal right over here. And these are alternate interior angles, and they are going to be congruent. Or you could say that, if you continue this transversal, you would have a corresponding angle with CDE right up here and that this one's just vertical.
Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles and two of the corresponding angles are the same. And that by itself is enough to establish similarity. We actually could show that this angle and this angle are also congruent by alternate interior angles, but we don't have to.
So we already know that they are similar. And actually, we could just say it. Just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar, even before doing that. So we already know that triangle-- I'll color-code it so that we have the same corresponding vertices.
And that's really important-- to know what angles and what sides correspond to what side so that you don't mess up your, I guess, your ratios or so that you do know what's corresponding to what. It's similar to vertex E. And then, vertex B right over here corresponds to vertex D. Now, what does that do for us? Well, that tells us that the ratio of corresponding sides are going to be the same.
They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it in just the way that we've written down the similarity. If this is true, then BC is the corresponding side to DC. So we know that the length of BC over DC right over here is going to be equal to the length of-- well, we want to figure out what CE is.
That's what we care about.
Area Of Similar Triangles
The corresponding side over here is CA. It's going to be equal to CA over CE. This is last and the first. Last and the first.