Lesson video

Hello all and welcome to another maths lesson with me, Mr. Gratton.

During today's lesson, we will look at the construction of perpendicular bisectors, using the properties of a rhombus.

Pause here to check the definitions of these two keywords.

During this lesson, we will first of all look at the basics of constructing a perpendicular bisector and then make our method more efficient.

Jacob asks, what happens when you draw on the diagonals of a rhombus? Whilst some notices that the rhombus becomes two isosceles triangles that are congruent to each other, but a rhombus has two diagonals, what happens when you draw on the other? Well, Jacob thinks that the diagonals are perpendicular to each other, whilst Sam spots that each diagonal, splits the other diagonal into two equal parts.

Whose observation is correct, Jacob's or Sam's? Well, actually both observations are correct.

This is because the angle between the two diagonals is always going to be 90 degrees.

Furthermore, the length of each diagonal has cut the other diagonal into two equal halves.

One diagonal of a rhombus is the perpendicular bisector of the other diagonal, where perpendicular means to meet at right angles and bisector means to divide into two equal parts.

Therefore, the perpendicular bisector divides a line segment into two equal parts with a line that is perpendicular or at right angles to the other line segment.

Here's a quick check for understanding, about perpendicular bisectors and rhombi.

Which of these shows a perpendicular bisector on a rhombus? Pause now to look through all five shapes.

And the only one correct answer is D.

This is because the other quadrilaterals are not rhombi, because they do not have four equal sides, noted by the hash marks.

Furthermore, some of the incorrect answers also have diagonals that do not look like they meet at right angles, such as option B.

Jacob thinks that drawing on a perpendicular bisector is pretty easy because he can use a ruler to find the midpoint, the point of bisection and whilst this might be sometimes true, it is then far more difficult to guarantee that the right angle can be drawn, even when using a protractor.

You can construct a rhombus with a line segment as its diagonal in order to find the line segments perpendicular by sector.

It can be any rhombus as long as AB is one of its diagonals.

One vertex of the rhombus will be above the diagonal and the other one will be below it.

Each vertex must be an equal distance away from the two endpoints of the line segment that creates one of the diagonals of that rhombus.

Okay, let's go through the process to construct a rhombus given a line segment that is one of the diagonals of that rhombus.

By placing the compass needle at point B, I can find all the points that are a certain distance away from point B.

In this example, the circumference of this circle is always going to be four centimetres away from point B.

The second step is exactly the same as the first, but with your compass needle at point A instead.

This is because all of the sides of a rhombus are of equal length, and so both of the circles need to be exactly the same size as well.

Okay, quick question.

Where in the circles is the distance from point A, exactly the same as the distance from point B? Well, this point up here is exactly four centimetres away from A, but it's much further away from point B.

However, this point here is exactly four centimetres away from point B, but it's actually closer to point A.

It is the two intersections of those two circles that is an equal distance away from point A and point B.

And so those two points at those two intersections can be used as the final two vertices of the rhombus.

And by drawing on that second diagonal, between the pair of intersections of those two circles, we have created a pair of perpendicular bisectors, meaning we have bisected AB at an angle of 90 degrees.

Okay, here's a quick check.

Alex has started a construction to find the perpendicular bisector of line segment PQ.

What has Alex done wrong in their construction? Pause now to look through those four options and choose any of the correct ones.

And the three correct answers are A, C and D.

This is because if you only use, one of the two intersections of those circles, there is no guarantee that you will draw a line perpendicular to PQ, nor is it likely that you'll draw a line that bisects PQ perfectly.

It is essential that you use two intersections in order to create a perpendicular bisector.

in your construction of a rhombus, both circles have to be the same size, but they can be any size as long as the two circles intersect each other.

Another way of looking at this is that the radius of each circle has to be greater than half the length of the line segment in order for the two circles to intersect.

Okay, let's go through a demonstration to construct a rhombus from a line segment that is one of its diagonals.

After each set of steps that I show you, perform the same steps for the rhombus that you see on the right.

Here's step one.

Use a ruler to draw a line segment of five centimetres and pause now to draw a line segment of four centimetres where you label one end point A and the other end point B.

Okay, now place your compass needle on point A and make sure that the pencil end is closer to point B than point A, and then draw a full circle.

Pause now to try yourself to draw circle with the centre at point A.

Before we get onto the next step, ensure that your compass width is exactly the same as for the first circle that you drew.

And then draw a second circle this time with the compass needle at point B.

Try this yourself.

Place your compass needle at point B and draw a second circle.

Pause now to do this.

And then your last set of steps is to mark the two locations that the two circles intersect and then join these two intersections to the points A and B.

This completes your rhombus.

Pause now to do this for yourself.

On your completed rhombus, draw on the second diagonal to create the perpendicular bisector to the diagonal that is the line segment AB.

Pause now to try this yourself and complete your perpendicular bisector.

This is what your perpendicular bisector would look like and here are the full constructions, without the construction lines.

Onto the next check for understanding.

Alex has again attempted a construction to find the perpendicular bisector of PQ.

Why can't Alex complete this construction? Pause now to look at the construction and check which options make sense.

Alex's circles are too small, therefore a rhombus cannot be constructed, because the circles do not intersect each other.

And so what advice can we give to Alex so that the construction can be improved? Pause now to decide which bit of advice is suitable.

If you two circles do not intersect, make both of the circles larger as long as they are both the same size.

Okay, onto the practise for this cycle.

And for question number one, give an explanation as to why each construction, does not result in the correct perpendicular bisector, being drawn for line segment XY.

And furthermore, give some advice as to what could be done to make the constructions better.

Pause now to do that.

Question number two, find the perpendicular bisector for each line segment by constructing a rhombus.

Pause now to give yourself time to do this.

And for question number three, measure the length of each line segment and then construct a rhombus and use that rhombus to construct a perpendicular bisector.

After you've done that, measure the length of one part of the bisected line and check how close is it to a perfect half of the original line segment.

Furthermore, measure the angle at the intersection of the two diagonals.

This should always be 90 degrees.

Pause now to do this question.

And onto the answers for question 1A, both circles are too small and so a rhombus cannot be constructed.

The advice is to just make both circles bigger as long as they're both the same size and speaking of the same size, part B shows two circles that aren't.

After constructing the first circle, make sure that your compass width is exactly the same, so that the second circle is exactly the same size as the first.

And for part C, the compass needle was not placed at the endpoints of line segment XY.

Therefore, the two circles do not intersect through the midpoint of XY.

And so the construction, will not show you a perpendicular bisector, just a perpendicular.

And for question number two, here's what your construction lines should have looked like.

And for question number three, here's what your final perpendicular bisectors, should have looked like.

Now that we understand the key properties of a perpendicular bisector, that is to say we need two intersecting congruent circles with centres at the end points of a line segment.

How can we make the construction a little bit more efficient? Sam's point is valid.

Keeping the whole circle on the paper is challenging, especially when the circles can be very large, especially for very long line segments.

Laura's point is also valid.

We do not need to draw the entire circle.

It is possible to accurately construct a rhombus by only drawing the arcs of the circle, rather than the full circle.

To do this, we need to draw two arcs with the compass needle at one endpoint of the line segment and then another two arcs with the compass needle on the other endpoint.

The most important part is the four arcs need to create two intersections, one intersection on either either side of the line segment.

Notice, how each pair of arcs is a part of the full circle that we could have constructed previously.

It is still possible to construct a fully accurate rhombus using just these arcs.

However, you can construct the perpendicular bisector by directly drawing a line that passes through both intersections.

There is no need to draw the sides of the rhombus if you are only looking at constructing a perpendicular bisector.

Okay, here are a few quick checks.

Andeep has correctly drawn three of the four arcs needed to construct the perpendicular bisector to line segment UV.

Point A is seven centimetres away from point U.

How far away is point B from point V.

Pause now to have a think about what your answer could be.

And your answer is seven centimetres.

For Andeep's construction, which of these statements is correct? Pause now to look through all of them and make a decision.

The correct answer is C.

Point C is equidistant to U and V.

Which of these diagrams shows a full and accurate construction of a perpendicular bisector to the line segment KL? Pause now to look through all four and make your decision.

And the answers are A and C.

C has used four smaller arcs, whereas A has used two big arcs.

Both are absolutely okay.

And for D, whilst that might be a perpendicular bisector, there are no construction lines or construction arcs to show this.

In your constructions, keep every single construction line and arc to show how accurate your construction is.

And onto the next demonstration, I have drawn a line segment of 5.

5 centimetres.

Pause now to draw a line segment that is 4.

2 centimetres long and label each endpoint with A and B.

Next up, set your pair of compasses to any width over half the length of AB.

If in doubt, set your compass width to exactly the length AB and then place your compass needle onto the point B and draw two arcs, one on each side of line segment AB and pause here to draw two arcs of your own with the compass needle at point B.

We're about to construct some more arcs.

If you are unsure if your compass width has changed, you can reset it by placing the compass needle on point B and setting the pencil to touch one of the arcs that you have already drawn.

Next step is to construct two more arcs, now with the compass needle on point A like so.

Try this yourself with your construction.

Pause now to do so.

And the last step is to draw a line segment, through the two intersections of your four arcs like so.

And pause now to try this yourself for your two intersections.

Your perpendicular bisector should have divided AB into two equal halves.

For my diagram, the original length of AB was 5.

5 centimetres, and so the length of one half, should have been 2.

75 centimetres.

For your construction, see whether each half of the line is going to be 2.

1 centimetres long.

Pause now to check.

Constructing perpendicular bisects, using only arcs rather than full circles can be especially helpful when trying to bisect more than one line segment on a diagram.

For this diagram, let's bisect AB and BC.

When trying to bisect multiple line segments, it is important to focus on the constructions for only one line segment at a time.

For example, when bisecting AB, let's ignore the line segment BC and so we would do the full construction for the line segment AB, before then ignoring line segment AB and doing the construction for BC.

And onto the next check, when constructing a perpendicular bisector of this line segment that has a length of 12 centimetres, which of these construction methods are possible? Pause now to look through all five and choose the possible ones.

The three correct answers are C, D and E.

If you drew arcs of radius three centimetres, the pairs of arcs would never intersect each other and if you used a radius of six centimetres, the line segments would touch exactly on MP itself.

This means a rhombus cannot be constructed, because there will be only one intersection and not two, therefore, you might be able to bisect the line, but it would be impossible for you to create a perpendicular bisector.

Next, check of the possible construction methods, which of these is most sensible? Pause now to have a think.

And the correct answer is D, whilst C seven centimetres is over half the length of line segment MP, it is only just over half the length and therefore your construction lines, will be very close to MP.

And if your pencil isn't sharp enough, there may be some overlap and confusion as to some of the intersections.

And E, 20 centimetres is probably too big for you to construct practically with your pair of compasses and the arcs themselves may even intersect off the page that you're constructing on.

Nine centimetres is just about possible, given a standard pair of compasses.

And onto the final set of practise questions.

For question number one, complete the construction of each perpendicular bisector.

Pause now to do each construction.

And for question number two, the circumcenter of a triangle is a point that is the centre of a circle whose circumference intersects with all three vertices of a triangle.

The circumcenter can be found by the point where the perpendicular bisectors of all three sides of the triangle intersect.

By creating three perpendicular bisectors for the three sides of this triangle, find the circumcenter of that triangle.

Pause now to do so.

And here are the answers to question number one.

Pause now to compare these constructions to your own.

Question number two, the circumcenter of this triangle is in that location.

Pause here to check that the circumcenter of yours, matches roughly with the one on screen.

And are very well done on a great lesson on constructions where we have learned that a perpendicular bisector is a line that intersects a line segment at right angles and divides the line into two equal parts.

And that constructing a rhombus from the end points of that line segment, using either full circles or arcs is an effective way of bisecting a line segment, perpendicularly.

That is all for this lesson on constructions.

Thank you so much for joining me today, and until next time, have a great day.