Construct a quadrilateral ABCD in which AB = 4 cm,

Steps of construction:

Step 1: Draw $A B=4 \mathrm{~cm}$

Step 2: Make $\angle B=90^{\circ}$

Step 3: $A C^{2}=A B^{2}+B C^{2}$

$5^{2}=4^{2}+B C^{2}$

$25-16=B C^{2}$

$B C=3 \mathrm{~cm}$

With $B$ as the centre, draw an arc equal to $3 \mathrm{~cm}$.

Step 4: Make $\angle C=90^{\circ}$

Step 5: With $A$ as the centre and radius equal to $5.5 \mathrm{~cm}$, draw an arc and name that point as $D$.

Then, $A B C D$ is the required quadrilateral.