Construct a quadrilateral PQRS, where PQ = 3.5 cm,

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

i. e., $\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}+\angle \mathrm{S}=360^{\circ}$

$\Rightarrow \angle \mathrm{Q}=75^{\circ}$

Steps of construction:

Step I : Draw PQ $=3.5 \mathrm{~cm}$.

Step II : Construct $\angle \mathrm{XPQ}=105^{\circ}$ at P and $\angle \mathrm{PQY}=75^{\circ}$ at Q.

Step III : With Q as the centre and radius $6.5 \mathrm{~cm}$, cut off QR $=6.5$

Step IV : At R, draw $\angle \mathrm{QRZ}=105^{\circ}$ such that it meets PX at S.

The quadrilateral so obtained is the required quadrilateral.