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*Working rule to construct a trapezium*Let us discuss the working rule to construct a trapezium when the measure of all the four sides of a trapezium are given.

Example:

Construct a trapezium \(ABCD\) in which \(\overline{AB}\) is parallel to \(\overline{CD}\), \(AB\) \(=\) \(9\) \(cm\), \(BC\) \(=\) \(5\) \(cm\), \(CD\) \(=\) \(6\) \(cm\) and \(DA\) \(=\) \(6.5\) \(cm\). Also, find its area.

Construction:

*: Draw a line segment \(AB\) \(=\) \(9\) \(cm\).*

**Step 1***: Mark a point \(P\) on the line segment \(AB\) such that \(AP\) \(=\) \(6cm\).*

**Step 2***: With \(P\) and \(B\) as centres, draw two arcs of radii \(6.5\) \(cm\) and \(5\) \(cm\) respectively such that they intersect each other at \(C\).*

**Step 3***: Join \(PC\) and \(BC\).*

**Step 4***: With \(C\) and \(A\) as centres, draw two arcs of radii \(6\) \(cm\) and \(6.5\) \(cm\) respectively such that they intersect each other at \(D\).*

**Step 5***: Join \(AD\) and \(CD\).*

**Step 6***: \(ABCD\) is the required trapezium. The measure of \(DQ\) gives the height of the trapezium \(ABCD\).*

**Step 7**Area calculation:

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

\(=\) \(\frac{1}{2} \times 5.8 \times (9 + 6)\)

\(=\) \(\frac{1}{2} \times 5.8 \times 15\)

\(=\) \(\frac{87}{2}\)

\(=\) \(43.50\) \(cm^2\)