CLASS-9
AREA 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.)