Area of a trapezium – formula, derivation and examples

Introduction to trapezium

• A trapezium is a two-dimensional closed figure having four sides and four vertices.
• The four sides are such that one pair of opposite sides are parallel to each other.

Parts of a Trapezium (Refer Figure 1)

1. Bases:

The parallel sides – AB and CD.

2. Legs:

The sides AD and BC, that are non-parallel.

The length of these sides may or may not be equal.

3. Height:

AE is the height of this trapezium.

4. Diagonals:

AC and BD are the diagonals.

Types of Trapeziums

Based on the sides, it can be classified into 2 categories

1. Scalene Trapezium: The measure of the non-parallel sides is not equal.

2. Isosceles Trapezium: The measure of the non-parallel sides is equal.

Based on the angles, it can be classified into 3 categories.

1. Acute Trapezium: The measure of two adjacent angles is less than 90°.

Acute Trapezium

2. Obtuse Trapezium: The measure of two opposite angles is more than 90°.

Obtuse Trapezium

3. Right Trapezium: The measure of two adjacent angles is equal to 90°.

Right Trapezium

Area of a trapezium

It is defined as the measure of the region enclosed by the four sides of the trapezium.

In the figure given below, the yellow shaded area is the area of the given trapezium.

The formula for finding the area

In the above figure,

ABCD is a trapezium where AB and CD are the parallel sides

AB = 1^{\text{st}} base = a

CD = 2^{\text{nd}} base = b

AE = height = h

Area of the trapezium ABCD =\frac{1}{2} \times(\text { sum of the parallel sides }) \times \text { height }

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

Derivation of the above formula

There are many ways for deriving the above formula out of which we are going to try 1 easy way using two triangles.

Given:

ABCD is a trapezium having AB and CD as parallel sides.

AB = a

CD = b

AE = height = h

To Prove:

Area of ABCD = \frac{1}{2}× (a + b) × h

Constructions:

1. Draw a line joining A and C to divide the trapezium into two triangles.
2. Produce AB to meet a line from C at F such that CF is perpendicular to BF and parallel to AE.
   (This shows that AECF is a rectangle where AE = FC)

Proof:

We have two triangles after drawing AC i.e. triangle ABC and triangle ACD

In triangle ABC 

Base = AB = a 

Height = FC = h   

\text { Area }=\frac{1}{2} \times \text { base } \times \text { height}
 =\frac{1}{2} \times a \times h

In triangle ACD 

Base = CD = b 

Height = AE = h   

\text { Area }=\frac{1}{2} \times \text { base } \times \text { height}
=\frac{1}{2} \times b \times h

\text { Area of Trapezium } A B C D=\text { Area of } \triangle A B C+\text { Area of } \triangle A C D

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

                                                        =\frac{1}{2}(a+b) h

\text { Area of Trapezium } A B C D=\frac{1}{2} \times \text { height } \times \text { sum of parallel sides }

Hence the formula is derived.

Approach for solving

1. Try to draw the diagram if not given and analyse it.

2. Then we need to note the given data:

     (i) Length of the bases (a and b).

     (ii) Height of the trapezium (h).

3. Sometimes to make the questions a little tricky sum of the parallel sides is given. But this makes the approach even easier as we know the value of (a+b) directly from the question.

4. Now just put the values in the formula to calculate.

Examples

1) Find the area of the given trapezium ABCD whose height is 4 cm. The length of sides AB and CD are 4 cm and 8 cm respectively. 

Solution:

Given:

AB = a = 4 cm 

CD = b = 8 cm 

AE = height = h = 4 cm

Formula for area:

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

             =\frac{1}{2}\times(4+8) \times 4 cm²

             =24 \mathrm{~cm}^{2}

2) The area of an isosceles trapezium having parallel sides 8 cm and 12 cm is equal to 20 cm². Find the height of the trapezium.

Solution:

Sum of parallel sides = a + b = 8 cm + 12 cm = 20 cm

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

⇒ 20 = 12×(a+b)×h 

⇒ 20 = 12×(20)×h 

⇒ 20 = 10 h

⇒ h = 20/10 = 2 cm 

Hence the height is equal to 2 cm.

Real-life applications

This concept of the trapezium is not only limited to books but also can be seen in real life.

Let’s explore.

1)Popcorn Box

2) Roof of a house