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Algebraic expressionsUnderstanding and using formulae

Area of quadrilaterals Rearranging equations Pythagoras’ Theorem TrigonometryThis topic is relevant for:

Here we will learn about the **area of an isosceles triangle** including how to find the area of an isosceles with given lengths and how to calculate those lengths if they are not given.

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.

An **isosceles triangle** is a type of triangle with two equal sides. The base angles which are opposite to the equal sides are also equal.

There are different types of isosceles triangles:

Isosceles right triangle

-two equal angles (45° each)

-two equal side lengths

A special case of an isosceles triangle is an **equilateral **triangle where all three sides and angles of the triangle are equal.

We can identify a triangle by putting a capital letter on each vertex (corner).

We can then refer to each of the sides of the triangle by using two letters to describe where the line starts and ends.

We can refer to the entire triangle by using all three letters.

E.g.

Name of sides:**side AB**, **side AC**, **side BC**

Name of triangle:**triangle ABC**

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

To find the area of a rectangle you must multiply **adjacent **sides together.

The area of the rectangle below would be calculated by multiplying the base x height

(b x h).

We can create an isosceles triangle by drawing two sides from the midpoint of a side of the rectangle to the corners.

The rectangle has been split into an isosceles triangle and two congruent (identical) right angled triangles. If we combine the area of the two right angled triangles they will form the pink isosceles triangle.

The area of the isosceles triangle is exactly half the area of the rectangle.

\[\text { Area of a triangle }=\frac{\text { base } \times \text { height }}{2}\]

This can be shortened to

\[A=\frac{1}{2} b h\]

where

Your final answer must be given in units^{2} (cm^{2}, m^{2}, mm^{2}).

To find the area of an isosceles triangle when only the side lengths have been given, we will need to calculate the height of the triangle using Pythagoras’ Theorem

**Finding the height using Pythagoras’ Theorem**

Let the base of the isosceles triangle be b and the two equal sides be l.

Drawing a perpendicular line from the top vertex to the base forms the height, h, and also forms a right angled triangle with hypotenuse l and short sides h and \cfrac{1}{2} \, b.

h=\sqrt{{{l}^{2}}-{{\left( \cfrac{1}{2} \, b \right)}^{2}}}We can also find the area of the isosceles triangle using the area of a triangle formula

Area=\cfrac{1}{2} \, ab\sin CIf we know (or can find) the two equal sides, l and the angle between them, \theta , the formula would give the area as

A=\cfrac{1}{2} \, {{l}^{2}}\sin \left( \theta \right)In order to find the area of a isosceles triangle:

1**Identify the height and base length of your triangle if given. (You might need to calculate these values)**

2**Write the appropriate** **formula**

A=\cfrac{1}{2} \,b h or A=\frac{1}{2}{{l}^{2}}\sin \left( \theta \right)

3**Substitute the values into the formula**

4**Calculate **

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

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

DOWNLOAD NOW**Area of isosceles triangles** 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 a right angled triangle
- Area of an equilateral triangle
- Area of a triangle
- How to work out area
- Area of a rectangle
- Area of a hexagon
- Area of a pentagon

Find the area of the triangle below:

**Identify the height and base length of the triangle****if given.**

2**Write down the appropriate formula**

\[A=\frac{1}{2} b h\]

3**Substitute the values into the formula**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(6)(10)
\end{aligned}\]

4**Calculate**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(6)(10)\\
&=30cm^2
\end{aligned}\]

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

Calculate the area of the triangle below:

**Identify the height and base length of the triangle if given.**

h = 120cm
b = 4m

*Note: You have 2 different units here. You must convert them to a common unit*:

**Write down the appropriate** **formula**

\[A=\frac{1}{2} b h\]

**Substitute the values into the formula.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(4)(1.2)
\end{aligned}\]

**Calculate**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(4)(1.2)\\
&=2.4m^2
\end{aligned}\]

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

Shown below is a triangular shaped field. Each cow needs 2000m^{2} to graze. How many cows can fit into this field?

**Identify the height and base length of the triangle if given.**

h = 100m
b = 150m

*Note: The * 125m* is not actually needed in this question and is just there to confuse you.*

**Write down the appropriate formula**

\[A=\frac{1}{2} b h\]

**Substitute the values into the formula.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(150)(100)
\end{aligned}\]

**Calculate**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(150)(100)\\
&=7500m^2
\end{aligned}\]

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

Now to calculate how many cattle will fit into the field take 7500m^{2} and divide it by 2000m^{2} because each cattle needs that much space to graze on the field.

7500 \div 2000 = 3.753 cows will be able to fit in the field.

Find the area of the triangle below:

**Identify the height and base length of the triangle if given.**

b = 4cm
h = 6.71cm

We can find the height by splitting the isosceles triangle into two right triangles. We can then apply Pythagoras’ Theorem to one of the right triangles to calculate its height.

\[\begin{array}{l}
h^{2}=7^{2}-2^{2} \\
h^{2}=49-4 \\
h^{2}=45 \\
h=\sqrt{45} \\
h \approx 6.71 \mathrm{~cm}(2 \mathrm{~d.p.})
\end{array}\]

**Write down the appropriate formula **

\[A=\frac{1}{2} b h\]

**Substitute the values into the formula.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(4)(6.71)
\end{aligned}\]

**Calculate**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(4)(6.71)\\
&=13.42cm^2
\end{aligned}\]

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

Below is the floor plan for a new house. Calculate the area of the plan.

**Identify the height and base length of the triangle if given.**

Split the plan into 2 shapes. We now have a square and an isosceles triangle.

For the triangle:

b = 8m h = 11m**Write down the appropriate formula **

\[A=\frac{1}{2} b h\]

**Substitute the values into the formula.**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(8)(11)
\end{aligned}\]

**Calculate**

\[\begin{aligned}
A &=\frac{1}{2} b h \\
&=\frac{1}{2}(8)(11)\\
&=44m^2
\end{aligned}\]

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

**Now you must find the area of the square.**

Area of square = 8^{2}

=64m^{2}

Total Area = 44 + 64 = 108m^{2}

Below shows an isosceles triangle with two sides of length 10 \, cm and a base angle of 75^{\circ}.

Find the area.

**Identify the height and base length of your triangle if given. (You might need to calculate these values)**

We have the length of the two equal sides, 10\, cm and the base angle of 75^{\circ}.

To use the formula A=\frac{1}{2}{{l}^{2}}\sin \left( \theta \right) we need the angle between the equal sides.

The base angles are both 75^{\circ}, so 180^{\circ}-150^{\circ}=30^{\circ} for the remaining angle

**Write down the appropriate formula **

A=\cfrac{1}{2} \, {{l}^{2}}\sin \left( \theta \right)

**Substitute the values into the formula.**

A=\cfrac{1}{2} \, {{(10)}^{2}}\sin \left( 30 \right)

**Calculate**

\begin{aligned}
A&=\cfrac{1}{2} \, {{\left( 10 \right)}^{2}}\sin \left( 30 \right) \\\\
& =25 \, cm^2
\end{aligned}

Sometimes a question might give you the area and ask you to work out the height or missing length. In order to do this you must rearrange the formula.

To find a missing length given the area:

**1Rearrange the formula**

2**Substitute in the values you know**

**3Calculate**

**Step by step guide:** Rearranging equations

Triangle ABC is an isosceles triangle with an area of 18cm^{2}. The height of the triangle is 6cm. Find the length of the base of the triangle.

**Rearrange the formula**

\[A=\frac{1}{2} b h\]

to make

\[2A=bh\]

Next we need to divide both sides of the formula by h

\[\frac{2(A)}{h}=b\]

**Now substitute the given values.**

Area = 18cm^{2}

Height = 6cm

\[b=\frac{2(18)}{6}\]

**Calculate**

\[\begin{aligned}
b&=\frac{2(18)}{6} \\
b&=\frac{36}{6} \\
b&=6 \mathrm{~cm}
\end{aligned}\]

**Identifying the correct information to use**

A question may give extra information that is not needed to answer it. Carefully identify the relevant pieces of information.

E.g.

To calculate the area here we only need the base and height.

Base= 12cm,

Height = 8cm

\[Area = \frac{12 \times 8}{2} = 48cm^{2}\]

We can ignore the values of the other sides (10cm)

**Units**

It is a common error to forget the units for area in the final answer. When calculating area, your answer must always have units squared.

1. Find the area of the triangle below

90cm^{2}

45cm^{2}

180cm^{2}

21cm^{2}

Using the lengths for base and height, the calculation we must perform is \frac{1}{2} \times 15 \times 6

2. Find the area of the triangle below giving your answer in cm^{2}

225cm^{2}

2250cm^{2}

22500cm^{2}

45000cm^{2}

After converting the height to the appropriate units, the calculation becomes \frac{1}{2} \times 300 \times 150

3. Shown below is a triangular shaped chicken enclosure cage. Each chicken needs 8m^{2} to roam around. How many chickens can fit into this enclosure?

125 chickens

15 chickens

16 chickens

30 chickens

The area of the enclosure is \frac{1}{2} \times 20 \times 12.5 = 125m^{2}. We then consider multiples of 8 to work out how many chickens will fit.

Or use a division method: 125 \div 8 = 15.625

So 15 chickens will fit.

4. Find the area of the triangle below (give your answer to two decimal places)

18.00cm^{2}

9.00cm^{2}

8.71cm^{2}

17.42cm^{2}

The height of the triangle can be found using Pythagoras’ Theorem.

Height = \sqrt{6^{2}-1.5^{2}} Height = 5.809…

The area of the triangle is then given by \frac{1}{2} \times 3 \times 5.809…

5. Find the area of the shaded region below (round your answer to one decimal place)

24.6m^{2}

18.3m^{2}

12.3m^{2}

19.4m^{2}

The shape can be split into a rectangle and a triangle.

The area of the rectangle is 3 \times 4 = 12m^{2}

The area of the triangle is \frac{1}{2} \times 3 \times 4.2 = 6.3m^{2}

The total area is 12 + 6.3 = 18.3m^{2}

6. Triangle MNP is an isosceles triangle with an area of 20cm^{2} . The base length of the triangle is 4cm . Find the height of the triangle.

80cm

5cm

2.5cm

10cm

Starting with the formula and the information we know already,

Area = \frac{1}{2} \times base \times height

20 = \frac{1}{2} \times 4 \times height

20 = 2 \times height

So the height is 10cm

1. This pattern is made from three identical isosceles triangles. Find the total area.

**(3 marks)**

Show answer

10 \div 2 = 5

**(1)**

One triangle:

\begin{aligned} A&= \frac{1}{2} \times 5 \times 12\\ A&=30cm^{2} \end{aligned}**(1)**

Total area:

3 \times 30 = 90cm^{2}**(1)**

2. (a) Find the area of the following isosceles triangle:

(b) The isosceles triangle below has the same area as the triangle in

part (a).

Work out the height of this triangle.

**(3 marks)**

Show answer

\begin{aligned}
A&= \frac{1}{2} \times 8 \times 12\\
A&=48cm^{2}
\end{aligned}

**(1)**

Rearrange area of triangle:

\begin{aligned} A &= \frac{1}{2} bh \\ 2A &= bh\\ h & = \frac{2A}{b} \end{aligned}

Substitute in values:

h= \frac{2 \times 48}{16}**(1)**

**(1)**

3. Calculate the area of the isosceles triangle.

**(4 marks)**

Show answer

Find height of triangle using Pythagoras:

h^{2}=5^{2}-3^{2}**(1)**

**(1)**

**(1)**

**(1)**

You have now learned how to:

- Apply formula to calculate and solve problems involving the area of triangles
- Use Pythagoras’ Theorem to solve problems involving triangles

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