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Answer :
Given:
In \(\triangle{XYZ}\) in which \(\angle{Y}\) = \(30^\circ\), \(\angle{Z}\) = \(90^\circ\) and XY + YZ + ZX = 11cm.
Steps of construction:
1. Draw a line segment BC = XY + YZ + ZX = 11cm.
2. Make \(\angle{LBC}\) = \(\angle{Y}\) = 30° and \(\angle{MCB}\) = \(\angle{Z}\) = 90°.
3. Bisect \(\angle{LBC}\) and \(\angle{MCB}\). Let these bisectors meet at a point X.
4. Draw perpendicular bisectors DE of XB and FG of XC.
5. Let DE intersect BC at Y and FC intersect BC at Z.
6. Join XY and XZ.
Then, \(\triangle{XYZ}\) is the required triangle.