We have already seen how to find the area of rectangles, triangles and parallelograms.
There is another 2D shape whose area can be found by looking at the relationship the shape has with a rectangle.
A trapezium (sometimes called trapezoid) is a 2D shape with the specific geometric properties of
All of these are trapeziums
Let me show you something.
Move the point D anywhere you like to create a trapezium and then slide the green slider to the right.
From the interactive you can see you to flip the trapezium over and turn it into a parallelogram, and of course we already know that a parallelogram has the same area as a rectangle!
And guess what? Because we know $\text{Area of a rectangle }=L\times W$Area of a rectangle =L×W, then we can work out the area of any trapezium.
We tend to call the top side of the trapezium BASE 1 (a), and we call the bottom side BASE 2 (b) - because when we flip it both those sides becomes part of the base. We also need the HEIGHT of the trapezium.
This gives us the following formula:
$\text{Area of a Trapezium}=\frac{1}{2}\times\left(\text{Base 1 }+\text{Base 2 }\right)\times\text{Height }$Area of a Trapezium=12×(Base 1 +Base 2 )×Height
$A=\frac{1}{2}\times\left(a+b\right)\times h$A=12×(a+b)×h
You might wonder why we divide this by two. Well remember we flipped the trapezium over to turn it into a rectangle, well that means that 2 trapeziums is the same as the parallelogram - but we really only want the area of 1 trapezium.
Lets do an example using our new rule
Question: A new chocolate bar is to be made with the following dimensions, the graphic artist needs to know the area of the trapezium to begin working on a wrapping design. Find the area.
Think: I need to identify the base1, base2 and height. I can see these on the diagram that is given.
Do:
$\text{Area of a Trapezium}$Area of a Trapezium | $=$= | $\frac{1}{2}\times\left(\text{Base 1 }+\text{Base 2}\right)\times\text{Height }$12×(Base 1 +Base 2)×Height |
$=$= | $\frac{1}{2}\times\left(a+b\right)\times h$12×(a+b)×h | |
$=$= | $\frac{1}{2}\times\left(4+8\right)\times3$12×(4+8)×3 | |
$=$= | $\frac{1}{2}\times12\times3$12×12×3 | |
$=$= | $18$18 cm^{2} |
So now we can find the areas of rectangles, triangles (which are half of a rectangle), parallelograms (like a rectangle) and trapeziums (like half a rectangle).
Find the area of the trapezium shown.
Find the value of $x$x if the area of the trapezium shown is $65$65 cm^{2}.
Start by substituting the given values into the formula for the area of a trapezium.
$A=\frac{1}{2}\left(a+b\right)h$A=12(a+b)h
We have already seen how to find the area of rectangles, triangles and parallelograms, and trapeziums.
There are just a few more 2D shapes whose area we need to be able to find.
A kite is a 2D shape with the specific geometric properties of
Of course the kite you fly around on a windy day is named after the geometric shape it looks like.
Kites can taken on many different shapes and sizes. Try moving points $A$A, $O$O and $D$D on this mathlet to make many kinds of kites.
Let me show you something. Slide the slider. What shape is created? Now try showing diagonals and then sliding the slider.
From the interactive you'll notice that if you copy the inner triangles of the kite and rearrange them you can create - you guessed it a rectangle. Of course, like the trapezium, our original shape is only 1/2 of this rectangle.
Because we know the $\text{Area of a rectangle }=L\times W$Area of a rectangle =L×W, then we can work out the area of any kite.
We tend to call the long diagonal $x$x and we call the short diagonal of the kite $y$y. These give us the length and width of the rectangle that the kite fits inside.
$\text{Area of a Kite}=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Kite=12×diagonal 1×diagonal 2
$A=\frac{1}{2}\times x\times y$A=12×x×y
Let's do an example using our new rule.
Question: Find the area of this kite
Think: I need to identify the long diagonal length and the short diagonal length.
Do:
$\text{Area of a Kite }$Area of a Kite | $=$= | $\frac{1}{2}\times x\times y$12×x×y |
$=$= | $\frac{1}{2}\times4\times\left(2\times0.9\right)$12×4×(2×0.9) | |
$=$= | $\frac{1}{2}\times4\times1.8$12×4×1.8 | |
$=$= | $3.6$3.6 mm^{2} |
So now we can find the areas of rectangles , triangles (which are half of a rectangle), parallelograms (like a rectangle), trapeziums (like half a rectangle) and kites (like half a rectangle).
Find the area of the kite shown.
The area of a kite is $640$640 cm^{2} and one of the diagonals is $59$59 cm. If the length of the other diagonal is $y$y cm, what is the value of $y$y rounded to two decimal places?
We have already seen how to find the area of rectangles, triangles ,parallelograms, trapeziums and kites.
There is just one more 2D shapes whose area we need to be able to find- a rhombus!
A rhombus is a 2D shape with the specific geometric properties of
You can play with this mathlet to make many kinds of rhombuses, it also shows that you only need 1 side length and 1 angle to create one.
Let me show you something;
From this interactive you can see that as you copy the inner triangles of the rhombus and place them accordingly you can create - you guessed it a rectangle. Of course, like the trapezium and kite, our original shape is only $\frac{1}{2}$12 of this rectangle.
Because we know the $\text{Area of a rectangle }=L\times W$Area of a rectangle =L×W, then we can work out the area of any rhombus.
Lets call the diagonals $x$x and $y$y. These give us the length and width of the rectangle that the rhombus fits inside.
$\text{Area of a Rhombus }=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Rhombus =12×diagonal 1×diagonal 2
$A=\frac{1}{2}\times x\times y$A=12×x×y
So now lets do an example using our new rule
Question: A packing box with a square opening is squashed into the rhombus shown. What is the area of the opening of the box?
Think: I need to be able to identify the two diagonals.
Do:
$\text{Area of a Rhombus }$Area of a Rhombus | $=$= | $\frac{1}{2}\times\text{diagonal 1 }\times\text{diagonal 2 }$12×diagonal 1 ×diagonal 2 |
$=$= | $\frac{1}{2}\times x\times y$12×x×y | |
$=$= | $\frac{1}{2}\times16\times4$12×16×4 | |
$=$= | $32$32 cm^{2} |
So now we can find the areas of rectangles, triangles (which are half of a rectangle), parallelograms (like a rectangle), trapeziums (like half a rectangle), kites (like half a rectangle) and now rhombuses (like half a rectangle).
Find the shaded area shown in the figure.
Deduce and use formulae to find the perimeters and areas of polygons and the volumes of prisms