Example 1, StudySmarter Originals

As a matter of fact, we can use a general formula for the perimeter and area of a square to work out these measurements. Recall that a square is a type of quadrilateral, which is a polygon of four sides and four corners. Throughout this discussion, we shall look at the perimeter and area formulas of the six types of quadrilaterals mentioned in our previous topic: Quadrilaterals.

## Recap: Quadrilaterals

Before we begin, let us recount a quick review of quadrilaterals.

A **quadrilateral **is a polygon with four sides, four vertices and four angles.

It is also known as a tetragon or quadrangle. Quadrilaterals have two diagonals and the sum of all their interior angles equals 360^{o}. There are six types of quadrilaterals we should familiarise ourselves with, namely the square, the rectangle, the parallelogram, the trapezium, the rhombus and the kite. For a more detailed discussion regarding the characteristics of these mentioned quadrilaterals, you can refer to the article: Special Quadrilaterals.

## Perimeter of Quadrilaterals

We shall begin our topic with the perimeter formula for quadrilaterals. The perimeter of a quadrilateral is defined as the total length of its boundary. That is to say, it is **the sum of all its sides.** Thus, if we had a quadrilateral ABCD

The perimeter of quadrilaterals, StudySmarter Originals

with sides AB, BC, CD and DA, the perimeter, P is

$P=\overline{)AB}+\overline{)BC}+\overline{)CD}+\overline{)DA}$

or

$P=a+b+c+d$

Let us go through some worked examples involving this derivation.

Find the perimeter of the parallelogram below.

Example 2, StudySmarter Originals

**Solution**

Recall that a parallelogram has opposite sides of equal length. This means that PQ = SR and PS = QR. Thus, SR = 16 cm and QR = 10 cm.

In order to find the perimeter of this given shape, we simply add the total length of each side as mentioned.

$P=16+16+10+10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=2\left(16\right)+2\left(10\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=32+20\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=52cm$

Thus, the perimeter of this parallelogram is 52 cm.

Find the length of the missing sides of the kite below given that the perimeter is equal to 98 cm.

Example 3, StudySmarter Originals

**Solution **

First, note that a kite has two pairs of equal adjacent sides. This means that WZ = WX (and YZ = YX = 32 cm).

By the formula for the perimeter of a quadrilateral, we obtain

$P=32+32+WX+WZ\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 98=64+WX+WX\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 98=64+2WX$

Now rearranging this, we find that

$64+2WX=98\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2WX=98-64\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2WX=34$

Further simplifying,

$WX=\frac{34}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow WX=17cm$

Thus, the length of WX and WZ is 17 cm.

### Calculating the Perimeter of Quadrilaterals on a Plane

Say you were given a set of four points, (x, y), on a Cartesian plane. Joining these points together with four (separate) line segments, we find that it forms the shape of some quadrilateral. You are then told to find the perimeter of this shape using these coordinates. Is there a method we could use to accomplish this?

To tackle such a problem, we shall make use of the distance formula. This is introduced below.

Given two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}), the distance between A and B, denoted by D_{AB} is found using the formula below.

${D}_{AB}=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$

With that being said, we can find the perimeter of this quadrilateral by calculating the distance of these four line segments (formed by their corresponding pair of points) and adding them all up together.

**Note**: Given a set of four points, it may be helpful for you to sketch the outline of this quadrilateral so that we can roughly gauge the type of quadrilateral we are dealing with. By doing so, we would be able to notice its distinct properties and thus calculate its perimeter much more efficiently.

To get a better visual of this, let us look at the examples below.

Find the perimeter of a rectangle with vertices at A (1, 6), B (1, 2), C (4, 2) and D (4, 6).

**Solution**

Let us begin by sketching this quadrilateral on the Cartesian plane.

Example 4, StudySmarter Originals

Since we have a rectangle, AB = DC and AD = BC. Thus, we can use the distance formula for the lengths of AB and AD.

**Distance AB**, A (1, 6) and B (1, 2)

${D}_{AB}=\sqrt{{(1-1)}^{2}+{(2-6)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AB}=\sqrt{{\left(0\right)}^{2}+{(-4)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AB}=\sqrt{16}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AB}=4units$

**Distance AD**, A (1, 6) and D (4, 6)

${D}_{AD}=\sqrt{{(4-1)}^{2}+{(6-6)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AD}=\sqrt{{\left(3\right)}^{2}+{\left(0\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AD}=\sqrt{9}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AD}=3units$

**Perimeter ABCD**

$P=AB+DC+AD+BC\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=4+4+3+3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=14units$

Deduce the perimeter of a quadrilateral with vertices at A (–2, 8), B (0, 8), C (1, 4) and D (–1, 6).

**Solution**

Let us begin by sketching this quadrilateral on the Cartesian plane.

Example 5, StudySmarter Originals

Looking at the sketch above, we need to find the distance of AB, BC, CD and AD to calculate the perimeter of ABCD.

**Distance AB**, A (–2, 8) and B (0, 8)

${D}_{AB}=\sqrt{{(-2-0)}^{2}+{(8-8)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AB}=\sqrt{{(-2)}^{2}+{\left(0\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AB}=\sqrt{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AB}=2units$

**Distance BC**, B (0, 8) and C (1, 4)

${D}_{BC}=\sqrt{{(1-0)}^{2}+{(4-8)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{BC}=\sqrt{{\left(1\right)}^{2}+{(-4)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{BC}=\sqrt{1+16}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{BC}=\sqrt{17}units$

**Distance CD**, C (1, 4) and D (-1, 6)

${D}_{CD}=\sqrt{{(-1-1)}^{2}+{(6-4)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{CD}=\sqrt{{(-2)}^{2}+{\left(2\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{CD}=\sqrt{4+4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{CD}=\sqrt{8}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{CD}=2\sqrt{2}units$

**Distance AD**, A (-2, 8) and D (-1, 6)

${D}_{AD}=\sqrt{{(-1-2)}^{2}+{(6-8)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AD}=\sqrt{{(-3)}^{2}+{(-2)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AD}=\sqrt{9+4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AD}=\sqrt{13}units$

**Perimeter ABCD**

$P=AB+DC+AD+BC\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=2+\sqrt{17}+2\sqrt{2}+\sqrt{13}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=12.6units(correcttoonedecimalplace)$

## Area of Quadrilaterals

In this segment of our discussion, we shall move on to the area formula for quadrilaterals. The area of a quadrilateral is described by the space bounded by its boundary. Each of the six types of quadrilaterals we have mentioned previously has its own area formula.

Quadrilateral | Area |

Square Area of a square, StudySmarter Originals | $A=a\times a={a}^{2}$ |

Rectangle Area of a rectangle, StudySmarter Originals | $A=a\times b$ |

Parallelogram Area of a parallelogram, StudySmarter Originals | $A=a\times h$ |

Trapezium Area of a trapezium, StudySmarter Originals | $A=\frac{1}{2}\times \left(a+b\right)\times h$ |

Rhombus Area of a rhombus, StudySmarter Originals | $A=\frac{1}{2}\times {d}_{1}\times {d}_{2}$ |

Kite Area of a kite, StudySmarter Originals | $A=\frac{1}{2}\times {d}_{1}\times {d}_{2}$ |

Here are several worked examples that show how we can apply these formulas.

Calculate the area of the rhombus below given that PO = 7 cm and SO = 4cm. Point O is the point at which the two diagonals PR and SQ perpendicularly bisect each other.

Example 6, StudySmarter Originals

**Solution **

**Remember: **We need the measures of the diagonals, PR and SQ, of the rhombus to calculate its area. Since the diagonals of a rhombus are perpendicular and bisect each other, we find that PO = OR and SO = OQ and thus,

$PR=PO+OR=2PO\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}SQ=SO+OQ=2SO$

Solving this, we obtain

$PR=2\left(7\right)=14cm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}SQ=2\left(4\right)=8cm$

Thus, the vertical diagonal, PR is 14 cm and the horizontal diagonal, SQ is 8 cm. By the area formula for a rhombus,

$A=\frac{1}{2}\times 14\times 8\phantom{\rule{0ex}{0ex}}\Rightarrow A=\frac{112}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A=56c{m}^{2}$

Thus, the area of this rhombus is 56 cm^{2}.

What is the height of the trapezium below given that the area is 330 cm^{2}?

Example 7, StudySmarter Originals

**Solution **

Since AB is parallel to DC, the bases of this trapezium are given by AB = 13 cm and DC = 31 cm. The height is given by AD. By the formula for the area of a trapezium, we obtain

$A=\frac{1}{2}\times \left(13+31\right)\times AD\phantom{\rule{0ex}{0ex}}\Rightarrow 330=\frac{1}{2}\times \left(44\right)\times AD\phantom{\rule{0ex}{0ex}}\Rightarrow 330=22\times AD$

Rearranging this and simplifying our expression, we obtain

$22\times AD=330\phantom{\rule{0ex}{0ex}}\Rightarrow AD=\frac{330}{22}\phantom{\rule{0ex}{0ex}}\Rightarrow AD=15cm$

Thus, the height of this trapezium, AD is 15 cm.

### Calculating the Area of Quadrilaterals on a Plane

To find the area of a quadrilateral represented by a set of points on the Cartesian coordinate system, we would simply use the same technique as with the perimeter case. Yes, the distance formula applies here as well! However, we would need to be careful here as there are some area formulas that do not include the sides of a given quadrilateral but rather their diagonal or perpendicular height; such as the parallelogram, trapezium, rhombus and kite.

The examples below will give a clearer picture of this procedure.

Find the area of a kite with vertices at A (0, 4), B (1, 2), C (0, –4) and D (–1, 2).

**Solution**

Let us begin by sketching this quadrilateral on the Cartesian plane.

Example 8, StudySmarter Originals

Since we have a kite, we need the length of the diagonals to equate its area. The diagonals here are AC and BD.

**Distance AC**, A (0, 4) and C (0, –4)

${D}_{AC}=\sqrt{{(0-0)}^{2}+{(-4-4)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AC}=\sqrt{{\left(0\right)}^{2}+{(-8)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AC}=\sqrt{64}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AC}=8units$

**Distance BD**, B (1, 2) and D (–1, 2)

${D}_{BD}=\sqrt{{(-1-1)}^{2}+{(2-2)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{BD}=\sqrt{{(-2)}^{2}+{\left(0\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{BD}=\sqrt{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{BD}=2units$

**Area ABCD**

$A=\frac{1}{2}\times AC\times DB\phantom{\rule{0ex}{0ex}}\Rightarrow A=\frac{1}{2}\times 8\times 2\phantom{\rule{0ex}{0ex}}\Rightarrow A=8unit{s}^{2}$

Find the area of a square with vertices at A (2, 3), B (2, –3), C (–2, –3) and D (–2, 3).

**Solution **

Let us begin by sketching this quadrilateral on the Cartesian plane.

Example 9, StudySmarter Originals

As have a square, AB = BC = CD = AD. Thus, we can simply find one side to compute the area of this square. We shall choose to find AB.

**Distance AB**, A (2, 3) and B (2, –3)

${D}_{AB}=\sqrt{{(2-2)}^{2}+{(-3-3)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AB}=\sqrt{{\left(0\right)}^{2}+{(-6)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AB}=\sqrt{36}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {D}_{AB}=6units$

**Area ABCD**

$A=AB\times BC\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow A={\left(AB\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow A={6}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow A=36unit{s}^{2}$

## Examples of Perimeter and Area of Quadrilaterals

We shall end this topic with two worked examples involving the perimeter and area formulas for quadrilaterals. In the final example, we will be looking back at our first example at the beginning of this discussion.

Find the perimeter and area of the parallelogram MBND inscribed in the rectangle ABCD below. Here, AM = 6cm.

Example 10, StudySmarter Originals

**Solution **

The formula for the area of any parallelogram requires the length of its width and perpendicular height. The width is described by MB (or DN and MB = DN) while the perpendicular height is defined by MO. The length of MO is equal to the height of the rectangle ABCD. Thus, MO = AD = BC = 55 cm.

The width of the rectangle is AB = 84 cm. This is made up of both line segments AM and MB so

$AB=AM+MB\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 84=48+MB\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow MB=84-48\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow MB=36cm$

Thus, the length of MB is 36 cm. By the area formula for a parallelogram, we obtain

${A}_{MBND}=55\times 36\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {A}_{MBND}=1980c{m}^{2}$

Thus, the area of this parallelogram is 1980 cm^{2}.

Now we need to find the length of the side MD to calculate the perimeter of this parallelogram. Notice that MOD is a right-angle triangle. Since we have the lengths of MO = 55 cm and DO = AD = 48 cm, we can use Pythagoras' Theorem! Here, MD is the hypotenuse.

$M{D}^{2}={55}^{2}+{48}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow M{D}^{2}=3025+2304\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow M{D}^{2}=5329\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow MD=\sqrt{5329}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow MD=73cm$

Thus, the length of MD is 73 cm. Note that MD = BN. So the perimeter is equal to 218 cm since

${P}_{MBDN}=73+73+36+36\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {P}_{MBDN}=218cm$

### Real-world Example for the Perimeter and Area of Quadrilaterals

The length of each side of this square patch is 3.7 metres.

Example 11, StudySmarter Originals

To find the perimeter of this square patch, we simply add the total length of each side. Similarly, we could multiply this side length by 3.7 meters.

$P=3.7\times 4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow P=14.8m$

The area is found by squaring the side length of this square patch.

$A={\left(3.7\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow A=13.69{m}^{2}$

Thus, the perimeter of this square patch is 14.8 m and the area is 13.69 m^{2}.

## Area and Perimeter of Quadrilaterals - Key takeaways

- The perimeter of a quadrilateral is the sum of all its sides, i.e. P = a + b + c + d
- Area formula for quadrilaterals
Quadrilateral

Area

Square

$A=a\times a={a}^{2}$

Rectangle

$A=a\times b$

Parallelogram

$A=a\times h$

Trapezium

$A=\frac{1}{2}\times \left(a+b\right)\times h$

Rhombus

$A=\frac{1}{2}\times {d}_{1}\times {d}_{2}$

Kite

$A=\frac{1}{2}\times {d}_{1}\times {d}_{2}$

- We can find the perimeter and area of a quadrilateral given by a set of four points using the Distance Formula.${D}_{AB}=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}}$

###### Learn with 0 Area and Perimeter of Quadrilaterals flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Area and Perimeter of Quadrilaterals

How do you find the perimeter and area of a quadrilateral?

The perimeter is the sum of all its sides. The area is the product of its height and width.

What is the formula for the perimeter of a quadrilateral?

The sum of all its sides.

What are the area and perimeter of quadrilaterals?

The area is the space bounded by its boundary and the perimeter is the total length of its boundary.

What are the rules for calculating the area and perimeter of quadrilaterals?

Ensure that you identify the side lengths, diagonals and perpendicular height before calculating the area and perimeter of a quadrilateral.

How do you calculate the area of a quadrilateral with coordinates?

Using the distance formula.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more