Question
Construct a $\triangle \text{XYZ}$ in which $\angle\text{Y}=30^\circ,$ $\angle\text{Z}=90^\circ$ and $XY + YZ + ZX = 11\ cm$.
✓
Answer
Steps of construction:
$1.$ Draw a line segment $AB$ of $11\ cm$.
$2.$ Draw $\angle\text{DAB}=30^\circ$ and $\angle\text{FBA}=90^\circ.$
$3.$ Draw the angle bisectors of $\angle\text{DAB}$ and $\angle\text{EBA}$ which intersect each other at $A$.
$4.$ Draw the perpendicular bisector of $XA$ and $XB$ which intersect $AB$ at $Y$ and $Z$ respectively
$5.$ Join $XY$ and $XZ$.
Hence $\triangle\text{XYZ}$ is the required triangle.
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