Lesson video

Hello everyone.

Welcome to this lesson on constructions with me, Mr. Gratton.

Make sure to grab a pair of compasses and a ruler in this lesson where we will look at the properties of a rhombus, and then use these properties to help us construct a perpendicular to a line through any point either on or off that line.

Pause here to have a quick look at some of the keywords that we'll be using today.

Let's start by constructing perpendiculars to points on a line segment or on sides of a polygon.

Okay, to begin with, let's look at an important distinction.

A perpendicular bisector to AB is a line that intersects the exact midpoint of AB.

It does this at a right angle.

However, a perpendicular to AB is a line that is at right angles to AB because perpendicular means at right angles.

However, because it doesn't say bisector, it doesn't have to pass through the midpoint of AB, it can be a perpendicular line anywhere on AB.

Jacob thinks it's only possible to construct a perpendicular to a line segment through its midpoint using construction arcs or full circles that look like this.

This is because the two points of intersection between these arcs are the vertices of a rhombus.

We can construct a perpendicular bisector from a rhombus which divides a line segment exactly in half.

However, Laura thinks we can draw the perpendicular anywhere on AB by simply adjusting where you construct the rhombus.

The rhombus could be over here or it could be over there instead.

But Jacob wonders how it's possible to draw a rhombus and guarantee it passes through one specific point.

It is absolutely possible to construct a perpendicular through any point on line segment AB.

All we need to do is either extend or shorten the original line segment, or sometimes both, so that the point we want to draw the perpendicular through is at the centre of this new modified line segment.

Basically, change the line segment to create a new midpoint.

But how? We put a compass needle through the point we want the perpendicular to pass through, and then draw a circle with a centre at that point.

Any part of the old line segment beyond the circle is ignored, and if one end of the line segment doesn't intersect with the circle, we can extend it until it does, but then go no further.

Using a ruler, here we can check, yes, the extended line segment to the left is five centimetres, just like the shortened line segment to the right.

Okay, so, Andeep draws a circle with a centre at point C using a pair of compasses whose width has not changed.

Pause here to find the distances A and B.

Both A and B are 6.

3 centimetres long.

Next up, here we have Sofia who has incorrectly constructed a perpendicular through C.

Pause here to identify what is wrong with Sofia's construction.

That circle only intersects AB once.

We want the circle to intersect AB twice.

We are also unconfident as to whether the line segment drawn is perpendicular or not.

This is due to no construction arcs being shown as helpful information as to what has been drawn.

It is always important to keep all construction lines and construction arcs wherever possible when using compasses and rulers.

Pause here once more to advise Sofia on what she can do to improve her construction.

We want the circle to intersect AB twice.

Sofia can achieve this by either making the circle smaller or by extending AB to make it intersect the current circle a second time.

And here's Alex.

Alex has partially completed a construction for a perpendicular.

Pause here to identify what is incorrect about Alex's pair of compasses.

The compass width is too short.

This is because the width of the compass should be over half the diameter of the circle.

Or more visually, if that compass needle is at the intersection of AB and the circle, then the pencil end needs to be beyond point E, closer to point B.

And lastly, here's Jun.

Jun has tried to draw a circle with a centre at point F, but we can see that the circle has ended up looking incredibly wobbly.

Pause here to give Jun advice on how to improve his circle.

Jun's circle is simply too small.

It is tricky to use compasses to construct a really small circle accurately.

There's a sweet spot for the size of circle you can draw with a pair of compasses.

And since we can extend or shorten parts of this line segment, we can set the compass width to anything that is suitable for you and your tools.

However, what you cannot do is move the location of the circle, since the circle's centre must be at the point that you want to draw your perpendicular through.

Okay, I want to model a good construction on the left.

After each set of steps, perform those same steps for your own construction on the right.

Here, I have drawn a four centimetre line segment and marked on the end point A.

The perpendicular that we want to draw should pass through A.

Pause here to draw a five centimetre line segment for yourself and mark on a cross at point A, one of the endpoints of your line segment.

This is the line segment we want you to construct a perpendicular through.

Here's the first main step.

Place the compass needle at point A and draw a circle.

Try this for yourself, and the circle can be of any size, so make sure it's big enough to draw accurately.

A circle like this is absolutely fine.

Next up, place your ruler carefully over line segment AB, and extend the line segment until it intersects with your circle a second time.

Make sure the extension touches the circle, but it does not need to go any further, like this.

Pause now to give it a go for yourself.

We now have a circle and a line segment that intersects the circle twice.

Keep your focus on those two points of intersection.

Place your compass needle at one of the two points of intersection between the line segment and the circle, and then set the compass width to over half the diameter of that circle, like so, and draw two arcs.

For this construction, the compass pencil should go beyond point A, closer to point B.

The arcs should look something like this, so pause now and try it for yourself.

The arcs can go either inside or outside of the circle.

Both are fine.

It will all depend on what you set the width of your compasses to be.

Carefully, so that the compass width stays the same, repeat another pair of arcs with the compass needle on the other intersection between the line segment and the circle to get intersecting arcs, like so.

Pause now to do this for yourself.

A very well done if your two sets of arcs intersect twice.

The two intersecting arcs are two of the vertices of a rhombus, along with the two points on the line segment that intersected your original circle.

Basically, all four vertices of the rhombus are at the four intersections that you have just constructed.

Join these intersections up to create a rhombus.

Alternatively, we can go straight to drawing the perpendicular by drawing a line that passes through intersecting arcs.

Pause now to try this for yourself.

This line is your perpendicular through point A, and you can use a protractor to see how close to 90 degrees this angle is.

Okay, let's check what you've learned.

Pause here to identify whose partially completed construction is correct.

Jacobs' is correct because he extended the line segment so it intersects with the circle a second time.

Both Andeep and Alex have not done this, whilst Sofia has drawn a different line segment through C and not extended the already existing line segment.

Great stuff.

Let's put that method into practise.

For question one, complete each construction of a perpendicular through point Z.

And for question two, construct a perpendicular to side EF through point G.

If your construction is correct, then the perpendicular should also pass through the vertex H.

Pause now for these two questions.

And for question three, construct these three perpendiculars, and then these three perpendiculars as well as the line segment EF bound a quadrilateral.

Explain what type of quadrilateral it is.

Pause now for question three.

Great effort on all of those constructions.

Pause here to compare your constructions to those on screen for both question one and question two.

And pause here once more to compare your constructions to that in question 3A.

The quadrilateral EFIH is a trapezium, specifically a right angled trapezium.

This is because both EH and FI are parallel lines that are both perpendicular to EF.

Exactly one pair of parallel lines and two 90 degree angles created by those perpendiculars make a right angle trapezium along with IH.

Well done if you spotted this.

So far we've constructed perpendiculars to points on a line segment.

However, it is also possible to construct perpendiculars to any point anywhere above, below or to the side of a line segment.

Let's see how.

Laura's observation is very important to pay attention to.

No matter where that point is, that point our perpendicular should pass through, we want to draw a circle with a centre at that point, where the circle is of a size that means it intersects with our line segment twice.

A circle that intersects with the line segment twice.

This is true no matter where the point is.

The point can be on the line segment or it could be well off of the line segment as well.

However, the further away from the line segment, the more likely it'll be that we'll also have to extend the line segment, so that the line segment intersects with the circle that we draw twice.

We can construct the perpendicular to any line segment through any point anywhere on a plane.

If that point is off of the line segment, then it is more likely that we have to draw a bigger circle and extend the line segment.

This will help ensure that the circle and the line segment intersect twice.

From there, our method is identical to before.

We ignore any part of the line segment outside of the circle that we draw, and construct a perpendicular that passes through that point off of the line segment.

Okay, here's a check.

pause here to identify which of these circles will help in the construction of the perpendicular through point Z.

Both circles B and C will, but you'll also need to extend that line segment QP first before it intersects with those circles twice.

We can construct perpendiculars to sides of shapes, as well as random line segments that we've just done.

The method is no different.

So pause here to grab a ruler and draw any triangle with a horizontal base.

So we're going to construct a perpendicular to that base through the third vertex of that triangle.

For my triangle, the perpendicular will pass through vertex C.

It is helpful to extend the length of the side beyond the triangle.

This applies to the side we are drawing the perpendicular through, like this.

Pause now to do so with your triangle.

You should have something like this.

Vertex C of the triangle is the point that we want to pass the perpendicular through, so place the compass needle on that point, then set your compass width to long enough that the circle will intersect your extension of AB twice.

And then draw a circle and make sure that it intersects with your extended line twice.

Alternatively, you could have just drawn arcs at the two locations where the circle would've intersected AB.

Pause now to construct your own circle or arc with the compass needle at point C.

This circle or this set of arcs should be similar to what you got.

We now have a new line segment.

I'm gonna label it DE.

DE will be one of the diagonals of our rhombus that we can construct.

Set your compass width to greater than half the length of DE, and then construct two circles or a set of arcs with D and E as their centres.

So we put the needle at D and draw a circle, and then put the needle at E and draw a second circle.

Alternatively, you could have drawn these arcs instead.

Pause now to construct your own set of circles or arcs.

Here are some circles or arcs that you might have constructed.

Using the arc intersections, let's construct a rhombus.

Alternatively, we can just draw the perpendicular that passes through both arc intersections.

Pause once more to finish your construction.

So, you might have either drawn a rhombus first or gone straight to drawing the line segment.

Either way, we had to use those two pairs of intersecting arcs.

We have constructed a perpendicular, hence this right angle.

Notice that this perpendicular passes through the third vertex of our triangle.

This is brilliant.

This is exactly what we wanted.

Well done if yours did as well.

Notice how my construction on the left had a perpendicular that intersected an extension of the line segment AB.

Yours could have done that, or intersected the original part of line segment AB, part of the original side of the triangle.

Great, but what's the point of constructing a perpendicular through points off of a line segment? Well, sometimes the length of a perpendicular can help find the shortest distance from that point to the line segment.

However, this is only true if the perpendicular intersects part of the original line segment and not a part of its extension.

Here's an example.

With centre at C, we draw a circle, and we extend the line segment, so the line segment intersects the circle twice.

We construct four arcs, and here's the perpendicular.

As the perpendicular intersects the original line segment AB and not the extension to the right, the shortest distance from C to the line segment AB is 4.

3 centimetres.

I measured this using a ruler.

On the other hand, let's construct our perpendicular just like before to get the same 4.

3 centimetre shortest distance to the line.

For point D, however, 4.

3 centimetres is not the shortest distance to the line segment AB, because the perpendicular only passes through that extension of AB, not the original line segment AB.

The true shortest distance is 5.

1 centimetres, which is the distance from D to the endpoint, B, the closest endpoint to point D.

Okay, for this final check, pause here to complete each statement with a value from that diagram.

The shortest distance from Z to an extension of SR is seven centimetres.

However, since this is just an extension of SR, this is not the true shortest distance.

The true shortest distance to SR is 12 centimetres along the path from Z to R.

The radius of the circle is 10 centimetres.

Great effort and attention to detail, everyone.

Let's put that extra perpendicular knowledge into practise.

For question one, pause here to complete each of these three constructions of the perpendiculars through point Z.

For question two, we have three lakes, each with a lake front that is a straight line segment.

Construct a perpendicular from point F to each of the three lake fronts, and hence identify which lake is closest to point F.

Pause now to try question two.

And finally question three, an orthocenter of a triangle is the point inside or on or outside a triangle where the three perpendiculars on a triangle intersect.

Each perpendicular must pass through the opposite vertex of that triangle.

Pause here to construct the orthocenter for each of these three triangles.

Good stuff, great effort on tackling all of those questions.

Pause here to compare your constructions to question one with those on screen.

And a very well done if you spotted that you had to extend the line segments CD and EF.

For question two, Lake Concave was the closest.

Whilst the distance to an extension of Lake Convex's lakefront was shorter, this is an extension to the lakefront and not the actual lakefront, and so it does not count.

Pause here to compare your constructions with these on screen.

And finally, question three.

Orthocenters can exist either inside a triangle, like with A, outside a triangle, like with B, or even on one of the vertices of a triangle, like with C.

Pause here to think about or discuss any theories that you might have for the properties of a triangle that will result in these different orthocenter locations.

If the greatest or largest interior angle of a triangle is acute, then the orthocenter will lie inside the triangle.

If the greatest interior angle of a triangle is obtuse, then the ortho centre will lie outside of the triangle.

So if the greatest interior angle is a right angle, what do you think you can conclude? Since the right angle is neither acute or obtuse, it cannot be either inside or outside of the triangle.

What's the one remaining option? One and all, a superb effort on this constructions lesson where we have constructed perpendiculars to a line segment through any point, either on or off of a line segment.

And that most of the time, you'll need to either extend or shorten that line segment to make that perpendicular happen.

When constructing a perpendicular, we need to draw a circle with a centre at the point that the perpendicular must pass through.

Make sure to draw a circle large enough that it intersects the line segment twice, even if it intersects at an extension of the original line segment.

And lastly, we have seen that if a perpendicular through a point that is off of the line segment intersects the original line segment and not its extension, then the distance from that point to the line segment is the shortest distance from the point to line segment.

A fantastic job on all of these constructions, everyone.

Well done on your effort.

I have been Mr. Gratton, and you all have been absolutely brilliant.

So until next time, our next maths lesson together, take care and goodbye.