# CLASS-9AREA OF TRAPEZIUM

AREA OF TRAPEZIUM -

Area of Trapezium ABCD = Area of (ABC) X Area of (ACD)

1                       1

= ------- X AB X h + ------- X CD X h

2                       2

1

= ------- X (AB + CD) X h

2

1

= ------ X (Sum of Parallel Sides) X Distance between them

2

Example.1) Find the area of the trapezium whose parallel sides are 30 cm and 12 cm and the distance between them is 6 cm.

Ans.)   as per the formulae –

Area of Trapezium =

1

= ----- X Sum of parallel sides X Distance between them

2

1

= [------ X (30 + 12) X 6]  cm²

2

= 42 X 3 = 126 cm²         (Ans.)

Example.2) Find the area of a trapezium ABCD in which AB || DC, AB = 77 cm, BC = 25 cm, CD = 60 cm, and  DA = 26 cm.

Ans.) As per the above given condition, we would like to draw DE || CB and DF ⊥ AB in below figure

Now, DE = BC = 25 cm

AE = (AB – EB) = (AB – CD) = (77 – 60) cm

So, in DAE, we have –

AE = a = 17 cm, DE = b = 25 cm, and DA = c = 26 cm.

We have to find out the area of DAE

1                       1

So, s = ------ (a + b + c) = ------ (17 + 25 + 26) cm

2                       2

= 68/2 = 34 cm

Now, (s – a) = 34 – 17 = 17 cm

(s – b) =  34 – 25 = 9 cm

(s – c) = 34 – 26 = 8 cm

So, area DAE = √s (s – a)(s – b)(s – c)

= √34 X 17 X 9 X 8

= (17 X 3 X 4) cm² =  204 cm²

1                       1

Also, area (DAE) = ------ X AE X DF = ------ X 17 X DF

2                       2

1

So, ------ X 17 X DF = 204

2

204 X 2

So, DF = ----------- = 24 cm

17

1

Area of trap ABCD = ------ (AB + DC) X DF

2

77 + 60

= ----------- X 24

2

=  68.5 X 24 = 1644 cm²         (Ans.)