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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:

*: Draw a line segment \(PR\) \(=\) \(10\) \(cm\).*

**Step 1***: Mark the midpoint of the line segment \(PR\) as \(O\).*

**Step 2***: Draw a line \(XY\) through \(O\) such that \(\angle POQ\) \(=\) \(120^{\circ}\).*

**Step 3***: With \(O\) as centre, draw two arcs each of radii \(4\) \(cm\) on either side of \(PR\) intersecting \(XY\) at \(Q\) and \(S\).*

**Step 4***: Join \(PQ\), \(QR\), \(RS\) and \(SP\).*

**Step 5***: \(PQRS\) is the required parallelogram. The measure of \(SA\) gives the height of the parallelogram \(PQRS\).*

**Step 6**Area calculation:

Area of the parallelogram \(=\) \(base \times height\) square units.

\(=\) \(8 \times 4.3\)

\(=\) \(34.40\) \(cm^2\)