Construct $\Delta ABC$ with $AB = 3 \, cm$,$BC = 6 \, cm$,and $AC = 5 \, cm$. Then,construct $\Delta BPQ$ similar to $\Delta ABC$ such that each side of $\Delta BPQ$ is $\frac{3}{4}$ of the corresponding sides of $\Delta ABC$. Write the steps of construction.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 6 \, cm$.
$2$. With $B$ as the center and radius $3 \, cm$,draw an arc. With $C$ as the center and radius $5 \, cm$,draw another arc intersecting the first arc at $A$.
$3$. Join $AB$ and $AC$ to obtain $\Delta ABC$.
$4$. Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$.
$5$. Locate $4$ points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$6$. Join $B_4C$.
$7$. Draw a line through $B_3$ parallel to $B_4C$ intersecting $BC$ at $Q$.
$8$. Draw a line through $Q$ parallel to $AC$ intersecting $AB$ at $P$.
$9$. $\Delta BPQ$ is the required triangle.