The word "perimeter" comes from the Greek word perimetron. Peri - meaning around and metron - meaning measure. It has the following definition:
The distance around the external boundary of a two-dimensional shape.
A perimeter can be calculated by adding together the lengths of all the sides of the shape.
Here is a scalene triangle. To find its perimeter we add up the three side lengths. Notice that all sides are measured using the same units.
Perimeter | $=$= | $14+12+6$14+12+6 cm |
$=$= | $32$32 cm |
We can get clever with the way that we carry out calculations if the shape we are looking at has side lengths that are equal.
Here is a rectangle. We know that rectangles have opposite sides of equal length.
Perimeter | $=$= | $2\times13+2\times6$2×13+2×6 |
We have two $13$13 mm sides and two $6$6 mm sides. |
$=$= | $26+12$26+12 mm | ||
$=$= | $38$38 mm |
A square has $4$4 sides of the same length, so the perimeter of a square will be $4\times\text{side length}$4×side length.
Perimeter | $=$= | $4\times7.4$4×7.4 cm |
$=$= | $29.6$29.6 cm |
To find the perimeter of a shape, add up all the side lengths.
For finding the perimeter of a rectangle and a square we can use the following formulas:
All perimeters can be found by adding up each side length one at a time as we travel around the shape.
Here is a trapezium. To find its perimeter we add up the side lengths.
Perimeter | $=$= | $1.2+1.3+2.4+1.8$1.2+1.3+2.4+1.8 m |
$=$= | $6.7$6.7 m |
Also keep in mind that we can construct simple rules for other shapes as necessary, such as grouping together sides of the same lengths. For instance, let's calculate the perimeter of the figure below:
Perimeter | $=$= | $3\times1+2\times3.6+1.4+1.5+3.44$3×1+2×3.6+1.4+1.5+3.44 m |
$=$= | $3+7.2+2.9+3.44$3+7.2+2.9+3.44 m | |
$=$= | $16.54$16.54 m |
Find the perimeter of the shape pictured below, where the measurements are in metres:
Think: At first glance it might look like we don't have enough information to add up all the sides. If we think carefully, however, notice that the two unknown horizontal side lengths must add up to $13$13 m. Similarly, the two unknown vertical side lengths must add up to $8$8 m. We can picture this by moving the sides to the edge as follows:
This shape is a rectangle with exactly the same perimeter as our original shape (although a different area).
Do: So the perimeter of the composite shape can be calculated as:
Perimeter | $=$= | $2\times\left(8+13\right)$2×(8+13) m |
$=$= | $2\times21$2×21 m | |
$=$= | $42$42 m |
Perimeter is a measure of length. Recall from the previous lesson that common metric units for lengths are millimetres (mm), centimetres (cm), metres (m) and kilometres (km).
Find the perimeter of a rectangle with a length of $4$4 m, and a width of $23$23 m.
Find the perimeter of the isosceles triangle shown.
Consider the following figure.
Find the length $x$x.
Find the length $y$y.
Calculate the perimeter of the figure.
Given the perimeter of our shape, we can often work out how long any missing sides are by using the special features of that shape together with any other given measurements.
A $32$32 cm long piece of wire is bent into the shape of a square. What is the side length of the square?
Think: Since the piece of wire was $32$32 cm long, we know that the perimeter of the square will be $32$32 cm. We can reverse the formula for the perimeter of a square, $\text{Perimeter}=4\times\text{side length}$Perimeter=4×side length, to find a solution.
Do:
$P$P | $=$= | $4s$4s |
Write out the formula. |
$32$32 | $=$= | $4s$4s |
Substitute in the known value. |
$\frac{32}{4}$324 | $=$= | $\frac{4s}{4}$4s4 |
Divide both sides by $4$4. |
$\therefore s$∴s | $=$= | $8$8 |
So we find that the side length of the square is $8$8 cm.
A rectangle has length $8$8 cm and perimeter $26$26 cm. Find the width of the rectangle.
Think: We can try reversing the formula for the perimeter of a rectangle, $P=2L+2W$P=2L+2W, to find a solution.
Do:
$P$P | $=$= | $2L+2W$2L+2W |
Write out the formula. |
$26$26 | $=$= | $2\times8+2W$2×8+2W |
Substitute in the known values. |
$26$26 | $=$= | $16+2W$16+2W | |
$\therefore2W$∴2W | $=$= | $10$10 |
Take $16$16 from both sides. |
$W$W | $=$= | $5$5 |
Divide both sides by $2$2. |
So we find that the width of the rectangle is $5$5 cm.
Find the side length $d$d indicated on the diagram. The perimeter of the shape is $35$35 cm.
The length of a rectangle is twice its width and its perimeter is $30$30 cm.
Let $x$x be the width of the rectangle.
Write an expression for the perimeter of the rectangle in terms of $x$x.
Using the expression obtained from part (a), find the width of the rectangle.
Find the length of the rectangle.
Finding the boundary length of a shape lends itself to many practical applications such as pricing fencing to boarder a property or pool, finding length of skirting boards to surround a room, distance in a training circuit around the neighbourhood and so forth.
Andrea needs to buy fencing to keep he dog from digging up her vegetable garden. She has a rectangular garden bed that is $3$3 metres wide and $4$4 metres long and requires new fencing on all sides. If the fencing costs $\$25$$25 m, how much in total will the fencing cost?
Think: First find the length of fencing and then multiply by the cost.
Do:
Length of fence | $=$= | $2\left(L+W\right)$2(L+W) |
$=$= | $2\left(3+4\right)$2(3+4) m | |
$=$= | $14$14 m |
Thus the cost is:
Cost of fencing | $=$= | $14\times25$14×25 |
$=$= | $\$350$$350 |
A rectangular athletics field is $140$140 metres long and $40$40 metres wide. How far, in km, will an athlete run by completing $6$6 laps of along the edge of the field?
Find the length of wire needed to create the frame of this rectangular prism.
use metric units of length, their abbreviations, conversions between them, and appropriate levels of accuracy and choice of units
convert between metric units of length and other length units
calculate perimeters of familiar shapes, including triangles, squares, rectangles, and composites of these