Lesson video

Hello there.

My name is Dr.

Rowlandson and I'll be guiding you through this lesson.

Let's get started.

Welcome to today's lesson from the unit of circle theorems. This lesson is called "The angle in a semicircle is a right angle." And by the end of today's lesson we'll be able to derive that theorem and also use it.

Here are some previous keywords that'll be useful during today's lesson.

You may want to pause the video if you want to remind yourself what any of these words mean, and then press play when you're ready to continue.

The lesson is broken into two learning cycles.

In the first learning cycle we're going to be investigating angles using circular geoboards, this is so that we may observe specific cases where this theorem occurs.

And then in the second learning cycle we're going to introduce the theorem more formally.

Let starts off with investigating angles with circular geoboards.

A geoboard is a mathematical tool to explore aspects of geometry.

It's often a board that contains pegs laid out in a pattern, and the patterns can vary between boards.

Polygons can be made on geoboards by wrapping elastic bands around the pegs.

like you can see in these examples.

Here we have Jacob who is using a circular geoboard that contains 10 points, equally spaced around a circle and a point in the centre.

He makes an isosceles triangle.

He says, "I know that this is an isosceles triangle because to the sides are radii of the circle." Those are the sides that go from the centre to its circumference.

He then says, "I can work out the angles in this triangle," by doing 360 degrees divided by 10, for the 10 equally spaced points, that gives the 36 degrees in the centre, and you can subtract from 180 degrees divide by two to get the other two angles of 72 degrees.

Following this, Jacob makes different isosceles triangles and each time he works out the missing angles.

Here's one of them, here's another, here's another.

He says, "I wonder what would happen if I combine some of these isosceles triangles on the same geoboard." It looks something a bit like this.

We have a lot of information in this diagram now so it can be helpful to put some labels on here, so we can specify precisely what it is we're talking about as we go along.

Jacob says, "I'm going to label a point at the centre O, and then label a point at the circumference A, B, C and D.

And this will help me describe the line segments, angles and shapes in a much more precise way." One circle theorem tells us that the angle at the centre is twice the angle at the circumference, if the angles are subtended by the same arc.

Jacob says, "I can see this circle the in the diagram.

Angle BAD, which is the angle at the circumference, and angle BOD, which is angle at the centre, are subtended by the same arc.

The arc is BCD, the minor arc on the right." If we work out these angles, angle BAD, which is the angle at the circumference, that's the sum of 36 degrees and 18 degrees, which is 54 degrees.

Angle BOD, which is the angle at the centre, is the sum of 36 and 72, which gives 108 degrees.

And what we can see there is that angle BOD is double the angle of BAD, the angle at the centre is double the angle at circumference.

We could do the same here with this shape.

The reflex angle, BOD, is subtended by a different arc, that's the arc BAD, that's the major arc on the left, but this arc also subtends the angle BCD, that's the angle that circumference on the right.

If we work out these angles, angle BCD is 54 + 72, which is 126 degrees.

The reflex angle, BOD, is 108 + 144, which is 252 degrees.

And once again we can see the reflex angle, BOD, which is the one at the centre, is double the angle, BCD, which is the one at the circumference.

Let's check what we've learned so far.

Here we have the same diagram.

Jacob thinks that AC is a diameter of the circle.

Now this is correct, what you need to do is explain why it's correct.

Now you can see that AC goes to the.

0, which is at the centre, but what you're truly trying to see here is whether or not these are two radii which look like they might be straight or whether it is a straight line.

Pause the video while you write down justification and press play when you're ready for an answer.

Let's take a look and example answer.

Let's start by thinking about what the diameter is.

The diameter is a chord that goes through the centre of the circle.

So that means if we can show that AOC is a straight line, it would mean that AC is the diameter, and if we're thinking about straight lines we should think about the fact that angles which form a straight line sum to 180 degrees.

So angle AOB + angle BOC is 108 + 72, that is 180 degrees.

Or you could do angle AOD + DOC, that is 144 degrees + 36 degrees, that is also 180 degrees.

And angles along a straight line sum to 180 degrees, therefore AC is the diameter of the circle.

So let's take a look at this diagram in a little bit more detail, and remind ourselves that the angle at the centre is twice the angle at circumference, if the angles are subtended by the same arc.

Jacob says, "The angle at the centre, angle AOC, is 180 degrees; and this is double the angle at the circumference, angle ADC, which is a right angle." And we can see as a right angle because one angle is 18 degrees, the other angle is 72 degrees, and when we find the sum of those we get 90 degrees.

Jacob also says, "The diameter AC forms a semicircle," and the a semicircle is half a circle.

And we can see the same with the angle at the bottom as well.

Jacob says, "The angle at the centre, AOC, is 180 degrees, and this has double the angle at circumference, which is angle ABC, and that's a right angle." Again, we can see it's a right angle because part of it is 36 degrees, part of it's 54 degrees, and then when we add them together we get 90 degrees.

Jacob says, "The diameter AC is the hypotenuse of a right-angled triangle, ABC." So let's check what we've learned with that.

Here we have two circular geoboards, one with 10 points, one with nine points.

Now triangle EFG is a right-angled triangle, What you need to do is show that triangle HIJ is not a right-angled triangle.

Pause the video while you do this and press play when you're ready to see an answer.

Okay, let's start by looking at triangle EFG.

We can see that EF is the diameter of the circle, as it passes through the centre of the circle.

The angle at the centre is 180 degrees, because it's a straight line, so angle FGE, the angle at the circumference, must be 90 degrees, because the angle at the centre is twice the angle of the circumference; therefore EFG is a right-angled triangle.

If we look at this one, hmm, well we can see that HI is not the diameter of the circle, as it does not pass through the centre of the circle.

So we need to use a different method to work out whether or not it has a right angle in it.

We can create isosceles triangles by drawing in the radii for the circle.

We could then work out the angles in the triangle.

We can do that by working out the one the centre first by doing a fraction of 360, in this case we get 80 degrees, and then we can subtract from 180 degrees and divide by two to get the other two angles, in this case we have 50 degrees and 50 degrees.

We can do the same thing again in the other triangles, and we can see that angle IJH is the sum of 30 degrees + 50 degrees, which is 80 degrees, and that is not a right angle.

Therefore, HIJ is not a right-angled triangle.

Now an alternative method could have been just to use these two radii here, and work out the angle at the centre, IOH.

That angle is 160 degrees because it's four ninths of 360.

And then we can think about how the angle at the centre is double the angle of circumference, if it is subtended by the same arc, which in this case it is.

Therefore, the angle at the circumference is 80 degrees, then it's not a right-angled triangle.

Okay, it's over to you now for task A.

This task contains three questions, and here is question one.

You've got a circular geoboard that contains six points equally spaced around a circle, and a point in the centre.

Three points in a circumference can be joined to create a triangle, and your job is to work out how many different right-angled triangles can be drawn when you use three points on this circumference.

And by different it means incongruence or unique triangles.

And part B, calculate the interior angles of the triangles you find.

Pause the video while you do that and press play when you're ready for question two.

Here is question two.

It's the same questions again, but this time you have a 12 point circular geoboard instead.

Pause the video while you do that and press play when you're ready for question three.

And here is question three.

This time you have a 15 point circular geoboard, but this time you need to explain why a right-angled triangle cannot be created on this circular geoboard.

Pause the video while you do that and press play when you're ready for answers.

Okay, let's take a look at some answers.

In question one there is only one unique right-angled triangle that can be drawn on this circular geoboard, using points on a circumference.

All other triangles you make will be reflections or rotations of this, and will be congruent to this particular triangle.

The hypotenuse triangle is the diameter of the circle, and then you need to work out the interior angles and you should get 60 degrees and 30 degrees, and 90 degrees, of course.

Then question two, there are three unique right-angled triangles that you can make.

The angles are 90 degrees, 75 degrees and 15 degrees.

Another one is 90 degrees, 60 degrees and 30 degrees.

Another one is 90 degrees, 45 degrees and 45 degrees.

All other triangles you make will be reflections or rotations of these three triangles, and will be congruent.

And for a 15 point circle geoboard, a diameter cannot be drawn by joining points on circumference of a 15 point geoboard, as there is an odd number of points, the chord will not pass through the centre and a semicircle will not be created.

Therefore, a right-angled triangle cannot be drawn.

In other words, you can't create a semicircle when you have an odd number of points on your circular geoboard.

Well done so far, now let's move on to the second part of this lesson where we're going to introduce a circle theorem more formally.

Here we have Jacob who noticed that the angle at the circumference subtended by the diameter is always 90 degrees.

You can see an example of that on this circular geoboard here, but Jacob is a curious fellow and he says, "Does this only work for drawing angles on circular geoboards or will it work for any circle?" Well, one way to investigate this is by using dynamic geometry software.

Doing so allows you to move the points around as much or as little as you like, so you can observe lots and lots of different examples.

And if you have access to the slide here, you can click on the link of the bottom of the slide to take you to a geometry file which allow you to do just that.

If you do so, it'll look something a bit like this.

Here we have a circle with a diameter, AC, and another point B on there with an angle at that point subtended by the diameter.

And in these examples we are looking at the relationship between the angle AOC, which is the angle at the centre, and the angle ABC, which is the angle at the circumference.

Jacob says, "Angle AOC is always twice the angle of ABC, no matter where I move point B on the circumference." For example, if we move it here we can see that the angle at the centre is 180 degrees and the angle at circumference is 90 degrees.

Same here as well, and same here as well.

Now AC is the diameter of the circle, so no matter where I move point A on the circumference, point C will always move so that AC remains as the diameter.

For example, if I move point A like so, point C moves as well.

Point B hasn't moved there, but what we can see it's at the angles have stayed the same.

We have 180 degrees at the centre, because it's a straight line, and we have at 90 degrees at the circumference because the angle at the centre is twice the angle of the circumference.

And we can keep moving it like so and see that it remains the case.

Jacob then says, "The diameter AC cuts the circle into two semicircle.

The angle ABC is described as the angle in the semicircle, and is 90 degrees." And we can see that no matter where we move the diameter, that angle in the semi-circle always remains as 90 degrees.

So let's check what we've learned.

Here we have a diagram with four points around a circle and a point in the centre, and you can assume that the line segment SOQ is a straight line segment and is the diameter of the circle.

So, in that case, in the diagram.

which angles are right angles? Pause while you choose and press play when you're ready for answers.

The answers are angle SRQ and angle QPS, those are right angles because they are in a semicircle.

Here's another question.

In which diagram is the chord XY also the diameter of the circle? Now the centre of the circle is unmarked, so you might need to think about something else that might help you decide whether or not the chord is the diameter.

Pause while you do this and press play when you're ready for answers.

Well, like we said, the centre of the circle is not marked so we can't see whether the chord goes through the centre and is therefore diameter, but what we can see is we have two angles in a triangle and we can work out the third remaining angle, and if that angle is 90 degrees we know that the triangle is right-angled and, therefore, the hypotenuse is the diameter.

For A the angle is 88 degrees, so no.

for B the angle is 92 degrees, so no.

for C the angle is 90 degrees, which means XY must be the diameter, because angles in a semicircle are 90 degrees and for it to be a semicircle xy must be the diameter.

So far we've observed lots of cases where the angle in a semicircle is a right angle, let's now prove that that is true for all possible cases.

And we can prove this using algebra and angle facts.

Jacob says, "If I draw in the radius OB, I split up the angle ABC into two parts, like this.

I'll label these as X degrees and Y degrees." Now, I can see that by drawing on the radius OB I also have created two isosceles triangles, and I can mark the equal sides of my diagram.

These three lengths are all equal to each other because they're all radii of the same circle, so we have two isosceles triangles.

I can then mark the other angles in the diagram.

We know that isosceles triangles have a pair of angles which are equal to each other.

So for the triangle on the left with the X degree angle, we know that another angle is X degrees, so OCB is equal to angle OBC, which are both X degrees because base angles and iso these triangle are equal.

We also know that angles in a triangle sum to 180 degrees, so we can write an expression for the third angle here by doing 180, subtract the two Xs, and that gives 180 degrees - 2x degrees for an expression for that angle.

We can then do the same for the other triangle on the right, the one the Y degrees in.

Angle OAB = angle OBA, which are both Y degrees, because base angles and isosceles triangle are equal.

We can then subtract those two Y degrees from 180 degrees to get an expression of 180 - 2y degrees for the angle at the centre with that triangle.

Now AC is a diameter and a straight line, and angles that meet a point on a straight line sum to 180 degrees.

And what we have is an expression for two angles that meet at a point on a straight line; 180 - 2x and 180 - 2y.

Those two angles must sum to 180 degrees, so we can write an equation.

So we have the sum of those two angles equals 180 degrees, because our Y term and our X term are currently negative we could make this a bit nicer by adding 2y and 2x to both sides to make them positive, and we'd have this.

We could simplify it by subtracting 180 from both sides to get this, and then we can see 180 = 2x + 2y.

So if 2x + 2y = 180, then we could factorised it and divide by two to see that x + y = 90, and x + y is the angle at circumference in this semicircle; and that therefore is equal to 90 degrees.

Therefore angle ABC is a right angle.

Jacob says, "So this theorem works for any angle in the semicircle." Huh, here we have Lucas.

I bet he's going to say what some of you have been thinking.

He says, "Why have you done all this work, Jacob? We can also prove this using another circle theorem which will be much faster.

Let me show you how." The other circle theme what we can use is the angle at the centre of the circle is twice to angle at circumference, if the angles are subtended by the same arc.

Well AC is the diameter of the circle, so the angle at the centre, angle AOC, is always 180 degrees.

Now this is always double the angle circumference, which in this case is angle ABC, which will therefore always be 90 degrees.

So once again we've proven that the angle in a semicircle is always a right angle, and you can prove that in multiple different ways.

In this particular way we've used one circle theorem to prove another.

Let's check what we've learned.

Here we have a diagram of a circle,.

0 is at the centre of the circle, points, K, L, M and N are on the circumference and can be joined to create a quadrilateral, a bit like this.

Now KM and LN are straight lines.

So what type of quadrilateral is formed here and how do you know? Pause the video while you do this and press play when you're ready for an answer.

The quadrilateral KLMN is a rectangle.

Now, you might have thought a rectangle by how it looks, but the important part here is proven how we know.

Well one way you can explain is that KM and LN are diameters of the circle, therefore diagonals of the quadrilateral are the same length and they bisect each other.

Now, this is a property of a rectangle.

But we can also prove this using a circle theorem, because a rectangle is a quadrilateral that has four right angles.

So we can show it has four right angles, we've shown that it's a rectangle.

Well, KM is the diameter of a circle, angles in a semicircle with right angles, therefore, angle KLM and angle KNM are both 90 degrees.

We can also look at the other diameter as well.

LN is diameter of the circle, which means angles in a semicircle are right angles and LKN and LMN are both 90 degrees as well.

All angles are 90 degrees, which is a property of a right angle.

Let's now look at how we can use this circle theorem.

If we know that the angler semicircle is a right angle, we can use this circle theorem to help us solve problems. For example, with this diagram here, let's find the area of this circle.

A problem here is that we don't know the length of the radius or the diameter, but we are given the length of two chords.

I wonder how we can use this information in order to help us work out the area of the circle.

We're going to work through this together in a moment, but perhaps pause the video while you think about what might do in this situation, and then press play when you're ready to see how to do it.

Let's take a look at this together.

ST is a diameter, so we know that KNM is 90 degrees because angles in a semicircle are always 90 degrees.

If the angle is 90 degrees, that means we are working with a right-angled triangle.

We have two lengths of the right-angled triangle, but we don't have the length of the hypotenuse, which is the diameter of the circle.

We could work it out, though.

We could use Pythagoras' theorem to work out the length of the hypotenuse, which will be 26 centimetres.

Therefore, if the diameter is 26 centimetres the radius must be 13 centimetres, and then we have the information we need to find the area of the circle.

We can do pi X 13 squared to get 169 pi centimetres squared, or you can convert it to a decimal if you want to.

Okay, it's over to you now for task B.

This task contains four questions, and here are questions one and two.

Pause the video while you do these and press play when you're ready for more questions.

Okay, and here are questions three and four.

Once again, pause the video while you do 'em and press play when you're ready for answers.

Okay, let's take a look at some answers now.

For question one, here are all your missing angles and also the justifications for why they are what they are.

Pause the video while you check against your own and press play when you're ready for more answers.

Then in question two you have to explain why the chord RU is not the diameter of the circle.

Well, one way you can do that is by working out the size of the angle that is subtended by that chord, the angle RSU, because if RU is the diameter then angle RSU will be 90 degrees, as the angle in a semicircle is a right angle.

Now there are lots of different ways you can calculate RSU, but let's take a look at one together.

We could use that given the interior angles in a triangle sum to 180 degrees, we can perform this calculation to work out TUV to be 37 degrees, and then we can use the fact that we have 72 degrees and 37 degrees in the triangle SRU, and work out the angle RSU to get 71 degrees.

And now we can see it's not a right angle, therefore RU is not the diameter.

Then question three, how to use a pair of compasses to construct a circle with AB as a diameter, to show that ACB is a right angle.

And you're told that M is the midpoint of AB.

Once you draw your circle, you should notice that point C is on the circumference of that circle, and if it's on the circumference of the circle, that means it's an angle in, a, a semicircle.

That means angle ACB must be 90 because angles in a semicircle is a right angle, or you could say the angle at the centre is twice as size the angle of circumference, AB is a straight line so angle AMB is 180 degrees and, therefore, angle ACB is 90 degrees.

You can use either of those circle theorems. Then question four, you have to find the area of the shaded region given the information of the diagram, and the fact that the radius of the circle is six centimetres.

The shaded region is a semicircle with a triangle cut out of it, so if we find the area of the semicircle and find the area of the triangle, we could perform a subtraction to find the area of the shaded region.

So let's do that.

The angle in a semicircle is always a right angle, which means EFG is 90 degrees.

Now we have a right-angled triangle, we can think about how we can do half X base X perpendicular height to find its area, but the issue is we don't know the length of EF, which is what's perpendicular to FG.

So we need to find that length first.

Well, the radius is six centimetres, which means the diameter is 12 centimetres so we have a right-angled triangle with a hypotony to 12 centimetres, one of its shorter size is 10 centimetres, and we can use Pythagoras' theorem to work out the length of the shorter side, which is 6.

6332 and more decimals centimetres.

We can use that to find the area of the triangle to get 33.

1662 and more decimals centimetres squared, and we can find the air of the semicircle by doing a half X pi X six squared, which gives us 18 pi centimetres squared.

And then we can find the air of the shade region by doing a subtraction with those two areas, and would get 23.

4 centimetres squared when rounded to three significant figures.

Fantastic work today, now let's summarise what we've learned.

We've been focusing on one particular circle of theorem, but we've also been using another.

When subtended by the same arc, the angle at the centre of the circle is always twice the angle at any point on the circumference.

That's one of the circle theorems we've been using to help us understand and learn about, and prove another circle theorem.

That's because when the angle at the centre is 180 degrees, then the angle at circumference is a right angle.

Therefore we can state the circle theorem, the the angle in a semicircle is a right angle and in order to use this theorem, you may need to draw a diagram or add more information to an existing diagram in order to do so.

Well done today, have a great day.