# Calculating Area

See also: Properties of PolygonsArea is a measure of how much space there is inside a shape. Calculating the area of a shape or surface can be useful in everyday life – for example you may need to know how much paint to buy to cover a wall or how much grass seed you need to sow a lawn.

This page covers the essentials you need to know in order to understand and calculate the areas of common shapes including squares and rectangles, triangles and circles.

### Calculating Area Using the Grid Method

When a shape is drawn on a scaled grid you can find the area by counting the number of grid squares inside the shape.

In this example there are 10 grid squares inside the rectangle.

In order to find an area value using the grid method, we need to know the size that a grid square represents.

This example uses centimetres, but the same method applies for any unit of length or distance. You could, for example be using inches, metres, miles, feet etc.

In this example each grid square has a width of 1cm and a height of 1cm. In other words each grid square is one 'square centimetre'.

Count the grid squares inside the large square to find its area..

There are 16 small squares so the area of the large square is 16 square centimetres.

In mathematics we abbreviate 'square centimetres' to cm^{2}. The ^{2} means ‘squared’.

Each grid square is 1cm^{2}.

The area of the large square is 16cm^{2}.

**Counting squares on a grid to find the area works for all shapes – as long as the grid sizes are known. **However, this method becomes more challenging when shapes do not fit the grid exactly or when you need to count fractions of grid squares.

In this example the square does not fit exactly onto the grid.

We can still calculate the area by counting grid squares.

- There are 25 full grid squares (shaded in blue).
- 10 half grid squares (shaded in yellow) – 10 half squares is the same as 5 full squares.
- There is also 1 quarter square (shaded in green) – (¼ or 0.25 of a whole square).
- Add the whole squares and fractions together: 25 + 5 + 0.25 = 30.25.

The area of this square is therefore 30.25cm^{2} .

You can also write this as 30¼cm^{2}.

Although using a grid and counting squares within a shape is a very simple way of learning the concepts of area it is less useful for finding exact areas with more complex shapes, when there may be many fractions of grid squares to add together.

Area can be calculated using simple formulae, depending on the type of shape you are working with.

The remainder of this page explains and gives examples of how to calculate the area of a shape without using the grid system.

## Areas of Simple Quadrilaterals:

Squares and Rectangles and Parallelograms

The simplest (and most commonly used) area calculations are for squares and rectangles.

To find the area of a rectangle, multiply its height by its width.

Area of a rectangle = height × width

For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area. This is the same as saying length^{2} or length squared.

It is good practice to check that a shape is actually a square by measuring two sides. For example, the wall of a room may look like a square but when you measure it you find it is actually a rectangle.

Often, in real life, shapes can be more complex. For example, imagine you want to find the area of a floor, so that you can order the right amount of carpet.

A typical floor-plan of a room may not consist of a simple rectangle or square:

In this example, and other examples like it, the trick is to split the shape into several rectangles (or squares). It doesn’t matter how you split the shape - any of the three solutions will result in the same answer.

Solution 1 and 2 require that you make two shapes and add their areas together to find the total area.

For solution 3 you make a larger shape (A) and subtract the smaller shape (B) from it to find the area.

Another common problem is to the find the area of a border – a shape within another shape.

This example shows a path around a field – the path is 2m wide.

Again, there are several ways to work out the area of the path in this example.

You could view the path as four separate rectangles, calculate their dimensions and then their area and finally add the areas together to give a total.

A faster way would be to work out the area of the whole shape and the area of the internal rectangle. Subtract the internal rectangle area from the whole leaving the area of the path.

- The area of the whole shape is 16m × 10m = 160m
^{2}. - We can work out the dimensions of the middle section because we know the path around the edge is 2m wide.
- The width of the whole shape is 16m and the width of the path across the whole shape is 4m (2m on the left of the shape and 2m on the right). 16m - 4m = 12m
- We can do the same for the height: 10m - 2m - 2m = 6m
- So we have calculated that the middle rectangle is 12m × 6m.
- The area of the middle rectangle is therefore: 12m × 6m = 72m
^{2}. - Finally we take the area of the middle rectangle away from the area of the whole shape. 160 - 72 = 88m
^{2}.

The area of the path is 88m^{2}.

A **parallelogram** is a four-sided shape with two pairs of sides with equal length – by definition a rectangle is a type of parallelogram. However, most people tend to think of parallelograms as four-sided shapes with angled lines, as illustrated here.

The area of a parallelogram is calculated in the same way as for a rectangle (height × width) but it is important to understand that height does not mean the length of the vertical (or off vertical) sides but the distance between the sides.

From the diagram you can see that the height is the distance between the top and bottom sides of the shape - not the length of the side.

Think of an imaginary line, at right angles, between the top and bottom sides. This is the height.

## Calculating the Area of Triangles

It can be useful to think of a triangle as half of a square or parallelogram.

Assuming you know (or can measure) the dimensions of a triangle then you can quickly work out its area by using this formula:

Area of a triangle = (height × width) ÷ 2.

In other words you can work out the area of a triangle in the same way as the area for a square or parallelogram, then just divide your answer by 2.

The height of a triangle is measured as a right-angled line from the bottom line (base) to the ‘apex’ (top point) of the triangle.

**Here are some examples:**

The area of the three triangles in the diagram above is the same.

Each triangle has a width and height of 3cm.

The area is calculated:

(height × width) ÷ 2

3 × 3 = 9

9 ÷ 2 = 4.5

The area of each triangle is 4.5cm^{2}.

In real-life situations you may be faced with a problem that requires you to find the area of a triangle, such as:

You want to paint the gable end of a barn. You only want to visit the decorating store once to get the right amount of paint. You know that a litre of paint will cover 10m^{2} of wall. How much paint do you need to cover the gable end?

**You need three measurements:**

A - The total height to the apex of the roof.

B - The height of the vertical walls.

C - The width of the building.

**In this example the measurements are:**

A - 12.4m

B - 6.6m

C - 11.6m

The next stage requires some additional calculations. Think about the building as two shapes, a rectangle and a triangle. From the measurements you have you can calculate the additional measurement needed to work out the area of the gable end.

**Measurement D = 12.4 – 6.6**

**D = 5.8m**

You can now work out the area of the two parts of the wall:

Area of the rectangular part of the wall: 6.6 × 11.6 = 76.56m^{2}

Area of the triangular part of the wall: (5.8 × 11.6) ÷ 2 = 33.64m^{2}

Add these two areas together to find the total area:

**76.56 + 33.64 = 110.2m ^{2}**

As you know that one litre of paint covers 10m^{2} of wall so we can work out how many litres we need to buy:

110.2 ÷ 10 = 11.02 litres.

In reality you may find that paint is only sold in 5 litre or 1 litre cans, the result is just over 11 litres. You may be tempted to round down to 11 litres but, assuming we don’t water down the paint, that won’t be quite enough. So you will probably round up to the next whole litre and buy two 5 litre cans and two 1 litre cans making a total of 12 litres of paint. This will allow for any wastage and leave most of a litre left over for touching up at a later date. And don’t forget, if you need to apply more than one coat of paint, you must multiply the quantity of paint for one coat by the number of coats required!

## Areas of Circles

In order to calculate the area of a circle you need to know its **diameter** or **radius**.

The **diameter** of a circle is the length of a straight line from one side of the circle to the other that passes through the central point of the circle. The diameter is twice the length of the radius (diameter = radius × 2)

The **radius** of a circle is the length of a straight line from the central point of the circle to its edge. The radius is half of the diameter. (radius = diameter ÷ 2)

You can measure the diameter or radius at any point around the circle – the important thing is to measure using a straight line that passes through (diameter) or ends at (radius) the centre of the circle.

In practice, when measuring circles it is often easier to measure the diameter, then divide by 2 to find the radius.

You need the radius to work out the area of a circle, the formula is:

Area of a circle = πR^{2}.

This means:

π = Pi is a constant that equals 3.142.

R = is the radius of the circle.

R^{2} (radius squared) means radius × radius.

Therefore a **circle with a radius of 5cm** has an area of:

3.142 × 5 × 5 = 78.55cm^{2}.

A **circle with a diameter of 3m** has an area:

First, we work out the radius (3m ÷ 2 = 1.5m)

Then apply the formula:

πR^{2}

3.142 × 1.5 × 1.5 = 7.0695.

The area of a circle with a diameter of 3m is 7.0695m^{2}.

### Final Example

This example pulls on much of the content of this page for solving simple area problems.

This is the Ruben M. Benjamin House in Bloomington Illinois, listed on The United States National Register of Historic Places (Record Number: 376599).

This example involves finding the area of the front of the house, the wooden slatted part – excluding the door and windows. The measurements you need are:

A – 9.7m | B – 7.6m |

C – 8.8m | D – 4.5m |

E – 2.3m | F – 2.7m |

G – 1.2m | H – 1.0m |

Notes:

- All measurements are approximate.
- There is no need to worry about the border around the house – this has not been included in the measurements.
- We assume all rectangular windows are the same size.
- The round window measurement is the diameter of the window.
- The measurement for the door includes the steps.

What is the area of the wooden slatted part of the house?

Workings and answers below:

Answers to above example

First, work out the area of the main shape of the house – that is the rectangle and triangle that make up the shape.

The main rectangle (B × C) 7.6 × 8.8 = 66.88m^{2}.

The height of the triangle is (A – B) 9.7 – 7.6 = 2.1.

The area of the triangle is therefore (2.1 × C) ÷ 2.

2.1 × 8.8 = 18.48. 18.48 ÷ 2 = 9.24m^{2}.

The combined full area of the front of the house is the sum of the areas of the rectangle and triangle:

66.88 + 9.24 = 76.12m^{2}.

Next, work out the areas of the windows and doors, so they can be subtracted from the full area.

The area of the door and steps is (D × E) 4.5 × 2.3 = 10.35m^{2}.

The area of one rectangular window is (G × F) 1.2 × 2.7 = 3.24m^{2}.

There are five rectangular windows. Multiply the area of one window by 5.

3.24 × 5 = 16.2m2. (the total area of the rectangular windows).

The round window has a diameter of 1m its radius is therefore 0.5m.

Using πR^{2}, work out the area of the round window: 3.142 × 0.5 × 0.5 =. 0.7855m^{2}.

Next add up the areas of the door and windows.

(door area) 10.35 + (rectangle windows area) 16.2 + (round window area) 0.7855 = 27.3355

Finally, subtract the total area for the windows and doors from the full area.

76.12 – 27.3355 = 48.7845

The area of the wooden slatted front of the house, and the answer to the problem is: 48.7845m^{2}.

You may want to round the answer up to 48.8m^{2} or 49m^{2}.

See our page onEstimation, Approximation and Rounding.