Question

find the area of a trapezium ABCD; AB parallel to DC In which AB=9cm ,DC=4cm ,A=5cm and BC=5cm​

Answer 1

Answer:

28.15 cm²

Step-by-step explanation:

ABCD is an isosceles trapezium with AB= 9 cm, DC = 4 cm, AD = 5 cm and BC = 5 cm.

The distance between AB and CD = [5^2–{(9–4)/2}^2]^0.5

= [25–6.25]^0.5

= 4.33 cm

So the area of the trapezium, ABCD = (9+4)*4.33/2

= 28.15 sq cm.