Construct a rhombus whose diagonals are $4 \,cm$ and $6 \,cm$ in length. Provide the steps of construction and justification.

(N/A) Steps of construction:
$1.$ Draw a line segment $AC = 6 \,cm$.
$2.$ Construct the perpendicular bisector of $AC$,which intersects $AC$ at point $M$.
$3.$ From point $M$,mark points $B$ and $D$ on the perpendicular bisector such that $MB = MD = 2 \,cm$ (since the total length of the second diagonal is $4 \,cm$,each half is $2 \,cm$).
$4.$ Join $AB$,$BC$,$CD$,and $DA$.
Thus,$ABCD$ is the required rhombus.
Justification:
In quadrilateral $ABCD$,the diagonals $AC$ and $BD$ bisect each other at right angles at point $M$. Since $AC = 6 \,cm$ and $BD = MB + MD = 2 \,cm + 2 \,cm = 4 \,cm$,and the diagonals bisect each other at $90^\circ$,the quadrilateral $ABCD$ is a rhombus.