Lesson video

Good day, everyone.

I'm Mr. Gratton and welcome to another maths lesson.

In this lesson, we will be using knowledge of constructions to solve a range of problems and look at a range of construction patterns.

Pause here to check the definitions of some of the keywords that we'll be using during this lesson.

A wide variety of polygons and angles of different sizes can be found from both simple and complex constructions.

Let's have a look at some examples.

In this construction of a perpendicular bisector, there is a right angle.

This one is obvious since the core purpose of a perpendicular bisector is to construct a perpendicular and 90 degree right angle.

We also have an equilateral triangle.

This can be formed by joining one of the intersections of the two circles to both endpoints of AB.

A 60 degree angle is also obvious.

If you formed an equilateral triangle, then naturally you have a 60 degree angle as the interior angles of an equilateral triangle are all 60 degrees.

And lastly, we've got a rhombus, which is an alternative way at looking at the entirety of this construction.

Each line segment that connects intersections of two circles or an intersection to an endpoint of the line segment, or simply the line segment itself, all form either the sides or the diagonals of a rhombus whose interior angles are either 60 degrees or 120 degrees.

Which properties of this construction are important in defining these polygons and angles? Well, it is important to be aware that both circles have the same radius as the length of AB itself.

This means that the distance from either endpoint to an intersection is equal in length to the original line segment AB.

This creates an equilateral triangle, or a rhombus, depending on how many points you join together.

However, further polygons can also be formed by drawing on extra line segments, such as this perpendicular bisector, creating a right angle triangle, for example.

Okay, here's your first check.

Three congruent circles are drawn.

Two circles have centres at the end points of the line segment AB, and the other has centre at the midpoint of AB.

How many sides of this hexagon have length eight centimetres? Pause now to consider the answer to this, or wait a few seconds to receive a hint.

The hint is these two line segments are radii of the centre circle and create an equilateral triangle with the top side of the hexagon.

Pause now to consider the answer.

All six sides have a length of eight centimetres.

This is a regular hexagon.

This is because the shape can be made from six equilateral triangles, each with a vertex at the centre of the middle circle.

We know that each of these congruent triangles is equilateral.

This is because each triangle has two side lengths that are radii of the same circle, and so each triangle is at least isosceles.

However, each triangle also shares a common vertex that is also the centre of that middle circle.

And all six angles meet around a single point where the sum of angles around a point is always 360 degrees.

360 divided by six is 60 degrees for each triangle.

An isosceles triangle with one angle of 60 degrees is always equilateral as all three angles must therefore be 60 degrees each.

Onto the next check.

Two circles of different size have opposite endpoints of the line segment AB as their centres.

Which quadrilateral is created by joining the two endpoints of AB and the two intersections of the circles? Pause now to look at the construction and think which quadrilateral it looks like.

And the quadrilateral is a kite, but which of these reasons explains why the quadrilateral is a kite? Pause now to look at the construction and reason the properties that make it a kite.

This quadrilateral is a kite because two adjacent sides of the kite have the same length, the two radii of the larger circle.

Furthermore, the other two adjacent sides of the kite also have the same length because they are the radii of the smaller circle instead.

And onto the practise.

Here I have two of the same construction.

Each copy shows congruent circles.

The centre of the left and rightmost circles are the endpoints of the line segment FG.

On both copies of this construction, draw and label each of these polygons and angles.

Pause now to do this.

And for question number two, I've got two copies of the same construction again.

For each construction, the top two circles have the endpoints of the line segment FG as their centres.

And the bottom circle has a centre at one of the two intersections of the top two circles.

Pause now to draw on and label each of these polygons.

And again, for question number three, I've got two copies of the same construction.

In this construction, I have two circles of different size.

However, their centres are still the endpoints of the line segment IJ.

Pause now to draw and label all of these polygons.

And finally, for question number four, this construction shows seven congruent circles, each with their centres labelled with a cross.

The centres of six of the circles lies on the circumference of the seventh central circle.

By using multiple copies of this construction, find, draw, and label as many different polygons and angles as you can.

And for part b, suggest a type of quadrilateral that cannot be accurately drawn from this construction.

And can you explain why? Pause now to do this question.

And here are the answers for question number one.

On the left hand construction, we have a right angle created by the perpendicular bisector construction.

We have an equilateral triangle on the far right, and on the left we've got a regular hexagon.

For the second construction, we have a rectangle and a parallelogram drawn.

And for question 2a, two of the side lengths of the smaller equilateral triangle is shared with two of the side lengths of the larger equilateral triangle.

And for part b, by drawing the is isosceles trapezium and then creating a perpendicular bisector, the half trapezium is also a right angled trapezium.

For question 3a, first of all, I draw a kite and then by drawing the bisector, I naturally create two isosceles triangles, each which share the base side with the bisector itself.

And for question 3b, the kite is bisected using the line segment IJ.

And so one of the two triangles is scaling in that bisection.

And by drawing part of the bisector that we did in 3a, we also create a right angled scaling triangle.

And for question number four, pause here to see how many of these polygons and angles match the ones that you found in your investigation.

And for question 4a, a square cannot be accurately drawn.

This is because all lengths that are equal to each other are at an angle of either 30 or 60 degrees.

We've spotted several different polygons on constructions that have been given to us.

Let's now look at creating constructions for ourselves that allow us to create several different types of polygon, as well as several different types of more general shapes too, but won't bother constructing it for ourselves.

As Sofia says, "It's great to spot shapes on different constructions," but Laura is also correct.

Sometimes it is more satisfying to construct the shape from scratch rather than relying on pre-made constructions.

And this demonstration will show you how to construct this shape.

As you may have seen from a previous cycle, many, many different shapes and angles can be found from this construction.

I will show you the construction itself on the left with the instructions shown on the right.

After each set of steps, perform those same steps for yourself.

In the centre of your page, draw a line of four centimetres and label it AB.

This will be the radius of each of the seven congruent circles.

This means that each circle will be of equal size.

Check your compass width after each circle If you are unsure if its width has changed.

With the compass needle on A, set your compass to the length of AB and draw a circle.

With the compass needle on B, draw another circle of equal size.

Point A is the centre of the left circle whilst point B is the centre of the right circle.

We will treat the circle with centre A as the most central circle in the finished diagram.

Pause now to draw a four centimetre line segment and construct these two congruent circles also with a radius of four centimetres.

Label one of the intersections on your two circles point C.

I've done this for the bottom of the two intersections.

Place your pair of compasses on point C and construct a third circle.

Pause now to do this for your diagram.

Now that you've got three circles in total, place your compass needle on the point where your most recently drawn circle intersects the circle with centre A, the blue dashed circle.

Label it the next letter.

In this case, label this intersection D.

With your compass needle on that point, draw another circle.

Pause now to try this yourself and repeat this set of instructions three more times to complete the entire construction.

And your finished construction should look like this.

Well done in all your effort in constructing this six-petal flower construction.

In the construction we just did, each radius of each circle is four centimetres, and therefore the side lengths of many polygons which have side lengths as the radii of these circles will also be four centimetres.

If we drew an original line segment AB to be five centimetres instead, this would increase the size of each circle, and therefore it would increase the side lengths of the polygons we can form from this construction.

And onto question one of the practise task.

By constructing either part or a whole six-petal flower, accurately construct the following shapes.

Pause now to do this for all four of the shapes.

And for question number two, many real world famous logos of companies are created using arcs and sectors of circles made from constructions.

By first of all using a template of the six-petal flower construction to help you plan, construct from scratch your own logo.

Pause now to get creative and create your own logo.

Okay, great work so far.

The answers for question one.

Well, for 1a, there's two possible answers.

You could have constructed this equal actual triangle from a construction with a line segment of four centimetres, or this bigger equilateral triangle from a smaller construction where the initial line segment was two centimetres instead.

And for question 1b, here is a parallelogram.

It has been formed from a construction with an initial line segment of five centimetres.

And for parts c and d, here are what your constructions could have looked like.

Note, there may have been other ways of representing these two shapes.

And for question number two, very well done if you were creative and made your own logo from a six-petal flower template, and even bigger well done if you were able to make it from your own construction.

We have looked at perpendicular bisectors, angle bisectors, and perpendiculars in a specific way, but there are other ways of achieving the same goal through a different method, or by looking at a familiar method in more detail.

Let's have a look.

To start off with, a perpendicular bisector can be constructed in many different ways.

In the construction of this perpendicular bisector, a line passes through the intersection of two congruent circles.

But what if there were multiple pairs of congruent circles? In this particular diagram, you have two intersecting smaller blue circles and two intersecting larger green circles that are dashed.

There are a total of four intersections of pairs of circles that are congruent to each other.

Two intersections for the smaller blue circles and two intersections for the larger dashed green circles.

Drawing a line through any two of these intersections will construct the perpendicular bisector of AB.

A perpendicular bisector can be drawn from these two intersections, even if the rest of the circle cannot be constructed or seen.

In this check, Alex says he has constructed a perpendicular bisector because his line passes through two intersections between pairs of circles.

Which of these statements explain why Alex's construction is incorrect? Pause now to give yourself some time to look through this diagram.

The answer is c.

The intersections should be between pairs of circles of equal size and not an intersection between two circles of different sizes.

An angle by sector can be constructed in many different ways.

This angle by sector involves constructing a rhombus from three congruent circles.

One of the diagonals of the rhombus bisects the angle.

However, this angle by sector involves constructing a kite.

This angle by sector only requires two circles with each circle being a different size.

This means it has a different radius and drawn with a different compass width.

However, each circle has the same centre, the vertex of the angle given.

Each of these two circles can be of any size, but it's best to make each circle fairly different in size so that the following steps are easier to work with.

Two line segments are drawn.

Each line segment joins an intersection between the smaller circle and leg, and an intersection between the larger circle and the other leg.

The fourth point on the kite is the intersection of these two line segments.

The diagonal that bisects the kite also bisects the angle.

Which two pairs of points need to be joined with a line segment to find the fourth point in a kite? Pause now to consider what the two line segments should look like and where they should intersect.

Your answer should be two pairs of two points.

The two answers are BF and CG.

The perpendicular to a line segment through a point off of a line sometimes shows the shortest distance from that point to the line segment.

This only happens if the perpendicular intersects the original line segment and not an extension of it.

If a perpendicular insects an extension of the line segment, then the shortest distance from that point to the line segment is the distance from the point to one of the endpoints of the line segment.

For this check, let's look at the reasoning behind this property.

Which of these statements help explain why the shortest distance from X to AB is seven centimetres? Pause to give yourself time to look through all six of these options.

And the answers are all of them except for option d.

Pause again to consider why this is the case.

This is because all points outside the circle are greater than seven centimetres away from the point X.

And all points on AB are outside the circle.

Therefore, every single point on AB is greater than seven centimetres away from X, except for one point.

And this point is point B, which is exactly seven centimetres away from point X because it is part of a radius of that circle, and therefore this distance that is seven centimetres away from X is closer than all other points on the line segment, or the extension of the line segment AB.

Okay, onto the final set of practise questions.

For question number one, accurately construct a perpendicular bisector to XY without any of your construction arcs and circles going outside the box.

Note, you are not allowed to cheat and rub out any construction lines.

And for question number two, accurately construct the angle bisector to this reflex 290 degree angle using the circle given.

You must use it as part of your construction.

After you have bisected that angle, measure the size of the bisected angle and you should get an angle of 145 degrees.

Pause now to attempt both of these questions.

And for question three, the final question of this lesson.

The area of a triangle can be found using the formula, area equals 1/2 base times perpendicular height.

By performing the appropriate construction, and I'm not gonna tell you which one, accurately measure the perpendicular height of each triangle and use that perpendicular height to find the area of each triangle.

Pause now to figure out which construction you need to do and answer each of these questions.

Here are the answers for questions one and two.

For question one, you needed to construct two pairs of arcs each of different size.

And for question number two, you needed to draw a second smaller circle still with a centre at the vertex of the angle given.

You should have measured your bisected angle to be 145 degrees.

And for question number three, you needed to draw a perpendicular to the base side through the opposite vertex of each triangle.

The measurements for each triangle will depend on the scale with which your triangles have been printed.

However, the perpendicular height of A is equal to its base.

The perpendicular height of B is half its base, and the perpendicular height of C is 1/3 of its base.

Well done if you managed to figure out that a perpendicular construction was required for this question.

Furthermore, well done on this entire lesson on challenging problem solving constructions where we have identified and constructed many polygons, such as equilateral triangles, rhombi, parallelograms, and trapezia, where some of the angles of these polygons are either 60 degrees or 120 degrees.

We've also identified that kites can be constructed from two non-congruent circles.

We have seen alternative methods for constructing angle and perpendicular bisectors, and seen the rationale behind why a perpendicular sometimes shows the shortest distance from a point to a line segment.

Thank you so much for all of your effort in this lesson.

Have an amazing rest of your day, and I look forward to seeing you again soon for some more maths.

Have a great day.