Construct a triangle $XYZ$ in which $\angle Y = 30^{\circ}, \angle Z = 90^{\circ}$ and $XY + YZ + ZX = 11 \,cm.$

(N/A) $I.$ Draw a line segment $AB = 11 \,cm = (XY + YZ + ZX)$.
$II.$ Construct $\angle BAP = 30^{\circ} = \angle Y$.
$III.$ Construct $\angle ABQ = 90^{\circ} = \angle Z$.
$IV.$ Draw $\overrightarrow{AR}$,the bisector of $\angle BAP$.
$V.$ Draw $\overrightarrow{BS}$,the bisector of $\angle ABQ$,such that $\overrightarrow{AR}$ and $\overrightarrow{BS}$ intersect each other at $X$.
$VI.$ Draw the perpendicular bisector of $AX$,which intersects $AB$ at $Y$.
$VII.$ Draw the perpendicular bisector of $XB$,which intersects $AB$ at $Z$.
$VIII.$ Join $XY$ and $XZ$.
Thus,$XYZ$ is the required triangle.