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Hello, I'm Mrs. Lashley and I'm looking forward to working with you as we go through this lesson.

I really hope you're looking forward to it and you're ready to try your best.

So our learning outcome today is to be able to use our knowledge of loci and constructions to solve problems. On the screen, there are some keywords that you've used before in your learning and I'm gonna be using them during today's lesson.

So it might be that you wish to pause the video and read through them again before we make a start.

So the lesson on problem solving with loci and constructions is split into two learning cycles.

The first learning cycle is going to be looking at loci, constructions, and bearings, and the second one is loci, constructions, and areas.

So let's make a start looking at bearings.

A boat is leaving a port on a bearing of 130 degrees.

So we can see the port is located on the screen and we know that the boat is on a bearing of 130 from that port.

Laura says, using a protractor and measuring from north, we can locate the boat.

So she's got her protractor and she's measuring clockwise 'cause bearings are measured clockwise.

She's made a mark at 130 degrees.

And that's where she thinks that the boat is.

Lucas says that is only a possible location for the boat.

So why is Lucas correct? Have a think about that.

So Lucas is correct because actually the boat could be anywhere along the ray line.

We know it's on a bearing from the port, but we don't have any distances from the port to give it an exact location.

Laura recognises what her mistake was and says, oh, it's the locus of points but not the exact location.

So more information is needed to find the exact location.

Can you think about what information we could be given so that we would know exactly where the boat is in comparison to the port? So here's a check for you.

So what bearing is the boat travelling on if this is the locus of points that a boat is on having left the port? Pause the video and when you're ready to check, press play.

So the boat is travelling on a bearing of 110 degrees.

So remembering that bearings are an angle measured from north in the clockwise direction and given with three digits.

This happens to have three digits anyway, which is 110 degrees.

So if we continue thinking about what other information do we need in order to find the location of the boat? So here we've got some further information, which is the speed it is travelling.

So it's travelling at a speed of 40 kilometres per hour.

So where is the boat? Well, we know the boat is along the locus of points that is 130 degrees from the port.

So along that ray line.

We have a scale which is one square is 10 kilometres.

So where's the boat? Well, Lucas says, well actually the location will depend on how long it has travelled for.

So just because we have the speed it is travelling still doesn't give us the boat's location.

So after one hour the boat will have covered 40 kilometres, that's what a speed of 40 kilometres per hour means.

And we can use our squares because that's our scale to know that it will be four squares away from the port.

But not just four squares in any direction but along that ray line.

And we can use our pair of compasses to mark an arc that is four squares away from boat B, that would be our radius would be 4.

And where it intersects would be the location of the boat after one hour of travel.

If it was two hours of travel, then it would now be 80 kilometres from the port.

And so we could draw an arc that has a radius of eight squares or 80 kilometres so that we know that moment in time the boat would be located there.

So here's a check for you.

A boat leaves on a bearing of 55 degrees at a speed of 30 kilometres per hour.

How long has the boat been travelling for? So you can see the location of the boat, how long has it been travelling for? Pause the video and when you're ready to check, press play.

Well it's travelled 60 kilometres and we know it's travelled 60 kilometres 'cause we can see that the radius of that arc is six squares, which equates to 60 kilometres.

So 60 kilometres has travelled at a speed of 30 kilometres per hour.

So the distance has doubled, so therefore the time will have doubled too.

So it's travelling for two hours.

The same boat that we've been trying to locate is on a bearing of 130 degrees from the port.

But another boat which we're gonna call C is due east of the port and picks up boat B on its radar system on a bearing of 235 degrees.

Where is boat B? So we can see the location of boat C.

And where is boat B? Well, we can draw our bearing.

The locus points that are on a bearing of 130 degrees from point B and the locus points that are on a bearing of 235 degrees from boat C where they intersect is where the boat would be.

So this further information allows us to find the exact location of boat B.

So here's a check.

Which of the following describes the location of the boat B? So you can see the accurate construction on the screen and which of the descriptions matches it.

So pause the video and then when you're ready to check, press play.

So description b.

And description a is very similar, but the two bearings are the wrong way round.

So it's important that we get those correct.

And part c, the bearing or the angle that's given is the angle between the two ray lines.

However, that is not a bearing, because a bearing needs to be measured from north in a clockwise direction.

So we're gonna look at a different scenario for loci, constructions, and bearings.

So this time there are three pylons in a local area.

Pylon B is at a bearing of 150 degrees from pylon A.

Pylon B and C are both 500 metres from pylon A.

Pylon C is due west of pylon B.

And a diagram may be helpful here.

You can see that there is a scale in the top right corner.

So that distance, that length represents 100 metres.

So I'm gonna start with the first information point.

But you don't need to start with the first information point, but that's the one I'm gonna start with.

So I'm gonna place pylon A on my diagram.

And then I'm going to find the location of pylon B and C relative to that point.

So pylon B is at a bearing of 150 degrees from pylon A.

So I can measure that angle of 150 degrees.

And I know that that doesn't give me the exact location of pylon B, but it tells me a locus of points that pylon B could be.

Then I can move to the second point.

And that says that pylon B and C are 500 metres from pylon A.

Pylon A is a fixed point.

So think about what the locus of points looks like when it's a fixed point.

And this one I could have done first before my bearing.

So I need to set up my pair of compasses to 500 metres.

And using that scale, so it'll be five times that length and I can draw my locus of points.

So all points along the circumference of that circle are 500 metres from pylon A.

Because I know that pylon B is along the ray line somewhere and I also know it needs to be 500 metres from pylon A.

The point of intersection of those two locusts is where pylon B is located.

Now I'm onto the final information point, which is pylon C is due west of pylon B.

Pylon C is due west means it is to the left of B.

And if we think about our compass directions, north and west are perpendicular directions.

So we now need to think about how we can construct a perpendicular line that passes through B that is perpendicular to the north.

So we're gonna do that by thinking about perpendicular bisectors, the skill, the knowledge we have.

So if I extend the north line, so that is still in the direction of north I've just extended it, I now want a perpendicular line that passes through B, perpendicular to that north line.

So think about how we construct that.

So the steps are we're gonna create a line segment where B, our points off of the line is equidistant to either end.

So we're gonna use pair of compasses to do that because that's a fixed distance.

That distance is your choice.

So you can see my arc here that crosses through the north line.

I'm now gonna do a perpendicular bisector on the line segment between those two intersection points.

So I would be putting my pair of compasses at one of those arcs which meets the north line and drawing two arcs.

And without changing the radius of my pair of compasses, do it from the other intersection point and draw my line.

So I've got a perpendicular bisector between that line segment that I created.

I created it such that the point B was equidistant from both ends of it.

So now I know where pylon C is.

It is where the perpendicular bisector or the perpendicular line that passes through B also intersects with the circle.

So a road has been planned that runs equidistant to pylon A and B.

Will pylon C need to be moved? So remember these are electricity pylons in a town and we now know that a road is gonna run equidistant from pylon A and pylon B.

So if we remove our construction lines, just so we've got our locations only, we need to find the road and we know it's equidistant to A and B.

So what does that mean? Well, it means that this needs a perpendicular bisector.

Because a perpendicular bisector gives us the locus of points that is equidistant from both A and B.

So I'm gonna put a line segment, a faint line segment on and construct a perpendicular bisector.

And if you draw the perpendicular bisector, which indicates our road, the road would actually be passing through point C which are pylon C.

So yes, the pylon will need to be moved or they will need to reroute the road.

So here's a check for you.

So for this check you need to label the pylons A, B, and C given the information on the screen.

So pause the video, read through it, and when you're ready to check, press play.

So C, then A, then B.

I think it's probably.

2 that you can start with.

And that's the pylon B and C are both 600 metres from pylon A.

So pylon A has a circle around it and that's the locus of points that are 600 metres from it.

And therefore B and C are on the circumference, which means that they are both 600 metres from A.

And then you needed to work out which one was B and which one was C.

And you could use either.

1 or.

3 to work that out.

So going back to ships or boats, things in the water.

A ship needs to pass through a point that is midway between two buoys.

So the buoys are B1 and B2 and the ship is the S.

Aisha and Jim both do some constructions to find the correct point to pass through.

So you can see their constructions.

Can you name them? And now I want you to think about who has constructed the correct bisector.

So Aisha has done an angle bisector and Jun has done a perpendicular bisector.

So who has chosen the correct bisector for this scenario? Jun has.

And that's because it is two single points.

So the equidistant line of two points is a perpendicular bisector.

Aisha has constructed an angle bisector where she has joined the three points together with line segments, which isn't necessary in this scenario.

So continuing with this, using Jun's correct construction, what bearing should the ship travel? So you can see the ship, that's the one with the S.

It needs to pass through the points that is midway between the two buoys.

So what's the bearing? So where is that point? That construction allows us to find that point, which point is it? Well, it's the point that's halfway the bisector part of the two buoys, the midway.

So what's that bearing? Well we're gonna measure using a protractor.

And in practical terms, you're probably gonna measure the obtuse angle and subtract from 360 degrees to get the reflex bearing.

But do remember that the bearing needs to be measured in the clockwise direction.

So 246 degrees is the bearing that the ship would need to travel from the location it's at in order to pass through the midway point of the two buoys.

So here's a check for you.

Which point represents the location of the point midway between the points A and B? So there are four points that you could choose from, but which one is the location of the midway? Pause the video and then when you're ready to check, press play.

So I'm hoping you went for d.

And that's because of the perpendicular bisector allows us to bisect this line segment between A and B.

So that finds the mid point of the line.

If you went for a, that's actually just a point that is on the construction of the angle bisector.

It doesn't really mean too much.

It's just used to construct an angle bisector, Part b is the point on the line AB, which is equidistant from the lines SA and SB.

So the angle bisector is showing a locus of points that is equidistant from both of those line segments.

And that just happens to be a point on AB as well.

And point c is the point where the line bisector and the angle bisector intersect.

So we're now onto the first task of the lesson.

So question one, a yacht is sailing on a bearing of 80 degrees from a port.

Draw the locus of the yacht's path is part a.

And then you've told the speed and it's travelled for 30 minutes, so locate the yacht.

And then for part c, after that 30 minutes, another yacht is 10 miles away on a bearing of 220 degrees from the first yacht.

So mark its location.

So pause the video and then when you press play, we'll move to question two.

Here's question two and this is about three phone mast in a local area.

There are four information points.

You need to use those to find the location of mast A and mast C.

So pause the video and when you're finished with that, press play and we'll move to question three.

Here's question three.

So the scenario here is a remote controlled boat race is taking place.

We can see the start and the finish.

The boat needs to pass through the checkpoints, which are pairs of flags, such that it's equidistant to both.

On the diagram you can see C1 and C1, so that's checkpoint 1 and they're the pair of flags that go together.

And then you've got checkpoint 2 and they're another pair of flags.

So for part a, you need to construct the appropriate bisectors to show the boat's path through the course checkpoints.

And then for part b, you need to write down the bearing of the finish point from the start point.

So pause the video.

When you finish, we're gonna go through the answers to task A.

So here's the answers to question one.

All the constructions are on the diagram and hopefully yours looks very similar to this.

So part a, you needed to draw the locus of the yacht's path.

So that was you measuring the bearing 80 degrees from the port.

So if using the north line to get your 80 degree angle and drawing a ray line.

Then for part b, you were told the yacht was travelling at 7 miles per hour and you needed to show the location of the yacht after 30 minutes.

So you need to think about 30 minutes as half an hour and therefore the distance travelled would be 3 1/2 miles.

Using the scale, that would be 3 1/2 squares.

So you could draw a circle with a radius of 3 1/2 squares and look at where it intersects your ray line from part A and that would be the location of the yacht after 30 minutes.

For part c, it was after 30 minutes, so we're using the location of A that there is this second yacht that we're gonna call B and we know it's 10 miles away and it's on a bearing of 220 degrees.

So mark it's location.

So it doesn't matter which order you did these two constructions, you needed to draw a circle.

We can only see an arc, part of the circumference on the diagram.

But that needed to be 10 squares of a radius because it was 10 miles.

And then you also needed to measure the bearing 220 degrees.

And then when those two intersect, would be where the location of the yacht B is.

Question two, you needed to read through the four information points, decide a appropriate order in order to locate A and C.

B was already given to you.

So using the first two information points you can say that mast A is on a bearing of 140 degrees.

So we can measure that angle.

And the ray line indicates where mast A will be located somewhere along it.

The same for the second point, mast C is on a bearing of 50 degrees.

So again, if the 50 degree angle is overlapping with 140 degree, the angle between the two ray lines is 90.

Then we've got mast A is 1,100 metres from mast C and mast B is 900 metres from mast C.

So the third point you can't use of at this point because you don't know where A is and you don't know where C is.

But the fourth point includes mast B and we do know where mast B is located.

So mast B is 900 metres from mast C is the same as saying mast C is 900 metres from mast B.

So we can use the location of B, draw our locus points that are 900 metres, which would be nine squares using the scale.

And where that intersects with the bearing is where mast C is.

So that's how you could have found where mast C is located.

And now that you have mast C, you know that the distance between C and A is 1,100 metres.

So you can draw your locus of points that is 1,100 metres from C.

And where that intersects with the bearing of 140 degrees from B will be the location of A.

Onto question three.

You needed to construct the appropriate bisectors.

In both cases, it was perpendicular bisectors.

So you needed to use perpendicular bisectors for checkpoint 1 and a perpendicular bisector for checkpoint 2.

And then basically, you've shown the route that the boat needs to take.

It will start at the start, it will travel equidistant through the checkpoint 1.

And at the point at which that perpendicular bisector meets the other perpendicular bisector is where the boat would need to spin round, change its direction of travel so that it now travels along the perpendicular bisector through checkpoint 2, keeping it equidistant from both flags.

And then that takes it to the finish line.

For part b, you need to write down the bearing of the finish from the start.

So if we draw on a line segment between the start and finish a north line in order to measure the bearing, use a protractor, that's 153 degrees.

So the second learning cycle is loci, constructions, and areas.

So three motion sensors are set up for security.

So the rectangle represents the area of land that is being protected and we've got three sensors.

So sensor 1 has a range of 15 metres, sensor 2 has a range of 12 metres, and sensor 3 has a range of 20 metres.

And you can see where the sensors are located.

What area does sensor 1 cover? Well sensor 1 is just a point, so we're gonna draw a circle for our locus.

And so it's the locus of a single point.

Some of the circle is external to the area we're trying to protect.

So we've only got part of a circle that we can actually see.

But we'd use our pair of compasses and we'd fix the radius using the scale and make sure it's got a radius of 15 metres.

What about the area for sensor 2? Well that's gonna, again, this one's a segment because part of it's being cut by a chord.

But using the scale we're gonna draw our circle because it's a locus of a single point.

And lastly, what about sensor 3? Well, similar idea.

That sensor 3 is a single point, so we need to show the region of points that are up to and include in 20 metres from that single point.

So we'd use a pair of compasses.

What areas are not covered by the sensors? So remember these emotion sensors.

So where could you move, where could you be on this piece of land, and then none of those sensors would be tripped, none of them would activate? Well, it's gonna be the regions that are outside of range for all of the sensors.

The circles show the boundary where the range they go out of range.

So outside of all three of them.

So here's a check for you.

Sensor 1 and sensor 2 are activated, so they have picked up motion, where might the intruder be? So pause the video.

When you're ready to check, press play.

So it's gonna be this region here and that's the area that is covered by both sensor 1 and the sensor 2 only.

So moving on and continue with thinking about loci, constructions, and areas.

We've got a scenario here where a dog is tied to the wall of an outbuilding.

The outbuilding is 5 metres square and the lead is 8 metres long.

The region that the dog can go is shown here.

How far along the wall is the dog tethered? So that cross is where the dog is tied, how far along that wall is it? I'm gonna use this part of the diagram to help me get to an answer.

And so I know that the outbuilding is 5 metres square and the dog lead would be tethered at that cross.

When it got to that corner and turns around the corner, it gets 5 metres and the lead is 8 metres in length, which means that this part here would be 3 metres in order to total 8.

So we could describe it as 3 metres from the left corner or 2 metres from the right corner depending on how you describe it.

So how much space can the dog reach? So we can see that this is the region it can go, but how much space is that? So what's its actual area? Well, we need to total the semicircle, the quarter circles, to find that area.

This is a compound shape made of different sectors.

So let's look at this a bit closer.

So if we start with the semicircle, this is when the lead is at maximum length of 8 metres and it's not caught along or going round the corner.

So it's on this one side of the building.

So it's a semicircle.

So it's 180 out of 360 degrees times 5 times 8 squared, which is 1/2 times pi times 8 squared.

So the 1/2 is what fraction of the full circle is it using the angle between the two radii? And that simplifies to 32 pi.

Now if we look at this quarter circle, it has a radius of 5 metres.

We know that because once the dog gets to that corner and turns around that side of the outbuilding, the lead would get trapped.

3 metres of it would be taut to the wall and therefore there's only 5 metres of like free lead left.

So we need 1/4 of a full circle with a radius of 5.

Why 1/4? Well, because the angle between the two radii is 90 degrees and that's out of 360 degrees for a full circle.

So 1/4 of pi times 5 squared simplifies to 25 over 4 pi square metres.

We're leaving it in terms of pi so that we don't have any rounding errors as we work through the solution, especially because there's multiple calculations.

Now if you look at this quarter circle, well 2 metres of the lead would take us from the tether point to the corner.

And that leaves 6 metres that can wrap around that side of the building.

So it's a quarter circle with a radius of 6.

So 1/4 of pi times 6 squared simplifies to 9 pi.

Lastly, we've got this final quarter circle.

And so if 2 metres of the lead gets us to the corner and then the outbuilding is 5 metres long, when it turns around that corner, there's only 1 metre of lead left.

A quarter circle of radius 1 metre simplifies to 1/4 pi.

So what space can the dog reach? Well it's the total of all four parts.

And so that comes to 95 over 2 pi, as an exact answer, square metres.

149.

2 square metres to one decimal place or you could say 149 square metres, just three significant figures.

So here's a check.

A dog is tied to the wall of an outbuilding.

So the same scenario.

The region that the dog can go is shown here.

What are the missing lengths? So pause the video and then when you're ready to check, press play.

So a would be 12 foot.

So that's the length of the lead or the rope that this dog is tied to.

And you can get that using the 3 foot and the 9 foot.

B would also be 9 foot.

So because it's a radius, it's a fixed length throughout the sector.

So 9 foot.

C is 7.

5 feet.

There's a couple of ways you could do this.

You could use the 4.

5 feet and think about how much more lead there would be to get to 12 feet.

Alternatively, it's the length of the outbuilding and you can see that the top edge is 4.

5 and 3 feet, which has a total of 7 1/2 feet.

And lastly, there'd be 1 1/2 feet of the lead that could wrap around to that third edge.

So if the lead has gone 3 feet to the corner along the 7 1/2 feet of the building, that's 10 1/2 feet of the lead used, which leaves you 1 1/2 feet.

Another scenario is a goat is attached to a 12 metre bar with a 2 metre rope on a ring, so that it can graze either side.

So because it's on this ring, the ring will be able to slide along the bar and also the goat could go underneath the bar and go onto the other side, et cetera.

So what will the locus of the goat be? Is it gonna be this one where the middle line is the bar? Is it gonna be this one, this one, this one, or this one? So pause the video and think about if the goat is attached to this bar that's 12 metres long on a 2 metre rope which is on a ring which means it can go along the bar and around to the other side, what would the locus look like? It's gonna be c.

So at the end of the bar it would then make a semicircle.

Because that rope is a fixed length when it is extended to its maximum.

And so it would draw, if you'd imagine like a chalk line, it's gonna draw this semicircle.

So Sofia says that the area the goat can graze is 64 square metres.

Does this seem a sensible answer? So pause the video and think about that.

Does that seem sensible? So no, I'd say that doesn't seem sensible because of the circular nature.

So we've got semicircle at either end and so we'd expect the answer to be in terms of pi, if it was in its exact form, or if it's been rounded, then the degree of accuracy should be stated.

So the fact that this area is an integer value doesn't feel sensible when there is circular parts to this area.

So what is the actual area? Well, we're gonna think about it as two congruent semicircles and the rectangle between them.

And those two congruent semicircle can be thought of to form one circle.

So we've got a circle with a radius of 2 and a rectangle that's length is 12 and width is 4.

So the areas are 4 pi for the circle pi r squared.

Add length times width, which is 48.

So if we find the total of those two, 'cause we've broken it into its parts, component parts would be 4 pi plus 48 square metres.

And that's the exact form.

So here's a check.

Goat is tethered to a pole by a rope with a ring attached, complete the locus of the goat.

So pause the video and then when you're ready to check, press play.

So we needed those semicircle at the end 'cause if you think about the ends of the bar as single fixed points, what is the locus of a single fixed point? It is a circle.

So here we've got semicircle because it actually meets a different locus.

And so we've got loci that builds this.

Now we've told that the pole is 6 metres long and the rope is 3 metres long.

So what is the area? So pause the video and calculate the area.

Press play when you're ready to check.

Well the area would be 9 pi plus 36 or you could factorised out the factor of 9 and write it as 9 lots of pi plus 4.

So 9 pi came from thinking of the two semicircle as one circle.

A radius of 3, 3 squared is 9 times by pi.

The rectangle part of the locust would have a length of 6, which was the pole.

And it was also 6 wide, it's actually a square.

So 6 times 6 is 36.

So onto the last task of the lesson.

So question one, three sensors, these are those motion sensors again, are set up for security.

Sensor 1 ranges 15 metres, sensors 2 is 10 metres, and sensor 3 is 12 metres.

For part a, you need the use this scale and draw the locus of each sensor.

Part b is shade the area that's covered by all three sensors.

Part c is use a different colour, shade the area that's not covered by any of the sensors.

And part d, in a third colour, shade the area that is only covered by sensor 2.

Press pause and then when you finish your question one, press play for the last question.

Here's question two.

So an external plug is located on the outside of a house, 2 metres from a corner.

And that's marked there with the x.

The house is 10 metres square.

A corded lawnmower is plugged in.

The cord is 14 metres long.

What area of the lawn can the lawnmower mow? So you can see the region that it can mow.

So what is the area? Pause the video and when you're ready to check the answers to task B, press play.

So here's question one, part a.

You needed to use the scale to draw the locus of each sensor.

So that was circles, but as much as you could draw within the region.

Part b was shade the area that is covered by all three sensors.

And so that would be this region here.

It's within each of the three circles.

Part c, using a different colour, shade the area that's not covered by any of the sensors.

And it would be these four regions that you should have coloured in.

And part d, using a different colour, shade the area that is only covered by sensor 2.

So it needs to be the part that is within the area of sensor 2 but not within sensor 1 and not within sensor 3's regions or locus.

And that would be this region here.

Question two is gonna go over a few slides just for spacing.

So we needed to work out the area which was the total of the semicircle and the three quarter circles.

So the semicircle would be 14 metres, which is the maximum length of the cord.

And so it's a semicircle.

So 1/2 times pi times 14 squared is 98 pi.

Moving on to this quarter circle, we we need to work out that the radius is 6 and that's by using that the tether point was 2 metres from the corner and the house was 10 metres square, which means it was 8 metres to the top corner.

And that would leave 6 metres of the cord to wrap around onto that side of the house.

So it's a quarter circle with a radius of 6.

And that is 9 pi when you evaluate it.

Part C is gonna be this quarter circle, which has a radius of 12 metres.

And that's because we know that the plug is 2 metres from the corner and so there is 12 metres of cord to mow on that side of the house.

So a quarter circle with a radius of 12 comes out as 36 pi.

And lastly we've got the final quarter circle, which would have a radius of 2.

Because if the cord has gone 2 metres on that side of the building where the plug is, then 10 metres along the house edge there would be 2 metres left.

And that comes out as pi square metres.

So the total area is all four of those added together, the sum of those four, and that is 144 pi square metres.

I've left that in terms of pi.

So you can check, we use an calculator if you've given it as a rounded value.

So to summarise today's lesson, which was problem solving with loci and constructions.

Constructions such as perpendicular bisectors and angle bisectors can sometimes be used in problems that also involve bearings.

And problems that involve area can sometimes require loci to identify the correct region in question.

So well done today and I look forward to working with you again in the future.