# Areas of Shapes

## Areas of Shapes Revision

**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

**Area of a Parallelogram**

The area of a **parallelogram** is base \times vertical height

\text{Area} = b\times h

**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.

**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.

**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.

**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}

**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}

## Areas of Shapes Example Questions

**Question 1:** The triangle below has a base of 11.5 cm and a perpendicular height of 12 cm.

Calculate its area.

**[2 marks]**

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

**Question 2:** Below is a trapezium with sides of length 8cm, 5cm, and 5cm as shown below.

Calculate the perpendicular height and use it to find the total area.

**[2 marks]**

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

**Question 3:** Calculate the area of the parallelogram shown below.

**[2 marks]**

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

**HIGHER ONLY**

Question 4:** The triangle below has area** 1.47\text{m}^2.

Work out the length of x to 2 decimal places.

**[3 marks]**

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.)}

## Areas of Shapes Worksheet and Example Questions

### (NEW) Areas of Shapes Exam Style Questions - MME

Level 4-5Level 6-7GCSENewOfficial MME## Areas of Shapes Drill Questions

### Area and Volume - Drill Questions

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