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

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.