Common Core Geometry Unit 4 Lesson 4 The Circumscribed Circle

Hello and welcome to another common core geometry by E math instruction. My name is Kirk weiler and today we'll be doing unit number four lesson number four on the circumscribed circle. That that is a mouthful. The circumscribed circle. So without any further ado, I guess other than saying, make sure that you've got your straight edge and your compass out, right? Let's get into it. And let's talk about what a circumscribed circle is. One of the great great mysteries and really kind of interesting things about geometry is how connected circles and triangles are. All right? And the first connection that we're going to really see with that is today in what's called the circumscribed circle. A circumscribed circle is a circle which passes through all three vertices of a triangle. And that's it. Okay? Every triangle has only one circumscribed circle. So right here in this Georgia brew widget, we can already see, given this triangle, we've got this circle, which is hitting each one of the three vertices. And give me a minute, because I'm going to kind of futz with a little bit right now. So what's amazing about this is that again, without really doing anything at all, other than just moving the vertices around. What you can see is that no matter how I move them, there is only one circle that passes through those three points. It's really rather cool. So what we're going to be doing today is we're going to be learning how to construct it. Now I will say this is a rather challenging construction. In theory, according to the common core standards, you could have to do this construction on a test. But it's rather lengthy. All right. Still, strangely enough, you have all the tools right now to do it, but you probably don't realize that. Let's start with a pretty easy question, though, just to make sure you understand what a circumscribed circle is. All right, exercise number one. Which of the following shows a circle circumscribed about a triangle? Take a moment, pause the video if you need to. All right. Well, again, I'm sure that this term circumscribed is new to you. And again, it's a little tricky because circumscribed really means to surround. All right? And what we've really got here is we've got choice two. The idea of a circumscribed circle is a circle where the three points of the triangle are on the exterior of the circle, right on the circumference of the circle around the circle itself. Remember, the circle is not the space inside of it, but actually the points that lie on its perimeter or circumference. So now, let's start to build what we need in order to create that circle. Okay, here we go. Starting off. Make sure again, straight edge, right? Compass, we're all set. Letter a, asks me to construct the perpendicular bisectors of sides AB and BC. Mark their intersection point as D all right, now the perpendicular bisector of a segment is something that we've seen in multiple lessons at this point. But let me review how to do that for one of the two segments, and then you can practice on the other segment. So let's do it. Perpendicular bisector of AB and BC. Maybe I'll do the more challenging one. I'll do the perpendicular bisector of BC, the one that's kind of slanted. I think AB will be a little bit easier for you. But let's go ahead and talk about how to do that. Remember to create the perpendicular bisector, what I'm going to do is I'm going to put the point of my compass at one of the two end points of the segment. And I'm going to stretch the compass out so that it's more than half the length. The radius of the compass is more than half the length of segment BC. What I'm then going to do is I'm going to draw an arc and I try to draw an arc still not happening. All right, maybe we'll get the arc up here. All right, I'm going to draw an arc here. And that's not working. All right. Let's draw it not the video over, but we're going to start the construction over. Let me lock this thing down. That's going to help quite a bit. Lock in place. One more time. We've already got the compass at a good radius. There we go. Thank you, compass for working. Maybe if I thank it more. It'll work for me better. All right. Now I'm going to switch, keep that radius the same. Okay? When you draw those two arcs from B, then keep the radius the same as you move over to C and I'll draw this arc. And again, you can make those a complete arc, and this arc. There we have it. Now, X marks the spot, so we'll put our ruler. And I'll slide it a little bit. And draw that. And that's my perpendicular bisector, a little bit of a struggle, but there it is. Hopefully it won't be as problematic. The next time. Now, what I'd like you to do is pause the video and draw the perpendicular bisector of AB. And then mark its intersection point with this thing as point D take a moment to do that. All right, I'm going to go through it. Hopefully I'll have more success than before. But I'm not going to hold my breath. Okay. So just like before, what I'm going to do is I'm going to bring my compass down, set it at point a, stretch its length out so that it's more than half the length of AB. Bring it up. This is where the picture can start to get kind of messy, right? So I've got my arc there, rotate it down. Of course, now you're going to work for me, compass. Bring it back up. Switch, bring it over here. Down, arc up. Arc, get it out of there, back up. Grab my ruler. Bring my ruler over, get it lined up, and. Okay. So perpendicular bisector of AB. And then point D I think that's what I said to label it. Let's just make sure, mark their intersection point D okay. So so far, this is what we've got. All we've done is the perpendicular bisector of one side, the perpendicular bisector of the other side, and their intersection point. Put that intersection point is key. And let's see why. Letter B, explain why point D must be equidistant from all three vertices of the triangle. All three vertices of the triangle. Okay? So take a moment and think about that. Why is point D equidistant from all three vertices of the triangle? Pause the video now and think about this. All right, let's talk about it. I'm going to extend the page a little bit. Since D lies on the perpendicular. Bisector of AB, it is equidistant. From a and B now make sure you understand what that means. What that means is if I measure the distance that D was from a and from B, that would be the same distance. Maybe it would be four inches each, right? So again, if I measured the distance that D was from a and B four inches, if I measure the distance D was from B, it would be four inches. They'd be the same distance, no matter what, maybe not four. Obviously. But likewise, I'll say also sense D lives. On the. Perpendicular bisector. Of BC, DY is in the perpendicular bisector of BC. It lies equidistant from C and B now again, think about what that means. That means that if the distance from B to D is four inches, then the distance from C to D is four inches. So again, keep this in mind. It's kind of cool, right? This length, and this length must be the same. But this length, and this length must be the same, and therefore all three lengths must be the same. Literally, to be equidistant from point a and B means that AB, I'm sorry, AD, AD is equal to BD. But being equidistant from B and C means that BD must be equal to BC. Okay. And therefore, this means that that point is equidistant from all three points. But this is really cool. Draw the circle that circumscribes the triangle. Well, we kind of have it now, and let me explain why. You see, this length is congruent to this length, which is congruent to that length. And therefore, if I use D as my center, and any one of those three lengths, which are all the same, as the length of the radius of the circle, I've got the circle. Take a look. Now it's not likely to turn out perfect. And that's because this is a fairly challenging construction, and also we're in the world's point D, you know what I mean? It's all sorts of messy at this point. But let me stretch it out. So right now. As good as I can up at the board, I've got my center at D and my radius set to that length, which is again the same as that length and the same as that length. And now. I've got that circle. Now again, notice it's a little bit off up here. Right? It more or less hit B and a perfectly. And the point is, there's almost zero chance that if I'm doing this with a hand ruler and encompass or electronic ones up here, it's very doubtful that I'm going to nail it perfectly. And a little bit, I will show you a perfect construction using geo gebra. But this is good enough. Now again, all we have to do to draw this circle. All we have to do is construct the perpendicular bisector of one side, construct the perpendicular bisector of the other side, and find the intersection point. This point, let me put this in red because it's important. This point right here is the center of the circle. It's got a fancy name. It's called the circumcenter, the circumcenter. You don't particularly need to know that piece of terminology, as far as I know, for an exam. But your teacher may want you to know it. The circumcenter, and then any one of these three, and they're all the same, or they should be all the same. Any one of those three, the length of which you can use as the radius of your circle. How cool is that? All right, we're going to do this construction a couple more times. And see how bad I get. All right. And also see how good you get. So let's keep going. All right, exercise number three. For triangle EFG shown, construct the circumscribing circle, and mark the circumcenter there it is. That piece of terminology mark the circumcenter as point N as always leave all construction marks. All right, well, let's see what you remember. Okay. Take a little bit of time, construct the perpendicular bisector of two of the sides, whichever two you find the easiest if that's EF and FG grade, if it includes EG, great, whatever. Try to locate that point, that circumcenter M, the intersection point of the two perpendicular bisectors, and then draw in the circle. Don't worry if it's not perfect, okay? Again, there are different shades of perfection and imperfection, but don't worry about that. Just try to go through the procedure. Take a few minutes, it will. All right, let's go through it. And as always, the amount of time that I pause is not the amount of time that you should take on this construction. It should take longer than that. So let me go through it. I'm going to switch back to my blue. All right. I think I am going to first perpendicularly bisect FG. I think that's going to be easiest for me. So I'll come in. Widen this out. That looks good. I'll make an arc up here. I'll make an arc down here. I will bring it up, switch over all right, bring this up. Oh my goodness, I didn't even come close enough with that arc. I'll have to bring that back. That's probably a mistake some of you will make on occasion. It's not really a mistake so much in the math as a mistake in my ability to see what I'm doing. So now it's not going to be the best, but it'll be better. Let's get that there we go. I now have those two. Whoops. Get this out of here. Bring this down. Rotate. And rotate some more. Let's try that one more time. Thank you. Excess mark the spot. All right, nice and long. Make sure that you make those perpendicular bisectors nice and long. Otherwise, they may not intersect with each other. Don't get me wrong, you could always go back later on and draw it longer because you still have the two X's that you need. Let me now go in and perpendicularly bisect, I think, EF. I think I'm going to stay away from EG because of how long the segment is. It doesn't really matter. Let's go with something like that. Again, that's probably good, but I think I'm going to make it a bit shorter. Okay, good. Let me make an arc. You got bigger arc this time. You could sweep that arc all the way through if you want. To go up here a little bit, bring it down. Make the other arc, you had it this way. You got that nice piece of paper in front of you much easier to work with. At F. And you probably have a real compass. I mean, don't get me wrong, I like my virtual compass, but we've been having arguments lately. So here we go. I don't actually argue with my compass. Well, not often. All right, a little bit more in the rotation there, while they're. It's going to be a little tricky. I've got less on the rotation. All right. And got to make sure that they intersect, still don't have a rotation quite right. There we go. Okay. I take a little while, but I've got the two perpendicular bisectors now, right? This then, strangely enough, is the center of the circle. I say what to label it. Yep, as point M now I need to get that circle drawn. Let me kind of do this. I need to put the center of my circle. Or the point there. Okay. Let me rotate this up and out a little bit more. A little bit more. Remember, I can go to any one of the three vertices now. Any one of the three. Okay. And draw that circle. Well, I'm not going to get much better than that. That's pretty good, right? Because it went through points F, E, and G, right? I find my circumcenter right there. And I've got my circumscribing circle. Now, little kind of question, right? In the previous exercise, oh, isn't that beautiful? There's my circumcenter. Right there in the middle of the triangle. In the middle of the triangle. It's on the inside of the triangle. On the other hand, in this particular one, the circumcenter is on the outside of the triangle. And why do you think, why do you think the circumcenter for EFG, right? Wide outside of it, whereas in exercise two, it lied inside. Do you have any speculation on that? Take a moment. This is pure conjecture. Why is that? Well, it has to do with whether or not the triangle is acute or obtuse. Little reminder from middle school. A triangle is an acute triangle. If it has all acute angles, angles that are less than 90°. A triangle is an obtuse triangle if one angle of it is greater than 90°. And of course, then you have right triangles. Which have a single 90° angle. But it's because. The first. Triangle. Was acute. And the second. Obtuse. Now, what's really cool about this is it's really kind of a neat one to demonstrate on geo gebra. So let's take a look at that really quick, all right? I'm going to do the construction really fast on this. Because I'm going to let I'm going to let geobra actually create those perpendicular bisectors for me. But then I'll create the circumcenter and the circle itself. So let's take a look. In this particular picture, what we've got is we've got triangle ABC. For some reason, I have one of these things. Not quite as highlighted as the other one, so let me make that a little bit of a thicker line. There we go. And remember, what essentially I do is I go in and I construct the perpendicular bisector of a segment like so and like so. And then I find their intersection point, like so. Okay, that's now my circumstance. And now I can draw a circle with a center and a point, and there's my circum circle. Boy, that was a lot faster, wasn't it? But now, watch what happens as I start to move these vertices around. Suddenly, I've got an acute triangle here. All those angles are less than 90°. Here now I have an obtuse triangle, and that circumstance is falling outside of it. You know, I can do all sorts of things. Like if I have an isosceles triangle like that, where the circumcenter is. And what we'll be looking at in a moment is what happens when I have a right triangle. So it's kind of neat. It used to be required knowledge that you knew, okay, well, if the triangle was acute, then the circumcenter lied inside of it, if it was obtuse, it lied outside of it. And if it was a right triangle, it lied, um. Well, let's get to that now. Let's talk about our right triangle. Okay? And one of the things with the right triangle is finding the circumscribed circumscribing circle. And the circumcenter is a lot easier than it is with any of the other triangles. And let's take a look why. Okay, exercise number 5. For a right triangle, PQR, shown below, answer the following. Construct the perpendicular bisector of the hypotenuse QR label the midpoint of QR as point C all right, so now of course we've done the perpendicular bisector repeatedly today. I'd like you to pause the video now. Do the perpendicular bisector, but only only of hypotenuse QR. Take a moment to do that. All right, let's go through it. Oh, let's see how quick we can do it. All right, put the sharp end at R hope like mad that the compass works. And that the ruler doesn't get in our way. That looks good. Let's draw an arc up here. Let's draw another one. Down here, make sure they're nice and long. All right, let's switch the pointed end. So that's over here. It's good enough. I think oh boy. Did I get it? Yeah, just barely. Notice that intersection point. The two are just intersected. Not by much, I'll tell you. Now I've got to draw that line segment. Always a tricky business. Let me pan down a little bit. I'm going to make sure I've really got it. Little bit more. Come on. All right. Oh, what I would give for just a normal ruler. Good enough. Okay. The perpendicular bisector. I also want to label its midpoint, I'm going to call that point C so simple enough. The perpendicular bisector. All right. Now, part B of this problem, let me extend the page a little bit, so we get it up. It says draw a circle with a center of C and a radius that's the length of RC. Is this a circle? Is this circumscribing? Triangle PQR. So in other words, what I'd like you to do now is draw a circle that has that as its center and has RC or QC, it doesn't matter because they're equal in length as the radius. Go ahead and do that really quick. All right, let's go for it. Ready, we're going to rotate this up a little bit, maybe shrink this down, bring it up to here. And out to there. Rotate a bit more. Looks pretty good. Oh, and look at that. Indeed. It is the circumscribing circle. We didn't even have to do two of them now. We just did that one. One of the remarkable things about right triangles is that the circumcenter, all right, again, let me go to red just to really emphasize this. The circumcenter lies at the midpoint of the hypotenuse. It's something very, very specific with right triangles. Okay? And we're not really going to delve too deeply into why that is right now. But it's kind of cool, and it really has to do with the fact. That rectangles that right triangles are really kind of half of rectangles. So we have to sort of wait until we get into that quadrilateral unit to understand this a little bit more. But let's finish the problem up and wrap up the lesson because there is one last part. It says draw a line segment CP in. So do that really fast. Draw in line segment CP. And then it says explain why triangle CPQ must be isosceles. Why don't you go ahead and do that? All right, let's go through it. Real fast. Easy enough to actually draw the segment in, especially for you. Let's do this. I think I'll leave it in red that way will really highlight it. Nice. And then it asks us to explain why triangle CPQ must be isosceles. Well, it's pretty simple, right? C. Is the center. Of a circle. With radii, a level word with two I's in a row with radii, CP. And cq, which must be the same length. Already, I have a given circle of the same length. So triangle CPQ. Has two congruent sides. And maybe I should say at least two congruent sides. It may have three, but it doesn't appear at least in this diagram. It doesn't appear in this diagram that PQ is the same length as these two. All right, that's it. Let me bring this back up. Now, one of the very important things about this exercise, I don't want it to get away from us again, is that if I ask you to find the circumscribed circle for a right triangle, all you have to do, all you have to do is construct the midpoint of the hypotenuse, and that becomes the circumcenter. That becomes where you're going to put the sharp end of the compass, right? And then you can use either end point, either end point of the hypotenuse to then set the radius. And then you get that nice circumscribing circle. All right, let's wrap this up. Oh, hello. All right. So what we saw today was a very special relationship between triangles and circles. For every triangle, there is one and only one circle that passes through the three vertices of the triangle. It's called the circumscribing circle, and its center is called the circumcenter. Defined that triangle to find that circle or to construct that circle, so to speak. Means that we take the triangle and we find the perpendicular bisector of two of its sides. When we do that, the intersection point of those two perpendicular bisectors is the circumcenter, because it is equidistant, the same distance from all three vertices. And so the three vertices now become points on the exterior of a circle, or lying on a circle. Okay? Now we're going to see another relationship between triangles and circles in a later lesson. But for now, I just want to thank you for joining me for another common core geometry lesson by E math instruction. My name is Kirk weiler, and Intel next time, keep thinking. And keep solving problems.