Sentence
stringlengths 50
2.16k
| video_title
stringlengths 16
104
|
|---|---|
So B's slope is negative 1 half. So we know that B's equation has to be y is equal to its slope m times x plus some y-intercept. We still don't know what the y-intercept of B is, but we can use this information to figure it out. We know that y is equal to negative 7 when x is equal to 6, negative 1 half times 6 plus b. I just know that this is on the point. So this point must satisfy the equation of line B. So let's work out what B must be. So in this, or what the B, the y-intercept, this is the lowercase b, not the line b.
|
Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3
|
We know that y is equal to negative 7 when x is equal to 6, negative 1 half times 6 plus b. I just know that this is on the point. So this point must satisfy the equation of line B. So let's work out what B must be. So in this, or what the B, the y-intercept, this is the lowercase b, not the line b. So we have negative 7 is equal to what's negative 1 half times 6? That's not a b there, that's a 6. What's negative 1 half times 6?
|
Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3
|
So in this, or what the B, the y-intercept, this is the lowercase b, not the line b. So we have negative 7 is equal to what's negative 1 half times 6? That's not a b there, that's a 6. What's negative 1 half times 6? It's negative 3 is equal to negative 3 plus our y-intercept. Let's add 3 to both sides of this equation. So if we add 3 to both sides, I just want to get rid of this 3 right here, what do we get?
|
Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3
|
What's negative 1 half times 6? It's negative 3 is equal to negative 3 plus our y-intercept. Let's add 3 to both sides of this equation. So if we add 3 to both sides, I just want to get rid of this 3 right here, what do we get? The left-hand side, negative 7 plus 3 is negative 4. And that's going to be equal to, these guys cancel out, that's equal to b, our y-intercept. So this right here is a negative 4.
|
Writing equations of perpendicular lines Mathematics I High School Math Khan Academy.mp3
|
Two of the points that define a certain quadrilateral are zero comma nine and three comma four. The quadrilateral is left unchanged by a reflection over the line y is equal to three minus x. Draw and classify the quadrilateral. Now I encourage you to pause this video and try to draw and classify it on your own before I'm about to explain it. So let's at least plot the information they give us. So the point zero comma nine, that's one of the vertices of the quadrilateral. So zero comma nine, that's that point right over there.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
Now I encourage you to pause this video and try to draw and classify it on your own before I'm about to explain it. So let's at least plot the information they give us. So the point zero comma nine, that's one of the vertices of the quadrilateral. So zero comma nine, that's that point right over there. And another one of the vertices is three comma four. Three comma four, that's that right over there. And then they tell us that the quadrilateral is left unchanged by a reflection over the line y is equal to three minus x.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
So zero comma nine, that's that point right over there. And another one of the vertices is three comma four. Three comma four, that's that right over there. And then they tell us that the quadrilateral is left unchanged by a reflection over the line y is equal to three minus x. So when x is zero, y is three, that's our y-intercept. And it has a slope of negative one. You could view this as three minus one x.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
And then they tell us that the quadrilateral is left unchanged by a reflection over the line y is equal to three minus x. So when x is zero, y is three, that's our y-intercept. And it has a slope of negative one. You could view this as three minus one x. So it has a slope of negative one. So the line looks like this. So every time we increase our x by one, we decrease our y by one.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
You could view this as three minus one x. So it has a slope of negative one. So the line looks like this. So every time we increase our x by one, we decrease our y by one. So the line looks something like this. Y is equal to three minus x. Try to draw it relatively, pretty carefully.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
So every time we increase our x by one, we decrease our y by one. So the line looks something like this. Y is equal to three minus x. Try to draw it relatively, pretty carefully. Pretty carefully, so that's what it looks like. Y is equal to three minus x. So that's my best attempt at drawing it.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
Try to draw it relatively, pretty carefully. Pretty carefully, so that's what it looks like. Y is equal to three minus x. So that's my best attempt at drawing it. Y is equal to three minus x. So the quadrilateral is left unchanged by a reflection over this. So that means if I were to reflect each of these vertices, I would essentially end up with one of the other vertices on it, and if those get reflected, you're gonna end up with one of these, so the thing is not going to be different.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
So that's my best attempt at drawing it. Y is equal to three minus x. So the quadrilateral is left unchanged by a reflection over this. So that means if I were to reflect each of these vertices, I would essentially end up with one of the other vertices on it, and if those get reflected, you're gonna end up with one of these, so the thing is not going to be different. So let's think about where these other two vertices of this quadrilateral need to be. So this point, let's just reflect it over this line, over y is equal to three minus x. So if we were to try to drop a perpendicular to this line, notice we have gone diagonally across one, two, three of these squares.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
So that means if I were to reflect each of these vertices, I would essentially end up with one of the other vertices on it, and if those get reflected, you're gonna end up with one of these, so the thing is not going to be different. So let's think about where these other two vertices of this quadrilateral need to be. So this point, let's just reflect it over this line, over y is equal to three minus x. So if we were to try to drop a perpendicular to this line, notice we have gone diagonally across one, two, three of these squares. We need to go diagonally across three of them on the left-hand side. So one, two, three gets us right over there. That's the reflection of this point.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
So if we were to try to drop a perpendicular to this line, notice we have gone diagonally across one, two, three of these squares. We need to go diagonally across three of them on the left-hand side. So one, two, three gets us right over there. That's the reflection of this point. This is the reflection of this point across that line. Now let's do the same thing for this blue point. To go, to drop a perpendicular to this line, we have to go diagonally across two of these squares.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
That's the reflection of this point. This is the reflection of this point across that line. Now let's do the same thing for this blue point. To go, to drop a perpendicular to this line, we have to go diagonally across two of these squares. So let's go diagonally across two more of these squares, just like that, to get to that point right over there, and now we've defined our quadrilateral. Our quadrilateral looks like this. Our quadrilateral looks like this.
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
To go, to drop a perpendicular to this line, we have to go diagonally across two of these squares. So let's go diagonally across two more of these squares, just like that, to get to that point right over there, and now we've defined our quadrilateral. Our quadrilateral looks like this. Our quadrilateral looks like this. Both of these lines are perpendicular to that original line, so they're going to have the same slope, so that line is parallel to that line over there, and then we have this line. We have this line, and then we have this line. So what type of quadrilateral is this?
|
Constructing quadrilateral based on symmetry Transformations Geometry Khan Academy.mp3
|
So this would be a circle that's inside this triangle where each of the sides of the triangle are tangent to the circle. And one way to, or probably the easiest way to think about it is the center of that circle is going to be at the in-center of the triangle. Now, what is the in-center of the triangle? The in-center of the triangle is the intersection of the angle bisectors. So if I were to make a line that perfectly splits an angle in two, so I'm eyeballing it right over here, this would be an angle bisector. But to be a little bit more precise about angle bisectors, I could actually use a compass. So let me make this a little bit smaller.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
The in-center of the triangle is the intersection of the angle bisectors. So if I were to make a line that perfectly splits an angle in two, so I'm eyeballing it right over here, this would be an angle bisector. But to be a little bit more precise about angle bisectors, I could actually use a compass. So let me make this a little bit smaller. And what I can do is I could put this, the center of the circle, on one of the sides of this angle right over here. Now let me get another circle. And I want to make it the same size.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
So let me make this a little bit smaller. And what I can do is I could put this, the center of the circle, on one of the sides of this angle right over here. Now let me get another circle. And I want to make it the same size. So let me center it there. I want to make it the exact same size. And now let me put it on the other one, on the other side of this angle.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
And I want to make it the same size. So let me center it there. I want to make it the exact same size. And now let me put it on the other one, on the other side of this angle. I'll put it right over here. And I want to put it so that the center of the circle is on the other side of the angle. And the circle itself, or the vertex, sits on the circle itself.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
And now let me put it on the other one, on the other side of this angle. I'll put it right over here. And I want to put it so that the center of the circle is on the other side of the angle. And the circle itself, or the vertex, sits on the circle itself. And what this does is I can now look at the intersection of this point, the vertex, and this point. And that's going to be an angle bisector, or the angle bisector. So let me go.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
And the circle itself, or the vertex, sits on the circle itself. And what this does is I can now look at the intersection of this point, the vertex, and this point. And that's going to be an angle bisector, or the angle bisector. So let me go. I'm going to go through there. And I'm going to go through there. Now let me move these circles over to here so I can take the angle bisector of this side as well.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
So let me go. I'm going to go through there. And I'm going to go through there. Now let me move these circles over to here so I can take the angle bisector of this side as well. So I could put this one over here. And I could put this one. See, I want to be on the side of the angle.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
Now let me move these circles over to here so I can take the angle bisector of this side as well. So I could put this one over here. And I could put this one. See, I want to be on the side of the angle. And I want to go right through. I want the circle to go right through the vertex. And let me add another straight edge here.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
See, I want to be on the side of the angle. And I want to go right through. I want the circle to go right through the vertex. And let me add another straight edge here. So I want to go through this point. And I want to bisect the angle. So I want to bisect the angle, go right through the other point of intersection of these two circles.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
And let me add another straight edge here. So I want to go through this point. And I want to bisect the angle. So I want to bisect the angle, go right through the other point of intersection of these two circles. Now let me get rid of one of these two circles. I don't need that anymore. And let me use this one to actually construct the circle inscribing the triangle.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
So I want to bisect the angle, go right through the other point of intersection of these two circles. Now let me get rid of one of these two circles. I don't need that anymore. And let me use this one to actually construct the circle inscribing the triangle. So I'm going to put it at the center right over there. And actually, this one's already pretty close in terms of dimensions. And with this tool, you don't have to be 100% precise.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
And let me use this one to actually construct the circle inscribing the triangle. So I'm going to put it at the center right over there. And actually, this one's already pretty close in terms of dimensions. And with this tool, you don't have to be 100% precise. It has some margin for error. It has some margin for error. And so let's just go with this.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
And with this tool, you don't have to be 100% precise. It has some margin for error. It has some margin for error. And so let's just go with this. This actually should be touching. But this has some margin for error. Let's see if this was good enough.
|
Constructing circle inscribing triangle Geometric constructions Geometry Khan Academy.mp3
|
I encourage you to pause the video and try this out on your own. Well, let's think about what's going on right over here. AB is definitely the A diameter of the circle. It's a straight line. It's going through the center of the circle. O is the center of the circle right over here. And so what do we know?
|
Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3
|
It's a straight line. It's going through the center of the circle. O is the center of the circle right over here. And so what do we know? Well, we could look at this angle right over here, angle C, and think about it as an inscribed angle, and think about the arc that it intercepts. It intercepts this arc right over here. This arc is exactly half of the circle.
|
Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3
|
And so what do we know? Well, we could look at this angle right over here, angle C, and think about it as an inscribed angle, and think about the arc that it intercepts. It intercepts this arc right over here. This arc is exactly half of the circle. This arc right over here is exactly half of the circle. Angle C is inscribed. If you take these two sides, or the two sides of the angle, it intercepts at A and B, and so it intercepts an arc, this green arc right over here.
|
Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3
|
This arc is exactly half of the circle. This arc right over here is exactly half of the circle. Angle C is inscribed. If you take these two sides, or the two sides of the angle, it intercepts at A and B, and so it intercepts an arc, this green arc right over here. So the central angle right over here is 180 degrees, and the inscribed angle is going to be half of that. It's going to be 90 degrees. Or, another way of thinking about it, it's going to be a right angle.
|
Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3
|
If you take these two sides, or the two sides of the angle, it intercepts at A and B, and so it intercepts an arc, this green arc right over here. So the central angle right over here is 180 degrees, and the inscribed angle is going to be half of that. It's going to be 90 degrees. Or, another way of thinking about it, it's going to be a right angle. And what that does for us is it tells us that triangle ACB is a right triangle. This is a right triangle, and the diameter is its hypotenuse. So we can just apply the Pythagorean theorem here.
|
Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3
|
Or, another way of thinking about it, it's going to be a right angle. And what that does for us is it tells us that triangle ACB is a right triangle. This is a right triangle, and the diameter is its hypotenuse. So we can just apply the Pythagorean theorem here. 15 squared plus 8 squared is going to be the length of side AB squared. So let me just call this side right over here, let me just call that x. That's going to be equal to x squared.
|
Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3
|
So we can just apply the Pythagorean theorem here. 15 squared plus 8 squared is going to be the length of side AB squared. So let me just call this side right over here, let me just call that x. That's going to be equal to x squared. So 15 squared, that's 225. 8 squared is 64, plus 64 is equal to x squared. I want to do that in green.
|
Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3
|
That's going to be equal to x squared. So 15 squared, that's 225. 8 squared is 64, plus 64 is equal to x squared. I want to do that in green. Is equal to x squared. 225 plus 64 is 289, is equal to x squared. And then 289 is 17 squared.
|
Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3
|
I want to do that in green. Is equal to x squared. 225 plus 64 is 289, is equal to x squared. And then 289 is 17 squared. And you could try out a few numbers if you're unsure about that. So x is equal to 17. So the diameter of the circle right over here is 17.
|
Hypotenuse of right triangle inscribed in circle Circles Geometry Khan Academy.mp3
|
So each of these is one unit. Which of the following sequences of transformations maps triangle PQR onto triangle ABC? So we have four different sequences of transformations. And so why don't you pause this video and figure out which of these actually does map triangle PQR, so this is PQR, onto ABC. And it could be more than one of these. So pause this video and have a go at that. All right, now let's do this together.
|
Mapping shapes.mp3
|
And so why don't you pause this video and figure out which of these actually does map triangle PQR, so this is PQR, onto ABC. And it could be more than one of these. So pause this video and have a go at that. All right, now let's do this together. So let's first think about sequence A. And I will do sequence A in this purple color. So remember, we're starting with triangle PQR.
|
Mapping shapes.mp3
|
All right, now let's do this together. So let's first think about sequence A. And I will do sequence A in this purple color. So remember, we're starting with triangle PQR. So first it says a rotation 90 degrees about the point R, so let's do that. And then we'll do the rest of this sequence. So if we rotate this 90 degrees, so one way to think about it is a line like that is then going to be like that.
|
Mapping shapes.mp3
|
So remember, we're starting with triangle PQR. So first it says a rotation 90 degrees about the point R, so let's do that. And then we'll do the rest of this sequence. So if we rotate this 90 degrees, so one way to think about it is a line like that is then going to be like that. So we're gonna go like that. And so R is going to stay where it is. You're rotating about it.
|
Mapping shapes.mp3
|
So if we rotate this 90 degrees, so one way to think about it is a line like that is then going to be like that. So we're gonna go like that. And so R is going to stay where it is. You're rotating about it. But P is now going to be right over here. One way to think about it is to go from R to P, we went down one and three to the right. Now when you do the rotation, you're going to go to the right one and then up three.
|
Mapping shapes.mp3
|
You're rotating about it. But P is now going to be right over here. One way to think about it is to go from R to P, we went down one and three to the right. Now when you do the rotation, you're going to go to the right one and then up three. So P is going to be there, and you could see that. That's the rotation. So that side will look like this.
|
Mapping shapes.mp3
|
Now when you do the rotation, you're going to go to the right one and then up three. So P is going to be there, and you could see that. That's the rotation. So that side will look like this. So that is P. And then Q is going to go right over here. It's going to once again also do a 90 degree rotation about R. And so after you do the 90 degree rotation, PQR is going to look like this. So that is Q.
|
Mapping shapes.mp3
|
So that side will look like this. So that is P. And then Q is going to go right over here. It's going to once again also do a 90 degree rotation about R. And so after you do the 90 degree rotation, PQR is going to look like this. So that is Q. So we've done that first part. Then a translation six units to the left and seven units up. So each of these points are gonna go six units to the left and seven up.
|
Mapping shapes.mp3
|
So that is Q. So we've done that first part. Then a translation six units to the left and seven units up. So each of these points are gonna go six units to the left and seven up. So if we take point P, six to the left, one, two, three, four, five, six, seven units up. One, two, three, four, five, six, seven. It'll put it right over there.
|
Mapping shapes.mp3
|
So each of these points are gonna go six units to the left and seven up. So if we take point P, six to the left, one, two, three, four, five, six, seven units up. One, two, three, four, five, six, seven. It'll put it right over there. So that is point P. If we take point R, we take six units to the left, one, two, three, four, five, six, seven up. One, two, three, four, five, six, seven. It gets us right over there.
|
Mapping shapes.mp3
|
It'll put it right over there. So that is point P. If we take point R, we take six units to the left, one, two, three, four, five, six, seven up. One, two, three, four, five, six, seven. It gets us right over there. And then point Q, if we go six units to the left, one, two, three, four, five, six, seven up is one, two, three, four, five, six, seven. Puts us right over there. So this looks like it worked.
|
Mapping shapes.mp3
|
It gets us right over there. And then point Q, if we go six units to the left, one, two, three, four, five, six, seven up is one, two, three, four, five, six, seven. Puts us right over there. So this looks like it worked. Sequence A is good. It maps PQR onto ABC. This last one isn't an R. This is a Q right over here.
|
Mapping shapes.mp3
|
So this looks like it worked. Sequence A is good. It maps PQR onto ABC. This last one isn't an R. This is a Q right over here. So that worked, sequence A. Now let's work on sequence B. I'll do this in a different color. A translation, eight units to the left and three up.
|
Mapping shapes.mp3
|
This last one isn't an R. This is a Q right over here. So that worked, sequence A. Now let's work on sequence B. I'll do this in a different color. A translation, eight units to the left and three up. So let's do that first. So if we take point Q, eight to the left and three up. One, two, three, four, five, six, seven, eight.
|
Mapping shapes.mp3
|
A translation, eight units to the left and three up. So let's do that first. So if we take point Q, eight to the left and three up. One, two, three, four, five, six, seven, eight. Three up, one, two, three. So this will be my red Q for now. And now if I do this point R, one, two, three, four, five, six, seven, eight.
|
Mapping shapes.mp3
|
One, two, three, four, five, six, seven, eight. Three up, one, two, three. So this will be my red Q for now. And now if I do this point R, one, two, three, four, five, six, seven, eight. Let me make sure I did that right. One, two, three, four, five, six, seven, eight. Three up, one, two, three.
|
Mapping shapes.mp3
|
And now if I do this point R, one, two, three, four, five, six, seven, eight. Let me make sure I did that right. One, two, three, four, five, six, seven, eight. Three up, one, two, three. So my new R is going to be there. And then last but not least, point P, eight to the left. One, two, three, four, five, six, seven, eight.
|
Mapping shapes.mp3
|
Three up, one, two, three. So my new R is going to be there. And then last but not least, point P, eight to the left. One, two, three, four, five, six, seven, eight. Three up, one, two, three, goes right there. So just that translation will get us to this point. It'll get us to that point.
|
Mapping shapes.mp3
|
One, two, three, four, five, six, seven, eight. Three up, one, two, three, goes right there. So just that translation will get us to this point. It'll get us to that point. So we're clearly not done mapping yet, but there's more transformation to be done. So it looks something like that. It says, then a reflection over the horizontal line through point A.
|
Mapping shapes.mp3
|
It'll get us to that point. So we're clearly not done mapping yet, but there's more transformation to be done. So it looks something like that. It says, then a reflection over the horizontal line through point A. So point A is right over here. The horizontal line is right like that. So if I were to reflect, point A wouldn't change.
|
Mapping shapes.mp3
|
It says, then a reflection over the horizontal line through point A. So point A is right over here. The horizontal line is right like that. So if I were to reflect, point A wouldn't change. Point R right now is three below that horizontal line. Point R will then be three above that horizontal line. So point R will then go right over there.
|
Mapping shapes.mp3
|
So if I were to reflect, point A wouldn't change. Point R right now is three below that horizontal line. Point R will then be three above that horizontal line. So point R will then go right over there. Just from that, I can see that this sequence of transformations is not going to work. It's putting R in the wrong place. So I'm going to rule out sequence B. Sequence C, let me do that with another color.
|
Mapping shapes.mp3
|
So point R will then go right over there. Just from that, I can see that this sequence of transformations is not going to work. It's putting R in the wrong place. So I'm going to rule out sequence B. Sequence C, let me do that with another color. I don't know how I will do it with this orange color. A reflection over the vertical point through point Q. Sorry, a reflection over the vertical line through point Q.
|
Mapping shapes.mp3
|
So I'm going to rule out sequence B. Sequence C, let me do that with another color. I don't know how I will do it with this orange color. A reflection over the vertical point through point Q. Sorry, a reflection over the vertical line through point Q. So let me do that. So the vertical line through point Q looks like this. Just gonna draw that vertical line.
|
Mapping shapes.mp3
|
Sorry, a reflection over the vertical line through point Q. So let me do that. So the vertical line through point Q looks like this. Just gonna draw that vertical line. So if you reflect it, Q is going to stay in place. R is one to the right of that, so now it's going to be one to the left once you do the reflection. And point P is four to the right, so now it's going to be four to the left.
|
Mapping shapes.mp3
|
Just gonna draw that vertical line. So if you reflect it, Q is going to stay in place. R is one to the right of that, so now it's going to be one to the left once you do the reflection. And point P is four to the right, so now it's going to be four to the left. One, two, three, four. So P is going to be there after the reflection. And so it's going to look something like this after that first transformation.
|
Mapping shapes.mp3
|
And point P is four to the right, so now it's going to be four to the left. One, two, three, four. So P is going to be there after the reflection. And so it's going to look something like this after that first transformation. I know this is getting a little bit messy, but this is what you probably have to go through as well. So I'll go through it with you. All right, so we did that first part, the reflection.
|
Mapping shapes.mp3
|
And so it's going to look something like this after that first transformation. I know this is getting a little bit messy, but this is what you probably have to go through as well. So I'll go through it with you. All right, so we did that first part, the reflection. Then a translation four to the left and seven units up. So four to the left and seven up. So let me try that.
|
Mapping shapes.mp3
|
All right, so we did that first part, the reflection. Then a translation four to the left and seven units up. So four to the left and seven up. So let me try that. So four to the left, one, two, three, four, seven up. One, two, three, four, five, six, seven. So it's putting Q right over here.
|
Mapping shapes.mp3
|
So let me try that. So four to the left, one, two, three, four, seven up. One, two, three, four, five, six, seven. So it's putting Q right over here. I'm already suspicious of it because sequence A worked where we put P right over there. So I'm already suspicious of this, but let's keep trying. So four to the left and seven up.
|
Mapping shapes.mp3
|
So it's putting Q right over here. I'm already suspicious of it because sequence A worked where we put P right over there. So I'm already suspicious of this, but let's keep trying. So four to the left and seven up. One, two, three, four, seven up. One, two, three, four, five, six, seven. So R is going to the same place that sequence A put it.
|
Mapping shapes.mp3
|
So four to the left and seven up. One, two, three, four, seven up. One, two, three, four, five, six, seven. So R is going to the same place that sequence A put it. And then point P, one, two, three, four. One, two, three, four, five, six, seven. Actually, it worked.
|
Mapping shapes.mp3
|
So R is going to the same place that sequence A put it. And then point P, one, two, three, four. One, two, three, four, five, six, seven. Actually, it worked. So it works because this is actually an isosceles triangle. So this one actually worked out. We were able to map PQR onto ABC with sequence C. So I like this one as well.
|
Mapping shapes.mp3
|
Actually, it worked. So it works because this is actually an isosceles triangle. So this one actually worked out. We were able to map PQR onto ABC with sequence C. So I like this one as well. And then last but not least, let's try sequence D. I'll do that in black so that we can see it. So first we do a translation. Eight units to the left and three up.
|
Mapping shapes.mp3
|
We were able to map PQR onto ABC with sequence C. So I like this one as well. And then last but not least, let's try sequence D. I'll do that in black so that we can see it. So first we do a translation. Eight units to the left and three up. Eight to the left and three up. So we start here. One, two, three, four, five, six, seven, eight.
|
Mapping shapes.mp3
|
Eight units to the left and three up. Eight to the left and three up. So we start here. One, two, three, four, five, six, seven, eight. Three up, one, two, three. So I'll put my black Q right over there. So eight to the left.
|
Mapping shapes.mp3
|
One, two, three, four, five, six, seven, eight. Three up, one, two, three. So I'll put my black Q right over there. So eight to the left. One, two, three, four, five, six, seven, eight. Three up, one, two, three. I'll put my black R right over there.
|
Mapping shapes.mp3
|
So eight to the left. One, two, three, four, five, six, seven, eight. Three up, one, two, three. I'll put my black R right over there. That's actually exactly what we did in sequence B the first time. So P is going to show up right over there. So after that translation, sequence, that first translation in sequence D, it gets us right over there.
|
Mapping shapes.mp3
|
I'll put my black R right over there. That's actually exactly what we did in sequence B the first time. So P is going to show up right over there. So after that translation, sequence, that first translation in sequence D, it gets us right over there. Then it says a rotation negative 270 degrees about point A. So this is point A right over here. And negative 270 degrees.
|
Mapping shapes.mp3
|
So after that translation, sequence, that first translation in sequence D, it gets us right over there. Then it says a rotation negative 270 degrees about point A. So this is point A right over here. And negative 270 degrees. It's negative so it's going to go clockwise. And let's see, 180 degrees. Let's say if we were to take this line right over here.
|
Mapping shapes.mp3
|
And negative 270 degrees. It's negative so it's going to go clockwise. And let's see, 180 degrees. Let's say if we were to take this line right over here. If we were to go 180 degrees, it would go to this line like that. And then if we were to go another 90 degrees, it actually does look like it would map onto that. So this is actually looking pretty good.
|
Mapping shapes.mp3
|
Let's say if we were to take this line right over here. If we were to go 180 degrees, it would go to this line like that. And then if we were to go another 90 degrees, it actually does look like it would map onto that. So this is actually looking pretty good. If you were to, this line right over here, well then, if you go negative 270 degrees, will map onto this right over here. And then that point R will kind of go along for the ride is one way to think about it. And so it'll go right over there as well.
|
Mapping shapes.mp3
|
Let's do some solid geometry volume problems. So they tell us shown is a triangular prism. And so there's a couple of types of three-dimensional figures that deal with triangles. This is what a triangular prism looks like, where it has a triangle on one, two faces. And they're kind of separated. They kind of have rectangles in between. The other types of triangular three-dimensional figures is you might see pyramids.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
This is what a triangular prism looks like, where it has a triangle on one, two faces. And they're kind of separated. They kind of have rectangles in between. The other types of triangular three-dimensional figures is you might see pyramids. This would be a rectangular pyramid because it has a rectangular or it has a square base, just like that. You could also have a triangular pyramid, where just literally every side is a triangle, so stuff like that. But this over here is a triangular prism.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
The other types of triangular three-dimensional figures is you might see pyramids. This would be a rectangular pyramid because it has a rectangular or it has a square base, just like that. You could also have a triangular pyramid, where just literally every side is a triangle, so stuff like that. But this over here is a triangular prism. I don't want to get too much into the shape classification. If the base of the triangle B is equal to 7, the height of the triangle H is equal to 3, and the length of the prism L is equal to 4, what is the total volume of the prism? So they're saying that the base is equal to 7.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
But this over here is a triangular prism. I don't want to get too much into the shape classification. If the base of the triangle B is equal to 7, the height of the triangle H is equal to 3, and the length of the prism L is equal to 4, what is the total volume of the prism? So they're saying that the base is equal to 7. So this right over here is equal to 7. The height of the triangle is equal to 3. So this right over here, this distance right over here, H is equal to 3.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
So they're saying that the base is equal to 7. So this right over here is equal to 7. The height of the triangle is equal to 3. So this right over here, this distance right over here, H is equal to 3. And the length of the prism is equal to 4. So I'm assuming it's this dimension over here is equal to 4. So length is equal to 4.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
So this right over here, this distance right over here, H is equal to 3. And the length of the prism is equal to 4. So I'm assuming it's this dimension over here is equal to 4. So length is equal to 4. So in this situation, what you really just have to do is figure out the area of this triangle right over here. We could figure out the area of this triangle, and then multiply it by how much you go deep. So multiply it by this length.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
So length is equal to 4. So in this situation, what you really just have to do is figure out the area of this triangle right over here. We could figure out the area of this triangle, and then multiply it by how much you go deep. So multiply it by this length. So the volume is going to be the area of this triangle. Let me do it in pink. The area of this triangle.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
So multiply it by this length. So the volume is going to be the area of this triangle. Let me do it in pink. The area of this triangle. We know that the area of a triangle is 1 half times the base times the height. So this area right over here is going to be 1 half times the base times the height. And then we're going to multiply it by kind of our depth of this triangular prism.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
The area of this triangle. We know that the area of a triangle is 1 half times the base times the height. So this area right over here is going to be 1 half times the base times the height. And then we're going to multiply it by kind of our depth of this triangular prism. So we have a depth of 4. So then we're going to multiply that times the 4. Times this depth.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
And then we're going to multiply it by kind of our depth of this triangular prism. So we have a depth of 4. So then we're going to multiply that times the 4. Times this depth. Times the 4. And we get, let's see, 1 half times 4 is 2. So these guys cancel out.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
Times this depth. Times the 4. And we get, let's see, 1 half times 4 is 2. So these guys cancel out. You'll just have a 2. And then 2 times 3 is 6. 6 times 7 is 42.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
So these guys cancel out. You'll just have a 2. And then 2 times 3 is 6. 6 times 7 is 42. And it would be in some type of cubic unit. So if these were in, I don't know, centimeters, it would be centimeters cubed. But they're not making us focus on the units in this problem.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
6 times 7 is 42. And it would be in some type of cubic unit. So if these were in, I don't know, centimeters, it would be centimeters cubed. But they're not making us focus on the units in this problem. Let's do another one. Shown is a cube. If each side of the cube, or if each side is of equal length, x equals 3, what is the total volume of the cube?
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
But they're not making us focus on the units in this problem. Let's do another one. Shown is a cube. If each side of the cube, or if each side is of equal length, x equals 3, what is the total volume of the cube? So each side is equal length x, which happens to equal 3. So this side is 3. This side over here, x is equal to 3.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
If each side of the cube, or if each side is of equal length, x equals 3, what is the total volume of the cube? So each side is equal length x, which happens to equal 3. So this side is 3. This side over here, x is equal to 3. Every side, x is equal to 3. So it's actually the same exercise as the triangular prism. It's actually a little bit easier when you're dealing with a cube, where you really just want to find the area of this surface right over here.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
This side over here, x is equal to 3. Every side, x is equal to 3. So it's actually the same exercise as the triangular prism. It's actually a little bit easier when you're dealing with a cube, where you really just want to find the area of this surface right over here. Now this is pretty straightforward. This is just a square. Or it would be the base times the height.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
It's actually a little bit easier when you're dealing with a cube, where you really just want to find the area of this surface right over here. Now this is pretty straightforward. This is just a square. Or it would be the base times the height. Or since they're the same, it's just 3 times 3. So the volume is going to be the area of this surface, 3 times 3, times the depth. And so we go 3 deep.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
Or it would be the base times the height. Or since they're the same, it's just 3 times 3. So the volume is going to be the area of this surface, 3 times 3, times the depth. And so we go 3 deep. So times 3. And so we get 3 times 3 times 3. This is 27.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
And so we go 3 deep. So times 3. And so we get 3 times 3 times 3. This is 27. Or you might recognize this from exponents. This is the same thing as 3 to the third power. And that's why sometimes, if you have something to the third power, they'll say you cubed it.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
This is 27. Or you might recognize this from exponents. This is the same thing as 3 to the third power. And that's why sometimes, if you have something to the third power, they'll say you cubed it. Because literally, to find the volume of a cube, you take the length of one side, and you multiply that number by itself three times. One for each dimension. One for the length, the width, and or I guess the height, the length, and the depth, depending on how you want to define them.
|
Find the volume of a triangular prism and cube Geometry Khan Academy.mp3
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.