Geometry (all content)
- Challenge problems: perimeter & area
- Challenging perimeter problem
- CA Geometry: Deductive reasoning
- CA Geometry: Proof by contradiction
- CA Geometry: More proofs
- CA Geometry: Similar triangles 1
- CA Geometry: More on congruent and similar triangles
- CA Geometry: Triangles and parallelograms
- CA Geometry: Area, pythagorean theorem
- CA Geometry: Area, circumference, volume
- CA Geometry: Pythagorean theorem, area
- CA Geometry: Exterior angles
- CA Geometry: Pythagorean theorem, compass constructions
- CA Geometry: Compass construction
- CA Geometry: Basic trigonometry
- CA Geometry: More trig
- CA Geometry: Circle area chords tangent
- Speed translation
51-55, Pythagorean Theorem, compass constructions. Created by Sal Khan.
Want to join the conversation?
- Is there any videos to do with loci? or the locus of a point?(3 votes)
- I have no particular idea, but it makes me think of Linear algebra. Maybe you should try that. I hope that there are some, and I will try to find them if there are any. Hope this helps some.(1 vote)
- First, I came up with a^2+2ab+b^2-c^2 = (area of 4 the small triangles)
Then, I visually rearranged the 4 triangles into a rectangle of side b by side 2a = 2ab for its area.
a^2+2ab+b^2-c^2 = 2ab....
Would that be considered legitimate as proof?(2 votes)
- i'm confused how do i do the puthogorean theorom on a graph?
question applies to freshman geometry
- at9:40(problem 54)
Sal says that she is trying to draw a line parallel to line L. isn't it also possible to connect the two points and make a line parallel to the line that passes through p?(1 vote)
- Why are the things that you use for drawing circles and the things that you use for finding north both called compasses? This is confusing.(1 vote)
- why is the volume for pryamid 1/3Bh(0 votes)
- because its is in the 3rd dimension and if u put 3 pyramids in a prism 2/3 of it would be filled and 1/3 wouldnt(2 votes)
- This video got me thinking. If you have a perfect square and a smaller perfect square inside it, the angles of which hit the sides of the outer square exactly at their centre, is there anything you can say about the relationship between them that would be true for all cases?
I'm assuming that a quadrilateral hitting an outer square at the middle of its sides would also be a square. It just seems intuitively that that would be the case.(0 votes)
- In question 52, why would the old route be the hypotenuse i thought it would be the opposite because it's opposite of the angle?(0 votes)
- Um, I'm trying to find out about Construction of Loci, does anyone know if there is a video for anything to do with Loci? I cant find anything!(0 votes)
We're on problem 51. And they say, a diagram from a proof of the Pythagorean theorem is pictured below. And they they say, which statement would not be used in the proof of the Pythagorean theorem? So since they have drawn this diagram out, I think we might as well just kind of do the proof and then we can look at their choices and see which ones kind of match up to what we did. Hopefully, they do it the same way. And this is a pretty neat proof of the Pythagorean theorem. I don't think I've done it yet. So I might as well do it now. Well, let's figure out what the area of this large square is right. Well there's two ways to think about it you could just say, OK, this is a square. That's a, that's b. Well this is going to be b as well. This is going to be a as well. So the area of the square is going to be the length of one of its sides squared. So we could say the whole square's area is a plus b squared. And that's equal to a squared plus 2ab plus b squared. Fair enough. Now we can also say that the area of this larger square, and it's a bit of an optical illusion, it looks like it's tilted to the left because of the way it's drawn. But anyway, that the area of this larger square is also the area of these four triangles plus the area of this smaller square. So this, the area of the larger square, which we figured out just by taking one side of it and squaring it, that should be equal to the area of the four smaller triangles. So there's four of them. And what's the area of each of them. Let's see, let's just pick this one. 1/2 base times height. So it's 1/2 times a times b. So 1/2 ab is one of these and I multiply by 4 to get all four of these triangles. And then we want to add the area of this inside square. And that's just going to be c squared. So plus c squared. Let's see if we can simplify this. So you get a squared plus 2ab plus b squared is equal to 4 times 1/2 is 2ab plus c squared. Well, we could subtract 2ab from both sides of this equation. The top and the bottom of this equation the way I've written it. But if we do that, subtract 2ab from there, subtract 2ab from there, and you're left with a squared plus b squared is equal to c squared, which is the Pythagorean theorem. And we've proved it. So let's see which of their choices matches what we did. OK, which statement would not be used in the proof of the Pythagorean theorem. The area of a triangle equals 1/2 ab. We used that. The four right triangles are congruent. No, we used that. The area of the inner square is equal to half of the area of the larger square. We didn't use that. I think this is the one that would not be used in the proof. Choice D, the area of the larger square is equal to the sum of the squares of the smaller square and the four congruent triangles. No, that that was the crux of the proof. So we definitely used that. So C is our answer. That's the statement that would not be used in the proof. I'm learning to copy and paste ahead of time. So I don't waste your time. All right, a right triangle's hypotenuse has length 5. If one leg has length 2, what is the length of the other leg? Pythagorean theorem, x squared plus 2 squared is equal to 5 squared, because 5 is the hypotenuse. x squared plus 4 is equal to 25. Subtract 4 from both sides. x squared is equal to 21. So x is equal to the square root of 21. So choice B. Next question. A new pipeline is being constructed to reroute oil flow around the exterior of a national wildlife preserve. I guess that's the national wildlife preserve. The plan showing the old pipeline and the new route is shown below. OK, how many extra miles will the oil flow once the new route is establised. So the new route is going to be 60 miles plus 32 miles. So the new route is 92 miles. So what was the old route? Well the old route was the hypotenuse of this triangle. So we could say, let's call that x. 60 squared plus 32 squared is equal to x squared. Because that's the hypotenuse. And these numbers, that's a bit of a pain to deal with. Maybe if I can factor out something here I can make it more interesting. So I don't have to multiply out 60 squared and 32 squared and all of the rest. Well, let me see. Both of those are divisible by 4. So then I would have 15 and 8. Yeah, that still doesn't make it that useful. So I'll just multiply them out. So this is 3600. At 32 squared, let's see, 32 times 32. 2 times 32 is 64. 3 times 2 is 6. 3 times 3 is 9. So it's 1024. Plus 1024 is equal to x squared. So let me just switch both sides. x squared is equal to 3600 plus 1024 is 4624. Let me see if I can get an approximate. So x is going to be the square root of this thing right here. So let's see if I can get a handle at least on the magnitude of where this would be. So 20 times 20 is 400. So this is way too small. 60 times 60 is 3600. So 68 times 68, this looks right. Especially because 8 times 8 should end in a 4. Let me try that out. 68 times 68. 8 times 8 is 64. 8 times 6 is 48 plus 6 is 54. 6 times 8, 48. 6 times 6, 36 plus 4 is 40. 4624. So x is 68. Oh, I used 68, I shouldn't have. Because they don't want to know how long was the old pipeline. That's 68. It just happened to be one of the choices. That's just to make sure that you read the question properly. But they want to know how much longer is going to be the new pipeline. So the new one was 92. And the old one is 68. Good thing they had that number there so I could try it out. That was the square root of 4624. So how much longer is the new one? Well 92 minus 68 that's 24 miles. So choice A. Not choice B. B is how long the old pipeline was. We want to know how much longer the new route is. That was tricky. Well not tricky, but I kind of fell for it by forgetting what the question was about. Anyway, next question. Marcia is using a straightedge and compass to do the construction below. Interesting. Which best describes the construction Marcia is doing. So, I assume when they say construction she's drawing something. Let's see what it looks like. It looks she's taking her compass, she's probably putting one of the points here, she put one of the points there and then she kind of drew this arc. And then it looks like she put the point there and then she drew that arc. And then she put the point here and drew that arc. And then put the point there and drew that arc. And the end result, it seems like the reason why she picked this point here is it goes through this line L. So she's probably trying to find another point here, so that she can draw another line. Because they say she has a straightedge. A straightedge is to draw these lines. A compass is to draw these curves. So if she were to draw another line between these two points, it looks something like that, then she would have parallel lines. The reason why she would have parallel lines is because these would be corresponding angles and they would be congruent. And so if you have a transversal the corresponding angles are congruent, you're dealing with parallel lines. So my read of this question is that she's probably trying to draw a line that is parallel to L. A line through P parallel to line L. Yeah, that's what I think she's trying to do. All right, choice A. 55. Given angle A. So given this angle. What is the first step in constructing the angle bisector of angle A? OK, well actually I've never done this. But I can assume that if I have a compass. You know what a compass is, it has those two points. One of them is like a pivot point. It looks something like this. It looks like it has a little pivot point, and then on the other side you can stick your pencil. And you can adjust it up here. And the bottom line, you pivot around this and then you can draw circles of arbitrary radiuses. It seems like that's what they did here. So if I want to draw the angle bisector of a, just thinking about it, it seems I could put the pivot point here, and then I can put the pencil and I can draw this circle. And really, as long as I just find the two points that it intersects those two lines or those two rays, then I'll be fine. And I could have done it anywhere. I could have done it here. I could have done it out here. They just picked points B and C. And then from each of those points, you can put your pivot here. If you put your pivot here, and then you were to draw a circle around that, you would have gotten this one right here. And then if you were to put your pivot point right here, draw a circle, you would be able to draw that. And then where they interect, that would that would give you an indication of where the angle bisector is. And you could then draw that line to where they intersect. So let's see, they say what is the first step in constructing the angle bisector of angle a. So they say draw ray AD. Well that seems like that would be the last step. Then you're done. Draw AD, that is the angle biector. Draw a line segment connecting points B and C. No, that's useless. You don't need a line segment. I mean even what they have drawn, that's an arc. It's not a line. From points B and C, draw equal arcs that intersect at D. That was the second step. You have to have points B and C before you can draw those equal arcs. From point A, draw an arc that intersects the side of the angle at points B and C. Yeah, that's what we said. That was the first step. Put your pivot here, and use your pencil to draw the arc. You say OK, this point and this point. So that would be the first step. D. And I'm all out of problems and I'm out of time. See you in the next video.