PDF chapter test TRY NOW

- Depending upon the lengths of the sides and diagonal, create a triangle of \(PQR\) based on \(SSS\) construction.

- Make an arc (point \(S\)) at a certain distance from \(P\).

- Make an arc at a certain distance from point \(R\) on the earlier arc on \(S\). Name the two points intersection as \(S\).

- \(P\) and \(R\) join \(S\). The quadrilateral \(PQRS\) will be achieved.

Example:

**Construct a quadrilateral**\(ABCD\)

**with the following measurements**.

\(AB =\) \(4.5 cm\), \(BC =\) \(5.5 cm\), \(CD = 4cm\), \(AD = 6cm\), \(AC = 7cm\).

**Step 1**:Draw side \(BC = 5.5 cm\) and cut arcs above it from \(B\) (\(4.5 cm\)) and \(C\) (\(7 cm\)). Mark the intersection as \(A\). Join \(AB\) and \(AC\).

**Step 2**: Draw and arc from \(A\) equal to \(6 cm\) which is the length of \(AD\).

**Step 3**: Draw and arc from \(C\) equal to \(4 cm\) which is the length of \(CD\). Mark the intersection as \(D\) and join \(AD\) and \(CD\).

Thus, the \(ABCD\) is a required quadrilateral.

**Calculate Area of quadrilateral**:

Area of the quadrilateral \(ABCD\) \(=\) \(\frac {1}{2}\) \(\times\) \(d\) \(\times\) \(h_1 + h_2\) sq. units

\(=\) \(\frac{1}{2} × 10 (1.9 +2.3)\)

\(=\) \(5\times 4.2\)

\(=\) \(21 cm²\).