Find the area of trapeizium ABCD;AB||DC i which AB =9cm, DC=4cm, AD=5cm and BC=5cm.
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Hey Mate,
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. Answer
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