Lesson video

Hello, welcome one and all to this lesson on constructions with me Mr. Gratton.

Grab a pair of compasses and a ruler in this lesson where we will use properties of constructing a rhombus in order to construct a perpendicular bisector of a line segment.

Pause here to check the definitions of important keywords, including both perpendicular and bisector.

First up, let's look at how we can construct a rhombus and analyse its properties.

Okay, here we have a rhombus.

We'll have a look at the properties of rhombus a little bit later.

Jacob says, if we need to draw a rhombus accurately, we can use a protractor and a ruler.

However, Sofia disagrees with this.

Sofia says it's easier to use a pair of compasses and a ruler instead.

Jacob is intrigued by Sofia's suggestion, but it is accurate.

A pet of compasses are useful for much more than just drawing a circle.

We can use the properties of rhombus in order to figure out why using a pair of compasses is an effective way of constructing a rhombus.

A rhombus is a quadrilateral with two pairs of parallel sides.

All four of its edges or sides are equal in length and each pair of opposite angles are equal.

It has two lines of symmetry across its two diagonals and its two lines of symmetry meet at right angles with each other.

If we used one line of symmetry to split the rhombus into two triangles, both of those triangles will be congruent with each other.

Notice that one triangle shares two of its sides with two of the four sides of the rhombus.

Also, every side of a rhombus is equal in length.

That means two of the sides of the triangle are also equal in length.

What type of triangle has two sides of equal length, an isosceles triangle.

That means a rhombus can always be split into two isosceles triangles.

Right, what does that mean when trying to construct a rhombus? Well, to construct a rhombus, we need to construct two congruent isosceles triangles where these two isosceles triangles share a side that is also one diagonal of the rhombus.

So if you are able to construct an isosceles triangle, you'll be halfway there to constructing a rhombus as well.

Okay, let's your knowledge of the properties of a rhombus with some checks.

Here we have a rhombus with a perimeter of 48 centimetres.

Pause here to find the length, X, centimetres.

X centimetres is 12 centimetres because all four sides of a rhombus are equal in length and 12 times four is 48.

Onto the next check.

Pause here to find the value of Y.

Y degrees is 133 degrees because opposite angles in a rhombus are equal.

And onto the last check for now.

Pause here to identify which of these quadrilaterals are made from two congruent isosceles triangles.

The key here is to have a look at the hash marks.

B and C are both rhombi, A and D are kites whilst E is a parallelogram.

Right, get a pair of compasses and a ruler at the ready.

Let's construct a rhombus with side lengths of seven centimetres.

We can use the protractor to construct either circular arcs of varying lengths or we can draw a full circle like so.

Sometimes it'll be helpful to draw full circles, but other times only arcs are necessary.

However, there's rarely one correct choice either way.

Practising different constructions will help you identify where circles or arcs are more helpful at any given moment.

Okay, onto step one.

We want to construct two congruent isosceles triangles.

For this example, each triangle will have two sides that are seven centimetres long.

We start by drawing this shared base edge like so.

We can also draw a seven centimetre line segment at the bottom of the page.

This might help ensure your pair of compasses are correct in their width later on.

Using your pair of compasses, we can then draw a full circle.

To do this, set your pair of compasses to the seven centimetre line segment that we drew earlier.

Once your compasses are the correct width, place the compass needle on one of the two endpoints of your base and then draw a full circle.

Pause now to do this.

every point on the circumference of the circle is exactly seven centimetres from the endpoint that the compass needle was placed on.

Let's now do exactly the same again, but this time, draw a second circle with the needle on the other open endpoint.

You can draw a full circle if you want, but only one arc is necessary.

So let's place the compass needle on the other open endpoint and draw one arc that intersects the circle like this.

Pause now to do that.

That point of intersection is the third vertex of an isosceles triangle.

The other two vertices of that triangle are the two endpoints of the original baseline segment that we drew.

So using a ruler, let's join each of the vertices together in order to construct this isosceles triangle with two sides of length seven centimetres.

Pause now to do that.

Remember, a rhombus Is just two congruent isosceles triangles that share a side.

If we've already drawn one triangle, then surely drawing a second triangle is just as easy.

Let's use that same base line segment and construct a second isosceles triangle that's underneath the first triangle.

If I place my compass needle on this particular endpoint, you may notice that it'll just end up drawing the exact same full circle that we constructed before.

However, if I place the compass needle on this other endpoint, then I can create an arc that intersects the circle once more.

This time underneath the triangle that we drew earlier, once again, let's use this ruler to join each pair of vertices together, creating another isosceles triangle with two sides of length, seven centimetres.

Pause now to do that.

And we're done.

These two congruent isosceles triangles make a rhombus where each of its four sides have length seven centimetres.

Very well done if you manage to construct that rhombus as well.

However, Sam quite sensibly asks, "How big does that base line segment of the triangle need to be?" That diagonal of the rhombus.

Well, let's have a look.

Even if the base length that diagonal of the rhombus that we first drew, even if that changes in length, we can still guarantee that all the sides of the rhombus will remain exactly seven centimetres long.

If we check our compass width is set to exactly seven centimetres using that reference line at the bottom of the page that we drew quite early on.

So once again, let's set our compass width to seven centimetres and then place the needle on one of the two endpoints and draw an arc.

Then place the compass needle on the other open endpoint and draw a second arc that intersects the first.

If your two arcs do not intersect each other, do it again, but make the arcs a little bit longer this time.

This point of intersection then becomes the final vertex of a different isosceles triangle with a pair of sides of seven centimetres.

We can do this exact same thing again a second time to create a second congruent isosceles triangle like so.

As Sam says, the initial base length doesn't matter.

However, there is a limit to how long this line segment can be.

Let's have a look at what that limit is.

Yes, we can be flexible over the length of the base, however, some base lengths simply do not result in a rhombus.

We make the sides of our rhombus using either circles or arcs.

The radius of each circle or arc has to be greater than half the length of the line segment in order for the two circles or arcs to intersect.

For this base length of 10 centimetres, the radius needs to be over five centimetres each.

In conclusion, for a rhombus with a side length of X units, the base length has to be less than two x units long.

Okay, here are a few checks.

Lucas starts to construct a rhombus with side lengths of 5.

5 centimetres.

Pause here to identify which of these constructions are correct when trying to construct the first isosceles triangle.

Both A and C are valid.

This is because circles or arcs, they both work, as long as we have a point of intersection between both circles or arcs.

B is clearly incorrect because the two circles are different in size, you always need to make sure that both circles or arcs have the same radius because your pair of compasses have been set to the exact same compass width.

Lucas has correctly drawn the constructions for the first isosceles triangle.

Pause here once more to identify which of these lengths are definitely 5.

5 centimetres long.

A, D and E are radii of the arc and the circle.

Both radii must be equal in length for this to be an isosceles triangle.

D is one of the two sides of the isosceles triangle that is also exactly 5.

5 centimetres long.

Lucas now makes a mistake and ends up constructing a general kite and not specifically a rhombus.

Here to identify possible explanations for Lucas's mistake.

The width of the compasses has changed, meaning that the bottom iso triangle is different to the top one.

And lastly, here we have Sam.

Sam attempts to construct a rhombus by drawing two congruent circles with their centres on the endpoints of a base line segment.

Pause here to explain why Sam cannot construct a rhombus.

The circles are too small because each circle has a radius that is less than half the length of the base line segment itself, the diagonal of the rhombus.

Brilliant stuff, let's practise those constructions and properties of rhombi.

For question one, label the rhombus with all of its lengths and angle sizes as well as appropriate hash marks.

And for question two, here's an isosceles triangle.

Using that triangle, construct a rhombus from it.

Pause here for these two questions.

Onto question three.

Construct two different rombi by using combinations of the lengths given and explain why some rombi by are impossible to construct.

And finally, for question four, here's a rhombus for you.

Use pyrosis theorem to predict the length of the other diagonal of the rhombus, knowing that one diagonal has length eight centimetres.

And then actually construct this rhombus from scratch to verify the answer that you found from Pyrosis theorem.

Pause here for these two questions.

A very well done on all of your effort on those four questions.

For question one, the rhombus has side lengths of 8.

5 units.

We have two pairs of equal opposite angles with one pair at 99 degrees and the other pair at 81 degrees.

And all sides should have the same number of hash marks as all sides are equal in length.

Pause here to check your rhombus constructions for question two versus those on screen.

And for question three, We have a rhombus with a diagonal of Z units and a side length of A units that is impossible to construct because A is less than half of Z, meaning that circles with radius of A will not intersect each other.

Pause here to check the other two constructions for the two rhombi against those on screen.

And for question four, Pythagoras's Theorem would lead to the other diagonal of the rhombus being exactly six centimetres long.

Well done if you constructed a rhombus to verify this.

You've done all so well at constructing a range of different rombi.

But what's the point? Well, we can use a rhombus construction to construct a perpendicular bisector.

Let's find out what that is.

Here we have Jacob once more who asks what happens when you draw on the diagonals of a rhombus? And here's Sam who reminds Jacob of what we've just learnt.

If we draw on one diagonal, the rhombus is cut into two congruent isosceles triangles.

But what happens when you draw on the second diagonal? Well, when both diagonals of a rhombus are drawn on, the angle between the two diagonals is always 90 degrees.

The two diagonals are always perpendicular to each other and the length of each diagonal has been cut in half.

One diagonal of the rhombus is the perpendicular bisector of the other diagonal.

This means that the two diagonals meet at right angles with each other where each diagonal is bisected, meaning divided into two equal parts.

In general, the purpose of a perpendicular bisector is to divide a line segment into two equal parts at a right angle.

Right, here's one quick check, pause here to identify which of these show a perpendicular bisector on a rhombus.

only D, all the other quadrilaterals are not rhombi 'cause they do not have four equal sides.

Many of the incorrect answers also do not look like their diagonals meet at right angles.

Okay, now we're a bit more familiar with what a perpendicular bisector is, let's construct one.

Let's construct the perpendicular bisector of line segment a AB.

And Jacob thinks this is really easy because we can just measure the length of AB using a ruler and then find its midpoint.

So AB here is 12 centimetres long and so its midpoint is at six centimetres.

However, whilst Jacob found that bisecting might be easy, making sure that the bisector is also perpendicular to AB is a lot more tricky.

Can you convince me that this line segment, this line segment or this line segment is exactly perpendicular to AB? It's pretty tricky to tell.

So how do we accurately construct a bisector that is perpendicular to line segment AB? With the rhombus of course, you can construct a rhombus with a line segment as its diagonal in order to find the perpendicular bisector of that line segment.

It can be any rhombus as long as AB is its diagonal.

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

Each vertex must be an equal distance away from the two endpoints of the diagonal line segment.

As we know, the second diagonal of the rhombus will perpendicularly bisect the first.

It will perpendicularly bisect line segment AB, which is what we want.

And remember, when constructing the rhombi, the circles have to be exactly the same size because they have to have exactly the same radius.

However, they can be any size as long as they are two circles that intersect each other.

So they cannot be these two circles because their radii are too small.

This is also true if you choose to draw arcs rather than full circles.

These four arcs have to intersect twice, but must still have the same radius as each and every other arc.

This set of arcs does not work because the radii are again too small.

So here's our full highly impressive rhombus construction and what it should look like.

Drawing on the second diagonal will perpendicularly bisect the line segment PQ.

However, you can construct the perpendicular bisector directly by drawing a line that passes through both intersections, which means there's no need to draw the four sides of the rhombus itself, as long as those intersecting arcs or circles have themselves been drawn.

Okay, here we have Andeep who has correctly drawn three of the four arcs needed to construct the perpendicular bisector of line segment UV.

Pause here to identify the distance between point B and point V.

The distance is seven centimetres.

We know this distance because the distance or the radius of that arc is the same as the radius from U to A, which is labelled at seven centimetres.

Remember, all arcs or circles must have the exact same radius.

For and Andeep\s correct construction, pause here wants more to consider which of these statements is correct.

Point C is equidistant to U and V.

This is because C is also the third vertex of an isosceles triangle, UCV.

And finally, pause here to identify which of these diagrams shows an accurate construction of a perpendicular bisector of the line segment KL.

only A and C are fully convincing constructions.

B does not look like a bisector as that perpendicular line is much closer to K than it is to L.

D does not have any construction lines.

So we cannot assume that the vertical line drawn has been drawn in the correct place.

Remember, it is always oh so helpful to keep your construction lines, your workings as it were, in order for others to know how you managed to make your perpendicular bisector itself.

Lovely, down to you now for some independent practise.

For question one, draw two different perpendicular bisectors.

One for a line segment that is 12 centimetres long and the other that is five centimetres long.

For each one, set your compasses to a width of exactly 6.

5 centimetres.

And for question two, you have free reign over your compass width, and well everything else, construct a perpendicular bisector for those three line segments.

Pause here for these two questions.

And finally, brace yourself for something, well really interesting, a circumcentre, that sounds like circumference and centre.

The circumcentre of a triangle can be found by constructing three perpendicular bisectors.

One for each of the three sides of that triangle.

And then we find where each of the three perpendicular bisectors intersect.

And they will always intersect at the same single point.

by constructing a bunch of different perpendicular bisectors, pause here to find the circumcenter of these two triangles.

Fantastic work everyone.

Here are the answers to question one.

For the two rhombi that you constructed, the diagonals are the same length and therefore the sides of the rhombi must also have been exactly the same length.

Therefore, both rhombi are congruent to each other.

Just one is a rotation of the other.

And for question two, pause here to check that your constructions match those on screen.

And for question three, here are the locations of the circumcenters for those two triangles, we can draw the circum circle in order to check that the circumcenter is in the correct place.

The radius of the circum circle will be the distance from the circumcenter to any one of the vertices of the triangle.

They will all be exactly the same distance.

A superb effort everyone on this really detailed lesson on constructions where we looked at rhombi and how they can be constructed from two congruent isosceles triangles.

Furthermore, the diagonals of a rhombus meet at right angles and bisect each other.

It is possible to construct the perpendicular bisector of a line segment without drawing the rhombus itself, as long as the constructions for the rhombus where there are two points of intersection between arcs and circles are drawn.

We can construct a perpendicular bisect of a line segment that is also part of a shape, such as a diagonal of a rhombus or a side of a triangle or other polygon.

Again, great effort everyone on this lesson.

I've been your teacher, Mr. Gratton, and you have been absolutely brilliant.

Take care everyone and have an amazing rest of your day.