Construct a triangle $XYZ$ in which $\angle Y = 30^{\circ}$,$\angle Z = 90^{\circ}$ and $XY + YZ + ZX = 8 \, cm$.

(N/A) Steps of construction:
$(1)$ Draw a line segment $AB = 8 \, cm$.
$(2)$ Construct $\angle LAB = \frac{1}{2} \times 30^{\circ} = 15^{\circ}$ at point $A$.
$(3)$ Construct $\angle MBA = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$ at point $B$.
$(4)$ Let the rays $AL$ and $BM$ intersect at point $X$.
$(5)$ Draw the perpendicular bisector of $XA$ to intersect $AB$ at $Y$.
$(6)$ Draw the perpendicular bisector of $XB$ to intersect $AB$ at $Z$.
$(7)$ Join $XY$ and $XZ$. Then $\Delta XYZ$ is the required triangle.