From the information given in the figure, the length of all sides of the ABCD. The area of the trapezium is 33cm 2and side AB parallar side DC.
Answers
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.
Proof: 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)
Formula of Area of a trapezium = ¹/₂ × (sum of parallel sides) × (distance between them)
Solved Examples of Area of a Trapezium
1. Two parallel sides of a trapezium are of lengths 27 cm and 19 cm respectively, and the distance between them is 14 cm. Find the area of the trapezium.
Solution:
Area of the trapezium
= ¹/₂ × (sum of parallel sides) × (distance between them)
= {¹/₂ × (27 + 19) × 14} cm²
= 322 cm²