- Books Name
- class 8 th Mathematics Book

- Publication
- ReginaTagebücher

- Course
- CBSE Class 8

- Subject
- Mathmatics

**Area of Quadrilaterals and Polygons**

**Area of a general quadrilateral**

We know that a quadrilateral is made up of four sides. Any quadrilateral can be divided into two triangles with the help of a diagonal. This process of dividing a quadrilateral into triangles is called triangulation.

Let us look at the figure given below to know how a quadrilateral would look like after triangulation.

Here, △ABD and △BCD are formed after triangulation.

To find the area of the quadrilateral, we should know the values of h1 and h2. h1 and h2 are the perpendiculars drawn from the diagonal to the vertices. Let us look at the figure given below for a better understanding.

From the figure given above, we can come to the following inferences.

ABCD is a quadrilateral with d as the diagonal and h1 and h2 as its heights.

Area of quadrilateral ABCD= (Area of△ABD) + (Area of△BCD)

[Since the diagonal is divided into two triangles after triangulation]

= (12×BD×h1) + (12×BD×h2)

[Since,Area of a triangle=12×b×h,where b and h are its base and height respectively]

= 12×BD×(h1+h2)

= 12×d×(h1+h2)

Therefore,** **Area of a quadrilateral = 12d(h1+h2) square units.

**Area of a special quadrilateral (Rhombus),**

A special quadrilateral is nothing but a rhombus. To find the area of a rhombus, we can use the same triangulation method as used in finding the area of a general quadrilateral.

A rhombus after triangulation is given below.

From the figure given above, we can come to the following inferences.

ABCD is a rhombus with diagonals AC and BD.

Let AC be d1 and BD be d2. The diagonals d1 and d2 intersect at O.

Also, as per the properties of a rhombus, the diagonals bisect each other.

Area of the rhombus ABCD= (Area of △ABD)) + (Area of △BCD)

= 12×BD×OA + 12×BD×OC

= 12×BD×(OA+OC)

= 12×BD×AC

[Since AC=OA+OC]

= 12×d1×d2

[Since we have assumed that AC=d1, and BD=d2]

Therefore, area of a rhombus=12×d1×d2 square units or the area of the rhombus is half the product of its diagonals.

**Area of a polygon**

A polygon is any shape with a minimum of three sides.

Let us consider the following pentagon ABCDE and try to find its area.

Area of a pentagon can be found in two ways.

**Type** 1

Let us use the method of triangulation using two diagonals.

Let us look at the figure given below for a better understanding.

Area of a pentagon=Area of △AED+Area of △ABD+Area of △BCD

**Type** 2

Another way of obtaining the area of a pentagon is by drawing one diagonal and drawing two perpendiculars to that diagonal.

Let us look at the figure given below for a better understanding.

In the figure given above, BD is the diagonal, EO and AN are the two perpendiculars to BD.

Here, Area of a pentagon=Area of △ODE+Area of △BCD+Area of trapezium AEON+Area of △ABN

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## Batch List

###### EDUCARE

Course : CBSE Class 8

Start Date : 09.02.2023

End Date : 01.03.2023

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###### Medha Sharma

Course : CBSE Class 8

Start Date : 08.02.2023

End Date : 01.09.2023

Types of Batch : Classroom