Lesson video

Hello, I'm Mrs. Lashley, and I'm gonna be working with you as we work through the lessons today.

I'm hoping you're ready to give this your best shot and try your hardest as we move through some quite difficult mathematics.

So our lesson outcome today is to be able to form and solve equations from complex shapes.

So it's gonna bring algebra and geometry together.

On the screen, there's some keywords that we will use during the lesson.

So you may wish to pause the video now so that you can re-familiarize yourself before we make a start.

And then, as I say, we'll be using those during the lesson.

So the lesson's in two parts.

The first learning cycle is to focus on forming equations from one shape.

In the second learning cycle, we'll move on to forming equations from two shapes.

So let's make a start of forming equations from one shape and going on to solve them where appropriate.

So if we're told that the perimeter of this scalene triangle is 42 centimetres, then what are the three edge lengths? Well, what do we have here? We've got expressions for each edge length.

We need to know the definition of a perimeter.

And the perimeter is a value of 42 centimetres.

It's scalene, which tells us that they should be different.

So this perimeter is the sum of the edges.

And we can sum together algebraic expressions.

We'll sum our algebraic expressions that we have for the edges of this scalene triangle.

We might need to expand brackets before we can collect together like terms. And so that's 2x plus six when you expand our brackets.

And then simplifying, we get 12x plus six.

So we now have an expression, an algebraic expression for the perimeter, which can be equated to its numerical given value.

So 12x plus six, which is the sum of our three edges, which is how you would work out the perimeter of a shape, is equal to 42.

So this is (indistinct) forming an equation based on the information given.

We can then go on to solve it to find the value of x.

So we're gonna subtract six from both sides and divide by 12 to find that one x is equal to three.

By substituting that value of x into each expression, we'll be able to get our numerical value, which is what we want to do.

So six times three minus one is 17.

Four times three, add one is 13.

And two times three, add three is 12.

And we've come out with three different values, which is what we expected because it's a scalene triangle.

We can do a quick check by summing those up.

And they equal 42, which is what was given to us as the perimeter.

So we can feel quite confident that nothing we have done has gone wrong.

So here's a check for you.

Given that the perimeter of the rectangle is 54 centimetres, what are the dimensions of the rectangle? So pause the video, you wish to form the equation and solve the equation to get to the answer.

Press play when you're ready to check.

The working out is on the slide here.

Firstly, you needed to get the perimeter as an algebraic expression and equate that to the numerical value.

The way that I've done it here is I've added the length and the width together and then multiplied by two because I know that the perimeter of a rectangle can be calculated that way.

If you did a sum of all four edges, that's absolutely fine.

You should still simplify to 30x plus six.

And that's when we go through the point of solving.

We then find that x is worth 1.

6.

And we can substitute that back into our algebraic terms and expressions to get the dimensions.

So this rectangle was eight centimetres by 19 centimetres.

And we're using the algebra to enable us not to do trial and improvement, not just trying numbers out to give us the correct perimeter.

We can use the algebra to get us there efficiently without any guesswork.

So, here, we have a quadrilateral.

We know that because there are four straight edges.

And the dimensions are written as algebraic expressions.

If you were told that this is actually a rectangle, what do we know about the edges? So just think about that for a moment.

What are your properties of a rectangle? Well, it means the opposite edges are equal.

So we now can say that some of these algebraic expressions are equal to each other because we know it is a rectangle.

And that means two different equations can be formed because of the equality of the opposite edges by being a rectangle.

So 8x is equal to 3y minus 10 and 3x plus 50 is equal to 6x minus 34.

Which one of those two equations that we formed can we solve? We can solve the second one.

And that's because there is only one variable.

On the first one, we have both x and y.

There are two unknowns.

Whereas in the second one there is the one unknown, so we can solve it.

It's an unknown on both sides, so we would want to collect together our x terms. And then using our inverse operations, step by step, solve it.

So x is worth 28, x is equal to 28.

By substituting this value of x into one of the expressions, we can find the lengths of the edges.

So 134.

By substituting it only into one, we know that they are equal to each other.

So if you put it into one, or the other, you'd get the same value.

By substituting that value x, which is 28, into the expression 8x, we can find the length of the shorter edge by the looks of the diagram, which is 224 centimetres.

So actually it's not shorter, the diagram is a little misleading in that way.

But we are just using the maths and using the algebra and the notation to allow us to calculate these values.

As the opposite edges are true, it's also true that 224, which we now know is the numerical length of that edge, is equal to 3y minus 10.

And we can, therefore, go on to solve that to work out the y is 78.

If you substitute 78 into that expression, we'd get 224.

The perimeter and area could now be calculated.

When it was algebraic, we could get algebraic expressions for the perimeter and the area.

Now that we have numerical values, we can get a numerical value for the perimeter and the area.

So the area would be 224 multiplied by 134.

So rectangle, remember.

So base times perpendicular height or length times width.

And the perimeter, length plus width times by two, or length plus width plus length plus width.

And look at the units.

We've got unit for area and unit for perimeter.

So a check based on this one.

Which of these equations are true for this square? All lengths are in centimetres.

So pause the video and decide which of the equations are true for this square.

Press play when you're ready to check.

They're actually all true.

We're using the properties of the shape.

A square, all edges are equal.

So the top one suggests that one edge is equal to another edge.

That is true for a square.

Part B also suggests that two edges are equal, which we know is true for a square.

This is the two opposite edges, but they would be equal in a square.

Part C, part C has said that four lots of 3x plus 14, that would be the perimeter of the square, would be equal to four lots of 9x minus 58 because that would be another expression for the perimeter of this square.

And the last one would be for the area.

How do you work out the area of a square? You do length squared.

So 3x plus 14 is one of the algebraic expressions for the length of this square.

So if we square it, we'd have an expression for the area.

Alternatively, we could multiply together the length by another length, in which case we've just chosen two algebraic expressions to multiply together to find the product.

And that would be equal to the other algebraic expression for a square.

So this is using those properties of a square to create and form equations.

Here we have a combination really of what we've seen so far.

An isosceles trapezium, so by being an isosceles trapezium, we've got symmetry, but we've also got two equal edges, and they're marked with the hash lines, has a perimeter of 78 centimetres.

So we've been given the numerical value of the perimeter.

The question is asking us to find the edge length.

So we want numerical values for the edge lengths of this isosceles trapezium.

So we need to start by thinking, "Where can we form equations?" Two equations can be formed.

Can you think where they're gonna come from? So one of the equations is formed by using the property that those opposite edges, the ones with the hash mark, are equal.

We have equality which means we can have an equation.

So 6x minus one is equal to 7y plus three.

That is true because it's an isosceles trapezium.

The second equation is by using the fact that we have a value for the perimeter.

And we also know what the perimeter is and we have all four edges.

So we can find an algebraic expression for the perimeter.

And that's 18x plus 11y plus two.

You might wanna pause the video and check you're happy where that expression has come from.

And then we've equated it, we've done it equal to 78, the value of the perimeter.

This sets us up with a pair of simultaneous equations because we have two equations, both with two variables.

So we can't solve one like we did previously when we had two and substitute that value into the other one.

We need to use them together to solve and find the value of x and y.

I've labelled the first one, which, remember, came from the equality of the isosceles trapezium sides, equation one.

And the second one, which came from using the perimeter value, as equation two.

There are a few methods for solving simultaneous equations.

I'm gonna go through this way, but you may do it a different way.

And that's fine, we should end up with the same x and y values, but do follow through this method.

So equation one is equivalent to 6x minus 7y equals four.

So I've just done a little bit of rearranging on that one.

I've collected the x and the y terms on one side of the equation and the numerical term on the right-hand side.

Equation two, if I follow the same kind of thing, I want x and y terms on the left-hand side of the equation and the numerical term on the right.

And, again, I've taken away two to move it over, inverse operations.

So they are equivalent equations one and two.

So there they are again.

These are our pair of simultaneous equations.

Those x and the y are unknown values at this moment in time.

By multiplying equation one by three, I'm just scaling up the equation.

The relationship and the proportions of Xs to Ys and its value hasn't changed.

So I'm scaling the equation.

Means that I'd have 18x minus 21y equals 12.

And this allows us to eliminate the x term.

So if I do equation two, subtract that equation one times by three.

The x terms have disappeared, have become zero x, because I've subtracted 18x from 18x.

So that has eliminated one of the unknowns, one of the variables, which allows us now to solve this one.

So 32y is equal to 64, which then means that y is equal to two.

So now I have one of the numerical values for one of the variables.

Given that y equals two, we just solved and found that y equals two, I can substitute that into one of my simultaneous equations.

Here, I'm substituting it into equation two.

And so 11 times two is 22.

It's a linear equation with one variable.

I can solve this.

So I can subtract my 22 and divide by 18.

I now know that x equals three.

I formed two equations using isosceles trapezium properties and the perimeter.

I then solved the simultaneous equations to get the value of x and the value of y.

The last thing we need to do is calculate what the value of the edges, the numerical lengths of these edges are.

And that's done by substituting our values into the expression.

So 8x minus 5y is equal to 14 when x is three and y is two.

7y plus three is equal to 17 when y is two.

4x plus 9y is equal to 30 when x is three and y is two.

And unsurprisingly, being that it's an isosceles trapezium, 6x minus one is also equal to 17.

And so now we have got, we went from having algebraic expressions for the lengths of this isosceles trapezium and only knowing the perimeter numerically to knowing all four edges without any guesswork.

So now that the edge lengths are known, can the area be calculated? So Sofia said yes, but first we need to calculate the perpendicular height of the trapezium.

And she's correct because the formula to work out the area of a trapezium, whether it's isosceles or not, involves the perpendicular height.

So because it's isosceles though, it does mean it is symmetrical.

And so we know we can split the bottom edge in the way that Sofia has.

The perpendicular height, therefore, can be calculated using Pythagoras' theorem because we've got these right-angled triangles that we can use.

17 is the hypotonus.

So 17 squared equals our perpendicular height squared plus eight squared.

We've worked out the eight using the symmetry of the isosceles.

Rearranging that to make h squared, the subject is 225.

And square rooting that means it is 15.

And, therefore, we can now use the formula because we've got all of the components that we need.

We've got our parallel edges, we know their length and we have the perpendicular height.

And 330 square centimetres is the area of this trapezium.

So here's a check.

This isosceles trapezium has a perimeter of 90 centimetres.

Write two equations from this information.

So I just want you to form the equations, not solve them.

Press pause whilst you're doing that and then press play to check.

So using the isosceles trapezium, the fact that we've got a pair of equal edges, we can equate those two algebraic expressions.

And using the perimeter, we can get an algebraic expression for the perimeter and we can equate that to 90.

If you did any further rearranging, so perhaps you got the x and the y terms onto one side of the equation, they are equivalent as well.

So if you did do further rearranging, just check that they are equivalent.

So here's the first task of the lesson.

So, question one, I would like you to, part A, write an algebraic expression for the perimeter of the scalene triangle.

And then given that the perimeter is 60, I'd like you to form and solve an equation to find the value of x.

And then finally, I want you to discuss and decide if the scalene triangle is also right-angled.

So press pause whilst you're doing that, and then when you're ready to move on to the next question, press play.

So, question two, we've got here is a parallelogram.

So think about all the properties you know about parallelograms. All lengths are measured in centimetres.

Part A, I'd like you to find the value of x and y.

And part B is, "Hence, find the perimeter of this parallelogram." So pause the video, think about the properties and where that will get you some equations, and then how you will solve them.

Press play when you're ready for question three.

Question three, here is an isosceles triangle.

Given that the perimeter is 32 centimetres, find the area of the triangle.

So think about where you can start.

Think about if there's an equation that you can form.

And then work through the question step by step.

Press pause whilst you're working through it, and then when you're ready and you press play, we'll go through our answers to task A.

So, question one, firstly, you need to write an algebraic expression for the perimeter.

So that was summing together the three algebraic expressions.

There was one expression that was in factorised form.

So you needed to expand that before you could collect up your x terms and your numerical terms. You should've finished with 12x plus 24.

It might be that you did factorise that as well.

And that would be 12, open bracket x plus two, close bracket, in a factorised form.

Part B was now given that you've been told that the perimeter is actually equal to 60 centimetres.

Form and solve the equation.

So you're gonna take the algebraic expression for the perimeter, equate it to the numerical value for the perimeter, and solve it.

X was equal to three.

And part C, you needed to say if it was right-angled or not.

So if it is a right-angled triangle, then Pythagoras' theorem will hold true.

The three edges, we could use our value of x, which we now know is three, to work out the numerical values.

So the edges were 26, 24, and 10.

So does 26 squared, which we know the hypothenuse is the longest one, so the largest value, 26 squared equal 24 squared plus 10 squared? Yes.

676 equals 576 plus 100.

So it is right-angled as well as being scalene.

Question two, the parallelogram, you need to find the value of x and y and then the perimeter.

So, part A, you needed to find yourself or form two equations.

And that came from parallelograms, opposite edges are equal.

So 8x minus 17 equals 5y and 10y equals 7x plus two.

So we've got a pair of simultaneous equations, both involving x and y.

Once again, if I multiply equation one by two, then I'll equate it to 10y.

So then I'd have 10y in both of the equations.

I can then substitute one into the other and make them equal to each other.

I've only got one variable now.

And I, therefore, can solve it.

So I work out that x equals four by going through the stages of solving, and then I can substitute the value of x being four back into an equation to work out that y, therefore, is three.

If I know that x equals four and y equals three, then I can find the perimeter of the parallelogram by firstly working out the length and the width or the longest edge and the shorter edge of the parallelogram, which is 15 and 30, and find the perimeter is 90 centimetres.

Lastly, question three, this question had quite a lot to it.

It was quite involved.

So, firstly, isosceles triangle, which means that we know there would be a pair of equal edges.

They were marked with a hash mark.

And they were the ones with the algebraic expression.

So we can equate those.

We can form an equation and solve them.

Solving that tells you that m is equal to two.

So then you could work out what the length of those two equal edges are, which was 10 each.

You were given the perimeter.

So because you knew the perimeter was 32 and you now know that 20 centimetres of the perimeter has been used by the equal edges, that means the base edge would be 12.

Now that we know the base edge is 12, we can use the symmetry of an isosceles triangle and the altitude to create a right-angled triangle to work out the perpendicular height.

So Pythagoras' theorem has come back in here.

Pythagoras' theorem allows us to work out that the perpendicular height of this isosceles is eight.

And then we can work out the area of the triangle by doing 1/2 times base times perpendicular height.

So our base is 12, our perpendicular height is eight, we just calculated that.

And so that is equal to 48 square centimetres.

Really well done if you managed to get through the whole of that task.

So we're now gonna move on to the second learning cycle where we are forming equations from two shapes.

So these two shapes have equal perimeters.

So we've got shape A and shape B.

Shape A is a compound shape and shape B is a triangle.

We've been told that they have the same perimeter.

So an expression for the perimeter of shape A is 6x plus 24.

That one is an L shape or rectilinear compound shape.

So the perimeter will be equal to the rectangle that would surround it because the vertical edges sum to the total height of 2x and the horizontal edges sum to the total length of x plus 12.

So that's why we can treat it like a rectangle because of it being a rectilinear L shape.

The perimeter on the triangle, we can get an expression for 4x plus 26.

What happens next? We've got two shapes.

We've worked out their algebraic expressions for their perimeter, we know that they are equal, and that's where the equation will come.

So because they are equal, we can form the equation, and we can solve it to say that x equals one.

Now that you know that x equals one, you can work out the individual edges.

Here's a check for you.

So this rectangle and a right-angled triangle and their areas are equal.

So I would like you to work out the value of x.

So pause the video whilst you're working through that one, think about what that means in terms of forming an equation, and then solve it.

Press play when you're ready to check.

So because you were told that their areas are equal, that's where we're gonna form an equation.

Area of A is gonna be equal to the area of B.

Now we need to use the algebraic expressions to get an algebraic expression for their areas, relative areas, and make them equal.

So a rectangle is length times width.

So we can multiply 5x by 2x plus seven.

And on a triangle, it's 1/2 times base times perpendicular height.

It's a right-angled triangle, so we know that they are perpendicular.

And we get 4x multiplied by 2x plus nine.

Now we need to manipulate the algebra so we can expand the brackets and then rearrange it.

And we have a quadratic equation to solve, 2x squared minus x equals zero.

We're not gonna divide by x, we're gonna factorise x and get it into its factorised form to find our solutions.

So x equals zero or x equals 0.

5, or maybe you wrote that as 1/2.

We're going to disregard x equals zero in this context because it cannot be zero.

The reason it cannot be zero is because we have two edges that are just multiples of x.

If they are multiples of zero, then that would be zero, and, therefore, there would be no edge.

So they wouldn't even have an area.

So you do need to, if you find yourself with a quadratic equation and you get two solutions, you need to decide if both of them are valid solutions or whether in the context one of them can be disregarded.

So the area of this parallelogram and triangle are equal.

So you just had a check with areas being equal.

So we know how we start this.

But using this fact, an equation can be formed, which is the parallelogram's area and the area of the triangle.

And so a 8xh, base times perpendicular height.

And then for a triangle, 1/2 times base times perpendicular height.

So that's 2x, the half is evaluated onto the 4x, times 3x plus five.

We can express h in terms of x.

Because in your check you had only one variable of x, which allowed you to solve the equation to get a numerical value.

Here, we've got two variables, x and h.

So we can't go forward and solve it because we haven't got any numerical equivalencies in terms of we don't know the area is 20, for example.

But we can write one variable in terms of another.

So we can express h in terms of x.

And that comes with rearranging the formula and making something the subject.

I want to make h the subject.

So I can divide both sides by 8x, which both of these are equivalent form.

So one is the factorised form and one is the expanded form.

This check's for you.

So this is an equation showing that the area of two shapes are equal.

I'd like you to use it to express h in terms of x.

Press pause whilst you work that one out.

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

So you wanna make h the subject.

Divide both sides by 4x because there is a product of 4x and h on the left-hand side.

So we're gonna use division as our inverse.

And there are two forms that you may have left your answer in.

Continuing now and looking at two shapes and equations from two shapes.

The area of the square and the area of the rectangle are in the ratio one to two.

So we've got algebraic expressions for their lengths and their widths, if it's the rectangle.

And we've also got this information that their areas are not equal, but their areas are in a ratio of one to two.

This means that the area of the rectangle is twice the area of the square, or the area of the square is half the area of the rectangle.

Here is a check for you.

So area of shape A to the area of shape B is in the ratio three to two.

Which of these are true? So, part A, the area of shape A is equivalent to 1 1/2 times the area of shape B; part B, the area of shape B is 40% of the area of shape A; or part C, the area of shape B is 2/3 of the area of shape A.

Pause video whilst you make a decision on those statements, and when you're ready to move on and check, press, play.

So it's part A and part C that are true.

1 1/2 could be written as 3/2 and 3/2 and 2/3 are reciprocals of each other.

And the statement was written in the converse way.

Shape A is equivalent to shape B, et cetera.

So if we go back to the area of the square and the area of the rectangle having a ratio of one to two, we can form an equation using that relationship between the two areas.

So we can still form an equation.

So two lots of the area of the square is equal to the area of the rectangle.

So the area of the square as an expression is 2x squared.

And we need two lots of that.

The area of the rectangle is width times length.

And we get our algebraic expression for that product.

We can expand and manipulate the algebra.

So 2x squared.

And the brackets are really important around that 2x term because the two needs to be squared as well as the x.

So that gives you 4x squared.

The area of the square is 4x squared.

But we want two of them, so that's why it's 8x squared.

If we expand the double brackets, we end up with the algebraic expression 8x squared plus 10x minus 25.

So we can subtract 8x squared from both sides or (indistinct) the coefficients.

And zero equals 10x minus 25.

Therefore, 10x is 25.

And that means the x is equal to 2.

5.

So now that we know that x is equal to 2.

5, what can now be calculated if needed? So we could calculate the edge lengths for both of the shapes.

We could, therefore, calculate the perimeter of both of the shapes.

Another check.

So the perimeter of shape A is 4k plus seven.

The perimeter of shape B is 6k plus one.

Given that the perimeter of shape A to the perimeter of shape B is the ratio four to one, write an equation to show this.

So form the equation given that relationship between their perimeters.

Pause the video whilst you're doing that, and when you're ready to check, press play.

So 4k plus seven is equal to four lots of 6k plus one.

One to the last task of the lesson, so task B.

So, question one, these two shapes have equal areas.

I want you, part A, to find the value of d, part B, find the value of the combined area, and, part C, find the perimeter of each shape.

Pause the video whilst you're working through question one, and then when you're ready to move on, press play.

Okay, so here's question two.

Shape A is a parallelogram and shape B as a triangle.

These two shapes have the same areas.

So part A is form an equation from those two shapes, part B is to express h in terms of y, and part C is given the h is equal to six centimetres, calculate the value of y.

So pause video whilst you work on question two.

When you press play, you've got one more question in this task.

So here's question three.

The perimeter of shape A to the perimeter of shape B is in the ratio two to three.

Find the area of shape A.

Figure out where you can form an equation and then solve it.

And then think about what that additional information that has given you and whether that leads you closer to finding the area of shape A.

Press pause whilst you work it through, and then when you're ready to check the answers to task B, press play.

Here is question one part A.

You needed to find the value of d.

So, first of all, you needed deform the equation from the two shapes because they had equal areas.

So one was a square, we knew that from the hash marks.

So that was d squared.

And the other one was a rectangle.

So length times width.

We get to a quadratic of zero equals 70 squared minus 14d.

So we can factorise that because there is a common factor of 7d in both terms. And then d is two because d cannot equal zero.

Once again, in this context, d cannot equal zero because some of the edge lengths are multiples of d.

Part B was find the value of the combined area.

So because the areas are equal, the combined area is double the area of A or double the area of B.

It doesn't matter which way 'round you went for.

The area of A is just d squared.

So that's the most efficient or easiest way to get to the area.

We've just calculated that d equals two.

So the area is four.

And, therefore, hence the combined area is eight.

And then the perimeter for each shape, if you substitute d in to get their dimensions, the numerical values, and work out the perimeter, shape A has a perimeter of eight centimetres and shape B has a perimeter of 10.

Onto question two.

Part A was to form an equation.

So we were forming an equation using the fact we were told their areas were equal.

So the area of parallelogram is base times perpendicular height.

So that was 3hy.

And the area of the triangle is half times base times perpendicular height, which is 3y multiplied by 2y plus five.

And you may have gone on and expanded that right-hand side.

So you get 6y squared plus 15y.

It doesn't matter if you did or you didn't.

Part B was to express h in terms of y.

H is only in the one term, which is on the left-hand side of the equation.

So we're trying to make h the subject.

And h is equal to 2y plus five.

On part C, you were given that h was equal to six.

So calculate the value of y.

So using that we know that h is equal to 2y plus five, substitute h to be six and solve it.

Y comes out as a 1/2 or 0.

5.

Question three then.

The perimeters were in a ratio and we need to find the area of A.

So you could've got an algebraic expression for the area of A from what was given straight away, but we do need the numerical value.

The perimeter is what this question sort of was discussing.

So let's get some algebraic expressions for the perimeter.

So rectangle is shape A.

We can find the perimeter, 10x plus 28.

Part B was a triangle that sum the three edges to get the perimeter.

And that gives you an algebraic expression of 23x minus 22.

We're then gonna use the ratio to form an equation that three lots of the perimeter of A is equivalent to two lots of the perimeter of B.

Expand the brackets and solve it.

Then we find the x is equal to eight.

If we know that x is equal to eight, we can then find the dimensions of the rectangle numerically by substituting them into the expressions, which gives you 29 and 25.

So the area of shape A, a rectangle, is 725 square centimetres.

To summarise this lesson, which was about equations from complex shapes, we were using properties of shape to form equations, when the dimensions are given as algebraic expressions instead of numerical values.

If there is a relationship between two shapes, more equations can be formed.

So if you are told their perimeters are equal, if you are told their areas are equal, or as we've seen, if you've been told there's a ratio between them.

And then finally, the equations that you form may be linear, may be quadratic, or they might be simultaneous equations, depending on the question.

So you need to use all of those algebraic skills that you have to solve equations in this context of geometry.

Really well done today.

And I look forward to working with you again in the future.