The formula for the perimeter of a rectangle | List of Perimeter of Rectangle formulas - BYJUS

# Perimeter of Rectangles Formulas

In simple words, the perimeter is the length of the boundary of any closed figure. A rectangle is a four sided polygon, that is, a quadrilateral having equal opposite sides and the interior angles all measure 90 degrees. Here, we will focus on the application of the formula of the perimeter of a rectangle. ...Read MoreRead Less

### Perimeter of Rectangle formula

The perimeter of a rectangle is defined as the measure of its boundary. The boundary of a rectangle consists of two equal and opposite sides; the longer sides and two equal shorter sides. The perimeter of the rectangle is the sum of the measure of each of its sides. In other words, add the measures of the length and the width of the rectangle twice. This provides the perimeter of a rectangle.

Perimeter of a rectangle is given by the following formula,

$$P=(2\times l)+(2\times w)$$

Where,

$$P \rightarrow ~$$perimeter of the rectangle

$$l \rightarrow ~$$length of the rectangle

$$w \rightarrow ~$$width of the rectangle

### How to use the Perimeter of a Rectangle formula?

The perimeter of any shape is the measure of its boundary. Hence, for a rectangle the perimeter formula can be obtained as:

Perimeter of any figure = length of the boundary.

Perimeter of the rectangle = $$l+w+l+w$$

= $$l+l+w+w$$

= $$(2 \times l)+(2 \times w)$$

or

= $$2 \times (l+w)$$

Therefore, the perimeter of a rectangle is two times the sum of its length and width.

Perimeter is measured in the same units as that of the sides of the figure, that is; cm, mm, m, km and so on.

For example, the perimeter of the given rectangle can be calculated as:

Perimeter of the rectangle = $$(2 \times l)+(2 \times w)$$

= $$(2 \times 5)+(2 \times 2)$$      [Substitute values]

= (10+4)                        [Simplify]

Hence, the perimeter of the given rectangle is 14 cm.

### Solved Examples

Example 1:

Find the perimeter of the rectangle shown here.

Solution:

Perimeter of the rectangle, P = $$(2\times l ) + (2 \times w)$$

P = $$(2\times 5) + (2 \times 9 \frac{1}{2})$$    [Substituting the values]

P = $$10 + 2\times (9+\frac{1}{2})$$       [Simplify]

P = 10 + 18 + 1                   [Simplify further]

Hence the perimeter is 29 inches.

Example 2:

Find the length of the given rectangle if the perimeter is 25 m and the width is 5 m.

Solution:

Perimeter of the rectangle, P = $$(2\times l ) + (2 \times w)$$

25 = $$(2\times l) + (2 \times 5)$$     [Substituting the values]

25 = $$(2\times l) + (10)$$         [Simplify]

15 = $$(2\times l)$$                     [Subtract 10 from both sides]

$$l=\frac{15}{2}$$                            [Divide both sides by 2]

$$l=7.5$$

Hence, the length is 7.5 m.

Example 3: You wish to decorate these rectangular shaped cards by pasting colored ribbons along their boundary. Which card requires a greater quantity of ribbons?

Solution:

Perimeter of the rectangle, P = $$(2\times l ) + (2 \times w)$$

Both cards are rectangular in shape, hence,

Perimeter of the first card = $$(2\times 15 ) + (2 \times 6)$$     [Substituting values]

= (30) + (12)                                                              [Simplify]

Perimeter of the second card = $$(2\times 18 ) + (2 \times 2)$$      [Substituting values]

= (36) + (4)                                                                      [Simplify]

42 > 40

Hence, the perimeter of the first card is greater than the second card.

So, you will need a greater quantity of ribbons for the first rectangular shaped card.

Example 4: While delivering a package, a worker had to paste the address label on a box and stick a tape around it. He noticed that the width of the rectangular box is twice the width of the rectangular label. How much tape does the worker need?

Solution:

The length of tape required will be equal to the perimeter of the label.

Length of the label = 4 cm

It is said that the width of the label is twice the width of the package; this means that to find the width of the label we need to divide the width of the box by two.

Width of the label = $$\frac{\text{width of the box}}{2}=\frac{6}{2}=3~$$cm

Perimeter of the rectangular label = $$(2\times l ) + (2 \times w)$$

Substituting the values we get,

= $$(2\times 4 ) + (2 \times 3)$$

Simplifying this further we get

= (8) + (6)

= 14

Therefore, the worker needs 14 cm of tape

Example 5: The length of a rectangular patch of land is twice the width of the same. What would be the cost of fencing the entire boundary if the cost of fencing one metre is $12? Solution: The width of the rectangular patch of land = 12 m The length of the rectangular patch of land = $$12 \times 2=24$$ cm Perimeter of the rectangular patch = $$(2\times l ) + (2 \times w)$$ Substituting the values we get: = $$(2\times 24 ) + (2 \times 12)$$ Simplifying the equation we get the following: = 48 + 24 = 78 m The perimeter is 78 m. The cost of fencing the entire boundary can be found by multiplying the perimeter with the cost of fencing one metre. Cost of fencing the rectangular patch of land = $$78 \times 12$$ = $$78 \times 12$$ = 936 Therefore, the cost of fencing the rectangular patch of land is$936

Example 6: Using square toy blocks, Ryan wanted to make a boundary around  a rectangular figure that has a length four times its width. How many toy blocks will he use if the length of each block is 2 cm?

Solution:

Width of the rectangular figure  = 10 cm

Length of the rectangular figure = $$4 \times 10$$ cm

Perimeter of the rectangular figure = $$(2\times l ) + (2 \times w)$$

Substituting the values we get the following:

= $$(2\times 40) + (2 \times 10)$$

Simplifying the equation further we get the following:

= (80) + (20)

= 100 cm

The perimeter of the rectangle is 100 cm.

To find the number of toy blocks Ryan needs, we divide the perimeter of the figure by the side length of the toy block.

Number of blocks used to make the  boundary = $$\frac{\text{Perimeter of figure}}{\text{Side length of toy block}}$$

= $$\frac{100}{2}$$

= 50 blocks

Therefore, a total of 50 square toy blocks were used to make the boundary.