Lesson video

Hello everyone.

Welcome and thank you for joining me in today's lesson on 2D Shapes.

I am Mr. Gratton.

And today we will look at calculating the perimeter and area of shapes on coordinate grids.

Pause here to check the definition and formula for the equation of a line, which will come in handy later on in the lesson.

First up, let's have a look at perimeters of polygons defined by their vertices.

Let's start by looking at two coordinates J and K.

We can find the shortest distance between these two coordinates on a Cartesian coordinate grid by using Pythagoras' theorem, but how when there's no triangle? Well these two coordinates are treated as the vertices on the hypotenuse of the right-angle triangle that we can create ourselves using the coordinate grid.

We use the lines on the coordinate grid to draw a horizontal side of the triangle where the horizontal distance between J and K is eight units and the vertical side of the triangle where the vertical distance between J and K is four units.

We can draw this right-angle triangle incredibly accurately and with confidence because we are using the coordinate grid itself to inform us of how long the two shorter sides of the triangle are.

Now we have a right-angle triangle.

We can use Pythagoras' theorem where a squared is the horizontal distance.

b squared is the vertical distance, and c squared is that shortest distance along the hypotenuse of the right-angle triangle.

Note that the order of A and B can be swapped around, but I personally will be keeping all calculations this way around for consistency.

The shortest distance between points J and K is therefore eight squared plus four squared equals c squared, which is 80 equals c squared.

So C is the square root of 80, which is approximately 8.

9 units.

For this first check pause here to choose which equation can help find the shortest distance between points P and Q? Using Pythagoras' theorem we have the horizontal distance squared plus the vertical distance squared equals the shortest distance across the hypotenuse squared, which is two squared plus nine squared equals c squared, which evaluates to four plus 81 equals c squared.

Using this information, pause here again to calculate the shortest distance between P and Q rounded to one decimal place.

4 plus 81 is 85 and the square root of 85 is approximately 9.

2 units.

Next up, pause here to perform some calculations to check whether the shortest distance between UW or TV is longer? Using Pythagoras' theorem between each pair of coordinates shows us that the distance between T and V is longer, but not by much.

It is 0.

3 units longer than UW.

Multiple points on a Cartesian coordinate grid can be joined together with line segments to create a polygon.

The quadrilateral ABCD is created by joining up the following four line segments, A to B, B to C, C to D, then D back to A.

Since this is a polygon, we do not need to join other pairs of coordinates such as A to C along the diagonal of that quadrilateral.

To find the perimeter of this polygon, we can use Pythagoras' theorem multiple times to find the length of each side of the shape one by one starting with the length AB.

Actually, we don't need Pythagoras' theorem to find this length as it is a horizontal line segment.

We can just count how many squares across from point A to B at six squares, meaning that this side is six units long.

However, for length BC we can set up Pythagoras' theorem as one squared plus four squared equals a hypotenuse squared of the length BC squared.

Therefore the length BC is the square root of 17.

This is therefore approximately 4.

1 units long.

Next up length CD.

With Pythagoras' theorem, we have five squared plus one squared equals the length CD squared.

Therefore the length CD itself is the square root of 26 or 25 plus one.

This side is therefore 5.

1 units long after rounding.

And lastly, we have the length DA.

We have two squared plus five squared equals the length DA squared.

Therefore the length DA itself is the square root of 29 at approximately 5.

4 units long.

Adding these four sides together gives us 20.

6, so the perimeter of this quadrilateral is 20.

6 units in length.

Notice how after all our working on our polygon, we have this sort of rectangle shape made from all the right-angled triangles.

This will not be the last time you'll see a rectangle play a part in helping us find perimeters or indeed areas of shapes on a coordinate grid.

For this next check, match the side of the triangle to the correct equation that finds the length of that side.

Pause here to do this question.

And here are the answers, for example, side JK matches with equation C as J and K are four units apart horizontally and two units apart vertically.

And for this check pause here to calculate the length of each side of this quadrilateral and match up the lengths that you calculate to the ones on screen.

And here are the answers, root 45 is an intermediate step to the rounded length 6.

7.

Now use these answers to calculate the perimeter of this quadrilateral.

3.

6 plus 4.

1 plus 6.

7 plus 7 is 21.

4 units.

Therefore the perimeter of this quadrilateral is 21.

4 units.

Brilliant onto the practise questions.

For question one, calculate the shortest distance between each pair of points.

And for question two, find the perimeter of quadrilateral RSTU.

Some guidance has been given to find the length of the side to you.

Pause now to do these two questions.

For question three, which of the two triangles FGH or IJK has the largest perimeter? And for question four, find the perimeter of quadrilateral LMMP and explain how you know this quadrilateral is a parallelogram? Pause now for these two questions.

Great stuff.

Here are the answers to question one.

The shortest distance between G and H is 6.

3 units between I and J is 5.

8 units and H and I at 5.

4 units.

For question two, pause here to check your calculations for this quadrilateral whose perimeter is 26.

7 units.

For question three, the perimeter of IJK is 0.

47 units longer at 18 units in length.

And for question four, the perimeter of LMNP is 16.

72 units.

We can tell that this is a parallelogram because both opposite pairs of sides are equal in length at 4.

24 units or 4.

12 units in length.

Next up area, but this time for polygons bounded by straight line equations.

If multiple straight line equations are drawn onto one coordinate grid, then it is possible for them to bound a polygon.

Not all sets of straight lines will bound a polygon, but many will.

Here's an example, equation A: y equals 10 gives this horizontal line.

Equation B: y equals five subtract x, gives this line with a negative gradient.

Then equations C and D both give lines with a positive gradient.

Did you spot this quadrilateral bound by these four equations? Well, let's see if you can spot one for yourself.

For this quick little check, what type of quadrilateral is bound by these three straight lines and a different fourth line that is currently not drawn that has the equation y equals three x plus two.

Pause now to identify the quadrilateral.

The line with equation y equals three x plus two has a positive gradient like this.

We have a right-angled trapezium after we've plot this fourth line.

In some basic cases, the area of some polygons can be found by using an area formula or by first splitting the shape into triangles and rectangles whose areas can be easily calculated.

For example, the area of this parallelogram is the base of seven units multiplied by its vertical height of six units giving an area of 42 units squared.

However, if two vertices lie on the same horizontal or vertical plane, then it can be helpful to split the shape up across these two vertices.

For example, we can split this more odd looking quadrilateral into two triangles using this horizontal line segment across these two vertices.

The area of the top triangle is therefore a base of 8 times by a height of 3 divided by 2 equals 12.

The area of the bottom triangle shares the same horizontal base length as the top triangle of 8 units multiplied by 5, divided by 2 equals 20.

Therefore, the total area of this entire quadrilateral is 12 plus 20 equals 32 units squared.

For this check we plan to plot a fourth line with equation y equals x.

This line intersects lines A and B to form a quadrilateral.

Pause here to consider what the coordinates are where the line y equals x intersects these two other lines.

Here is the line with equation y equals x and the coordinates where this line intersects A and B are six, six and two, two.

Notice how each pair of coordinates have the same X and Y coordinate because the line itself is y equals x.

Now that we have the line y equals x drawn, and this quadrilateral bounded, pause here to find the area of the quadrilateral bounded by these four straight lines.

First of all, a very well done if you spotted the quadrilateral bounded by these four straight lines was a trapezium with this calculation to find its area of 24 units squared.

Line C has equation y equals four and line C intersects with A and B to form a triangle.

Pause here to consider the location of line C and calculate the area of this triangle.

Line C is this horizontal line and the area of the triangle is 30 units squared.

Pause here again to consider how much smaller the area of this triangle would be if line C instead had the equation y equals seven, not y equals four? The new location of line C is here, this horizontal line.

The new triangle has an area of 7.

5 units squared.

The difference is therefore 22.

5 units squared.

When finding the area of a polygon defined by straight line equations, sometimes it is important to identify the coordinates where the pairs of lines intersect first.

So for these four equations, A, B, C, and D, the coordinates of intersection are 0, 6.

4, 8; 10, 5 and 6, 3.

In order to find the area of this quadrilateral with vertices at these four points, we can circumscribe a rectangle around this shape.

What is a circumscribed rectangle? Well, it is this a rectangle whose perimeter contains the four coordinates of our original bounded shape.

It is important to note that this circumscribed rectangle must have horizontal and vertical sides that are parallel to the axes so that its area can be easily found.

In this case, the area of the circumscribed rectangle is 10 units across multiplied by 5 units up to give an area of 50 units squared.

After finding the area of this circumscribed rectangle, we can then find the areas of all of the triangles inside the circumscribed rectangle that are external to the bounded shape that we want to find the area of.

The areas of the four triangles in this shape are nine units squared, four units squared, nine units squared, and four units squared.

The area of our original bounded polygon can be found by subtracting the area of each of these triangles from the area of the circumscribed rectangle that we formed ourselves.

50 for the area of the circumscribed rectangle, subtract 9, 4, 9, and 4 for those four triangles gives 24.

The area of the polygon bounded by these four straight line equations is 24 units squared.

Right for this check pause here to find the area of triangle D that is bounded by these three straight lines.

By first finding the area of the circumscribed square and the areas of the other three triangles, A, B, and C.

Pause now to do this.

The area of the circumscribed square is 25 units squared.

The area of D, the triangle bounded by these three straight lines is therefore 11.

5 units squared.

Brilliant work on this complex topic so far.

For question one of this practise task, draw the graphs of these straight line equations and hence calculate the area of the triangle bounded by these three lines.

For question two, do the same, but this time find the area of the bounded quadrilateral.

Pause now to do these two questions.

Next up, draw a rectangle that circumscribes, this bounded quadrilateral, and then by finding the area of this circumscribed rectangle and the area of three triangles, find the area of this bounded quadrilateral, pause now to do this.

And for question four, these three equations bound a triangle.

Pause now to find the area of this bounded triangle.

Brilliant effort on all of these questions.

Let's have a look at some of the answers.

For question one, pause here to check the locations of the two lines that along with the third already given, bound this triangle with an area of 16 units squared.

Similar for question two.

Pause here to check the locations of the two lines that along with the two already given, bound a trapezium with an area of 40 units squared.

For question three, pause here to check the location of the circumscribed rectangle.

The area of the circumscribed rectangle is 60 units squared and the area of these three triangles are 4, 10 and 6 units squared, meaning that the area of the bounded quadrilateral is 40 units squared.

And finally, question four, the area of the bounded triangle is nine units squared.

Pause here to check the locations of lines, the circumscribed rectangle and any calculations.

And for our last cycle, how on earth do simultaneous equations fit in with finding the area and perimeter of polygons on a coordinate grid? Well, let's find out.

Sometimes identifying the coordinates where straight line equations intersect can be pretty tricky.

For example, this could be because the straight lines could intersect at non integer coordinates and in between the guidelines on the coordinate grid.

In this intersection we can see the X coordinate is somewhere between four and five, but it is ambiguous as to the exact value.

We cannot just assume it is 4.

5 the midpoint.

A different reason could be that the scale of the graph is so large that reading a coordinate accurately can be incredibly challenging.

If you know the equation of each straight line, you can find the coordinates of each point of intersection by solving pairs of simultaneous equations.

For example, with the line with equation y equals four x plus one.

The line with equation y equals a quarter x plus one and the line with equation y equals negative x plus nine.

We can see that one intersection is at the coordinates, zero one, the intersection of lines A and B.

However, the exact values of these two other points of intersection are unclear.

First up, let's focus on that intersection between lines A and C.

With equations y equals four x plus one and y equals negative x plus nine.

Because both lines are given in the form y = mx + c, it is possible to simultaneously solve pairs of them using substitution.

Substituting equation A into equation C gives four x plus one equals negative x plus nine.

Rearranging gives five x plus one equals nine, therefore x equals 1.

6.

The X coordinate of the point of a dissection between lines A and C is 1.

6.

We can then substitute X equals 1.

6 into either original equation.

This time I'm gonna choose equation C, giving this substitution, resulting in a Y coordinate of 7.

4.

The full coordinates where lines A and C intercept are 1.

6, 7.

4.

So we know two of the three points of intersection for this bounded triangle.

For this check, let's find out the coordinates where lines C and B intersect by solving this pair of simultaneous equations.

Pause here to find the coordinates where lines B and C intersect.

By solving this pair of simultaneous equations, we get the coordinate of 6.

4, 2.

6.

Now we know more precise coordinates for the vertices of the bounded triangle, we can circumscribe a rectangle so the vertices of the triangle lie on its perimeter.

Knowing the precise coordinates of the vertices of the triangle will allow us to more precisely find the area of this circumscribed rectangle.

Both the area and perimeter of this triangle can be found by first calculating the widths and heights of each right-angled triangle.

It can be helpful to store or relevant information for each triangle in a table like this.

The width is the difference between the two x coordinates in this case of zero and 1.

6.

The height is the between the two white coordinates of one and 7.

4.

We can then use Pythagoras' theorem to find the hypotenuse of approximately 6.

59 and consider the area of the triangle using the width and the height that we just found divided by two to get 5.

12.

Here are the calculations, not the final answers for each of the key bits of information for triangle E.

Okay, using the information in the tables for triangles, D and E to help you pause here to find the values of A to D in the table for triangle F.

For triangle F, its width is 6.

4.

Its height is 1.

6, hypotenuse is 6.

59 and its area is 5.

12.

Where do those values look quite familiar? Those are the exact same values as in triangle D except with the width and the height the other way around.

Triangles D and F are actually congruent to each other.

Blindy filling in all of this information for each of those three tables was a lot of work, but what information can we take from those tables in order to find the perimeter of our original bounders triangle? Well, we take the sum of the hypotenuses of triangles, D, E, and F.

So the perimeter of our original triangle is 19.

98 units.

The sum of the hypotenuse 6.

59, 6.

78 and 6.

59.

And for the area we first have to calculate the area of the circumscribed rectangle that defines the triangles, D, E, and F.

The area of this rectangle is 40.

96 units squared.

We then take away the area of each of the three triangles from the area of this circumscribed rectangle to leave only the area of our original bounded triangle.

So 40.

96 subtract these three areas for these three right-angle triangles means that the area of our original bounded triangle is 19.

2 units squared.

Okay, for this quick final check pause here to find the width, height and therefore area of triangle D.

The width is the difference between the two X coordinates.

The height is the difference between the two Y coordinates and the area is therefore 1.

8 times by 7.

3 divided by 2, which is an area of 6.

573 units squared.

Great work.

Get ready for this final practise task.

For question one, two sets of coordinates that define two of the vertices of triangle K have in integer parts, but one vertex has coordinates that are non integer.

By first simultaneously solving a pair of equations, state the coordinates of all three vertices of triangle K and hence fill in the table of values to help find the perimeter and area of triangle K.

Pause now to do question one.

And lastly question two, by first sketching the three equations given, find the perimeter and area of triangle Q that is bounded by these three equations.

Please note that the three sketches do not need to be accurate and only used to visualise what triangle Q may look like and how to circumscribe it with a rectangle.

Pause now for question two.

Amazing effort on some really challenging questions.

For question one, the three vertices of triangle K are 8, 7, 2, 13, and approximately 0.

36, 3.

18.

Pause here to compare the information that you filled in on this table with the information on screen.

You may have rounded your answers slightly differently.

The perimeter of triangle K is therefore 27 units and its area is 34.

4 units squared.

For question two, here is an example of what your sketches should look like.

The coordinates of the vertices of triangle Q are 0, 10; 11, 87 and 80, 18.

The perimeter of triangle Q is 255.

8 units and its area is 3036 units squared.

Take a deep breath after the incredible effort that you put into those questions in a lesson where we have considered that the shortest distance between two points on a coordinate grid can be found using Pythagoras' theorem and that the perimeter of a polygon on a coordinate grid can be found by calculating the length of each and every side of that polygon using Pythagoras' theorem as well.

The area of a polygon bounded by straight line equations can be found by circumscribing, a rectangle whose vertices of this polygon lie on the perimeter of this circumscribed rectangle.

The area of the polygon is then found by subtracting the areas of triangles from the area of this circumscribed rectangle.

And lastly, the coordinates of the vertices of a polygon bounded by straight line equations can be found by simultaneously solving pairs of equations.

Once again, thank you all so much for your hercule effort today.

Take care and hope to see you soon for some more incredible maths fun.