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*Working rule to construct a trapezium*Let us discuss the working rule to construct a trapezium when the measure of two sides and two angles of a trapezium are given.

Example:

Construct a trapezium \(PQRS\) in which \(\overline{PQ}\) is parallel to \(\overline{RS}\), \(PQ\) \(=\) \(8.5\) \(cm\), \(QR\) \(=\) \(6.5\) \(cm\), \(\angle PQR\) \(=\) \(75^{\circ}\) and \(\angle QPS\) \(=\) \(110^{\circ}\). Also, find its area.

Construction:

*: Draw a line segment \(PQ\) \(=\) \(8.5\) \(cm\).*

**Step 1***: With \(Q\) as centre, mark an angle \(75^{\circ}\) using a protractor and mark it as \(X\). Join \(QX\).*

**Step 2***: With \(Q\) as centre, draw an arc of radius \(6.5\) \(cm\) intersecting \(QX\) at \(R\).*

**Step 3***: Draw a line \(RY\) parallel to \(PQ\).*

**Step 4***: With \(P\) as centre, mark an angle \(110^{\circ}\) using a protractor and mark it as \(Z\). Join \(PZ\) intersecting \(QY\) at \(S\).*

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

**Step 6**Area calculation:

Area of the trapezium \(PQRS\) \(=\) \(\frac{1}{2} \times h \times (a + b)\) square units

\(=\) \(\frac{1}{2} \times 6.3 \times (8.5 + 9.1)\)

\(=\) \(\frac{1}{2} \times 6.3 \times 17.6\)

\(=\) \(\frac{110.88}{2}\)

\(=\) \(55.44\) \(cm^2\)