Question

Q30. Find the area of the polygon ABCD where AB is parallel to DC/AB.​

Answer 1

Find the area of the polygon ABCD where AB is parallel to....?

See the above attached attachment...! You will have your answer..! ⬆️⬆️⬆️⬆️✨

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Answer 2

Answer:

Area of a trapezium ABCD

Area of a trapezium ABCD = area (∆DFA) + area (rectangle DFEC) + area (∆CEB)

Area of a trapezium ABCD = area (∆DFA) + area (rectangle DFEC) + area (∆CEB) = (¹/₂ × AF × DF) + (FE × DF) + (¹/₂ × EB × CE)

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)

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)

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)

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)

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.

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)