Construct $\Delta PQR$ in which $QR = 6\,cm$,$\angle Q = 60^{\circ}$ and $PR - PQ = 1\,cm$. Write the steps of construction.

(N/A) Steps of construction:
$1$. Draw a line segment $QR = 6\,cm$.
$2$. At point $Q$,construct an angle $\angle XQR = 60^{\circ}$.
$3$. Cut a line segment $QS = 1\,cm$ from the ray $QX$ such that $QS = PR - PQ$.
$4$. Join $SR$.
$5$. Draw the perpendicular bisector of $SR$,let it intersect $QX$ at point $P$.
$6$. Join $PR$. Thus,$\Delta PQR$ is the required triangle.