What is the correct diagram to construct a parallelogram (EFDC) with one of the angle equal to 'x' and whose area is equal to that of the given triangle (ABC).

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Solution

The correct option is A

Rules of Construction
(1) Draw a certain triangle ABC.
(2) Draw PAQ through A parallel to BC.
(3) Bisect BC of the $Δ A B C$ at D.
(4) Draw $∠ C D E$ at D on BC equal to $∠ x$ the side DE of which intersect PQ at E.
(5) Cut EF from EQ equal to DC.
(6) Join points C and F.

DEFC is the required parallelogram.

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