One to one maths interventions built for KS4 success

Weekly online one to one GCSE maths revision lessons now available

In order to access this I need to be confident with:

Arithmetic Adding and subtracting fractions Multiplying and dividing fractions Adding and subtracting decimals Multiplying and dividing decimals Area of a rectangle Pythagoras theoremThis topic is relevant for:

Here we will learn about the area of a triangle, including how to find the area of a triangle with given dimensions.

There are also area of a triangle* *worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

The **area of a triangle** is the amount of enclosed **two-dimensional space** that three connected straight lines take up.

In order to derive the area of a triangle, we start with the area of a rectangle.

The area of the rectangle below would be calculated by multiplying the base by the height ( b \times h ). This is usually pronounced as “base times height”.

We can split a rectangle into two smaller rectangles by drawing a perpendicular line through any two parallel sides of the rectangle.

Splitting these two rectangles in half by joining the diagonals together, we can create two pairs of congruent (identical) right-angled triangles.

The area of the rectangle is now halved to get the area of the triangle, so the area of the triangle is exactly half the area of the rectangle, when we know the base and the perpendicular height of the triangle**.**

The formula for the area of a triangle is:

\text{Area of a triangle}=\frac{\text{base }\times \text{ height}}{2}This can be shortened to

A=\frac{1}{2}bhwhere b is the base length and h is the perpendicular height of the triangle.

(Perpendicular means that the base and the height are at right angles to each other.)

Your final answer must be given in square units, e.g. cm^2, \ m^2, or mm^2 .

**Higher** – There is an alternative method to find the area of a triangle which involves sine, two sides and the included angle. This is part of trigonometry.

**Step-by-step guide:** Area of a triangle ( \frac{1}{2}ab \sin C )

On this page we’ll look at the area of** scalene triangles**. A scalene triangle is a triangle where all of the sides and all of the angles are different.

This means that a scalene triangle can contain three acute angles, a right-angle, or one obtuse angle (and two acute angles).

**Remember:** All of the angles within any triangle must add up to 180^{\circ} .

The triangle below has the three sides labelled a, b, and c, and the three vertices labelled A, B, and C. The side a opposes the vertex A and so on.

You should also be familiar with the notation that this triangle is called triangle ABC .

Each side could also be labelled using the capital letters that denote the start and end point of the line. For example, the side c is also known as the side AB , the side a can be referred to as side BC , and side b can be expressed as side AC .

In order to calculate the area of a triangle:

**Identify the base and perpendicular height of the triangle.****Write the area formula.****Substitute known values into the area formula.****Solve the equation.****Write the answer, including the units.**

Get your free area of a triangle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free area of a triangle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Area of a triangle** is part of our series of lessons to support revision on **area**. You may find it helpful to start with the main area lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

- Area
- Area of a circle
- Area of a quadrilateral
- Area of a trapezium
- Area of a parallelogram
- Area of compound shapes
- Pi r squared
- Area of a rhombus
- Area of an isosceles triangle
- Area of an equilateral triangle
- Area of a right angled triangle
- How to work out area
- Area of a rectangle
- Area of a hexagon
- Area of a pentagon

Calculate the area of the right-angled triangle ABC . Write your answer in square centimetres.

**Identify the base and perpendicular height of the triangle.**

The sides AC and BC are perpendicular to each other (the angle at C is a right-angle) and so we can state the values for the base and the perpendicular height of the triangle to be:

The base: b=3.2 \ cm

The perpendicular height: h=5.3 \ cm

2**Write the area formula.**

3**Substitute known values into the area formula.**

4**Solve the equation.**

Work out the value for A by solving the equation:

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(3.2)(5.3) \\\\ &=\frac{1}{2}\times 3.2\times 5.3\\\\ &=8.48 \end{aligned}5**Write the answer, including the units.**

**Remember:** Your final answer must be in square units.

For triangle ABC , the angle at C is an obtuse angle, the length AC=12cm , and the height of the triangle is 9cm . Calculate the area of triangle ABC.

**Identify the base and perpendicular height of the triangle.**

As AC is perpendicular to the height stated for the triangle, we can state:

\begin{aligned}
&b=12 \ cm \\\\
&h=9 \ cm.
\end{aligned}

**Write the area formula.**

A=\frac{1}{2}bh

**Substitute known values into the area formula.**

\begin{aligned}
A &=\frac{1}{2}bh \\\\
&= \frac{1}{2}(12)(9)
\end{aligned}

**Solve the equation.**

Solving for A , we have

\begin{aligned}
A &=\frac{1}{2}bh \\\\
&= \frac{1}{2}(12)(9) \\\\
&=\frac{1}{2}\times 12\times 9\\\\
&=54
\end{aligned}

**Write the answer, including the units.**

A=54\text{ cm}^2

Calculate the area of triangle ABC below. Write your answer in square metres.

**Identify the base and perpendicular height of the triangle.**

Currently we have 2 different units here. We must convert them to a common unit (here we will use metres as the question requires the answer to be in square metres):

240 \ cm = 2.4 \ m.

So the base and the perpendicular height of the triangle are:

\begin{aligned}
&b=2.4 \ m \\\\
&h= 2\ m
\end{aligned}

**Write the area formula.**

A=\frac{1}{2}bh

**Substitute known values into the area formula.**

\begin{aligned}
A &=\frac{1}{2}bh \\\\
&= \frac{1}{2}(2)(2.4)
\end{aligned}

**Solve the equation.**

\begin{aligned} A &=\frac{1}{2}bh \\\\
&= \frac{1}{2}(2)(2.4) \\\\
&=\frac{1}{2}\times 2\times 2.4\\\\
&=2.4 \end{aligned}

**Write the answer, including the units.**

A=2.4\text{ m}^2

PQR is an obtuse scalene triangle. The point S lies on the line PQ such that angle PSR=90^{\circ} .

Calculate the area of the triangle PQR below, correct to 2 decimal places.

**Identify the base and perpendicular height of the triangle.**

Here, we need to distinguish which two lengths are perpendicular to one another as we can use these as the base and the height of the triangle PQR .

As the question states that angle PSR=90^{\circ} , the side of the triangle PQ is perpendicular to the length RS which is the height of the triangle. We therefore can state:

\begin{aligned} &b= 11.02 \ cm \\\\ &h= 4.48\ cm. \end{aligned}

**Write the area formula.**

A=\frac{1}{2}bh

**Substitute known values into the area formula.**

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(11.02)(4.48) \end{aligned}

**Solve the equation.**

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(11.02)(4.48)\\\\ &=24.6848 \end{aligned}

**Write the answer, including the units.**

A=24.68 \text{ cm}^2 \text{ (2dp)}

A rectangle and a triangle are placed together to make a compound shape ABCDE .

Calculate the area of the compound shape to 1 decimal places.

**Identify the base and perpendicular height of the triangle.**

If we split the compound shape into two polygons, we have a rectangle and a triangle.

As AB and DE are two parallel sides of a rectangle, AB=DE and so we can calculate the vertical height of the triangle by subtracting 4m from 6m. This gives us the following dimensions on the original diagram:

If we sketch another copy of the triangle and label the dimensions of it, we get:

We can now state the values for the base and perpendicular height:

\begin{aligned}
&b= 5 \ m \\\\
&h= 2 \ m.
\end{aligned}

**Write the area formula.**

A=\frac{1}{2}bh

**Substitute known values into the area formula.**

\begin{aligned}
A &=\frac{1}{2}bh \\\\
&= \frac{1}{2}(5)(2)
\end{aligned}

**Solve the equation.**

Work out the calculation

\begin{aligned} A &=\frac{1}{2}bh \\\\ &= \frac{1}{2}(5)(2)\\\\ &=\frac{1}{2}\times 5\times 2\\\\ &=5 \end{aligned}

Now we must find the area of the rectangle.

\begin{aligned} A &=l\times w \\\\ &= 5\times 4\\\\ &=20 \end{aligned}

\text{Total area}=5+20=25

**Write the answer, including the units.**

A=25\text{ m}^2.

Sometimes a question might give you the area and ask you to work out the missing length. In order to do this you must rearrange the formula to make the missing value the subject, and then solve for this letter.

In order to find a missing length given the area:

**Write the area formula.****Substitute the known values into the area formula.****Rearrange the equation and solve.****Write the answer, including the units.**

The area of a triangle is 45 \ cm^2 . The height of the triangle is 10 \ cm. Find the length of the base of the triangle.

**Write the area formula**

A=\frac{1}{2}bh

**Substitute the known values into the area formula**

\begin{aligned}
A&=\frac{1}{2}bh\\\\
45&=\frac{1}{2}\times b\times 10
\end{aligned}

**Rearrange the equation and solve.**

We can solve the equation by rearranging to find the unknown b.

First we can tidy up the right-hand-side

\begin{aligned} 45&=\frac{1}{2}\times b\times 10\\\\ 45&=b\times 5 \end{aligned}

Then we divide both sides by 5.

\begin{aligned}
45&=b\times 5\\\\
\frac{45}{5}&=b\\\\
b&=9
\end{aligned}

**Write the answer, including the units.**

b=9\text{ cm}

**Remember:** Your final answer must be in linear units only as it is a length.

**Identifying the correct information to use**

The base and the height **must** be at a right angle to each other. They must be perpendicular. You may be given additional sides of a triangle that are not needed. You may also need to rotate the triangle around so that you can identify the base and height more easily.

**Units**

It is common to forget the units for area in the final answer. When calculating area, your answer must always have units squared (or square units).

1. Calculate the area of the triangle:

8\text{ cm}^2

7.5\text{ cm}^2

15\text{ cm}^2

16\text{ cm}^2

The base and the perpendicular height are 5 \ cm and 3 \ cm respectively, so the calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(5)(3) \\\\ &=\frac{1}{2}\times 5\times 3 \\\\ &=7.5 \end{aligned}

The units will be cm^2.

2. Find the area of the triangle. Give your answer in cm^2:

126\text{ cm}^2

252\text{ cm}^2

12.6\text{ cm}^2

25.2\text{ cm}^2

The base is 63 \ mm which needs to be converted to 6.3 \ cm. The height is 4 \ cm. The calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(6.3)(4) \\\\ &=\frac{1}{2}\times 6.3\times 4 \\\\ &=12.6 \end{aligned}

The units will be cm^2.

3. Find the area of the triangle:

24\text{ cm}^2

21\text{ cm}^2

12\text{ cm}^2

10.5\text{ cm}^2

It may be easier to see the base and height if you turn the triangle. The base and the height are 3 \ cm and 7 \ cm. The 8 \ cm is not needed. So the calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(3)(7) \\\\ &=\frac{1}{2}\times 3\times 7 \\\\ &=10.5 \end{aligned}

The units will be cm^2.

4. Triangle ABC is an obtuse scalene triangle. The angle at A=120^{\circ} . Calculate the area of the triangle. Give your answer in m^2 :

13.8 \text{ m}^2

13 \ 800 \text{ cm}^2

6.9\text{ m}^2

69 \ 000\text{ cm}^2

The base is 460 \ cm which needs to be converted to 4.6 \ m. The height is 3 \ m. The calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(4.6)(3) \\\\ &=\frac{1}{2}\times 4.6 \times 3 \\\\ &=6.9 \end{aligned}

The units will be m^2.

5. Find the area of the compound shape. Give your answer in m^2 :

1250\text{ m}^2

500\text{ m}^2

750\text{ m}^2

2000\text{ m}^2

The compound shape is made up of a rectangle and a triangle.

The area of the rectangle is 10\times{50}=500\text{ m}^2

The area of the triangle is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ &=\frac{1}{2}(50)(30) \\\\ &=\frac{1}{2}\times 50 \times 30 \\\\ &=750 \end{aligned}

The total area will be 500+750=1250

The units will be m^2.

6. The area of a triangle is 72 \ cm^2 . It has a height of 12 \ cm . Find the base length of the triangle.

24 \ cm

12 \ cm

6 \ cm

18 \ cm

The area is 84 \ cm^2 and the height is 14 \ cm. The calculation we need to do is:

\begin{aligned} A&=\frac{1}{2}bh \\\\ 84&=\frac{1}{2}(b)(14) \\\\ 84&= b\times 7 \\\\ b&=84\div 7 \\\\ b&=12 \end{aligned}

The units will be cm.

1. A triangle has a base of 6 \ cm and perpendicular height 7.2 \ cm.

Work out the area of the triangle.

**(2 marks)**

Show answer

\frac{1}{2}\times 6\times 7.2

**(1)**

21.6 \ cm^2

**(1)**

2. The diagram shows a triangular lawn.

A gardener wants to buy grass seed for this triangular lawn.

A packet of grass seed covers 15 \ m^2 of ground.

A packets of grass seed costs £6.99.

Calculate the cost of buying grass seed to cover the lawn.

**(4 marks)**

Show answer

\frac{1}{2}\times 9.5\times 8.7=41.325

**(1)**

41.325\div 15=2.755\text{ or }3

**(1)**

3\times 6.99

**(1)**

£20.97

**(1)**

3. Here is a triangle and a rectangle.

The area of the rectangle is 5 times the area of the triangle.

Work out the width x of the rectangle.

**(4 marks)**

Show answer

\frac{1}{2}\times 6\times 4=12

**(1)**

12\times 5=60

**(1)**

60\div 8 = 7.5

**(1)**

7.5 \ cm

**(1)**

You have now learned how to:

- Calculate the area of triangles
- Derive and apply the formula to calculate and solve problems involving area of triangles

The ancient Greeks used a different method for finding the area of a triangle. This is a useful method as we can work out the area of a triangle if we know the side lengths of the triangle. It is based on the semi-perimeter of the triangle (half of the perimeter).

The semi-perimeter is s and the side lengths of the triangle are a, b, and c.

s=\frac{a+b+c}{2} \text{Area of triangle}=\sqrt{s(s-a)(s-b)(s-c)}This formula is known as Heron’s formula.

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths tuition programme.