Lesson video

Hello there.

You made a great choice with today's lesson.

It's gonna be a good one.

My name is Dr.

Rowlandson, and I'm gonna be supporting you through it.

Let's get started.

Welcome to today's lesson from the unit of 2D and 3D shape with compound shapes.

This lesson is called checking and securing understanding of area for standard shapes.

And by the end of today's lesson, we'll be able to calculate the area of rectangles, triangles, parallelograms, and trapeziums effectively.

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

So you might want to pause the video if you want to remind yourselves what these words mean before pressing play to continue.

This lesson contains three learning cycles.

In the first learning cycle, we're going to start by finding the areas of rectangles and parallelograms, and then we'll find the areas of triangles and trapezia.

And finally we're going to push ourselves further by looking at problems where we have to apply other aspects of mathematics before we can find the area.

Let's start off with areas of rectangles and parallelograms. Area is the amount of 2D space that is taken up by a shape and finding the area of a shape always involves multiplying together two perpendicular lengths somewhere within the calculation.

Now it might not always be obvious where that multiplication of perpendicular lengths is, but somewhere within the formula it will be there, even if it's quite subtle, and areas measured in square units.

Here are some examples of some units have found in the area.

The top row we have metric units, those are square kilometres, square metres, square centimetres and square millimetres.

But you may pronounce those as kilometres squared, metres squared, centimetres squared and millimetres squared means the same thing.

On the bottom row we have some imperial units for area, square miles, square feet and square inches, but there are other units as welfare for area.

Here we have a parallelogram and a rectangle, which are drawn on a grid.

Aisha, Jun and Sam are trying to decide which shape has the greatest area.

Let's hear from 'em.

Aisha thinks it's the parallelogram that has the greatest area, Jun thinks it's the rectangle, and Sam thinks that the areas are equal.

Who do you agree with? Pause the video while you think about this and press play when you're ready to continue.

It's Sam here who is correct.

These two shapes do have the same area.

Let's take a look at why.

The parallelogram and the rectangle both take up the same amount of 2D space.

And one way we can see that is by taking all of the squares from the parallelogram on the left and transferring them into the rectangle or taking some of the partial squares and seeing how they fit together in the rectangle.

Let's do that now and see what happens.

Both of these shapes take up the same amount of space.

They both take up six squares worth of space.

That means they have the same area.

The areas of the parallelogram and the rectangle can both be calculated by multiplying the base by its perpendicular height.

This can be written as a formula, so the form for calculating each area for these two shapes can be written as A equals bh.

So in this case we can see that both shapes have a length of two for its base, and a length of three units for its height.

So the area would be two multiplied by three, which would be six, and we don't know what the units are in particular here, but we can say they are square units.

Let's take a look at an example of finding the area of a parallelogram together, and then you'll go try one yourself in a very similar way.

Calculate the area of this parallelogram.

So we have a parallelogram where we've got three lengths given to us.

We've got the base which is 20 inches, the height which is six inches, and the length of its slope, which is eight inches.

Now, we will not use all three of those.

We need to choose the right ones when we find the area.

The formula is area equals base times perpendicular height.

Well the base is 20 inches, and the perpendicular height is six inches.

The eight inches, that's not the height, and it's not perpendicular to either of the other two lengths, so we're not using the eight.

The area is 120 square inches.

Here's one for you to try.

Calculate the area of this parallelogram, pause the video while you do this, and press play when you're ready for an answer.

Here's your answer.

You'll do 12 multiply by 12 to get 144 square metres or metres squared.

Here we have Lucas.

Lucas is trying to find the area of this parallelogram.

Lucas says, I can see that the base is 11 centimetres, but I don't know the height, so I'm not sure what to do.

Is it possible for Lucas to find the area of the parallelogram using the information that is provided here? Pause the video while you think about this and press play when your ready to continue.

Here's what Lucas saw.

Lucas creates a congruent parallelogram and rotates it so that the sides with length nine centimetres are horizontal.

Now these two parallelograms are congruent, so they both have the same area.

And now when we look at the one on the right, it's a bit more clear how we might go about finding the area of it.

Lucas says, "I could do nine multiplied by 10.

The lengths multiplied together do not need to be horizontal and vertical, they just need to be perpendicular to each other.

And if you're ever unsure, can always turn the page around so that you have lens which are horizontal and vertical if that helps you see it more clearly.

Let's do another example together and then you'll try one yourself.

Calculate the area of this parallelogram.

We have three measurements given to us.

We need to choose the right ones for our formula and what we need are two lengths which are perpendicular to each other.

Now one of the sides of this parallelogram is 20 feet long, but we do not have a perpendicular height in relation to that side.

The other side is 10 feet long and we do have a perpendicular height from that one, which is 15 feet.

That's 10 and a 15 are perpendicular to each other.

So let's use those.

The area is 10 multiplied by 15, which is 150 square feet.

Here's one for you to try.

Calculate the area of this parallelogram, pause the video while you do that and press play when you are ready for an answer.

Here is your answer.

Your two perpendicular length are seven and 10, and that will give you 70 square centimetres or centimetres squared.

Here's another question for you, which two calculations could be used to calculate the area of this parallelogram? Pause the video while you make a choice and press play when you're ready for answers.

The answers are A multiplied by C, and B multiplied by D.

In both of those situations, the two lengths that you're multiplying together are perpendicular to each other.

One of them is the length of a side, and the other one is the distance between that and its parallel side.

Over to you now for task A, this task contains two questions, and here is question one.

You need to calculate the area of each of these shapes and all the lengths are given in centimetres, so your units for area will be centimetres squared.

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

And here is question two.

You'll have a parallelogram where the area is 24 square centimetres and the perimeter is 22 centimetres.

Lengths b, h and L are all integers.

And what you need to do is find a possible set of values for b, h and L.

Now there is more than one solution to this problem, so if you do find one set of answers, maybe push yourself to try and find another.

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

Here are your answers to question one, pause and check against your own and then press play to continue.

And question two, we defined possible values for b, h and l where they are integers.

Well, here are two possible solutions you can have.

Great work so far.

Now let's move on to the next part of the lesson, which is all about finding the areas of triangles and trapezia.

Here we have that parallelogram and rectangle again, which both have areas of six square units.

Sophia cuts each shape along its diagonal to make triangles, and it looks a bit like this.

Sophia says, "Each triangle takes up half the amount of space as its original quadrilateral.

That means the area of each triangle is three square units.

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

This can be written as a formula A equals bh over two, and in this case we would have two multiplied by three divided by two to give three square units.

This foam can also be written as A equals half bh.

They both give the same answer.

Once again, the lengths multiplied do not need to be horizontal and vertical, but they do need to be perpendicular to each other.

Here are some examples.

On the left you have the base which is horizontal and the height which is vertical.

But on the right you have a base and height where neither 'em are horizontal or vertical, but they are perpendicular to each other.

And if you're ever unsure you can always turn the page around so that the base is horizontal from your perspective and the height will be vertical in that situation.

Let's find the area of a triangle together and then you'll try one yourself.

Calculate the area of this triangle.

We have three different measurements given to us.

We will not use all three of those.

The formula is half times base times height and we need to find two lengths which are perpendicular to each other for the base and the height.

Well we have that seven metres on the bottom of that triangle and that usually is where our eyes are drawn when it comes to the base, but we don't know the height in relation to that base.

So let's look at the five metres, which is a different side of that triangle.

We do have a height that is in relation to that one.

Six metres is perpendicular to the five metres, so let's use the five and the six.

We have half times five times six which gives 15 square metres.

Here's one for you to try.

Calculate the area of this triangle, pause the video while you do this, and press play when you are ready for an answer.

Here is your answer.

You do a half multiplied by eight multiplied by 10, 'cause those two lamps are perpendicular to each other and that will give 40 centimetres squared.

Here's another question for you, which calculations give the area of this triangle? You've got four options to choose from, and it may be more than one correct answer.

Pause the video while you do this and press play when you are ready to see what the answers are.

The answers are A and C.

You could either do 15 multiplied by six and that comes from doing a half times 15 times 12 where we've half the 12 or you could do nine multiplied by 10, and that comes from doing a half times 18 times 10 where you have halved the 18 this time to get nine.

They both give the same answer and that'll be 90 millimetres squared.

So what about areas of trapezium? Well, one way you can find the area of a trapezium is by splitting it into other shapes.

For example, you could split it into three shapes a bit like this, a triangle, a rectangle, a triangle.

We could find the area of each shape.

Shape A.

The base is one unit long, so the area would be a half times one times three, which is 1.

5.

The area of shape B is a rectangle where the base would be two, so we do two times three, now get six.

And the area of shape C, it's a triangle, that base would be three units, so it'd be a half times three times three, which is 4.

5.

Now, we have the area of all three parts of the trapezium.

We can add them together to find the total area, and that would give 12 square units.

We don't have to split it this way though.

There are plenty of different ways we could split the trapezium up.

We could split it into a parallelogram and a triangle a bit like this.

We can then find the area these two shapes.

The parallelogram would be two multiplied by three to get six.

The triangle would have a base of four units, so we'll do a half times four times three to get six as well.

Add those together and we get the same answer again, 12 square units.

We could even split the trapezium up like this into two triangles.

The area of triangle A would be a base of six multiplied by a height of three, multiplied by a half to get nine.

And the area of shape B would be two multiplied by three and multiplied by a half, and it would give you three, and once again, if we add those together, we get the same answer again, 12 square units.

Now for each of those methods we used, we could generalise our calculations by using algebra instead of specific numbers.

For example, let's generalise the last method that we used, which was this way of splitting up the trapezium here.

Both of these triangles have the same height, it's the height of the trapezium, so let's call that h.

Now each triangle has a different base, so we can't use the same letter for each base, let's call one base A for shape A and other base B for shape B.

And then if we find the area of triangle A, we'll do a half multiplied by a, the base, multiplied by h, the height to give half ah.

If we find the area of shape B, we'll do a half multiplied by its base B multiplied by once again the height, h, and that would give a half bh.

So to find the area of the trapezium, we would add those two answers together to get half ah plus half bh and then we have a half and ah is a factor of both of those terms. So we could factorised it if we want to neaten this up and that would give a half multiplied by the sum of a and b multiplied by h.

You could write the h between the half and the brackets if you want to, but the way you can see it there is the most conventional way of seeing a formula for the area of trapezium.

The area of a trapezium can be found by using the formula A equals half multiplied by the sum of a and b multiplied by h where A and B are the lengths of the parallel sides.

For example, if were to use this formula on this trapezium would do a equals half multiplied by the sum of a and b multiplied by H as our formula.

If we substitute in our numbers, we'd have half multiplied by the sum of six and two multiplied by three where six and two are our parallel sides and three is the height or the distance between those parallel sides and that would give 12 square units once again.

So let's have a go at an example together and then you'll try one yourself.

Calculate the area of this trapezium.

Well here's our formula, let's choose the right numbers now.

a and b represent the parallel sides.

That will be eight and three and h is the height which is perpendicular to those sides, which would be four.

So here's our formula with the numbers substituted in and our answer is 22 centimetres squared.

Here's one for you to try, pause the video while you do this and press play when you are ready for an answer.

Here's your answer.

Your parallel sides are three and seven centimetres, the height is five centimetres, so you do half times a sum of three and seven times five to get 25 centimetres squared.

Here's another question for you, which calculations give the area of the trapezium.

You can see on the screen, and there may be more than one answer.

Pause a video while you choose and then press play when you're ready to go through it together.

Let's take a look at these together.

A would not work, eight and six are not parallel to each other and seven is not perpendicular to the eight.

B would work seven and two are parallel to each other and six is the perpendicular distance between them.

The calculations in C would work, and they are equivalent to if we split the trapezium up into two triangles like you can see on the screen here, one part of the calculation is the area one triangle, the other part is the area of the other triangle, but if you factorised the half and six out of those two, you would have the same calculation as you have in B, a half multiplied by six, multiplied by the sum of seven and two.

And D would also work, it's equivalent to if we split the trapezium up into a rectangle and a triangle.

Okay, it's over to you now for task B, this task contains one question and here it is.

All these shapes have the same height and the lengths which are labelled are given in centimetres.

You need to calculate the areas of the shapes and then consider the question in part B, pause the video while you do this, and press play when you're ready to go through some answers.

Let's check how we got on.

Here are your answers to part A, pause and check against your own, and then press play to continue.

In part B.

Alex thinks it's possible to use the formula A equals half a plus b times h, which is the formula for trapezium to calculate the area of all the shapes.

Is Alex correct? Explain your reasoning.

Well this formula is for the area of a trapezium, and we can see we have three trapezia up in part A, so that formula could be used for those three shapes in the middle.

On the right we have a parallelogram.

Now this formula could be used to calculate the area of the parallelogram by substituting four for both a and b.

On the left we have a triangle.

This formula could also be used to calculate the area of the triangle by substituting either a equals zero or b equals zero.

Fantastic work so far.

Now let's push ourselves even further by looking at some problems where we may need to apply other aspects of mathematics before we can find the area.

Sometimes you may need to use knowledge from other aspects of mathematics to help you find the area and here's an example.

Laura's trying to find the area of this parallelogram.

She says, "I can see the base is 15 centimetres but I don't know the height, so I'm not sure what to do.

We do have an angle there though, that might help us.

How could Laura obtain the information she needs to find the area? Maybe pause the video while you think about this and press play when you're ready to work through it together.

The problem here is that we don't know the perpendicular height in relation to any of those sides.

So Laura draws a perpendicular height up to the vertex in the top left of the parallelogram and if she knew how long that new line segment was, that would be the height of the parallelogram and she'd be able to find the area.

Laura says, "I now have a right angle triangle with an angle and the hypotenuse, so I could use trigonometry to find the height." If we did that, what we're trying to do is find this length here which we can call X.

If we look at that right angle triangle on the left of the parallelogram.

We can see that the 12 centimetres would be the hypotenuse and the X would be the opposite, because it's opposite the 60 degree angle, and because we have the opposite and hypotenuse, we could use the formula sin theta equals opposite divided by hypotenuse and we substitute our numbers in.

We'd have sin 60 degrees equals x over 12.

We could rearrange that to get x equals six, route three.

We could write that as a decimal but it'd be more accurate to keep it in form until we get to our final answer.

Laura says, "Now I know the base, which is 15 centimetres and the perpendicular height which is six route three centimetres.

I can calculate the area.

She can do 15 multiplied by six route three to get 155.

9 centimetres squared if we round it to one decimal place.

Laura says, "I didn't necessarily need to find the value of x, I could have just used 12 sin 60 degrees for the height 'cause 12 sim 60 degrees is the calculation that gives you the height.

So instead of those last two steps, what Laura could have done is multiply 15, which is the base by 12 sin 60, \which is the calculation that gives the height and that would give the same answer of 155.

9 centimetres squared.

Let's do an example together, and then you can try one yourself.

Calculate the area of a parallelogram accurate to one decimal place.

Well we don't have a perpendicular height in relation to either of those sides, but we do have that 40 degree angle so we could draw a line segment that is perpendicular to the 60 metres and goes up to that top right vertex of the parallelogram.

We can now see a right angle triangle.

The hypotenuse is six and the opposite is the unknown we want, x.

We could substitute those into our formula to get sin 40 degrees equals X over six.

Rearrange it to get x equals six sin 40 degrees and we could work that out and multiply it by 16 or we can just do 16 multiplied by six sin 40 and that would give us the area of 61.

7 metres squared to one decimal place.

Here's one for you to try.

Pause the video while you have a go at this and press play when you are ready for an answer.

Here's your answer.

You would use eight sin 70 degrees to get the height, which is x, and multiply it by 10 to get 75.

2 centimetres squared to one decimal place.

Here's another example.

Sophia is trying to calculate the area of this triangle and once again we don't have the height but we do have an angle.

Sophia says, "I can see that the base is 15 centimetres but I don't know the height so I'm not sure what to do." How could Sophia obtain the information that she needs in order to find the area? Pause the video while you think about this, and press play when your ready to continue.

Well, Sophia could do a similar thing to Laura earlier.

She could draw a perpendicular height from the base to the vertex, the top of the triangle.

Now what we can see here is we have two right angle triangles and within one of those triangles we have a few measurements.

Sophia says I have a right angle triangle with an angle and the hypotenuse, that's a triangle on the left so I could use trigonometry to find the height.

If we did that, we could label the height x, and in that left hand right angle triangle, the hypotenuse would be 12 centimetres, the opposite would be X centimetres, and we have an angle of 60 degrees so we could do a similar thing again.

Sin of 60 degrees equals x over 12, rearrange it to get six route three, so our height of our triangle is six route three, and Sophia says, now I know that perpendicular height I can calculate the area.

That would be by doing a half times base times height, so half times 15 times six route three is 77.

9 centimetres squared to one decimal place.

Now similar to earlier, Sophia says I didn't necessarily need to calculate the value of x, I could have just used 12 sin 60 degrees for the height.

So instead of those last two steps, we could have done a half times 15, which is the base, multiplied by 12 sin 60, which is the calculation that gives the height, and then the whole calculation would give 77.

9 centimetres squared again.

Here's Jacob.

Jacob is trying to find the area of this triangle and this one is slightly different.

Jacob says, "I don't know the height but I also don't know any of the angles, so trigonometry won't help me here." He's not sure what to do.

How could Jacob obtain the information he needs here to find the area, and if he's not going to use trigonometry in this case because we don't have any angles, maybe something else from another aspect of mathematics might help him.

Pause the video while you think about this, and press play when you are ready to continue.

Similar to the others, Jacob draws a perpendicular height from the base to the vertex at the top of the triangle.

We want to know how long that line segment is because that's the height of the triangle.

Jacob says, "My new line segment splits the base in half because the triangle is isosceles." This triangle is symmetrical around that line segment we just drew, which means the length of the base on the left triangle is the same as the length of the base on the right triangle.

That only works because it's is isosceles, so that means the base of the triangle on the left is half of the total base, so that's 7.

5 centimetres.

Jacob looks at triangle on the left hand side, and says, "I have a right angle triangle where two of the lengths are known so I could use Pythagoras's theorem to find the height." It does it like this.

We have x squared plus 7.

5 squared equals 12 squared.

We simplify and rearrange, and we get x equals 9.

3674, and there are more decimals.

Jacob's not go around it here.

He's gonna use the entire of that number on his calculator in his next calculation.

He says, "Now I know the base and perpendicular height, I can calculate the area." It does half times 15 times the entire of that number 9.

3674 and so on, and that gives 70.

3 centimetres squared to one decimal place.

Let's do an example and then you can try one yourself.

Calculate the area of the triangle, and give your answer accurate to one decimal place.

Now this triangle is isosceles, which means if we draw the perpendicular height on, we would have a base of seven centimetres for each of those two right angle triangles.

And then if we want to find the height in that right hand triangle, we have two lengths, we have seven centimetres and eight centimetres, we could use Pythagoras's theorem.

If we rearrange and solve, we get a value of route 15 for our perpendicular height.

That means we could do half multiplied by 14 multiplied by route 15 to give 27.

1 centimetres squared to one decimal place.

Here's one for you to try.

Pause the video while you give it a go and press play when you're ready for an answer.

Here's your answer.

You would use Pythagoras to get a height of six centimetres and then you would do a half times 16 times six to get 48 centimetres squared.

Let's work through another example together and then you can have a similar one to try yourself.

Calculate the area of this triangle, give your answer accurate to one decimal place.

Now this is not isosceles, which means when we draw our perpendicular height and split it into two right angle triangles, we don't actually know how long the base of either those triangles are, because it's not splitting that base exactly in half, but what we do have is that 50 degree angle, which can help us, because we have a right angle triangle with an angle and a length, and we're trying to find another length.

We could use trigonometry, we could do sin of 50 degrees equals x over eight, rearrange it to get X equals eight sin 50 degrees.

Now you could work that out or you could substitute that calculation into the formula for the area of a triangle and do a half multiplied by the base 14 multiplied by the calculation for the height eight sin 50 degrees and that will give you 42.

9 centimetres squared.

Here's one for you to try, pause a video while you have a go, and press play when you are ready for an answer.

Here's the answer.

You would do 10 sin 30 degrees to get the height, and then for the area a half multiplied by 16 multiplied by 10 sin 30 degrees where 16 is the base and 10 sin 30 degrees is the height and that'll give you 40 centimetres squared.

Okay, it's over to you for one last time now for task C, this task contains one question, and here it is, find the area of each shape, and round any decimal answers to one decimal place.

Pause the video while you work through this, and press play when you're ready for answers.

Right, let's see how we've done.

Here are the answers, pause and check these against your own and then press play to continue.

Fantastic work today.

Now let's summarise what we've learned in this lesson.

Formulae and logic can be used to find the area of standard shapes.

By logic we mean splitting the shapes into smaller shapes are easier to find the area of and then adding them together.

The formula will always include two perpendicular lengths being multiplied together, and it might not be obvious where that is, but it is in there somewhere, even if it's quite subtle.

The formula for the area of a trapezium is A equals half multiplied by the sum of a plus b multiplied by h, where a and b are the parallel sides, and h is the distance between 'em, which is perpendicular to them.

The formula for the area of a parallelogram is A equals bh, base times height.

The formula for the area of a rectangle is the same.

A equals b times h, where it's base and height again.

The form of the area if a triangle is the same as the ones above apart from you are also multiplying by a half and sometimes you may not initially have the information you need to calculate the area, but you may be able to obtain that information by using other aspects of mathematics such as Pythagoras's theorem and trigonometry like we see in today's lesson.

Great work today.

Thank you very much.