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Working rule to construct a parallelogram:

Let us discuss the working rule to construct a parallelogram when the measure of two diagonals and one included angle of a parallelogram are given.
Example:
Construct a parallelogram $$PQRS$$ with $$PR$$ $$=$$ $$10$$ $$cm$$, $$QS$$ $$=$$ $$8$$ $$cm$$ and $$\angle POQ$$ $$=$$ $$120^{\circ}$$. Also, find its area.

Construction:

Step 1: Draw a line segment $$PR$$ $$=$$ $$10$$ $$cm$$.

Step 2: Mark the midpoint of the line segment $$PR$$ as $$O$$.

Step 3: Draw a line $$XY$$ through $$O$$ such that $$\angle POQ$$ $$=$$ $$120^{\circ}$$.

Step 4: With $$O$$ as centre, draw two arcs each of radii $$4$$ $$cm$$ on either side of $$PR$$ intersecting $$XY$$ at $$Q$$ and $$S$$.

Step 5: Join $$PQ$$, $$QR$$, $$RS$$ and $$SP$$.

Step 6: $$PQRS$$ is the required parallelogram. The measure of $$SA$$ gives the height of the parallelogram $$PQRS$$.

Area calculation:

Area of the parallelogram $$=$$ $$base \times height$$ square units

$$=$$ $$8 \times 4.3$$

$$=$$ $$34.40$$ $$cm^2$$