Construct a quadrilateral ABCD in which AB = BC = 6 cm,

Steps of construction:

Step I: Draw $\mathrm{AB}=6 \mathrm{~cm} .$

Step II : Construct $\angle \mathrm{ABC}=120^{\circ}$.

Step III : With B as the centre and radius $6 \mathrm{~cm}$, cut off BC $=6 \mathrm{~cm} .$

Now, we can see that AC is about $10.3 \mathrm{~cm}$ which is greater than AD $+\mathrm{CD}=4.5+4.5=9 \mathrm{~cm} .$

We know that sum of the lengths of two sides of triangle is always greater than the third side but here, the sum of AD and CD is less than AC.

So, construction of the given quadrilateral is not possible.