Construct a triangle $XYZ$ in which $\angle Y = 30^o , \angle Z = 90^o$ and $XY +YZ + ZX = 11\ cm$.
Given:
A triangle $XYZ$ in which $\angle Y = 30^o , \angle Z = 90^o$ and $XY +YZ + ZX = 11\ cm$.
To do:
We have to construct the given triangle.
Solution:
Steps of construction:
(i) Draw a line segment $PQ =11\ cm$.
(ii) At $P$, draw a ray $PL$ making an angle of $30^o$ and $Q$, draw another ray $QM$ making an angle of $90^o$.
(iii) Draw the angle bisector of $\angle P$ and $\angle Q$ intersecting each other at $X$.
(iv) Draw the perpendicular bisectors of $XP$ and $XQ$ which intersect $PQ$ at $Y$ and $Z$ respectively.
(v) Join $XY$ and $XZ$.
Therefore,
$\triangle XYZ$ is the required triangle.
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