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## Class 10 math (India)

### Course: Class 10 math (India) > Unit 6

Lesson 2: Problems involving similar triangles# Solving similar triangles: same side plays different roles

CCSS.Math:

Sal finds a missing side length in a problem where the same side plays different roles in two similar triangles. Created by Sal Khan.

## Want to join the conversation?

- is there a practice for similar triangles like this because i could use extra practice for this and if i could have the name for the practice that would be great thanks(84 votes)
- yes there are go here to see: https://www.khanacademy.org/math/geometry/similarity/triangle_similarlity/e/solving_similar_triangles_2 and http://www.mathopolis.com/questions/q.php?id=1806&site=1&ref=/geometry/triangles-similar-theorems.html&qs=1806_1807_1808_1809_1810_1811(4 votes)

- At8:40, is principal root same as the square root of any number?(8 votes)
- The principal square root is the nonnegative square root -- that means the principal square root is the square root that is either 0 or positive.(16 votes)

- I understand all of this video.. But then I try the practice problems and I dont understand them.. How do you know where to draw another triangle to make them similar? Is there a video to learn how to do this? I never remember studying it.(11 votes)
- Keep reviewing, ask your parents, maybe a tutor? Try to apply it to daily things. Hope this helps.(0 votes)

- Is it algebraically possible for a triangle to have negative sides?(3 votes)
- No because distance is a scalar value and cannot be negative. This is also why we only consider the principal root in the distance formula.(14 votes)

- How do you know that angle B is congruent to angle D?(1 vote)
- I don't get the cross multiplication?(1 vote)
- Cross Multiplication is a method of proving that a proportion is valid, and exactly how it is valid. It can also be used to find a missing value in an otherwise known proportion.

An example of a proportion:

(a/b) = (x/y)

When cross multiplying a proportion such as this, you would take the top term of the first relationship (in this case, it would be a) and multiply it with the term that is down diagonally from it (in this case, y), then multiply the remaining terms (b and x).

The outcome should be similar to this:

a * y = b * x

Now, say that we knew the following:

a=1

b=2

x=2

We wished to find the value of y.

Simply solve out for y as follows.

a*y = b*x

(1) * y = (2) * (2)

1 * y = 4

divide both sides by 1, in order to eliminate the 1 from the problem.

y = 4

Hope that helped!(11 votes)

- is there a website also where i could practice this like very repetitively(2 votes)
- You could always do the 'Solving Similar Triangles 2' over and over!(7 votes)

- when u label the similarity between the two triangles ABC and BDC they do not share the same vertex. why is B equaled to D(4 votes)
- In the first triangle that he was setting up the proportions, he labeled it as ABC, if you look at how angle B in ABC has the right angle, so does angle D in triangle BDC.(2 votes)

- At2:30, how can we know that triangle ABC is similar to triangle BDC if we know 2 angles in one triangle and only 1 angle on the other?(3 votes)
- In △ABC & △BDC

∠ABC = ∠BDC {90°}

∠BCA = ∠BCD {common ∠}

So with AA similarity criterion, △ABC ~ △BDC(3 votes)

- I have watched this video over and over again. I have also attempted the exercise after this as well many times, but I can't seem to understand and have become extremely frustrated.

Any videos other than that will help for exercise coming afterwards?(2 votes)- try going slowly. write the problem that sal did in the video down, and do it with sal as he speaks in the video. find some worksheets online- there are plenty-and if you still don't under stand, go to other math websites, or just google up the subject.(4 votes)

## Video transcript

In this problem, we're asked
to figure out the length of BC. We have a bunch
of triangles here, and some lengths of sides,
and a couple of right angles. And so maybe we can
establish similarity between some of the triangles. There's actually three different
triangles that I can see here. This triangle, this triangle,
and this larger triangle. If we can establish
some similarity here, maybe we can use
ratios between sides somehow to figure
out what BC is. So when you look at it, you have
a right angle right over here. So in triangle BDC, you
have one right angle. In triangle ABC, you
have another right angle. If we can show that they have
another corresponding set of angles are congruent
to each other, then we can show
that they're similar. And actually, both of those
triangles, both BDC and ABC, both share this angle
right over here. So if they share that
angle, then they definitely share two angles. So they both share that
angle right over there. Let me do that in
a different color just to make it different
than those right angles. They both share
that angle there. And so we know that
two triangles that have at least two
congruent angles, they're going to be
similar triangles. So we know that
triangle ABC-- We went from the unlabeled angle,
to the yellow right angle, to the orange angle. So let me write it this way. ABC. And we want to do
this very carefully here because the same
points, or the same vertices, might not play the same
role in both triangles. So we want to make sure we're
getting the similarity right. White vertex to the
90 degree angle vertex to the orange vertex. That is going to be similar
to triangle-- so which is the one that is neither a
right angle-- so we're looking at the smaller triangle
right over here. Which is the one that
is neither a right angle or the orange angle? Well it's going to be
vertex B. Vertex B had the right angle when you think
about the larger triangle. But we haven't thought
about just that little angle right over there. So we start at
vertex B, then we're going to go to the right angle. The right angle is
vertex D. And then we go to vertex C,
which is in orange. So we have shown that
they are similar. And now that we know
that they are similar, we can attempt to take
ratios between the sides. And so let's think about it. We know what the
length of AC is. AC is going to be equal to 8. 6 plus 2. So we know that AC-- what's
the corresponding side on this triangle
right over here? So you could literally
look at the letters. A and C is going to
correspond to BC. The first and the third,
first and the third. AC is going to correspond to BC. And so this is interesting
because we're already involving BC. And so what is it
going to correspond to? And then if we look at BC
on the larger triangle, BC is going to correspond to
what on the smaller triangle? It's going to correspond to DC. And it's good because
we know what AC, is and we know it DC is. And so we can solve for BC. So I want to take one more
step to show you what we just did here, because BC is
playing two different roles. On this first statement
right over here, we're thinking of BC. BC on our smaller
triangle corresponds to AC on our larger triangle. And then in the
second statement, BC on our larger
triangle corresponds to DC on our smaller triangle. So in both of these cases. So these are larger
triangles and then this is from the smaller
triangle right over here. Corresponding sides. And this is a cool
problem because BC plays two different
roles in both triangles. But now we have enough
information to solve for BC. We know that AC is equal to 8. 6 plus 2 is 8. And we know the
DC is equal to 2. That's given. And now we can cross multiply. 8 times 2 is 16 is
equal to BC times BC-- is equal to BC squared. And so BC is going to be equal
to the principal root of 16, which is 4. BC is equal to 4. And we're done. And the hardest part
about this problem is just realizing that BC
plays two different roles and just keeping your
head straight on those two different roles. And just to make it
clear, let me actually draw these two
triangles separately. So if I drew ABC separately,
it would look like this. So this is my triangle, ABC. And then this is a right angle. This is our orange angle. We know the length of this
side right over here is 8. And we know that the length of
this side, which we figured out through this problem is 4. Then if we wanted to draw BDC,
we would draw it like this. So BDC looks like this. So this is BDC. That's a little bit
easier to visualize because we've already--
This is our right angle. This is our orange angle. And this is 4, and this
right over here is 2. And I did it this
way to show you that you have to flip this
triangle over and rotate it just to have a
similar orientation. And then it might make it
look a little bit clearer. So if you found
this part confusing, I encourage you to try to flip
and rotate BDC in such a way that it seems to
look a lot like ABC. And then this ratio
should hopefully make a lot more sense.