Question

Prove that: 

Area of a trapezium ABCD = {¹/₂ × (AB + DC) × h} square units.

Answer 1

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Area of a trapezium ABCD 

            = area (∆DFA) + area (rectangle DFEC) + area (∆CEB) 

            = (¹/₂ × AF × DF) + (FE × DF) + (¹/₂ × EB × CE) 

           = (¹/₂ × AF × h) + (FE × h) + (¹/₂ × EB × h) 



            = ¹/₂ × h × (AF + 2FE + EB) 

            = ¹/₂ × h × (AF + FE + EB + FE) 

            = ¹/₂ × h × (AB + FE) 

            = ¹/₂ × h × (AB + DC) square units.

            = ¹/₂ × (sum of parallel sides) × (distance between them) 

Answer 2

Let ABCD be a trapezium in which AB ∥ DC, CE ⊥ AB, DF ⊥ AB and CE = DF = h. Prove that: Area of a trapezium ABCD = {¹/₂ × (AB + DC) × h} square units.