Construct a quadrilateral ABCD, where ∠A = 65°,

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

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

$\Rightarrow \angle \mathrm{D}=115^{\circ}$

Steps of Construction:

Step I : Draw BC $=5.7 \mathrm{~cm} .$

Step II : Construct $\angle \mathrm{XBC}=105^{\circ}$ at B and $\angle \mathrm{BCY}=105^{\circ}$ at C.

Step III : With C as the centre and radius $6.8 \mathrm{~cm}$, cut off CD $=6.8 \mathrm{~cm}$.

Step IV : At D, draw $\angle \mathrm{CDZ}=115^{\circ}$ such that it meets BY at A.

The quadrilateral so obtained is the required quadrilateral.