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# 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.