Draw $\Delta ABC$ with $AB = 5 \text{ cm}$,$BC = 6 \text{ cm}$,and $m\angle B = 90^{\circ}$. Then,construct $\Delta BPQ$ similar to $\Delta ABC$ such that its sides are $\frac{3}{2}$ times the corresponding sides of $\Delta ABC$. Write the steps of construction.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 6 \text{ cm}$.
$2$. At point $B$,construct an angle of $90^{\circ}$ using a protractor or compass.
$3$. Cut an arc of $5 \text{ cm}$ on the ray from $B$ to mark point $A$. Join $AC$ to complete $\Delta ABC$.
$4$. Draw an acute angle $\angle CBX$ below $BC$.
$5$. Mark $3$ points $B_1, B_2, B_3$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3$.
$6$. Join $B_2$ to $C$. Draw a line through $B_3$ parallel to $B_2C$ intersecting the extended line $BC$ at $Q$.
$7$. Draw a line through $Q$ parallel to $AC$ intersecting the extended line $BA$ at $P$.
$8$. $\Delta BPQ$ is the required triangle.