Perimeter

Perimeter is the distance around any two-dimensional shape. To calculate the perimeter of a shape we simply add together the values of the various sides.

Example:

Calculate the perimeter of the shape below.

An L shape. The bottom is labelled 4.2 metres, the left-hand side is labelled 4 metres, and the top is labelled 2.6 metres.

Solution:

To calculate the perimeter we need to add up the value of each side. To do this for the shape above, we first need to determine the values of the three sides that are not labelled.

The same L shape from above, with the three unknown sides labelled A, B, and C.

From the image above, we can see that sides A and B together will give us another \(4m\). We can find side C by subtracting \(2.6\) from \(4.2\), therefore C is \(1.6m\).

So the perimeter of the shape will be:

\[P = 2.6 + 4 + 4.2 + 4 + 1.6 = 16.4m\]

Check your understanding of perimeter by working through the questions below.

Shapes with two perimeters

Sometimes you will encounter shapes with both an internal and an external perimeter, such as the external walls or concrete footings of a building:

An image of a concrete footing, with interior and exterior perimeter marked

To find one perimeter given the other you need to know the thickness of the wall or footing. For example, to find the external perimeter we will need to know this thickness measurement as well as the internal perimeter, and to find the internal perimeter we will need to know this thickness measurement as well as the external perimeter.

The two perimeters of a shape will be:

\[\begin{aligned} External\;perimeter &= internal\;perimeter\;+ \\\ & 2 \times the\;wall\;thickness \times the\;number\;of\;corners \\\ \end{aligned}\] \[\begin{aligned} Internal\;perimeter &= external\;perimeter\;- \\\ & 2 \times the\;wall\;thickness \times the\;number\;of\;corners \\\ \end{aligned}\]

Note: For a circular shape, you would use the radius of the external circle and the thickness of the wall to find the radius of the internal circle to calculate the circumference.

You may also need to find the average of the two perimeters, or the mean girth. How you do this will depend on the information you have.

If you know the values of both the external and internal perimeters:

\(Mean\;girth = (external\;perimeter + internal\;perimeter) \div 2\)

If you know the thickness of the wall and the external perimeter:

\[\begin{aligned} Mean\;girth &= external\;perimeter\;-\\\ & the\;wall\;thickness \times the\;number\;of\;corners \\\ \end{aligned}\]

If you know the thickness of the wall and the internal perimeter:

\[\begin{aligned} Mean\;girth &= internal\;perimeter\;+\\\ & the\;wall\;thickness \times the\;number\;of\;corners \\\ \end{aligned}\]

Example 1:

A rectangular brick wall has thickness \(120\,mm\) and internal perimeter \(48\,m\). Calculate the external perimeter.

Solution:

\[\begin{aligned} 120\,mm &= 0.12\,m \\\ External\;perimeter &= internal\;perimeter\;+ \\\ & 2 \times the\;wall\;thickness \times the\;number\;of\;corners \\\ \therefore External\;perimeter &= 48 + 2 \times 0.12 \times 4 \\\ \therefore External\;perimeter &= 48.96\,m \\\ \end{aligned}\]

Example 2:

A square room has an external perimeter of \(33.4\,m\) and a thickness of \(0.18\,m\). Calculate the mean girth of the perimeters of this room.

Solution:

\[\begin{aligned} Mean\;girth &= external\;perimeter\;- \\\ & the\;wall\;thickness \times the\;number\;of\;corners \\\ \therefore Mean\;girth &= 33.4 - 0.18 \times 4 \\\ \therefore Mean\;girth &= 32.68\,m \\\ \end{aligned}\]

Check your understanding of two perimeter shapes by working through the following questions.

Circumference

The perimeter of a circle is called the circumference of the circle.

We can calculate the circumference of a circle using the formula:

\[C = 2 \pi r\]

Where \(r\) is the radius, or half the diameter, of the circle.

A circle with the diameter marked as the distance from one side of the circle to the other when passing through the middle, the radius marked as the distance from the middle to the outside of the circle, and the circumference marked as the distance around the outside of the circle.

Example:

If a circle has a diameter of 34cm, calculate it’s circumference to two decimal places.

Solution:

A circle with the diameter labelled 34cm.

\[\begin{aligned} C &= 2 \pi r \\\ C &= 2 \times \pi \times 17 \\\ C &= 106.81cm \\\ \end{aligned}\]

Check your understanding of circumference by working through these questions.

Sectors

We can also find the perimeter of a sector of a circle. To do this we need to find the length of the arc and then add the two radii that form the sides of the sector.

To find the arc length of a sector we use:

\[Arc\;length = \frac{\theta}{360} \times 2\pi r\]

So to find the perimeter of a sector we have:

\[Perimeter = \frac{\theta}{360} \times 2\pi r + 2r\]

A circle with a major and minor sector labelled. The radius of the circle is also labelled, and the angle in the minor sector is labelled theta.

Example:

Find the perimeter of the shape below. Give the answer to one decimal place.

A sector of a circle with the radius labelled 32cm and the angle labelled 125 degrees.

Solution:

\[\begin{aligned} P &= \frac{\theta}{360} \times 2\pi r + 2r \\\ P &= \frac{125}{360} \times 2 \times \pi \times 32 + 2 \times 32 \\\ P &= 133.8cm \\\ \end{aligned}\]

Check your understanding of sector perimeters by working through the questions below.

Area

The area of a two-dimensional shape is how much space it takes up on a surface. It is measured in \(m^2\) (as the base unit of measurement; other derived units of measurement such as \(cm^2\) or \(mm^2\) may also be used as appropriate). The formula used to calculate the area of a shape will depend on the type of shape in question.

Triangles

To calculate the area of a triangle we use the formula:

\[Area = \frac{b \times h}{2}\]

which can also be written as:

\[Area = \frac{1}{2} b \times h\]

Where b is the value of the base of the triangle, and h is the height of the triangle.

A triangle with it's base labelled b and its height labelled h.

Example 1:

Calculate the area of the triangle.

A triangle with it's base labelled 18.2cm and its height labelled 7cm.

Solution:

\[\begin{aligned} Area &= \frac{1}{2} b \times h \\\ Area &= \frac{1}{2} \times 18.2 \times 7 \\\ A &= 63.7cm^2 \\\ \end{aligned}\]

Example 2:

Calculate the area of the triangle below.

A triangle with it's base labelled 52cm and its height labelled 61cm.

Solution:

\[\begin{aligned} Area &= \frac{b \times h}{2} \\\ Area &= \frac{52 \times 61}{2} \\\ A &= 1\,586cm^2 \\\ \end{aligned}\]

Check your understanding of the area of triangles by working through the questions below.

Quadrilaterals

A quadrilateral is any shape with four sides. However, not all quadrilaterals use the same formula for area.

For a rectangle we use:

\[Area = l \times w\]

Where l is the length of the rectangle, and w is the width.

A rectangle with it's base labelled l and its width labelled w.

Example:

Calculate the area of the rectangle.

A rectangle with it's base labelled 15.1m and its width labelled 6.9m.

Solution:

\[\begin{aligned} Area &= l \times w \\\ A &= 15.1 \times 6.9 \\\ A &= 104.19m^2 \\\ \end{aligned}\]

A square is similar, however the length and the width is the same, so we use the formula:

\[Area = l^2\]

As l is the length on all sides.

A square with two of its four equal sides labelled l.

Example:

Calculate the area of a square with a length of \(6.5\,m\)

A square with one of its sides labelled 6.5m.

Solution:

\[\begin{aligned} Area &= l^2 \\\ A &= 6.5^2 \\\ A &= 42.25\,m^2 \\\ \end{aligned}\]

To find the area of a parallelogram we use:

\[Area = l \times h\]

Where l is the length of the shape, and h is the height.

A parallelogram with its base labelled l and its height labelled h.

Example:

Calculate the area of the parallelogram.

A parallelogram with its base labelled 2.3m and its height labelled 6.1m.

Solution:

\[\begin{aligned} Area &= l \times h \\\ A &= 2.3 \times 6.1 \\\ A &= 14.03\,m^2 \\\ \end{aligned}\]

For the area of a trapezium we use the formula:

\[Area = \frac{1}{2} (a + b) \times h\]

Where a is the length of the top side, b is the length of the bottom, and h is the height of the trapezium.

A trapezium with its base labelled b, its top side labelled a, and its height labelled h.

Example:

Calculate the area of the trapezium. Give the answer to one decimal place.

A trapezium with its base labelled 2.9, its top side labelled 2.2, and its height labelled 1.7.

Solution:

\[\begin{aligned} Area &= \frac{1}{2} (a + b) \times h \\\ A &= \frac{1}{2} (2.2 + 2.9) \times 1.7 \\\ A &= 4.3 m^2 \\\ \end{aligned}\]

Check your knowledge of the area of quadrilaterals by working though the following questions.

Circles

The area of a circle is calculated using the formula:

\[Area = \pi r^2\]

Where r is the radius of the circle.

A circle with its radius labelled r.

Example:

Calculate the area of a circle that \(7.3\,m\) wide to two decimal places.

A circle with a dotted line from one side to the other, crossing the middle of the circle. This line is labelled 7.3mm.

Solution:

We know the diameter of the circle is \(7.3\,m\), therefore we first need to halve this to find the radius.

\[\begin{aligned} r &= 7.3 \div 2 = 3.65 \\\ Area &= \pi\;r^2 \\\ A &= \pi \times 3.65^2 \\\ A &= 41.85\,m^2 \\\ \end{aligned}\]

Sectors

We can also find the area of sectors within a circle using the formula:

\[Area = \pi r^2 \times \frac{\theta}{360}\]

Where r is again the radius of the circle, and \(\theta\) is the angle of the sector.

A circle with a minor sector sectioned off and labelled, along with the major sector shown. The minor sector has an angle labelled theta degrees.

Example:

Calculate the area of the shape below. Give the answer to one decimal place.

A circle with a minor sector cut out of it. The shape resembles pacman, with a radius of 4.2mm and an angle labelled 291 degrees.

Solution:

\[\begin{aligned} Area &= \pi r^2 \times \frac{\theta}{360} \\\ A &= \pi \times 4.2^2 \times \frac{291}{360} \\\ A &= 44.8\,mm^2 \\\ \end{aligned}\]

Check your knowledge of the area of circles and sectors by working through the following questions.