Construct a quadrilateral ABCD, where AB = 5.5 cm,

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

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

$\Rightarrow \angle \mathrm{C}=105^{\circ}$

Steps of construction:

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

Step II : Construct $\angle \mathrm{XAB}=60^{\circ}$ at A and $\angle \mathrm{ABY}=105^{\circ}$.

Step III: With B as the centre and radius $3.7 \mathrm{~cm}$, cut off BC $=3.7 \mathrm{~cm}$.

Step IV : At C, draw $\angle \mathrm{BCZ}=105^{\circ}$ such that it meets AX at D.

The quadrilateral so obtained is the required quadrilateral.