Lesson video

Hello, I'm Mrs. Lashley and I'm gonna be talking you through the lesson today.

I really hope you're willing to try your best.

And even if it gets challenging, remember I'm here to support you.

So our lesson outcome today is to be able to derive the formula for the area of any triangle.

So key words that we should be familiar with from previous learning are area, which is the area is the size of the surface and states the number of unit squares needed to completely cover that surface.

And also the sign of an angle.

And the sign of the angle is the Y coordinate of point P on the triangle formed inside the unit circle.

So this lesson on the area of any triangle is in two learning cycles.

In the first learning cycle, we're gonna be looking at the standard formula for the area of a triangle, and that's one you've used many times previously.

But when we get to the second learning cycle, we're going to look at using sign to find the area of a triangle instead.

So let's make a start at recalling and reviewing the standard formula for the area of a triangle.

The area of a triangle can be found by multiplying the base and perpendicular height of the triangle and then dividing by two.

And you've been using that formula for years and years and years.

But let's have a look at it just because it's gonna help us as we move through the lesson.

So if we've got this right angle triangle, we've got two perpendicular lengths.

So we can call one of them the base and one of them the perpendicular height and work out the area, pause the video and work out the area of this right angle triangle and then press play when you're ready to check that you've got that correct.

So the area would be three times four, which is our base times our perpendicular height, and then we're gonna divide it by two and that gives us six square units.

So it's sometimes easier to show the perpendicular height outside of the triangle.

So can you think of examples of triangles where you tend to see the perpendicular height noted outside of it? Pause the video, maybe sketch yourself a couple of different triangles.

And when would we normally label the perpendicular height outside of it? Press play when you're ready to look at a few.

So when we have an obtuse angle within a triangle and it's orientated in this way, we tend to draw our perpendicular height outside, but the area is still calculated in the same way.

So base times perpendicular height divided by two or half times base times perpendicular height.

And so for this triangle, the area is 17.

5 square units, but the base and the height just have to be perpendicular to each other.

And we tend to get ourselves into a bit of a pickle sometimes by using the word base and considering there always has to be the edge that is at the bottom of the triangle when actually in fact we just need two perpendicular lengths.

So if we orientate our triangle differently or if we look at it from a different, then our base and our perpendicular height can be two different values.

So here on the left hand side, we've got the base as that edge that the triangle is sort of sat upon.

And our perpendicular height is the height that's from the base to the highest point, that vertex.

Whereas if we know the edge length, if you look at the second copy of the diagram, if we know that other edge length, we can call that the base.

And then our perpendicular height is that distance from the base to the highest point which is the vertex opposite it directly.

And so if you can imagine just rotating that second image, our triangle will look slightly different despite the fact that it's exactly the same triangle.

And so we need to just be mindful that when we're calculating the area using half times base times perpendicular height, the base doesn't always have to be the edge that the triangle is sort of orientated to sit on.

So you just are looking for two perpendicular lengths.

So here is a triangle with four different measures.

So the base and height of this triangle measure to be which pair? Pause the video and when you're ready to check, press play.

So it's B, 14 and 17 are perpendicular measures to each other.

So we are gonna think of the 17 edge as our base and our perpendicular height is that distance from the base to the vertex.

12 millimetres is a perpendicular height of that triangle, but without the edge that it's perpendicular to, we wouldn't be able to use it for our area calculation.

18 millimetres is at one of the edges of that triangle, which we could think of as the base, but we do not have the perpendicular height to that base.

So we have to make sure we've got two lengths that are perpendicular.

So let's look at this example together.

If we are trying to calculate the area of this triangle, we've got four different measures, but which two can we use in our equation? Well, we need to have perpendicular ones.

So if I choose the 10, what measure is perpendicular to it? Well the five.

So our area will be 10 times five divided by two, which is 25 square metres.

Our base is that 10 metre edge, despite the fact that it's sitting on the nine metre edge without knowing the perpendicular height from that nine metre edge, we cannot use it.

So 10 and five are perpendicular.

So that is what we're going to use in our base times perpendicular height, just imagine rotating it.

So here's one for you, pause the video, work out the area of this triangle, and then when you're ready to check, press play.

The answer was 30 square metres.

Base, 10, perpendicular height, six.

So we can also use other strategies to help us find measures for the base and height when they're not that obvious.

So here we have an equilateral triangle, we know it's equilateral because of the hash marks.

We also have the edge length of the equilateral triangle, which is 50.

So we know that all of those edges are 50.

So Alex has added some additional information to our diagram.

He's added the altitude, which in an equilateral triangle is also a line of symmetry.

So we know it bisects the edge that it meets.

So that's where the 25 centimetre edges come from.

The 60 degree angle comes from the definition and the properties of equilateral triangles, which is that all angles are equal and therefore is 60 degrees.

So now we have this information, remember we're trying to calculate the area of the triangle.

Alex says, what methods could I use here? Pause the video and think about what methods that he could use in order to get the right information to calculate the area.

Press play when you're ready to look at what Alex decides to do.

So Alex decides to use Pythagoras' theorem because we can see a right angle triangle.

We have the hypotenuse and we have a shorter edge.

So we can make use of Pythagoras' theorem to work out the third edge of the right angle triangle, which in this case is our perpendicular height.

But is that the only method that Alex could use? Alex has also said sine of 60 equals H over 50.

Where's that come from? Well this is using our trigonometric ratios that using that 60 degree angle, the perpendicular height is the opposite.

The 50 is the hypotenuse.

So we can look at using the sine ratio to calculate the hypotenuse.

Sine of 60 is also one of our exact trigonometric values.

Can you remember what it is? And actually it was an equilateral triangle where we derived it from.

Sine of 60 is equal to root three over two.

So if we make sine of 60 equal root three over two, we can solve this to find H and we'll get 25 root three.

Alternatively, we could have thought about cosine.

Cosine of 30 degrees because we can calculate that the third angle in that right angle triangle is 30 degrees and then the 25 is the opposite to the 30, but the perpendicular height is the adjacent to the 30.

So then it would become cosine.

So cosine of 30 is equal to H over 50 and sine of 60 is actually equal to cosine of 30.

They're both root three over two.

So regardless of which method Alex chooses to use, we can work out that our perpendicular height is 25 root three as an exact value.

Once he's got the perpendicular height, then he can calculate the area of the equilateral.

We know that the base is 50, the perpendicular height is 25 root three, so we can do 50 times 25 root three divided by two, and then we'd get 625 root three square centimetres.

So here's the first question of task A.

And for question one, you need to write an expression for the area of each triangle.

So pause the video and write down on expressions.

You're not calculating the area as a numerical value, you're writing an algebraic expression for the area.

Press pause, work out question one and when you're ready for the next question in task A, press play.

So here's the next question, question two, and you need to calculate the area of each triangle on the screens.

There are four triangles that you need to calculate the area for, pause the video and do that.

When you press play, you've got question three to work on next.

So question three, calculate the area of each triangle.

So consider what you know in each diagram, what information is being told to you explicitly and what also is implicit there, what information you can work out.

Press pause whilst you do that.

And then when we press play, we'll go through our answers to task A.

So on question one, you were writing an expression for the area.

And so this was identifying which two lengths were perpendicular.

What was your base and what was your perpendicular height? So on part A, it was a right angle triangle, so you needed to choose the perpendicular edges.

So B times D divided by two is an expression for the area.

If you've written half BD, that's equivalent, so correct.

And if you've written DB divided by two or half DB, then remember algebraically we tend to put things in alphabetical order, but that is an equivalent expression.

On question part B, we needed to identify our base and our perpendicular height.

So our perpendicular height was eight units and the base or the edge that was perpendicular to that measure was 9X.

So 9X times eight divided by two.

So the expression in simplified form would be 36X.

Question two in that purple box is gonna be the answer for each question as we go through them.

So on part A, the two perpendicular measures were five and eight.

So a half times five times eight gives us 20 square centimetres.

On B, it was the 10 and the four, so a half times 10 times four is 20 square centimetres.

On C, right angle triangle.

So it was the six and the eight, the two perpendicular edges, a half times six times eight is 24 square centimetres.

And on D, it was seven and the 8.

4, the 2.

4 was a perpendicular measure to the 8.

4 edge.

However, it was not the perpendicular height, it didn't go to the highest point of the triangle.

So a half times seven times 8.

4 is 29.

4 square centimetres.

Question three, the first two triangles are here on the slide.

So question three, the first one was an equilateral triangle.

So remember you had quite a few options in how to get the perpendicular height.

Pythagoras' theorem using sine of 60 or using cosine of 30 were the ways that Alex sort of spoke us through.

So here I've used Pythagoras' theorem to get the perpendicular height.

So that would be the square root of 10 squared minus five squared and then I'm gonna multiply it by 10 and divide it by two.

And that gives us 25 root three.

So the perpendicular height was five root three, however you may have calculated that.

And then we've got our base, our perpendicular height and divide it by two.

On the second one it was an isosceles triangle.

So we can split the triangle into two right angle congruent triangles by again dropping the altitude, which would be the line of symmetry because it's an isosceles, that means that the base edge could be bisected into two four centimetres.

So we can think about one of the right angle triangles to get the perpendicular height.

A three, four, five triangle is very famous for Pythagorean triple and you can use Pythagoras to calculate that.

But the perpendicular height would be three.

So then the base times the perpendicular height divided by two gives you an area of 12.

The final one was inside of a circle, so it was a triangle that would also be isosceles.

And we know it's an isosceles triangle because two of its edges are radii of the same circle because we were told that top angle of the triangle was 120 degrees and we can deduce that it's an isosceles, then we can split it once again into two right angle triangles.

So the answer is important.

So did you get four root three as an exact value? Maybe you rounded that.

But four root three is the exact value.

If you've got that, well done.

But there's a chance that you didn't do it in the same way that the solution is on the screen.

There's often multiple ways to do mathematical questions.

So this solution on the screen is using base times perpendicular height divided by two where the perpendicular height is the two and the base is two times the square root of four squared minus two squared.

And that is the length of the chord on that circle.

You may be thinking, why is there the equilateral triangle? And the equilateral triangle is useful as we've already established to work out the length of half of the cord, which is the perpendicular height of that equilateral triangle.

So we can use Pythagoras' theorem to work out the perpendicular height and then double it to get the length of the cord.

And we can see that the perpendicular height of the isosceles triangle is two because it is half of the edge length of the equilateral triangle.

Alternative ways is you could have worked out the perpendicular height using the trigonometric ratios.

'Cause if you know that's 60 degrees and that's the adjacent to the 60 degree angle, you could use cosine of 60 degrees is equal to the perpendicular height over four and then that would come out as two.

You could then use Pythagoras' theorem to work out half of the cord length and then double it with the four and the two.

So multiple ways of working with trigonometric ratios and Pythagoras' theorem to calculate the area of the triangle.

So in this next learning cycle, we're going to look at using the sine function to find the area of a triangle instead.

So let's have a look at that.

So we're now looking at using sine to find the area of a triangle.

So for non-right angle triangles, we can sometimes use Pythagoras to find the area.

And we've seen that when we looked at the equilateral triangle.

So here we've got a non-right angle triangle.

So angle A, angle B and angle C are not a right angle, they're not 90 degrees and therefore we can't just do base times height divided by two using two of the edges.

So we're trying to look at a way of finding the area for non-right angle triangles.

So here we've got the edge AC is partitioned into an edge of three and an edge of nine.

So we know that the edge AC is 12 in total.

So that is the altitude that's been drawn there.

So it's from the vertex B and it's perpendicular to the edge AC.

We know that this is not an isosceles nor an equilateral because it does not bisect the edge that it meets.

And that's a common misconception to always think that when we drop a perpendicular in a triangle it will bisect the edge and that only happens if it's an equilateral or an isosceles triangle.

So if we focus on this right angle triangle that the altitude has created, then what can we say with Pythagoras' theorem? Because we have got that the edge is three, we can work out the length which is the perpendicular height.

So we can do five squared minus three squared and square root it, which will give us four.

We'd then be able to calculate the area because we'd say that the perpendicular height is four and we know that AC, the base edge is 12, so we can do 12 times four and divide it by two.

Sam though has pointed out what if we can't work out where B is partitioned? So if we drop the perpendicular from the vertex B, if we draw an altitude, as I've said, it doesn't always bisect, but how do we know where it has met that line? Sam recognises that we can actually make use of trigonometry to help us.

So we use Pythagoras when we did know where it bisect, we did know where it partitioned it, but if we don't know that we can make use of trigonometry.

So which formula calculates the area of triangle ABC? Don't overthink this, just look at the diagram, think about the information you need to be able to work out the area of triangle.

Which of those three would calculate the area? Press pause and then when you're ready to check, press play.

So it would be C.

We need two perpendicular lengths, the base and the perpendicular height.

We can say the base is the length B and the perpendicular height is the length P.

So it's a half times P times B.

We're gonna consider how the trigonometry can help us now.

Let's focus on the highlighted right angle triangle.

So what do we know about this right angle triangle that's been highlighted? Well we know the hypotenuse is A, we know that P is opposite the angle at C.

So we can say that sine of C, the angle at vertex C is equal to P over A because remember that the sine ratio is opposite over hypotenuse.

P is the opposite to that angle.

And A is the hypotenuse in the right angle triangle.

And we can rearrange this to say that P is equal to A times sine of C.

Why is P important? Well P is important because that is our perpendicular height of this triangle.

And to work out the area of a triangle, we need base times perpendicular height before we divide it by two.

So now we've got an expression for the perpendicular height.

So if we take our formula for the area of ABC that was part of your check a half times P times B, a half times base times perpendicular height, then we can substitute our expression for P and we get a new formula a half times A times B times sine of C.

So it's not a right angle triangle.

And as long as we've got the edge A and the edge B and the angle C, we can calculate the area.

So let's have a look at an example of this.

So we've got a seven centimetre edge, we've got an angle of 32 degrees and we've got an edge of 14 centimetres.

So using this new formula for the area of a triangle, we'd do half times seven times 14 sine of 32 and we'd evaluate that and calculate that on our calculator.

That's 25.

97 square centimetres to two decimal places.

So just be mindful that despite the formula being explicit with A, B and C, it just actually means that you need two adjacent sides and the angle between them.

So if you've got these two edge lengths and that angle that I've marked as theta, you can use this formula, you can label one of them to be lowercase A, one to be lowercase B, and the vertex would be C, as long as you've got two sides and the angle between them.

So if we had this side and the base edge and the angle that I've labelled as beta, you could still use the formula because you've got two adjacent sides and the angle between them and hopefully you can see that this means if you have this, you've got sufficient information to use this formula as well.

So two sides that are adjacent to each other and the angle that is between them is all you need to be able to use this formula even if the angles and the labelling is not A, B and C.

So for this check, for which triangles could you use the sine formula to calculate the area? Pause the video and have a look at that.

And then when you're ready to check, press play.

So you could only use the sine formula for triangle A because you've got two edges and the angle between them.

On B, you've got two edges and an angle but you haven't got the angle between them and that's the same for C.

On task B, you've got three questions, on question one, there are six parts.

So for which triangles can the area be calculated using the sine formula? You can calculate it if you want to, but for this question it's just identifying which triangles you can make use of it and which you cannot.

So pause the video and then when you're ready for the next question, press play.

So question two, find the area of each of these triangles to one decimal place and three of them there is a diagram to help you.

And for part D, it says using the sine formula, calculate the area of an equilateral triangle with side length of two.

So pause the video whilst you work through question two.

When you press play, you've got the last question of task B.

So on question three, what is the area of each of these triangles in a regular pentagon? Given that the edge of the pentagon is two centimetres.

So pause the video and then when you're ready to go through your answers to task B, press play and that's what we'll do.

So on question one, you just needed to identify out of the six how many you could use the sine formula to work out the area and that was B and F.

You were looking to have a pair of edges and the angle between them and it was only on B and F that you had that.

Question two, you needed to calculate their areas.

So for 2A, you've got two edges, the angle between them.

So you're gonna do a half times four times six times sine of 37 and that gave you 7.

2 to one decimal place.

On part B, you've got two edge lengths and the angle between them, so a half times five times seven times sine of 28 degrees gives us 8.

2 square centimetres to one decimal place.

And lastly, we've got three centimetres, 12 centimetres.

They are adjacent sides of the triangle.

The angle between them is 104 degrees, so a half times three times 12 times sine of 104 gives us 17.

5 square centimetres.

And then the last question was using the sine formula, calculate the area of an equilateral triangle with side length two centimetres.

And so we need an angle between the edges and we know that will be 60 degrees because it's an equilateral.

So a half times two, times two times sine of 60 gives us 1.

7 square centimetres to one decimal place.

So question three, what is the area of each of these triangles in a regular pentagon? Well the way that it was divided up gave us two congruent triangles and one different to that.

There was two different triangles in there.

We know that it's a regular pentagon and that information gives us quite a lot.

So there was one edge marked as two centimetres, but actually that means all of the edges are two centimetres.

Because it's a regular Pentagon, we can calculate that the interior angle would be 108 degrees.

So we know that the total interior angles is 540 and you could derive that by the triangles.

So each of those triangles has 180 degrees and there are three triangles, so that's 540.

Then we can divide that by five to share that evenly, which tells us that an interior angle is 108 degrees.

So if we look at the first triangle on the left, it's an isosceles triangle because it's got two of the edges, the external edges which we know are the same length because it's a regular pentagon.

And so we can work out the other two angles within that triangle are both 36 degrees.

You can use the sine formula to calculate that area.

So a half times two, times two, times sine of 108 gives us 1.

9 square centimetres to one decimal place.

And so each of the smaller triangles, the left one and the right one are identical.

So both of them are 1.

9.

Now we've got that triangle in the middle.

This triangle is also isosceles.

And we can again use the angles that the interior angle being 108, if we know that that part is 36 degrees, that means this other part is 72 degrees and we know that the other triangle is congruent.

So it would also be 72 degrees.

So it's an isosceles.

But at the moment, we do not have enough information to get the area because we need two adjacent side lengths.

So we've got all three angles, but we need two adjacent side lengths.

We've got one edge which is the two centimetre.

So really we're looking to try and calculate the third edge of the other triangle.

So this is where you may need to use a few strategies to get this.

So the central triangle, if we look at splitting our isosceles into two right angle triangles, we can then use trigonometry to get the length of half of that edge because two is the hypotonus and 36 is an angle within that right angle triangle.

So cosine of 36 equals the adjacent divided by two and the adjacent is half of the edge length of that central triangle, which means that we can rearrange that to say that the adjacent is two cosine 36 and the full edge length would therefore be 2A, two lots of that adjacent because we can see that the isosceles triangle, that first isosceles triangle has been split into two equal parts and now we have an value is an exact value of the edge length of that other triangle.

So we can calculate the area, half times two times four cosine 36 times sine of 72 gives us 3.

1 square centimetres to one decimal place.

So our two is one of our edge, the four cosine of 36 is the other edge and the angle of 72 degrees is between them.

So to summarise today's lesson on the area of any triangle, so the formula for the area of a right triangle is a half times base times perpendicular height.

With an equilateral or isosceles triangle, you can use Pythagoras' theorem to find the height because we can split it into two congruent right angle triangles.

If you know one of the base angles, you can use the sine ratio to find the perpendicular height.

And this leads to the formula half AB sine C.

So you can make use of the formula half AB sine C for any triangle, even a right angle triangle because sine of 90 is equal to one and then we just get a half AB where A and B or perpendicular.

Well done today and I look forward to working with you again in the future.