Let's call this the official beginning. Welcome to webinar number seven, 2D Geometry in the Level 5-6 Series. I'm Dr. Sarah Sagee. Your teaching assistant is Jacob. And we hope you enjoy the lesson today. Try your best. You should have some scratch paper. You should be working along with us. There was a handout. There's a folder of handout materials that's available for you to print each week. So hopefully you will have some of these figures with you. Which of the following geometric figures is not in the design to the right? So looking at this picture, which of these geometric shapes is not present? This is one of our three-point questions at the beginning of the contest. I think we can go ahead and start. I think the triangle is pretty obviously, there's triangles all over this, many, many triangles, so we can clearly say there is a triangle. How about the square? Well, I see the squares right next to those triangles. We have squares next to the triangles, so clearly there are multiple squares in the design. A regular hexagon. The hexagon seems to be in the center, six-sided figure in the center of these circular motifs. There's a hexagon. The octagon, well, there is a larger shape. Let's see how many sides it has. It has 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Well, the red shape has 12 sides, which makes it the dodecagon. I do see the dodecagon, but I do not see the regular octagon. Since the question asked, which is not there, the answer will have to be D. As you can see, we've gone through several different types of problems. We're giving you tools and some experience with the variety of problems that we have on Math Kangaroo Contests. Remember, these are contests at your level. There will be 30 questions, 10 worth 3 points, 10 worth 4 points, and 10 worth 5 points. Geometry questions can range in difficulty, so they could be a variety of points, but you'll find that sometimes there are difficult geometry questions on Math Kangaroo Contests. This is a good lesson to pay close attention. Today is 2D flat geometry, and next week we will move and do some three-dimensional geometry. Hopefully, you'll enjoy these two lessons. They somewhat go together. We assume by fifth and sixth grade that our students know their basic geometric shapes. Geometric shapes would include your triangle, squares, rectangle, circles, hexagons. We also hope that you understand that the perimeter is the distance around the outside of a shape, and that the area is the amount of space that a flat surface will cover. Okay, so you can cover or that you can put on top of it. You also will need to have some exploration of angles in the shapes. Squares and rectangles having right angles, equilateral triangles having the same angles in the triangle. The total degrees in a triangle is 180 if you add up the three angles. So things like that, we would expect you have already experienced in school, and we would expect that to be background knowledge for our Math Kangaroo Contests. Number one. Kathy draws a square with a side length of 10 centimeters. That is the large black square. She joins the midpoints of the sides to make a smaller square. That is the green square. What is the area of the smaller square? So it does tell you that she's joining midpoints. So that's my hint. And we want the area of the smaller square. The smaller square is the green. As you can see, the poll does have the image and the question in it, and if you scroll down along the side, you should be able to see the answer choices. Anybody else want to put an answer into the poll? Remember Math Kangaroo does not subtract points for incorrect guesses. So when you are not penalized for an incorrect answer, it is better to guess than to leave a blank. So you should know when you take a test or a contest or an evaluation, whether you get penalized. That will help you determine if guessing is okay or if you need to be more sure of yourself. Hey, I will share the results of the poll. So you can see that 64% of you think that the answer is E, 50 centimeters squared. That is correct. There were some other answers. So let's take a look and see how we can get there. So one of the things that some students have trouble with is they know, they're very familiar with the idea that the area of a square is the side length squared, right? It's equal side times side, side times side. But we don't have the side length here, do we? So we can find the side length of the black, the black large square. Is 10 times 10. So that equals 100 centimeters squared, correct? That's the large one. But we want to find this green one. So I have a bit of a trick that we can use to find the green one. We can compare, since we don't have the side length, you could use triangles to determine it, it is possible. But we can compare the green to the larger black square by cutting, because we know that these are midpoints, right? So we can see that if we take a look at this square here that I'm now gonna kind of outline in the blue color, we can see that it's half white and it's half green. And we can see that that goes for all four quarters. So half of the large square is green. So all we have to do is take the 100 and divide it by two or multiply it by a half. And we'll find that 50 centimeters squared is the area covered in green. So hopefully that makes sense just using some proportions on this, right? It's half of it, so we take half of the area. There is a square with line segments drawn inside it. The line segments are drawn either from the vertices or the midpoints of the other line segments. So midpoints or vertices. We colored 1 1⁄8 of the large square. Which one is our coloring? So that's another way of asking which of those squares, A through E, represents 1 1⁄8 of the large square. Has been colored. There is not a poll for this question because I didn't have a way of putting those figures as answer choices. So if that makes sense. Very good. I'm having overwhelmingly the students are replying to me. Hopefully, they are to you as well, Jacob, that they believe the answer is D. Is that what you see? Yeah, I'm getting a thumbs up from Jacob. If we take a look at it, we can do a very similar type of analysis that we did before. We can look at this in terms of quarters. That's a pretty good way to look at it. We can say that if we cut this into a quarter, this is a quarter is, of course, a fourth. Then one-eighth would be half of the fourth. This is one-eighth. That tells us right away that we have here is one quarter and one-eighth. D is our correct answer. Obviously, this little square here in red is very, very small. How do we deal with something like a diagonal? It's interesting because we can cut this. This would be a half. We cut that half into half. This is a fourth. But what is this? Is this actually cutting the fourth into a half? It's not. Because if we look at a triangle, the area of a triangle is one-half the base times the height. In this case, we're cutting the base of this triangle in half and the height in half. We would actually be doing, if you looked at the original, we'd have extra halves here because this is a half and this is a half. This is not going to give us half of a fourth. Here, of course, it's just like I showed in B, that's a fourth, and this is one-fourth. Our correct answer here is D. Each of the five neighbors owns a rectangular plot of land with the same area. The flower gardens of their land are fenced in. That is the solid lines in the picture. Who has the longest fence? I do have a poll for this one. The dotted lines show you the plot of land is the same rectangle for everybody, but they put their fencing in different ways. Who has the longest fence? Sorry. All right I'll end it here and share with you over 70% about three-fourths of you have said you think the answer is Mr. Jack. That is correct. Let's take a look at the problem. This is one of those problems where you're not actually be calculating lengths or perimeters of these fences but you're going to be comparing them anyway. So you can see that Mr. Adam uses the entire rectangle. Now if we try to do that look at Mr. John we see that this line would be the same length as that line and that the right-hand side would be the same as if we put it out to the left-hand side. So Mr. John's fence length is the same as Mr. Adams. So you can kind of what I what I call this is push it out. I don't know if there's another word but if you pushed it out you would see that you get the same as you do for the entire perimeter in A. We can do the same thing with Mr. Peter. We can push this out and we get this line here. We can push this out and we get this line here. And then if we push up and down we could get this comes up here and this one comes down here. So that's also going to form the rectangle. And you can see you get the sort the same sort of idea with Mark if we kind of push these out we would reconstruct the outside corners of the rectangle. But let's see what happens when we do that with Jack. With Mr. Jack if we push this side out we do fill in those dotted lines but we have the addition of this fencing in the red. That fencing does not push out and fill in the original perimeter. It is extra. So the red is additional fencing. So that is the correct answer here. The longest fence is in Mr. Jack's garden. How many convex angles with different measures are made by the rays with P as a starting point? Let me give you a picture. A convex angle because maybe you don't know convex angle but look they're giving you the definition here. The convex angle is an angle that is larger than zero and smaller than a straight angle. So straight angle would be you know just like this. It would be this whole angle. Right. So we're looking for anything that is in between like not an angle and all the way open into a straight line. There's a very important word here. Different measures. Most students get messed up by that right there. So I'll give you a minute to work on it. There is a poll for this one as well. Looks like some students are ready for the poll. The little picture of the angles is in the poll, so hopefully you can still see everything. You know, on one of my devices, I have to scroll up on the poll to see the answer choices. If you don't see the answers, try scrolling on your screen and see if you get to the bottom of the pop-up box. All right, I'm going to go ahead and solve this one with you. So here's the results. We have a mix of answers with the eight angles is winning, but it doesn't have the majority, it only has 46 percent of you. So let's take a look. Quite a few students also did not answer. So this is a medium difficulty problem. You can see it's number 16, so it's right in the middle of the math kangaroo contest. And I'm going to use a method that we teach in level 3.4. So if this is your first time taking math kangaroos this year, you might not have seen this approach. So it would be good to review it here as well. So the first thing I'm going to do is I'm going to kind of make an organized list or table of the different angle measurements that I have. So I'm going to take these angles one at a time. If I take them one at a time, I'm of course going to have 10 degrees, 20 degrees, 30 degrees, and 50 degrees. If I take them two at a time, two at a time, I can have 10 plus 20 is 30, 20 plus 30 is 50, 30 plus 50 is 80. Did I miss any? I think that's two at a time, right? Because I can take this one, this one, or this one, right? We can also take my angles three at a time. If I take them three at a time, I can get the top three, which is 10 plus 20 plus 30 is 60. Or I can take the bottom three, and I would get 80 plus 20 is 100. Or I can take all four of them together, in which case I get 110 degrees. So that is kind of my way of making a very organized list and make sure I haven't missed any of the angles. But this gives me 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 angles. But remember at the beginning, I said we have to read questions very, very carefully, and it says different measures. So if any of these measures are repeats, I want to cross them off the list. I see 50 appears twice and 30 appears twice. So I now have 10 minus 2, which is eight different angles. So my answer will be C, 8. So very important to read every word in the problem statement because you could have done all the angle measurements correct but missed the word different. And then you would have incorrectly marked answer choice D. The frame of a rectangular painting was made from wooden pieces of the same width. What is the width of these pieces if the outer perimeter of the frame is eight decimeters longer than the inner perimeter? Remember on one of my definition slides, perimeter was the distance around the outside. So we have an inner perimeter and an outer perimeter. And we know that the width is the same all along the frame. If you have a few minutes to work on this, there is a poll again for this problem. Just a reminder, we're asked for the width of the wooden pieces. So if you take a look at where we have the question mark, that's what we need to find is how wide are the frame pieces. And I will launch the poll now. This is the last poll in this lesson because the other questions didn't really work with the poll. I'm going to provide a hint. Try using some of the same method we used on the fence problems with the different gentlemen and their fences on the garden with this problem. What is the difference between the inside perimeter and the outside perimeter of the frame? The answer is on lines. Anybody else want to put a guess in the poll before I close that up? Remember the poll is anonymous. So, even if you get incorrectly, it's no big deal. Maybe you'll get the correct answer. All right on the poll and I'll share the results. It's a very interesting poll, so you'll notice we have a tie between 2 centimeter, 2 decimeters and 1 decimeter. We also have some students saying 4 and 8. So, let's take a look and see what we can do with this. I will tell you it is not easy. There's definitely some work that we can do here to figure this out. Okay, we'll clean that up a little bit. Let's go to what I was saying with the tracing it out. So, I can take this line here and I could put it right here. That would be the same. Maybe that color doesn't work so well. Let's try again. You see how I can take the pink top and bottom and I can put them on the outside frame and I can do that along the sides as well. So, that leaves me with some pieces here on the corners that are new perimeter in only the outside of the frame. So, the 4, the sum of the 4 green corners equals 8 decimeters. Because that's what it says. It says the frame is 8 decimeters longer on the outside. So, I'm looking at those corners so that I know there's 4 corners. All right, so I can take 8 divided by 4. And that's going to give me 2 decimeters. So, that might lead some of you to say, okay, the answer is B2, but if we carefully look at this. We can see that we have actually 2 pieces. So, there's kind of an X here and an X here. And each of those X's is the same as the width of the frame. Do you see that? We're trying to find the width of the frame. And you have 1 along the bottom of the corner and 1 along the side of the corner. So, each of these 2 decimeters is actually equal to 2 X's. So, therefore, X must be 1 decimeter. See, I hope that explains it. I hope it looks clearer now. Jacob, does that explanation work for you? Yeah, that's a good explanation. Okay, so if anyone is still confused, go ahead and ask Jacob some more questions about it. Okay, that's why he comes here every time. All right. Actually, if you're still confused, ask me because Jacob is going to explain and is going to help you through number 6. So, if you want more help on number 5, send me a chat. Jacob, go ahead and start them on number 6. Okay, so the rectangle shown in the picture with dividing the squares with different side lengths and the areas of some of the squares are given. And you want to find the area of the entire rectangle. So, I'll give you guys some time to think about this before I explain it. Okay, so we don't have too many correct answers, so I'll give you guys a hint that maybe you can try to use that the area of each of the squares is the side length squared, so hopefully that helps trying to find the rest of the side lengths of each of the squares. Okay, so another hint is that the area of the large rectangle is equal to the length of the rectangle times the width of the rectangle, so you need to determine those. And maybe you're trying to find another way, which is you can add up all the areas of the smaller squares, which will also give the same answer, so hopefully that helps. Okay, so we got a few more correct answers. So I think we're going to go ahead and explain it. So firstly, we should notice that we're given the areas of some of the squares. So what we can do is that we can find the side length of each of the squares. So for example, the area of the square with area 16, the side length of that would be four. So this would be four. For the square with area 196, the side length of that would be 14. The square with area 225, 225, the side length of that will be 15. And the side length of the square with area 81 would be nine. And this is all using the fact that the square, the side length of a square is the square root of the squares area. So now we can try to go ahead and try to find the side length of the other squares. So for this square here, notice that the side length of that square is sum of the square with side length, the square with area 196 and the square with area 16. So this would be 18. So now we know that one of the squares, the top right square has side length 18. So that means that the side length of this would be 18 plus 15. So that is 33. And similarly, we know that this edge here, well, we know that this is 14, because, well, it's a square, and all squares have equal side lengths. And then we also know that this is 18, because once again, all squares have equal side lengths. So the edge, the top edge would be 14 plus 18. So that's 32. And remember that the area of entire rectangle is the product of the side length of that rectangle. Yeah. So the area of the rectangle would be 32 times 33. And that comes out to be 1056. So that would be B. If any of the students found the areas or the lengths of all of the other squares, that is not incorrect. Okay, so that is another way to do it. But Jacob has found the area kind of using the minimum amount of steps. And because this is a timed contest, there is an advantage to using the minimum amount of steps, right? But certainly, if you had more time, you could check yourselves by doing some of the other steps as well by finding the other lengths of the sides of those squares. Okay. And yes, it is a no calculator contest, but there may be something where you have to kind of do a long multiplication like that, but it's not very common. So that won't be what holds you back in this contest. Okay, so there is officially a wrap up, but I made sure we had enough time to do our bonus problem because I really like it. So we'll do the bonus problem with Jacob, and then we'll go back and do the wrap up. All right, Jacob. Okay, so the shaded figure in the picture to the right consists of five identical Sotheby's right triangles. And you want to find the area of the shaded figure. So I'll give you guys some time to think about that. Read the problem statement a few times because there's important words in there about those triangles, right? Okay, so we have a lot of right answers, so we're gonna get started. So, notice that each of these right triangles are identical, and so that means that each of these links, so, sorry, but each of these links, each of these links here, so, yep, that link, this link, this link, this link, and this link, those links are all equal, and there's five of them, so that means that each of those links would be equal to 30 divided by 5, which is 6. Okay, so now the area of the shaded region is equal to 5 times each of these small triangles, so, and notice that, so now it just suffices to find the area of each of these small, I thought these triangles, and notice that we know one of the links, and we know it's a right triangle, and these links are equal, but notice that if we draw a perpendicular right here, then it splits the triangle into two triangles with equal area, and notice that that triangle is similar, so notice that this link would be 3, and this link would also be 3, but notice that this triangle is similar to, so like, this triangle here is similar to the blue triangle, so that means that this link and this link would be equal, so we know that the, that this side is 3, so that means that this side is also 3. Jacob, can you explain why they're similar? Some students might not, might not see that. So, so the blue triangle, notice it has angles, so this is a 90 angle, 90 degree angle, and notice that each of the, and notice that some of the angles in a right, in a triangle sum to 180, so that means that, and since it's isosceles, that means that we know that this angle and this angle are equal, and since the angles in a right triangle sum to 180, that means that, like, this angle would be 180 minus 90 divided by 2, that's 45. Now, notice that in the red triangle, the, when we draw the perpendicular, this perpendicular here, it forms a right angle right there, and this angle is still, once again, 45, because, well, this angle and this angle should be equal. So, they end up being similar because all the angles are the same. So, that means that this side would be 3, so we know that the area of each of the triangles is 6 times 3 over 2, because, remember, the area of the triangle can be modeled as base times height over 2, and 6 is the base and 3 is the height. So, the area of each of the triangles would be 9, so going back here, 5 times the, there's 5 triangles, so 5 times the area of each of the triangles is 5 times 9, so our answer would be 45. Thank you, Jacob. This problem works because they're a very special triangle. The isosceles right triangle is very special. Whenever you draw the height of the isosceles right triangle, you're getting new, smaller isosceles right triangles, so it's a very clever problem, and congratulations to some of you who got that correct. That's awesome. Okay, so in wrapping up today, I hope you like the 2D geometry, so you can see that sometimes you might have to explore a little bit. You might not have understood those right triangles that you could cut them in half, but if you tried exploring a little bit, you might see it like Jacob showed you. I hope that these were challenging. You can see we had some 5-point questions, but there also were some 3- and 4-point questions, so there's a varying level of difficulty. Don't forget how to find areas of generalized shapes. Don't forget how to find perimeters. Don't forget that you can add up all the areas, or you can multiply the side lengths on rectangles. Triangles are one half of rectangles, so all of those things that you've learned throughout elementary school and the beginning of middle school are going to come in really handy for these problems. Don't forget you can always add on to your drawings, right? Do you notice, like with the picture frame, when we were moving the perimeters out from the inside to the outside, don't be afraid to add some things to your diagrams. Jacob had to add the heights of triangles. It wasn't given to you, so don't be afraid to play around with them a little bit. I hope you liked this lesson. Remember that the best way to prepare is to be taking some practice contests. If you go into your registration, either for this course or for the contest, and look at the event info, there are some codes there to help you get some practice materials or to watch some videos of teachers solving contests. We hope you enjoy it. Bye, everyone!

2D geometry

Math Kangaroo

problem-solving

geometric shapes

area and perimeter

symmetry

geometric patterns

geometry problems

mathematical properties

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