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Area of a Trapezium
Trapezium:A trapezium is a quadrilateral having one pair of parallel opposite sides. In the given figure, ABCD is a trapezium in which AB ∥ DC. Area of a Trapezium: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) Formula of Area of a trapezium = ^{1}/_{2} × (sum of parallel sides) × (distance between them) Solved Examples of Area of a Trapezium1. 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} cm^{2} = 322 cm^{2} 2. The area of a trapezium is 352 cm^{2} and the distance between its parallel sides is16 cm. If one of the parallel sides is of length 25 cm, find the length of the other. Solution: Let the length of the required side be x cm. Then, area of the trapezium = {^{1}/_{2} × (25 + x) × 16} cm^{2} = (200 + 8x) cm^{2}. But, the area of the trapezium = 352 cm^{2} (given) Therefore, 200 + 8x = 352 ⇒ 8x = (352  200) ⇒ 8x = 152 ⇒ x = (152/8) ⇒ x = 19. Hence, the length of the other side is 19 cm. 3. The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm. Find the area of the trapezium. Solution: Let ABCD be the given trapezium in which AB = 25 cm, DC = 13 cm, BC = 10 cm and AD = 10 cm. Through C, draw CE ∥ AD, meeting AB at E. CE = AD = 28 cm and BC = 30 cm. Area of a Trapezium Area of a Trapezium  Worksheet
8th Grade Math Practice


