Construct a quadrilateral ABCD when BC = 5.5 cm,

We know that the sum of all the angles in a quadrilateral is 360 .

i. e., $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ}$

$\Rightarrow \angle \mathrm{C}=95^{\circ}$

Steps of construction:

Step I : Draw BC $=5.5 \mathrm{~cm}$.

Step II : Construct $\angle \mathrm{XBC}=110^{\circ}$ at A and $\angle \mathrm{BCY}=95^{\circ}$.

Step III : With C as the centre and radius $4.1 \mathrm{~cm}$, cut off CD $=4.1 \mathrm{~cm}$.

Step IV : At D, draw $\angle \mathrm{CDZ}=85^{\circ}$ such that it meets BY at A.

The quadrilateral so obtained is the required quadrilateral.