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

Given:

$\angle Y=30^o, \angle Z=90^o$ and $XY+YZ+ZX=11\ cm$.

To do:

We have to construct the $\triangle XYZ$.

Solution:

Steps of construction:

(i) Let us draw a line segment $AB$ of length $11\ cm$.

(ii) Now, construct an angle $LAB$ such that $\angle LAB=30^o$ from the point $A$

(iii) Similarly, construct an angle $MBA$ such that $\angle MBA=90^o$ from the point $B$

(v) Now, let us bisect $\angle LAB$ and $\angle MBA$ and they meet at the point $X$.

(vi) Then by taking compasses let us draw a perpendicular bisector of the line $XA$ and$XB$ respectively and mark the intersection points of the bisectors with $AB$ as $Y$ and $Z$ respectively.

(vii) Now, let us join $XY$ and $XZ$ respectively. Therefore, $XYZ$ is the required triangle.

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