Common Core Geometry Unit 6 Lesson 5 Rectangles

Hello and welcome to another common core geometry lesson by E math instruction. My name is Kirk weiler and today we'll be doing unit 6 lesson 5 on rectangles. And I know you've been studying rectangles, probably for as long as you've been in school, right? Recognizing them, looking at them, studying them, all of that kind of stuff. And really, in this lesson, in the next few, we're going to be walking down a path where we examine what are called special parallelograms. Sometimes the special quadrilaterals. All right? And what we're going to be doing is we're going to be seeing how each one of these is a parallelogram with additional properties. So let's jump right in and talk about what the definition of a rectangle is. A rectangle is any quadrilateral with four right angles. Now, one of the things that's sort of interesting about mathematics is that you can actually define a rectangle in a variety of different ways. So this is the definition that I hold with. A rectangle is any quadrilateral with four right angles. Some math teachers will define a rectangle as a parallelogram with at least one right angle, okay? But I'm going to just start with this. The defining characteristic of a rectangle. Well, it makes a rectangle a rectangle is all four of those interior angles are 90°. All right? So now let's move on and see what else we can figure out about a rectangle based on that definition. All right. The first thing that I'd like to do is I'd like to walk through an exercise that proves that based on the definition of a rectangle. Four right angles. It also has to be a parallelogram. All right? There's actually a variety of different ways to do that. But let's take a look at how we're going to do it in exercise one. Rectangle ABCD is formed below by lines N and P and Q give a justification for YM must be parallel to N and likewise P must be parallel to Q and that's basically going to prove it's a parallelogram. Because the definition of a parallelogram is a four sided figure with two pairs of parallel sides. So if I can prove that P is parallel to Q or E as parallel in I can see it's a parallelogram. So how can I prove that lines are parallel? There's going to be a variety of different ways I can do it. So let's first talk about why I can prove that M is parallel to N and again, lots of different ways to do it. One way to do it is by saying, well, let's talk about this angle. Let's talk about angle one. And let's call this angle two. All right. So angle one must be a right angle. And it's going to be a right angle because it this. Create a straight angle. But that means that angle one must be congruent to angle two, right? These two angles. Because that's a right angle. That's a writing. But if we have angle one to angle two, these are alternate interior angles that are being created by this line, and this line cut by this transversal. And so I can say that M is parallel to N because a pair of alternate. Interior angles. Are congruent. Now I can literally make the identical argument in order to show that P is parallel to Q, okay? But let me make actually just a slightly different argument. Just a slightly different argument, not much. Let me call this angle three over here. Now, obviously, I agree. So angle one is congruent to angle three. All right. These two angles are converted. Okay? But now what that's going to do is that's going to tell me that P is parallel to Q because a pair of corresponding angles. Are congruent. Again, we could have also gone with alternate interior angles. We could have talked about this angle, being 90 and that angle being 90, and then they would have been here and here, sorry. This one, and this one, and they would have been alternate interior angles that would have been established, P being parallel to queue. Either way, all those 90s ensure that we've got alternate interior angle pairs congruent, corresponding angle Paris congruent, all of which tell me that we've got two pairs of parallel sides, which leads me to be able to conclude a very important thing all rectangles are parallelograms. If you think about that for a minute, though, we know a lot about parallel grams, right? We know that in parallelograms, opposite angles are congruent, although I guess we already knew that because they were rectangles. But then therefore, the opposite angles are both 90°. But we also know things like opposite sides of a parallelogram are the same length, so those two must be the same length. And those two must be the same length. Okay? So there's a lot that comes along with the fact that every rectangle is a parallelogram. But to prove that rectangles are parallelograms, we really need to use those parallel line properties or the properties that establish that two lines are parallel. Let's move on to the next exercise. Okay. Take a look at exercise two. Given rectangle npq shown below. All right, so we know it's a rectangle. Do the following. Using a ruler or tracing paper, compare the lengths of the diagonals of the rectangle. And MP, are they the same? All right. So what I want to know is in this rectangle, all the two guys the same length. A few lessons ago, we looked at the question about whether or not the diagonals of a parallelogram in general are congruent or the same length. And we found that they weren't. But take a look at this. Let me take my room up. And I could actually measure the length itself. But I'm going to kind of do almost a tracing paper argument. Watch what I can do. And actually, I think this really works well in red. Very nice, thick red. I'm going to trace out diagonal MP. Okay? Not a perfect job, but good enough. And watch what I can do with it now. I can actually rotate it and what I see is that it is the same length as QN. So yes, the two diagonals of a rectangle, at least experimentally. At two diagonals. Are the same length. Now let's prove that. I want to somehow prove that the diagonals of a rectangle are congruent. Okay? Because it's one of the very special properties of a rectangle that most parallelograms don't have. Most parallelograms, there's a longer diagonal and a shorter diagonal. But in a rectangle, the two diagonals are the identical length. And the question is, why? Let's kind of sketch out the proof, and then I'm going to write it out in a paragraph form. Now think about this for a second, though. Ultimately what we want to prove is that NP is given to nq and that sounds a lot like a CPC TC proof. Sounds a lot like basically establishing that two parts of congruent triangles are also congruent. Well, let's talk about what we know based on the fact that this is a rectangle. All right? One thing I know based on the fact that it's a rectangle is that this angle and this angle are right in this. I also know the two up there are also right angles. But I'm just going to stick with those for the time being. Now because it's also the parallelogram. I know that this side length and this side length must be the same. All right, I'm getting somewhere. So I have a side equal to a side and angle equal to an angle. And finally, if I kind of concentrate on two triangles specifically, this triangle. Right? And this triangle. Right, I like to be able to prove that they're going to run with. Because if I can prove this to triangles are converted, and then I can just state that the two diagonals are congruent by CPC TC, and right now I've got 5%. Oh, nice. They share this side. Right? That whole reflexive property thing. So I'm going to be able to establish that this term is converted into this triangle by side angle side, and because those two triangles are congruent, and I can state that they're hypotenuses or by CPC TC. See if you can write all of that down in a paragraph type proof, and then we'll go through it. All right? All right, let's hit it. So because M and P Q is a rectangle. Angle nq is congruent to angle N pq. If you absolutely feel the need, you could say they are both right angles. Okay. And PQ is also a parallelogram. Since oops, look at that red. Since all rectangles are parallelograms. So MQ is congruent. To NP. Sense opposite sides. And a parallelogram. Are congruent. Finally. QP is congruent to QP. Bring this down a little bit. By the reflexive property. Actually, get ready to come in there. Period so triangle. Let me get this right. And cube P must be congruent to triangle. And PQ by side angle side. Therefore. Therefore, therefore, therefore, MP. Is congruent to. Phi C P C T, C there it is. Again, essentially what we're establishing is that when you draw the two diagonals of a rectangle in, you create two coherent right triangles. You can argue that you actually create four congruent right triangles. But as long as we've got these two right triangles here, then they're hypotenuses, are also congruent. No HL going on here. Just side angle side. And this is one of the extremely important properties of rectangles, okay? Generally speaking, if I've just got kind of a random random parallelogram, then what you can see is that the two diagonals do not have the same length. Generally speaking, right? Here are this dino 10.65 this diagonal 7.75. But the closer icon to having a rectangle, not quite a rectangle, not even 1.87° down here. But now, 14.59 1505, quite close, and there's no way I'm going to go get this to be exactly another. That's not too bad. 14.95, 14.83. Right? As soon as we lock down that 90° angle, then those two diagonals become exactly the same. A very important property of rectangles. And one that you've probably used in problems before is simple as this. Let's take a look at it. Exercise three. In rectangle RST U, it's known that RS is 12 and ST is 5. What is the length of the diagonal? SU. All right. So the point here is that if I have this rectangle, okay. R S T U RS is 12 ST is 5. And I want the length of diagonal SU, right? Well, that'll be the same as the length of diagonal RT. Don't get me wrong. You can also just invoke the fact that the sun is depending on you, which means if that's 5, this is fine. We know that this is 90°. So how can we actually solve for the length of diagonal SU? Well, hopefully what's completely obvious is that we can use the Pythagorean theorem to solve for that diagonal length, right? And we should be watching for the Pythagorean theorem, when we have situations that evolve a right angle all right triangle. So I could easily call that C and then I can just say, well, I've got 5 squared plus 12 squared is equal to C squared. 25 plus one 44. Is equal to C squared. C squared is one 69. Take the square root of both sides. And both diagonals will have a length of 13. All right? So nice, easy problem involving rectangles and the diagonals of rectangles. The Pythagorean theorem comes up quite a bit with rectangle work, simply because when you divide a rectangle using one of the diagonals, you always have a right triangle. Okay? All right. Let's go on and do some more challenging problems because that one was pretty easy. All right. The last little piece that we want to kind of work through today is something similar to the last lesson. We want to take a look at how, if we have a parallelogram with congruent diagonals, then it must be a rectangle. So we know that if we have a rectangle, it's got congruent diagonals. What we now want to show is if we know that it's a parallelogram with congruent diagnosis, this is important. Not just congruent diagonals. You can easily have a figure that's got two diagonals that are exactly the same length. But isn't the rectangle. We want to show that if we've got a parallel ground, both parallel diagonals. It's got to be a rectangle. So let's take a look at that in this exercise. Exercise number four. In the diagram below, all we know is that ABCD is a parallelogram whose diagonals are congruent. All right, first part of the problem. Explain why ADC must be congruent to BCD. Take a moment, pause the video and go through that. All right. Well, let's talk about it. We know it's a parallelogram. Which means opposite sides. Okay. We also know that the diagonals aren't congruent. So this side is converted to the side. It's a little bit hard to mark those segments, but the entire diagonals. And we also know that this third side is shared. So this kind of gets some of that noun. AD is congruent to BC. Because opposite sides congruent in a parallelogram. We know that AC is congruent to BD. That was simply a given. So we're doing kind of a little mini proof here. We know that CD is obviously congruent to itself, CD. That's the reflexive property. And therefore, triangle ADC is congruent to triangle BCD. By side side side. So just based on the fact that we've got a panel on the ground whose diagonals are congruent, those two triangles have to be congruent by the side side side triangle congruence theorem. All right, what does that then tell you about angle ADC and angle BCD? What do we know about those two angles based on the fact that the two triangles, this one and that one are congruent? Wow, they have to be congruent to each other. Angle ADC must be congruent to angle BCD. By that whole CPC TC thing, right? Those are congruent parts of congruent triangles. Corresponding parts of congruent triangles are congruent. Obviously, congruent parts of congruent triangles are congruent. But there are corresponding parts. Those two angles show up in the same place. Finally, how can we use this part in B to now conclude that angle ADC and BCD must be right angles. Why is it that knowing that this is congruent to this means that they must be 90s? All right, well, it's pretty simple. Since a, B, C D is a parallelogram. Remember, we knew that up front. Angle ADC is supplementary. To mystery red. Supplementary to angle BCD. Consecutive angles and a parallelogram are supplementary. They have to add up to one 80, right? But. Since they are equal. They must both be 90°. Right? So whatever they are, if this thing were a 50° angle, that would have to be 130° angle. These two angles in any parallelogram have to add it to a 180°. But the fact that they're equal angles means that both must be 90°. Any angle that is its own supplement must be a right angle. All right? So kind of neat. And that then really proves that a parallelogram that has congruent diagonals must be a rectangle. Because we can make the same argument up here. We can make the argument that angle a and angle B also have to be 90° angles, okay? Bernie nice. Let's move on to the last problem. Let's get the coordinate grid back in, okay? We've done a lot of work lately. Using Euclidean logic, CPC TC, things like that. Let's go back to the tools of coordinate geometry. Exercise number 5, quadrilateral EFG, has vertices at negative 6 two three 8 7 two and negative two negative four. Calculate the slope of all four sides of EFG, use the slopes to prove EFG, H is a rectangle. All right, there are many different ways to prove using coordinate geometry. That's something as a rectangle. Again, my preferred way is to basically establish the fact that you've got four right angles. And we can do that by using perpendicularity, and we can get it perpendicularity by using slopes. So what I'd like you to do is pause the video right now and also. A bit of GH in a slope of AH. Take a little bit of time. All right, let's go through it. Here we go. Nice and systematic. EF, let's take a look at its slope M equals 8 minus two in the numerator. Three minus negative 6 in the denominator. That gives me 6 divided by 9. And that reduces to two thirds. Let's take a look at FG. FG, the slope is going to be two -8 divided by 7 minus three. That's going to be negative 6 in the numerator. That's going to be four in the denominator, negative divided by a positive is a negative, and that reduces to three halves. So GH, let's take a look. Its slope is going to be negative four minus two, divided by negative two, -7. It's going to be negative 6 in the numerator, negative 9, and the denominator. Remember a negative divided by a negative is a positive and 6 ninths or reduced. Is two thirds. And finally, HE, or EH, however you want to call it the slope is going to be two minus a negative four in the numerator, divided by a negative 6 minus negative two, and the denominator. Two minus negative four is positive 6. Negative 6 minus a negative two is a negative four, a positive divided by a negative is a negative and 6 fourths reduces to three halves. Now let's take a look at those four slopes. Notice that two thirds and negative three halves are negative reciprocals. That means that EF must be perpendicular to FG. But notice negative. So if G must be perpendicular to GH. And likewise, two thirds is the negative reciprocal of negative three halves, so GH must be perpendicular to HE. And HE must be perpendicular to EF. Now, these are kind of like statements. You should justify them. And the justification for all of these because slopes are negative. I'm going to abbreviate this negative reciprocals. But now we have it, right? I mean, the time between perpendicularity and right angles is great enough to now say, give myself a little room here. EF GH is a rectangle. Because it contains. Four right angles. There we go. You have GH is a rectangle before it, because it contains four right angles. Now again, there are other ways to do this. You could also prove that it's a parallelogram, all right? You could then go on to establish that it's got a single right angle, and therefore a parallelogram with a single right angle must be a rectangle. You could prove that it's a parallelogram and then look at the diagonals. There's lots of different ways. To me, this is the most straightforward way of proving using coordinate geometry that a figure is a rectangle. Now let's take a look at letter B, it says calculate the midpoints of the diagonals of EFG H why does this show that EFG H must be a parallelogram? Now, we already know that all rectangles are parallelograms. But I want to kind of verify that by doing a little bit of midpoint work. So let's just make sure we've got E, I'm just going to read off the coordinates from here, which is at negative 6 positive two. We've got F which is at three comma 8, we've got G, which is at 7 two. And we've got H, which is at negative two negative four. Let's figure out what the midpoints of the diagonals are, right? That would be EG and FH. So let's do it. Let's do EG together. Remember by finding the midpoints all you're doing, all you're doing is averaging the X coordinates and averaging the Y coordinates. All right. It's a little weird because that's going to give me one half and two. Okay. And if I look at FH. All right, what do I have? I've got three plus negative two divided by two. And 8 plus negative four divided by two. And then I'll give you a midpoint of one half comma two. So what is this? Just this. Nothing else. Actually tells me that this must be a parallelogram. Why? Well, let's talk about it. Right? This means. Since they have the same midpoint. I'm going to abbreviate the diagonals bisect each other. And what we learned a couple of lessons ago. Is that any quadrilateral any quadrilateral I'm trying to make an R there it's not worth a very good. Any quadrilateral whose diagonals bisect each other must be a parallelogram. It's one of the many different ways you can prove something as a parallelogram. It must be a parallelogram. Kind of cool, right? And many times, teachers will suggest this as being the quickest method for proving in the coordinate plane. When you have current geometry, improving that a figure isn't a rectangle, but is a parallelogram. Simply find the midpoints of each of the diagonals, and if they're the same, you can conclude that that thing is a parallelogram, all right? I'm just going to need. Let's take a look at letter C it says calculate the lengths of diagonals. Why with B, does this show that EF GH is a rectangle? Now, I know. Letter a, we proved it was a rectangle already. But what we're trying to build up in B and C is an alternate way of showing something as a rectangle. So let's do it. Let's calculate the lengths of the diagonals. Okay, again, I'm going to just write out the coordinates really fast. Point E is at negative 6 comma two, point F is at three comma 8 point G is at 7 comma two point H is at negative two comma negative four. And I want to calculate the lengths of the diagonals. Now, something kind of funny about EG, okay? Let me actually draw that diagonal in EG is not a slanted line. All right? So you can use the distance formula. I'll even show you that. But you could also just count versus FH FH, you definitely have to use the distance formula to calculate its length. But let's take a look at EG. The distance formula should work whether or not that line segment is slanted or not. So for EG, we would have 7 minus negative 6 squared plus two minus two. Squared. And notice that's just going to end up being zero. Two minus two is zero. All right, but whatever. I mean, let's just kind of keep it on. Here we've got 7 minus negative 6, which is 13 squared plus zero squared. Which is the square root of a 169. Which is 13. All right, let's do FH. All right. For that, the distance formula a little bit more complicated. But we'll have negative two minus three. Squared. Plus negative four -8 squared. All right, that's going to be negative 5. Squared plus negative 12. Squared in that, again, turns out to be the square root of one 69 or 13. Now, why with this part B, does this prove that we have a rectangle? All right, well, that's pretty extend the page. So I have a little bit more room to write, okay? B proves right that E F G H is a parallelogram. Actually. Let me just say eh, they prove that it's a parallelogram. C prose it has congruent diagonals. And the plain fact is what we saw before. Any parallel ground. With congruent diagonals. Or diagonals that have the same length. Is a rectangle. And that's the way that a lot of students like to prove in the coordinate plane that you have a rectangle. They like to say, all right, well, I'm going to find the midpoints of the two lines. So it's all about diagonals then. You find the midpoint of the two diagonals. You find out that it's the same, and you know that it's a parallelogram, because the diagnosed bisect each other. Then you look at the length of the two diagonals, you find out that they're the same, and you conclude well. I've got a parallelogram with two diagonals that are the same length, and so it has to be a rectangle. Which is kind of fascinating because then all of that, there's no mention of 90° angles, which is of course the defining characteristic of a rectangle. A little strange, huh? Anyway, let me kind of step back, let you take a look at some of this. And then we'll wrap up. All right, let's do it. So what we saw today was a lot of different stuff. When we were looking at a rectangle. The defining characteristic of a rectangle, a four sided figure with all 90° angles. What's amazing about that is you can then take that property and prove that every rectangle has to be a parallelogram, and all the properties of parallelograms, opposite sides are congruent, opposite angles are congruent. The diagonals bisect each other, all that, then, gets to come along for the ride, if you will. But then we went on and we showed that rectangles have other important properties. The biggest one being that the diagonals are congruent to one another. We then looked in that last exercise at the fact that if you've got a parallelogram, with congruent diagonals, you must have a rectangle and how we can use that knowledge in the coordinate plane to prove something as a rectangle. And again, two very different but viable options in the coordinate plane for proving something is a rectangle. Either use all four slopes in the sides to just show that you've got four sets of perpendicular sides, or prove it's a parallelogram. By whatever method, including common midpoints, and then show it's a rectangle by showing that the diagonals are the same length. All right, well next time we're going to start getting into the rhombi or rhombuses, and then we'll move on to squares, et cetera, et cetera. 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 until next time, keep thinking. And keep solving problems.