Area of trapezium = 1 / 2 × (sum of parallel sides) × height ABCD is a trapezium AD II BC, AB = 5 cm, BC = 11 cm, AD = 7 cm, DA is produced to a point F such that AF = 3 cm and BF perpendicular to DF.In triangle BFA,
BF2=AB2−AF2 BF2=52−32 BF2=16cm BF=4cm Area of trapezium =1∕2×( sum of parallel sides )× height ⇒1∕2×(7+11)×4 ⇒1∕2×18×4 ⇒36cm2