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### Course: Geometry (all content) > Unit 17

Lesson 1: Worked examples- Challenge problems: perimeter & area
- Challenging perimeter problem
- CA Geometry: Deductive reasoning
- CA Geometry: Proof by contradiction
- CA Geometry: More proofs
- CA Geometry: Similar triangles 1
- CA Geometry: More on congruent and similar triangles
- CA Geometry: Triangles and parallelograms
- CA Geometry: Area, pythagorean theorem
- CA Geometry: Area, circumference, volume
- CA Geometry: Pythagorean theorem, area
- CA Geometry: Exterior angles
- CA Geometry: Pythagorean theorem, compass constructions
- CA Geometry: Compass construction
- CA Geometry: Basic trigonometry
- CA Geometry: More trig
- CA Geometry: Circle area chords tangent
- Speed translation

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# CA Geometry: Triangles and parallelograms

21-25, triangles and parallelograms. Created by Sal Khan.

## Want to join the conversation?

- I'm al little confused on Problem 23.(3 votes)
- Here is a little hint. Remember alternate interior angle theorem. I was confused for quite a while on this kind of stuff, but once I got the hang of it, it was easy. Don't worry, you'll get it.(1 vote)

- For #22, don't u have to follow pythagreom thereom?(2 votes)
- You can use pythagorean theorem to solve the problem if that is the method that you are most comfortable with, but the way Sal is showing, in which the sum of two of the sides must be greater than the length of the other side in order for the lengths to form a triangle, works just as well. In fact, I think Sal's method is the lesser time-consuming method between the two.(2 votes)

- I am confused on problem 21 can anyone clear it up for me(2 votes)
- There is this thing called the 'triangle inequality' which says, 'given two sides of a triangle a and b (a>=b), the third side x can be no smaller than (a-b) and no larger than (a+b). [ officially stated as (a-b) < x < (a+b) ]

Imagine a triangle has legs that are whole number lengths, one is 4 and the other 7. If the third side were 3 then it wouldn't be a triangle because it would have no middle... same problem if the third side were 11. And if the third side were 12 the 7 length side and the 4 length side wouldn't be able to touch, so the figure wouldn't be closed.

So in this problem, two legs are a and the third is 15, so (a - a) < 15 < (a + a). a - a is zero so that side doesn't matter but the other side is 15 < a + a or 15 < 2a or 7.5 < a. since a is known to be a whole number it must be 8 or greater. There is only one answer 8 or greater, so that is the right one.(1 vote)

- For what value of x is the given parallelogram a rhombus represented by (6x+9) and (9x-9)?(2 votes)
- The sides of a rhombus are all congruent, so 6x+9=9x-9. Then, 3x-9=9. Then, 3x=18, and finally x=6(1 vote)

- At6:19, Sal challenges us to draw a parallelogram where parallel sides aren't equal. Then, he says that you can't. But doesn't an isosceles trapezoid fit this requirement?

Help!(1 vote)- A trapezoid is not a parallelogram because it does not fit the properties of one. I made 2 documents with geometry properties, definitions, theorems, etc. You can look through and visualize the properties of different polygons.

https://docs.google.com/document/d/1s_uhlU1Fpg8xwhKfsv3hEBR7DjFlrW0K-m2GuIA0Jc4/edit

https://docs.google.com/document/d/1fY1RPGGYEKdp4_XObMANtUa5KcTD6ukyja6OKI7BDYk/edit(2 votes)

- I'm having trouble with a question and its "in triangle PQR. Angle P is congruent to Angle R and the measure of Angle Q is twice the measure of Angle R. Fine the measure of each angle."(1 vote)
- Triangles have measures that add up to 180 degrees. Because of so, the following system of equations can be modeled:

P+R+Q=180

Now we need to substitute everything relative to R, simply because using R is easier (You can use others if you want).

R+R+2R=180

Since P=R

R=R

and Q=2R

so simplify that and you get

4R=180

R=180/4

You can plug that into a calculator or solve it yourself for the decimal, or you can simplify the fraction, depending on what they ask for.

Hope this helps! Let me know if you have more questions. :D(2 votes)

- So if you have a rectangle, are all the sides automatically 90 degree angles?(1 vote)
- What is the area of the shape that's made up of one triangle and one parallelogram,and one side is 24km,one side is 8km,and the last side is 16km.(1 vote)
- i still think it is a little unclear.can you explain a little more.(1 vote)
- what program do you use to make these videos?(1 vote)

## Video transcript

We're on problem 21. In the figure below, n
is a whole number. All right. What is the smallest possible
value for n? OK, both of these sides are n. And so this is something that's
actually really good to get this intuition. Because this shows up on all
sorts of standardized tests. And so let's think about how
small can we make n. So the lower the top of
this pyramid becomes, the smaller n becomes. If we pushed this top of the
pyramid really high then n would have to be really big. For example, if we made the
triangle into that. Then clearly, this length is
shorter than that length. We want to keep lowering it to
get as small a possible n. But what happens if we lowered
it all the way? If we flattened this triangle. If we just flattened it
all the way down. So essentially this would be the
top of it and this would be n and this would be n. I hope you're visualizing
that properly. I've flattened the triangle. So these two sides would just
go flat with the base. And so this top if I
were to do it in magenta, is right here. So this is as small
as n could get. One could argue whether this is
a triangle at all anymore. It's really a line now, because
I've squished out all of the area in there. But even in this case, n would
have to be, in the smallest case, it would be 7.5. Each n would be half
of this 15. So, as we push this base down,
that's kind of the limit that n approaches. n cannot be any smaller
than 7.5. And they tell us that
n is a whole number. So n has to be greater
than 7.5 in order for this to be a triangle. And n is a whole number,
so n is equal to 8. That's choice C. And that's an important
intuition to have in general. That the third side of a
triangle can never be bigger than two of the other
sides combined. Then you're dealing with
something else. You're not dealing
with a triangle. Even if the third side is equal
to the other two sides, then you're actually dealing
with a line. Because you'd have to squish
out all of the area of the triangle in order
to get there. Anyway, I like that problem. Next question. I think, just eyeballing
it, they want us to do the same thing. Same type of intuition. And I had my rant in the last
video about how they weren't doing problems that give
you intuition or that test your intuition. But I'll take that back, because
I think that's what they are testing now. Which of the following sets of
numbers could represent the lengths of the sides
of a triangle. 2, and 2, and 5. So this is the same
thing again. How can I have two sides of
a triangle combined being shorter than the third side. If I had a side of length 2, and
then I had another side of length 2, there's no way that
this last side could be 5. Even if I completely flattened
this triangle, 2 and 2, the longest that this last
side, this third side, could be is 4. So that can't be a triangle. Same here, let's look
at the other ones. 3, 3, and 5. That's no reason why that
can't be a triangle. 3 and 3 is 6. So even if I flattened it out
a lot, then this side can be as long as 6. And obviously I could squeeze
them together like this. 3 and 3, and then this little
third side could be something really small. Because it's anything
in between 0 and 6. So obviously it can be 5. So that's going to
be the answer. 4, 4 and 8. Same problem. I would have to squish
the triangle so much. If those are two of the sides,
4 and 4, this last side is still going to be less than 8. The only way to get it to 8 is
if I push the top of this triangle all the way down and
essentially make it a line. Once again, one side of a
triangle can't be bigger than two of the other
sides combined. In fact, it can't be
equal to two of the other sides combined. Then we're dealing
with a line. So they're really testing
that same intuition. And the choice is B. 23. Copy and paste this. OK. I try to avoid talking when
I'm copying and pasting because I think it slows
down my computer. In the accompanying diagram,
parallel lines L and M are cut by transversal T. So it's a classic parallel
line with a transversal problem. And they're parallel. That's why I did those arrows. Which statements about angle
1 and 2 must be true? I don't know if you've seen the
Khan Academy videos of the angle game, but that's what
we're going to play here, the angle game. So angle 1, if you want to
look at its corresponding angle, its corresponding angle
on the other parallel line, or with the transversal and
the other parallel line, is right there. And they're going
to be congruent. Those two angles are going
to be congruent. So you could say this is equal
to the measure of angle 1. I've picked up their terminology
well, I think. So this is equal to the
measure of angle 1. And this is obviously angle 2. You see immediately that they
have to be supplementary. Because when you add them
together you get 180 degrees. Together they go all the
way around and they kind of form a line. So you know that if this angle
and this angle are supplementary. And this angle is congruent to
angle 1, then angle 1 and angle 2 must be supplementary. So what do they say? Angle 1 is definitely
not necessarily congruent to angle 2. It's congruent to
this angle here. Angle 1 is a complement
of angle 2. Complement means you
add up to 90. No, we're talking about
supplement. So it's not that. Angle 1 is a supplement
of angle 2. There you go. And there's nothing that says
that they're right angles, that's silly. All right, next problem. Let me copy and paste it. OK. And paste it. Ready to go. What values-- let me pick a good
color-- what values of A and B make the quadrilateral
MNOP a parallelogram. For this to be a parallelogram,
the opposite sides have to be equal. And I challenge you to
experiment to draw a parallelogram where opposite
sides are parallel where the opposite sides are
also not equal. If you make two of the sides not
equal, then the other two lines aren't going to
be parallel anymore. And you can play with
that if you like. But if opposite sides are going
to be equal that means 4A plus B is equal to 21. Because they're opposite sides,
so they should be equal to each other. Similarly, 3A minus 2B should be
equal to 13 because they're opposite sides. So 3A minus 2B is equal to 13. And now we have two linear
equations with two unknowns. So this is really an Algebra
1 problem in disguise. So let's see, they want
us to solve for both. Let's see if we can
cancel out B. So let's multiply this
top equation by 2. And I'm doing that to
cancel out the B's. So you get 8A plus 2B
is equal to 42. And I did that so that the 2B
and the minus 2B cancel out. So let's add these two
equations to each. The left-hand side,
3A plus 8A is 11A. The B's cancel out; minus 2B,
plus 2B, that's no B's. Is equal to 42 plus 13 is 55. That worked out well. Divide both sides by 11, you
get A is equal to 5. Now if A is equal
to 5, what's B? Let's substitute back into
that first equation. Because you picked either. So 4 times 5 plus B
is equal to 21. 20 plus B is equal to 21. So subtract 20 from both
sides, B is equal to 1. So A is 5, B is 1. That is choice B. Problem 25. I think I have to
clear this now. Quadrilateral ABCD is
a parallelogram. If adjacent angles are
congruent, which statement must be true? All right. So let me draw a
parallelogram. Well, I'll do it quick
and dirty. So a parallelogram says that
opposite sides are parallel. That's parallel to that and
that's parallel to that. But they give us another
statement. They say if the adjacent
sides are congruent. So they're saying that this
is congruent to that. They are saying if adjacent
angles are congruent. So I don't know if they're
saying it's just one of them or all of them. But it's the same
thing actually. If that angle is congruent
to that angle, something interesting has to happen. They both have to
be 90 degrees. And I want you to think about
that a little bit. So let me just draw the kind
of bottom part of that parellelogram. And I drew it more so if you say
that this line right here is a transversal. And that since it's a
parallelogram, we know that this line is parallel
to that line. So we have a transversal
between parallel lines. This angle is congruent
to that angle, they're corresponding angles. And this angle is a supplement
to this angle. They have to add up to 180. So this angle, this red angle
and this brown angle have to be supplements. Or if we said angle
1 and angle 2. They have to add up to 180. Measure of angle 1 plus measure
of angle 2 have to be equal to 180. But then they tell
us even more. They tell you that they're
congruent. These are adjacent
angles, right? That's angle 1 and
that's angle 2. They're adjacent. So if both of these are
congruent, if their measures equal each other, they both
have to be equal to 90. So if adjacent angles
are congruent in the parallelogram, then these
angles are 90. And if these angles are 90,
those are going to be 90 by the same argument. So now we're not dealing with
just a parallelogram. Well, I did want to draw
it all filled in. It looks tacky. It's a rectangle. They say quadrilateral
ABCD is a square. It could be a square, but in
order for it to be a square they'd have to tell us that
all the sides are equal to each other. So it's not A. That could happen, it
doesn't have to be. ABCD is a rhombus. No, in a rhombus all the
sides have to be equal to each other. They didn't tell us that. ABCD is a rectangle. Sure. Because we know all the angles
now are 90 degrees. And that is choice C. See you in the next video.