Multiple examples where Sal analyzes the similarity of triangles. Created by Sal Khan.
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- At6:11would that be considered an "improper" fraction because there are square roots in the denominators?(75 votes)
- No, because he assumes that you know to multiply the square roots by themselves to cancel out. An example : the square root of 3 times the square root of 3 is 3. That's how he cancels out the square root in the denominator.(21 votes)
- Why is it important to write the names of similar triangles in a certain order? Isn't triangle ABC the same as triangle CAB? Would similarity be affected?(22 votes)
- I think it's a good habit to arrange your triangle names. So if I told you that △ABC ≅ △XYZ, you don't have to look at a diagram to know that A and X are corresponding vertices in that congruence since they are both the first letters. If you do that, it's much easier to double check your CPCTC conclusions and that ASA or whatever was valid based on the evidence you gave.(61 votes)
- So we can't flip the figure... right? For example, the last question. It seems so tempting to be similar but it's not, unless we flip it. Is it allowed?!(13 votes)
- even if u flipped it the angle is not the included angle (btwn the 2 sides) they are not similar(2 votes)
- What would that squiggly line be called? ∆ ABC ~ ∆XYZ (~) <---- Would it be called similar?(1 vote)
- That does mean similar, which I thought that Sal Khan had mentioned it in the video, when he says "∆ABC ~ ∆XYZ is called similar."(4 votes)
- is AA rule a subset of AAA rule?(6 votes)
- They are essentially the same thing. This is because if you know 2 of the angles in a triangle, then you also can determine the third (because they must add up to 180˚).(5 votes)
- if it is proved by sss similarity postulate that
triangle ABC is similar to triangle EDF
then does it mean that angle A is equal to angle E?(2 votes)
- if it is proved by sss similarity postulate that
triangle ABC is similar to triangle EDF
then does it mean that angle A is equal to angle E?(4 votes)
- Yes, if both are SSS congruent triangles, and when put onto one another through angle preserving transformations, point A is on Point E, then yes. (aka If I can put point A on E and have the other points match up, then yes._(0 votes)
i am having trouble understanding the second example because why did he take the ratio's of longer sides and then shorter sides , how did get that formula or postulate ?(3 votes)
- The angle PQR and DEF for the correspondence PQR and EDF.The perimeter of angle
ABC is 15 and perimeter of angle PQR is 27.if BC=7and QR=9,find PR and AC.(3 votes)
aka if another triangle is double of another triangle does that mean their similar 100%?(1 vote)
- Rueben, your argument is confusing because he uses the similar language, not congruent and you support his conclusion.
If there is a scale factor of 2 (that is ALL sides are doubled) as appears to be the original question, then yes the two triangles are similar just as you noted - I just do not know what 100% means, because either two triangles are similar (I guess that would be 100%) or they are not (I guess 0%) - there is no such thing as 50% similar, so the 100% does not add to the question(4 votes)
What I want to do in this video is see if we can identify similar triangles here and prove to ourselves that they really are similar, using some of the postulates that we've set up. So over here, I have triangle BDC. It's inside of triangle AEC. They both share this angle right over there, so that gives us one angle. We need two to get to angle-angle, which gives us similarity. And we know that these two lines are parallel. We know if two lines are parallel and we have a transversal that corresponding angles are going to be congruent. So that angle is going to correspond to that angle right over there. And we're done. We have one angle in triangle AEC that is congruent to another angle in BDC, and then we have this angle that's obviously congruent to itself that's in both triangles. So both triangles have a pair of corresponding angles that are congruent, so they must be similar. 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. 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. This is two. XT is 3 plus 1 is 4. XZ is 3, and XS is 6. So you have 2 over 4, which is 1/2, which is the same thing as 3/6. 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. XYZ is similar 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. 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. So let's just figure out if the ratios between corresponding sides are a constant. So let's start with the short side. So the short side here is 3. The shortest side here is 9 square roots of 3. So we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here, is 3 square roots of 3 over the next longest side over here, which is 27. And then see if that's going to be equal to the ratio of the longest side. So the longest side here is 6, and then the longest side over here is 18 square roots of 3. So this is going to give us-- let's see, this is 3. Let me do this in a neutral color. So this becomes 1 over 3 square roots of 3. This becomes 1 over root 3 over 9, which seems like a different number, but we want to be careful here. And then this right over here-- if you divide the numerator and denominator by 6, this becomes a 1 and this becomes 3 square roots of 3. So 1 over 3 root 3 needs to be equal to square root of 3 over 9, which needs to be equal to 1 over 3 square roots of 3. At first they don't look equal, but we can actually rationalize this denominator right over here. We can show that 1 over 3 square roots of 3, if you multiply it by square root of 3 over square root of 3, this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3, times 3 is 9. So these actually are all the same. This is actually saying, this is 1 over 3 root 3, which is the same thing as square root of 3 over 9, which is this right over here, which is the same thing as 1 over 3 root 3. So actually, these are similar triangles. So we can actually say it, and I'll make sure I get the order right. So let's start with E, which is between the blue and the magenta side. So that's between the blue and the magenta side. That is H, right over here. I'll do it like this. Triangle E, and then I'll go along the blue side, F. Actually, let me just write it this way. Triangle EFG, we know is similar to triangle-- So E is between the blue and the magenta side. Blue and magenta side-- that is H. And then we go along the blue side to F, go along the blue side to I, and then you went along the orange side to G, and then you go along the orange side to J. So triangle EFJ-- EFG is similar to triangle HIJ by side-side-side similarity. They're not congruent sides. They all have just the same ratio or the same scaling factor. Now let's do this last one, right over here. Let's see. We have an angle that's congruent to another angle right over there, and we have two sides. And so it might be tempting to use side-angle-side, because we have side-angle-side here. And even the ratios look kind of tempting, because 4 times 2 is 8. 5 times 2 is 10. But it's tricky here, because they aren't the same corresponding sides. In order to use side-angle-side, the two sides that have the same corresponding ratios, they have to be on either side of the angle. So in this case, they are on either side of the angle. In this case, the 4 is on one side of the angle, but the 5 is not. So because if this 5 was over here, then we could make an argument for similarity, but with this 5 not being on the other side of the angle-- it's not sandwiching the angle with the 4-- we can't use side-angle-side. And frankly, there's nothing that we can do over here. So we can't make some strong statement about similarity for this last one.