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CA Geometry: Triangles and parallelograms

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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.