Similarity example problems Two example problems involving similarity
Similarity example problems
- 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 as 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.
- The other thing that might jump out at you
- Is that angle CDE is an alternate interior angle with CBA.
- So we have this transveral 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 this one is just vertical.
- Either way, this angle and this angle are going to be congruent.
- So we've established that we have two triangles.
- And they have two corresponding angles which are the same.
- And that, by itself, is enough to establish similarity.
- You can actually –
- 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.
- 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're similar,
- even before doing that.
- So we already know that triangle.
- I'll try to write it color coded, so that we have
- the same corresponding vertices.
- And it's really important to know what angles
- and what sides correspond to what side,
- so that you don't mess up your ratios,
- or so that you do know what is corresponding to what.
- So we know triangle ABC is similar to triangle –
- So A – this vertex A – corresponds to vertex E over here –
- Is similar to vertex E.
- And then vertex B right over here
- corresponds to vertex D, EDC.
- Now what does that do for us?
- Well that tells us that the ratio of corresponding sides
- are going to have the same.
- They're 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
- Just for 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 BC, the length of BC over DC
- Right over here
- Is going to be equal to the length of,
- Well, we wanna figure out what CE is,
- That's what we care about
- And I'm using BC and DC cause we know those values
- So BC over DC is going to be equal to
- A coresponding side to CE
- The corrresponding side over here is CA
- Is going to be equal to CA over CE
- Corresponding sides
- This is the last and the first, last and the first
- CA over CE
- And we know what BC is
- BC right over here is 5
- We know what DC is
- It is 3
- We know what CA or AC is right over here
- CA is 4
- And now we can just solve for CE
- So we can,
- There's multiple ways that you can think about this
- You could cross multiply and you get,
- Which is really just multiplying both sides by both denominators
- So you get 5 times the length of CE
- Is equal to 3 times 4
- Which is just going to be equal to 12
- And then we get CE
- CE is equal to 12 over 5
- Which is the same thing as 2 and 2 fifths
- 2 and 2 fifths or 2 4
- This is gonna be 2 and 2 fifths
- And we're done
- We're able to use similarity to figure out this side
- Just knowing that the ratio between corresponding sides
- Are going to be the same
- Now let's do this problem right over here
- Let's do this one.
- Let me draw a little line here.
- So this is a different problem now.
- So in this problem we need to figure out what DE is
- And we all once again
- We have these 2 parallel lines like this
- And so we know corresponding angles are congruent
- So we know that angle is going to be congruent to that angle
- Cause you could view this as a transversal,
- We also know that this angle right over here
- Is going to be congruent to that angle right over there
- Once again, corresponding angles for transversal
- And also in both triangles,
- So I'm looking at triangles CBD and triangles CAE,
- They both share this angle up here
- So we've actually shown,
- Once again, we could have stopped at two angles
- But we've actually shown
- that all three angles of these two triangles,
- all three of the corresponding angles
- are congruent to each other
- So we now know,
- And once again this is an important thing to do,
- Is to make sure that you get
- Is that you write it in the right order
- when you write your similarity,
- We now know that triangle CBD is similar
- Not congruent,
- It is similar to triangle CAE
- Which means that the ratio of corresponding sides
- Are going to be constant
- So we know for example, that the ratio between CD
- Is going to be the same
- That the ratio between CB to CA,
- Lets write this down,
- We know that the ratio of CB over CA
- Is going to be equal to the ratio of CD over CE
- And we know what CB is,
- CB over here is 5
- We know what CA is
- And we have to be careful here
- It's not 3
- CA this entire side is gonna be 5 plus 3
- So this is going to be 8
- And we know what CD is
- CD is going to be 4
- And so once again, we can cross multiply
- We have 5 CE, 5 times CE is equal to
- 8 times 4
- 8 times 4 is 32
- And so CE is equal to 32 over 5
- Or this is another way to think about that
- 6 and 2 fifths
- Now we're not done because they didn't ask for what CE is
- They're asking for just this part right over here
- They're asking for DE
- So we know that this entire length, CE right over here
- This is 6 and two fifths,
- And so DE right over here,
- what we actually have to figure out
- It's going to be this entire length 6 and 2 fifths minus 4
- Minus CD right over here
- So it's going to be 2 and 2 fifths
- 6 and 2 fifths minus 4 is 2 and 2 fifths
- So we're done
- DE is 2 and 2 fifths
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