Area of Shapes Worksheets | Questions and Revision | MME

Areas of Shapes Worksheets, Questions and Revision

Level 1-3
GCSE Maths Revision Cards

Areas of Shapes

The area of a 2D shape is the amount of space it takes up in 2 dimensions, and its units are always squared, e.g

\text{cm}^2,\hspace{1mm}\text{m}^2

You need to know the formulas to calculate the areas of some common shapes and be able to rearrange them. Revising rearranging formulae will help with this topic. 

Area of Rectangle 

The area of a rectangle is length \times width

\text{Area} = L \times W 

Level 4-5

Area of a Parallelogram

The area of a parallelogram is base \times vertical height

\text{Area} = b\times h

Level 4-5

Area of a Trapezium

The formula to calculate the area of a trapezium is:

\text{Area} = \dfrac{1}{2}(a+b)h

where a and b are the lengths of the parallel sides and h is the vertical height. 

Level 4-5

Area of a Triangle 1

The equation to calculate the area of a triangle is:

\text{Area} = \dfrac{1}{2} \times b \times h

Where b is the base width of the triangle and h is the vertical height. 

Level 4-5

Area of a Triangle 2

Another way to calculate the area of a triangle is as follows:

\dfrac{1}{2} \times a \times b \times \sin(C)

Where a and b are side lengths and C is the angle between the side lengths. 

Level 6-7

Example 1: Finding the Area of a Trapezium

The shape is a trapezium with a perpendicular height of 4mm. 

Calculate the area of the trapezium.

[2 marks]

Formula: \text{Area}=\frac{1}{2}(a+b)h,

where a = 8,\hspace{1mm}b = 12.5, and h = 4 .

\begin{aligned}\text{Area } &= \dfrac{1}{2}(8+12.5) \times 4 \\ &=\dfrac{1}{2} \times 20.5 \times 4 \\ &= 41\text{ mm}^2 \end{aligned}

Level 4-5

Example 2: Area of a Triangle

The triangle has a base of 6cm and an area equal to 24\text{ cm}^2.

Calculate its perpendicular height.

[2 marks]

Formula: \text{Area} = \dfrac{1}{2}\times b \times h

So we can add the numbers we know into the equation and solve for h:

24= \dfrac{1}{2}\times 6 \times h

24= 3 \times h

\dfrac{24}{3}= h

h = 8\text{cm}

Level 4-5
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Example Questions

The formula for the area is \frac{1}{2} \times b \times h, where b = 11.5 \text{ and } h = 12. So, we get:

 

\text{Area } = \dfrac{1}{2} \times 11.5 \times 12 = 69\text{cm}^2

To work out the area, we will need to find the perpendicular height by forming a right-angled triangle,

 

The hypotenuse of the right-angled triangle is 5cm and the base is 3cm (8cm-5cm=3cm), so we get the perpendicular height of the trapezium as,

 

\text{Perpendicular height} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\text{cm}

 

Now we know the perpendicular height, we can calculate the area.

 

\text{Area} = \dfrac{1}{2}(a + b)h = \dfrac{1}{2}(5 + 8) \times 4 = 26\text{cm}^2

Area of a parallelogram is given by the formula, 

 

\text{Area}=\text{base}\times\text{height}.

 

Therefore,

 

\text{Area}=8 \times 15 =120\text{cm}^2 

As we need to find a missing side-length rather than the area, we’re going to have to set up an equation and rearrange it to find x. The formula for the area we’ll need here is

 

\dfrac{1}{2}ab \sin(C),

 

so, our equation is

 

\dfrac{1}{2} \times 2.15 \times x \times \sin(26) = 1.47

 

Simplifying the left-hand-side, we get

 

1.075 \sin (26) \times x = 1.47

 

Finally, dividing through by 1.075\sin(26), and putting it into a calculator, we get

 

x = \dfrac{1.47}{1.075\sin(26)} = 3.12\text{m (2 d.p.)}

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