You can try this warm-up questions while waiting for others. Today, our topic is perimeter, area, and length. We are doing geometry. So this is a regular hexagon, A, B, C, D, E, F, has perimeter of 36. Each vertex is the center of a circle, this one. The radius of each circle is equal to, of course, a half of the length of its size. And the question asks, what is perimeter of this shaded shape? So we, I think Sveta and I sent the email with a link to the handouts, with an attachment to the handouts that you could have printed, but even not, I think it's a big trouble. So I will, you know, quickly. Okay. Oh, thank you, David, for launching the poll because I was drawing, I have to switch between the two. Okay, so for example, I can just, you know, complete the circle here. It's one way, it's just, you only need to draw one. Maybe this one is better. You just need to complete one circle to see if we can. Of course, the thick arcs, but they're all identical, right? So I only need to calculate one. Okay, so if, I think probably you remember that the, for a regular hexagon, this angle is 120 degrees, right? But suppose you don't remember, then just remember for an n-gon, the sum of all angles n minus two times 180. So for a hexagon to be 720 divides by 6 gives 120, but 120 is exactly a third of the circle, right? Because a full circle is 260. So for each circle, it's going to be 2 pi r, 2 pi r, and then we have to divide by 3, right? Because it's only a third, but then we have to multiply by 6 because there's 6 arcs. So in the end, that's going to be 4 pi r, and we know that r, oh perimeter is 36 divided by 6, so that's going to be, each side is 6, so r is going to be equal to 3, right? So in the end, you would have the answer b equal to 12, 12 pi. So a lot of the questions you see would just, you know, use symmetry. We can help you need to calculate a small part of it. Okay, so our topic, now we go to a totally different topics from what we've learned before. We're going to have the last three lessons on geometry, two lessons on 2D, today do with perimeter, area, and next week we're going to be doing angles. So let's review the basics of 2D geometry. First, congruent triangles, right? The test that we should be familiar with, psi, psi, psi. If two angles, there's pairs of psi, three pairs of psi that are the same, then the two triangles are congruent. The next test is side, angle, side. If you have an angle that's sandwiched between two pairs of equal sides like this, then it's also, they're also congruent. Or angle, side, angle. In this case, we have a side that is sandwiched between two pairs of angles, equal angles, then they are also congruent and we'll be using a lot of these tests in solving problems. And then it gets more exciting when we move to similar triangles, right? Those, they are the same shape, but just different by scaling factors. So we also have a psi, psi, psi test, but the sides are not equal, but the three ratios are equal, right? AB over A prime, B prime, and AC over A prime, C prime, and BC over B prime, C prime. Then we also have a side, angle, side test. Also, angles, two angles that are equal, sandwiched between the two sides, but they're not equal, the ratios. This is replaced the equal side by the equal side of two ratios. And finally, angle, angle. If two angles are the same, then the third angle is also the same and the two triangles are similar. And a lot of the time, you're gonna just use similar triangles when you have one triangle and you have a line that's parallel to this, right? So this small triangle, A prime, E, F's gonna be similar to A prime, B prime, C prime. I think that would be one of the most common situations when you have similar triangles. And the thing, one thing with similar triangles that we use a lot, that's the ratio of areas, right? If two triangles are similar, then the ratio of their areas is equal to the square of the ratio of one, of the length of side of, length of two sides. And this, this formula is actually very general. It applies to anything because we know that in general, area is square of length. So it applies to circles as well. It applies to any two polygons. If they are similar, if one is a scale version of the other, if you want to find the ratio of areas, you just need to find ratio of one length in any dimensions and then square that ratio. So that's also a very useful formula. Next thing, we have circles, right? I mean, it's a large topic, so I'm just trying to, you know, review the things that I hope would be the most useful. If we have any, here we have a circle, right? And if we count here, then we choose AB on the circle, then you would have OA because A equal to OB, then immediately ABO is an isosceles triangle. And then we know that C here, if C is the midpoint of AB, then C is the height of triangle AOB, OC perpendicular AB, then it also the median, it also the bisector, right? It's also the bisector. And then if a line is tangent to the ratio, to the circle at one point P here, then OP is perpendicular to that line. And you have two circles that are tangent to each other, that means they meet at exactly one point, then almost always you want to connect, to draw the lines that connect the two circles. And you want to use a relationship that the distance between the two centers is equal to the sum of the two radii. So I think almost always you need to use it. And of course, if we have a diameter of the, have a triangles that are inscribed in a circle, and one side is a diameter of the circle that goes through the center here, then this angle is 90 degrees. We're going to do a lot of angle next year, but I'm sorry, next lesson, but at least this is a very popular result that we want to use. Then the next thing is Pythagorean theorems we use a lot, right? We're going to go through today, so I'm going to stop here. Finally, some special triangles, 90-60-30 triangle is the hub of equilateral triangles. So we know this one is 30. I know that if I draw a midpoint over here, E for example, then I know that AE equals to EC, ME is also equal to that, and EC is also an equilateral triangle. And another thing I want to memorize is that MC, if I want equal to one, then the length of AC equal to two, and using the Pythagorean theorem, we know that the length of AM equal to square root of two, right? So that's because of three. So those are the few things that you want to remember by heart. So when solving geometry problems, lots of the time, we need to draw additional points, additional line segments, trying to label point segments, areas, and have to keep track of things, observe how shapes interact, intersect and meet, and then you have to imagine when shapes move, flip, and rotate. And a lot of the time, we actually use algebra to describe and solve geometry questions, just because we have a lot of ratios, and you know, we have formulas like Pythagorean formula, so you're going to see a lot of algebra as well. Okay, that's, I think, hopefully the most useful things. So let's see. First question, the center of a square with dimensions of two by two overlaps with the center of another square, which also has dimension two by two. What is the area of the common part of these two squares? So two equal congruent squares, side length is two. This one is the center of, something that we want to label it. Yeah. There's several different ways to do that. If you want to get the answer, you can you can use some tricks to get the answer right away. If you want you to prove in general, you probably need to draw some actual lines. That's always, almost always the case with some sort of funky area, with add or subtract, or use some tricks that we will discuss in this question. Let's put B and C. OK, the A as well. So one trick you can use to get the solution that they don't tell exactly where to intersect, you can find some special position when they intersect each other in a special position to get the answer. And in geometry, whatever information they give you, you have to use it. So if you have three pieces of information, you have to use all, using two won't help. So in this case, because E is a center, the center of the circle, so I may want to draw maybe this line and notice that E is the point of DB. Or I can try AC as well and make use of the fact that E is the center of this intersection of these two segments. So what do we notice? And don't forget that this angle is 90 degrees, right? So lots of time I find useful that we can draw all the notations that we can in the pictures. OK, so I may want to name this M and this N. And when I draw it, do you notice anything? These two triangles, E and C and EBM, they look like they are congruent. So you can suspect it. And then can we prove it? Remember, when two triangles are congruent, you need at least two sides have to be the same. So in this case, when I look at that, I saw that EB is already equal to MC, so EC. So I already have two sides that are equal. And this is 45 degrees, right? Because it's half of 90 degrees. And same, this one also 45. So I only need one more thing, which is this angle. So usually you can wish they were congruent. You guess that and they do it backward. And I need to prove that this angle equal to this angle. And are they equal? They are, right? Because they both, I think, supplementary or complementary. They both add up to this one. So you have this alpha in the middle, this one. And this angle and this angle add up to 90 degrees. And this angle and this angle, I go and go with 1 and 2. And this one is 3. So 1 and 3 is 90 degrees. And 2 plus 3 is also 90. So that means 1 and 2 are the same. And then that means EBm equal to ECn. And now we are all set, right? So then we know that we can move this piece here over here. And then the areas EMCn is just equal to BEC, which is a quarter of the square. And the square is a 4. So a quarter of that is B, which is 1. And in this case, in the exam, I think, as I said earlier, you can just use a special case. You can right away consider the special case in which the two squares, they're overlapped at EBC, which is a special situation. You just turn it that way. Or it can also overlap at this special position. It's called consider a special case or an extreme case. It's not a proof, but we can get away with that in the exam. You can also consider this case. When they go like this, they intersect at this square, which is also a quarter. And then if you have two cases, then you can, oh, maybe it's always equal to 1. Next one, tangent circles. Two circles are located inside a square with sides of length one. Okay, so this square as shown in the diagram. The circles are tangent to the side of the square and to each other, which is sort of obvious from the drawing. And now the question, what is the sum of the length of the radii of the circles? So you can start by drawing the square, drawing two circles, mark their centers, and because they are tangents, we discussed what we should do when you see two circles that are tangents, draw as many radii as you can, all the special radii, and see what we can do from there. And a lot of times in geometry, if you draw a perfect, like you suspect some points are on a line, you can prove it, you can use symmetry, or if you have a perfect drawing, usually you draw with a compass and a straight edge, then a lot of things you can just take for granted, sort of the result are obvious from the drawing. Okay, I have answer coming in, so let me just draw, right, I'm not doing anything, that's just because they say that the circles are tangent to each other and tangent to the square, that means I have to represent all of this in the picture, right, so tangent to the square means these two are the radii and they're perpendicular to a, b, and b, c. Similarly, these two are radii and perpendicular to this, right, I'm not doing anything fancy, I'm just, and finally, the two circles are tangent to each other, then I have to connect them, right, does it make sense, I connect them, and this is gonna be another radii, the radius is another one, and you know, you could, you might as well guess that due to symmetry, this thing actually is gonna lie on the diagonal with bd, bd, and that means this angle is 45 degrees, and then you may want to label it, like, for example, this is, you can call it a, this is a, a, so transfer as much as possible from words to symbols, points and segments, on the pictures, and, you know, still not enough, probably I have to do a little bit more, so how about I go all the way down here, and all the way down here, right, and now we know that this angle is 45 degrees, you can see it, this length here, I could call it h here, then you know that bc equal to 1, right, so this one's gonna be, fh gonna be 1 minus a plus b, because this is b as well, right, and same here, if you look at, let's look at the vertical direction, if you look at the horizontal directions, this is also a, and this b, and this 1, so it's gonna be 1 minus a plus b, right, and by now I think I've made use of everything, all the information, this to learn I got from the fact that this is a square, and then also now when I figure out this, I'm confident that this angle is 45 degrees, ehf is an isosceles right triangle, so let's wait for another 20 seconds to see if we can come up with the answer. we are asked to find what is a plus b, right? So, oh, when we were talking about Pythagorean, you also want to remember for an isosceles right triangle, what's the ratio if this one is, very good, if this one and one, so using Pythagorean, this one is square root of two, right? So if you do over here, we know that EF gonna be square root of two times one minus a plus b, right? And EF is also equal to a plus b. So again, we have to use algebra. So it'd be two, we're gonna have a square root of two equals to, if you put a plus b on the other side, we're gonna have a one plus square root of two times a plus b, right? So what's gonna be? So the answer is gonna be square root of two divided by one plus square root of two, is that correct? And then if we multiply by square root of two minus one, it's gonna be square root of two. And then if you multiply by this one, remember the identity we talked earlier, divided by one, so that's gonna be two minus square root of two, right? And multiply both the numerator and denominator by square root minus two minus one. And remember, we use this identity, a plus b times a minus b equal to a square minus b square. So lots of algebra in geometry, but it's good. We have some student who got the correct answer, which is d, okay? What's interesting about this problem, which really, I just wanna say this really quickly, is that when you have two circles, it seems no matter how big the circles are, your answer is two minus root two, right? And you think that that would mean that if you had only one circle, because the other circle is super tiny, it would be the same. But if you had a single circle that filled the whole square, its radius is only one half. Oh, that's a very interesting point. Yes, that's very, so it's like, you can sort of consider the extreme the special case and it didn't- The extreme case seems to not work. And there's a reason for that, is because if you have a single circle, you can still have a smaller circle in the tiny corner. Exactly, yeah. Oh, you need to use- Even this one, yeah, even this one occupies like the radius here, you still have a tiny one over here, right? Yeah, so sometimes special case works, sometimes not quite. Okay, thank you for that observation, Sue. Okay, so this is a tricky one. Again, a funky area. P is the area shaded with vertical lines. So you got to be careful with the vertical lines. And S is area shaded with horizontal lines. The diameters of the circles are 6, 4, 4, and 2. Therefore, how is P compared with S? And for this, I think I probably just vertical lines. So this one, yeah, this is P, just to make sure we don't get confused. So these three add up to P and this one to S. Yes, you may need to use the area of a circle, it's pi r squared. But remember when the area has a weird shape, you have to find a way to calculate it indirectly. Don't do it the direct way, sometimes it's sort of impossible. And this question actually, David, we can also solve it using the Stasher case. You can try the trick and find a Stasher, just consider a Stasher case to find the answer. Actually, when I look at this, it's pretty easy to see that all but one of the solutions just don't make sense to be always true. Because it just doesn't work for everything. And if you see what I'm saying, because I can't give too many details, then you know the answer has to be. Oh, good, I think we sort of figured it out, right? So we cannot calculate it, you know, we have no information of where these things can intersect with each other, so it's impossible to connect, to calculate them directly, but we notice that, yes, here, if we also, it's important of laboring, if I label these areas, this A, B, C, the white parts, and we notice right away that S plus A plus B plus C is equal to the big circle, right? The biggest one, which is pi, and times six square, which is 36 pi, right? So at least if I sort of use these complementary areas and I can calculate that, and similarly, if I add P with A plus B plus C, if I add them, like break P into three parts and add them, then I get the total areas of these three circles, which is equal to pi, right? And times, I have four square plus four square plus two square, and that's 16 plus 16 plus four, which is also 36 pi, right? So that means P equal to S. But really, because they do not tell us where the center of this one, you can consider the most sort of extreme case is when the big sphere and these two are, you know, just do not intersect the circle at all, right? And that means P, S is just the entire big circle, and P is just some of these three circles, and A, B, C equal to zero. Sort of cheating, but you can get the answer that way. If you want to prove it, you use it generally, but, and it's just very, very common method when you have all kind of weird shapes like this. Okay, good. So, more to similar triangles of David, we will do these questions. Okay, so very short statement, but what is the ratio of the gray area to the area of the whole rectangle? So as you can see, we have a gray triangle and the gray trapezoid, which is kind of ugly to calculate. And we have the dimensions of the rectangle. So this problem is harder than it looks. It probably has one of the longest solutions, really, out of any of these. So take your hand, and I'll be giving verbal hints because I have an annotation problem on this device. So I'm going to, I've taken a screenshot and I have to put all the solutions at the start, which I will do, but I'll give some verbal hints as the time goes on. Let me just make sure the poll. Oh yeah, you can just say verbal hint and I would type it up, David, if you want to say it, I can type it up. So I can even name points and things. Okay, sure. So I just need everything. So I'll just do that while you guys are working on this. Yeah, no, no worry. I can name, this name point I can call M and this one call N, maybe it's just, and this one is E and F. All right, that's pretty good. Yeah, thanks. Thank you, I appreciate that. Yeah, thank you. You're welcome. And then also you notice that these, as I say, one, one, right? So for example, you can sort of notice these are the same. All right. But the rest I can draw, I found a system. Bye-bye. 18. You can see that the question number from like only a year's that part of the geometry. Do you want to give some hint to start with, David? Yeah. So the first thing to realize is that you have two different shapes and each of them is ugly to calculate on its own. But what you can do to make it easier is if you realize that MEFB, so the shaded trapezoid, has the same area as the non-shaded trapezoid because they're symmetric, right? Your shaded region is just the big triangle ADC minus the corner triangle FNC, which means that the only thing you need to worry about is the triangle FNC, which makes your life a lot easier. Yes. So you see that this one, right? This area is just equal to this one, right? So we can move it over here and if we can figure out what's the area NFC, you are all set. That's one way to think about that. Another way is that do you see any other lengths that are equal over here? So because MD, notice that MD is parallel to DN, right? And they are equivalent over here. So that means MD is parallel to DN. I mean, you can see it. You can take it for granted from the drawing. So that means I can see that this length, DM, because ME is parallel to BF, and this one, so we can also mark this length here equal to this length, perhaps, right? And if you look over here, NF is parallel to DM, and you look at this way, NC equal to DM, so F also is a midpoint at EC, so we can make this way. Okay. I'm almost done with this annotating stuff. This works. Yes, we got the first correct answer. All right. So I'll begin if I can just, let me actually try to make this happen, share my screen on this. Okay. So, yeah, so I'm sharing this. All right. So here's the, if I can do this, yeah, here's the idea. There's a lot going on here. So first of all, You want to share the screen because I can't see it. Yeah. I will stop sharing so we can share yours. Oh, can you see now? No. It says I'm sharing. Can you guys see? I don't think so. So, okay, good. Okay. Now it works? Yes. Okay. Awesome. So here's what we have. The first step is to realize, yeah, that, you know, all of this, this is the same as the area we're looking for because this trapezoid and this trapezoid is the same area. So we just need ADC, the big triangle minus NFC, the small triangle. And since NFC, we know the base, the base of this is just one. What we need to find is this height page. That's enough for us to find the area. So the way we do this is that we realize that we have a bunch of similar triangles. So if there's a way to clear this, which apparently there isn't. No, there isn't. But the idea is we know that if we make this altitude here, FG, we find that this small triangle is similar to the big triangle ADC, right? Because they have two angles that are the same. This angle is shared and they have two 90 degree angles, which means that whatever this is, this is 2H because the ratio here is 1 over 2. So the ratio of FG and GC is also 1 over 2. So GC is twice FG, if that makes sense. That's step one. Step two is to realize that FBI, funny enough, and NBC, also funny enough, are also similar because they have a 45 degree angle and a 90 degree angle. That means that FH and BH are equal. These are equal segments in the same way that 1 and 1 are equal. That tells us that 1 minus H, GC, which we have over here, equals 2H, which we have over here. And then we solve to get the H equals 1 third. So we have to use two pairs of similar triangles. Finally, now that we have that, again, this is why this is a long solution, we can calculate the area of NFC. We know that this triangle here has an area of 1 times 1 over 2, which is 1 sixth, which means that the shaded region has an area of 1 minus 1 sixth, which is 5 sixth. And so your ratio is 5 sixth over 2 because the total area is 2, which gives you 5 twelfth. And the answer would be D. So conceptually, once we realized we only needed the area of this tiny triangle, the problem wasn't that hard. The tricky thing was the details because we needed to find two pairs of similar triangles and do a little bit of algebra to figure out what the height was. Yeah. Okay, thank you, David. So one thing I, where's my, I don't see where it is. Okay, so, okay, thank you. That's the real thing with geometry. We usually have a multiple solution depending on how you look at it. Let me just quickly, because while David was doing, I was thinking it this way, because all of this, you know, that this and this equal to this and this. So I just quickly mark it. If I mark this area equal to 1, then because this one is halfway, so this whole area is going to be 4, right? Because it's half. So the shaded one is 3. And then if I look on this side, because this length is half of this length, that means this length is half of this length. So that means this area is half of this one. So it's going to be 2. And then this is going to be 3. This is 2. This is 1. So it's going to be 5 divided by 12. That's how I did it. That's a good trick. The hard part is just realizing that there's so many of them. Yeah, no, but that's the beauty of geometry. You usually have a multiple solution. And sometimes the simple one is not necessarily the best one, because sometimes you go for more complicated solutions, but you will learn more. So thank you. This one, it's unclear. Actually, I'm not even sure. Wait, I'm curious now. If we go back, how do we know that the three segments OK, because this is OK. Because everything was lost when we showed A, B, C, D, right? These are parallel. Yeah, parallel, right? And that's why these lengths are the same. And you look on this side, this is equal to the same, right? And if this I... So really, there has to be drawn an altitude. An altitude from A to B, C has to be drawn. And then because that altitude makes a rectangle. So we still, technically, there are a few extra lines. Yeah, of course, but I mean, in the exam, you can just quickly do that. But yes, and I found that it's kind of easier. I like this one. Yeah. OK, let's go to this one. All right. So there is a related hint in the chat for this problem, for a reason, right? A, B, C, D is a square. What is the length of segment EC if AF equals 4 and FB equals 3? Which should hopefully start immediately some alarm bells in your head over how we would start this. And as you can probably guess, this is also a similar triangles type of problem because we have a lot of triangles. Yeah, so either congruent or similar, we can never get tired of this. So let's just put 4 and 3. And hope that rings a bell. And let's put it here. Oh, the button launched the poll. David, do you see the poll? Somehow it disappeared on my side. I see it. You have to, maybe you have to click on polls, but I see it and we already have three answers actually. Okay, good. Is that good? Yeah. So whenever you're ready, you can just say, and I can type. Yeah. So I'm almost. All right, so, um, Okay. I'm ready. Yeah. Oh, okay. Good. Yeah. It's time to get less cluttered. All right, let's make sure everyone can see. Does this work? Yes. Okay. So the first step is three, four, five, right? Triangle, right? This is a Pythagorean triple. You could also do this manually. You could do four squared plus three squared and get the 25, but this, this is one that should snap, especially as we do more of these types of problems. So that's the first step. We know that the square has side length five. That means that BC also has a side length of five, but we don't know what EC is yet, right? Because it's not a full side of the square. It's something smaller. We can guess from the fact that EC is in a right triangle BC. And by looking at the triangles that maybe we can use similar triangles to find a ratio between EC and BC. So I want to know what EC over BC is. And in fact, it turns out that ECB is similar to our original three, four, five right triangle by angle angle similarity. They both have a 90 degree angle, right here. Additionally, if we call this angle alpha, then the single B is 90 minus alpha because these two are complimentary. They sum up to 90. And because this here is a 90 degree angle, this angle and this angle are also complimentary, which means that these two angles are the same. That gives us similar triangles, which means that EC over BC equals to three over four, just like our original triangle. And since we know what BC is, now I just have to bash. What is three fourth of five? And if you do that math with decimals, you find that the answer is 3.75. So this time there's only one triangle, but it was probably a little bit harder to find because there was some angle chasing to do. Yeah. Thank you. It is. It's just when you rise the ratios and make sure it's the right one, like EC over BC and the other one is just the right order, FB over AF and not AF over FB. But that's good that you remind us of the Pythagorean troopers. So in this case, you may need to use it again in the convex. Oh, hello. Hello. Can you, can you see us? I can see you, but we can't see the screen. Oh, you can't see the screen. Okay. Sorry. Okay, good. So in the convex quadrilateral ABCD, the length of the sides are expressed with natural numbers and the perimeter is equal to 16. The interior angle C and D are right angles and angle B is obtuse. See the pictures. What is the length of side BC? So we know that the, the sides are all natural numbers, like whole numbers. And this one is obtuse. Let me just continue what David said about the Pfeiffer shipples, right? So we have 3, 4, 5, and of course we have a 6, 8, 10, that is just a scale burden of this, but it's good to know. And what's the next one? 7, okay, 5, 12, 13. Yeah, 5, 12, 13, also a good one to remember. And the last one, rarer, but it's cropped up in a couple of math kangaroos, 7, 24, 25. Very nice. Yeah, because 25 minus 24 squares 1 and 49, yeah, so. So, yeah, you do want to remember these by heart. These first three. Let's see. Bye-bye. Okay, so we have two, so I think the first thing we need to do is to draw some actual lines here. I draw a line like this. Sorry, I just lost control. So if I draw the actual lines like this, what do I see? Because this angle B is obtuse, I know H is going to lie inside the segment AD, right? And from here, what can I do? I have, I can use, let me just label this is like AB, also, you know, there's no escape here, algebra, ABDE. And we know that the sum of these are equal to, and this also C, right? Equal to 16. And then we have a constraint here, as you see, we have a very few conditions. So back to algebra, but we have a constraint that these numbers are all integers. So we know that from here, Pythagorean is A squared. How many students have got the answer, David? Because I lost the poll, I can't see. We have two, we only have two answers, three answers now, two which are correct. Okay, good. Okay, so we got the idea. So A squared is E squared plus C squared. So I started, this is really the only equations that I can write, right? But what I can learn from this, because these are integers, so I started guessing. These are Pythagorean triplets. If I start with 10 and 6 and 8, then they are just too big, right? Because then there's no way to get up to 15, right? So the only way I can have is 3, 4, 5. I have to use 3, 4, 5. So right away, I know that this guy A here equal to 5. The question now is that this one is 4 or 3. So let's try first, this one equals to 4 and this one is 3. Then we have a 3 and 4, that's 8, right? 8 and this 4 is 12. And these two sides are equal. Okay, I must be mistaken, this one must be equal to B as well. So B equals to 2, right? Really just guessing, 2. And that's the answer, B. Of course, I can try the other one, A equal to 5, I can have C equal to 4, and this one equal to, I'm sorry, B equal to 3, this equal to 4, but then it would not work out, right? Because 5 plus 4 is 9, 9 plus 6 is 15, 9 plus, did I mistaken here, 5? I think it would always work. I mean, because either way, you have the perimeter of A H B equals to 12, right? And then you're just missing H D and D C. Either way, it still gives 2, it can be anything you want. No, is this 5 and 4? So is this 5 and 4, that's 9. And then 9 plus 3 is 12. And this is actually the way, it should be 4 here and 3 here. 4 here and 3 here doesn't work because, no, I think it's the same, right? So 4 and 4, yeah. I think this whole problem is iffy because, okay, never mind, yeah. So either, the question is, the only question is whether you can have a side of length 2 and a side of length 5 making a side of the length A D. But anyway, it doesn't matter. No matter what you do, B has to be 2. So I think it's possible that A D equals to 6 and then 5 and 2 equals to 7. So you can still make a polygon like that. Yeah, I think it really doesn't matter. It does not matter. Either way, the point is the perimeter of the triangle in total is 12 and then there's 4 left. Exactly, yeah. So it doesn't matter, but really we have only one equation here. The trick is that we got to draw some actual lines and usually for geometries, that is really like a bottleneck. Once you get that step done, then you can move on. But always ask yourself what information you have and what tools I have. In this case, not so much, no similarity, no congruence, the only thing I have is the Pythagorean, right? And I go with that and use the constraints. Okay, next one, another one about tangents. A lot of tangents on circles over here, one in the middle and four of them. Oh, they also say that these are the diameters, right? The square 4 semicircles with center in the midpoint. So these are all semicircles. We are asked to find the radius of the small circle here. And as we said earlier, they are tangents to each other. You have to draw all the lines that represent the fact that they are all tangents on the picture. Would you please launch a poll, David? I lost it. I can't access it. Oh, it hasn't been ended, I see. That's confusing. Okay, good. There we go. So I draw this so this shows that these two circles are tangent and I mean symmetry so it goes like this and I think I should draw a few lines and see how it goes. You can only draw one side because the other side is the same and you want to give them names, right? So for example this is R, this is R, this is R, this is R, and this one oh I forgot one thing okay I haven't shown that these two are tangents so I also need to draw an actual one here this is R, this is small r, small r, I don't know if that makes sense to everyone. These are the same I don't need to draw. Oh good, I think you got it right. So this one over here, we don't know big R, right, but we know R plus R, is it equal to, because if you look at the sine length, they say that of the square, the sine and radii of R, radii of R, so R equal to one, right? And these are, as you can see, they are just isosceles triangles, right? Right triangles is 40 degrees and R, R, R plus R, R plus R plus over here. So isosceles right triangles, that means R plus R equal to one over square root of two of R plus R, which is two R, right? And R equal to one, so this guy is just equal to square root of two and big R equal to one, so small r is going to be square root of two minus one. So this actually could be one of the, maybe most straightforward question in today's class, right? You just remember, you have a tangent circle, just draw everything and make use of the fact that the distance from the two centers is equal to the sum of the two radii. And then lots of time you have to use Pythagorean theorems. So unfortunately we run out of time, even though we have a lot more questions. So what would be the takeaway for today's lessons? I would strongly suggest you go back to review some of the few slides we got in the beginning about congruent triangles, about similar triangles, the tools that we have with circles, with tangency, with Pythagorean and ratio. Super important how to relate ratio of areas to ratio of length, right? And also things like, simple things like if you have two triangles that share the same height and the ratio of area is just equal to the ratio of base. So simple thing like that, but if you remember it by heart and it's part of you, then you can save a lot of time in the exam and get to the answer very quickly. So today we do length and perimeter and area. Next week we're going to be doing angles. So I hope you had time to review today's lesson because whatever we use is going to be also be helpful for next week. And also we have a couple of things when we have to use algebra. I hope it's also a good opportunity for you to review it. We have only two more classes to go. I know we don't have a lot of interaction in this class, but you can come a little bit early in the exam, early before class and have any question, anything, please just send to the chat and David and I will try our best to answer.

geometry

perimeter

area

2D shapes

triangles

circle theorems

Pythagorean theorem

similar triangles

problem-solving

algebra