Question

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

Answer 1

$1.$Draw a line segment $AB$ equal to $XY+YZ+ZX=11$cm

$2.$Make angle equal to $\angle Y={30}^{\circ }$ from point $A$

Let the angle be $\angle LAB$

$3.$Make angle equal to $\angle Z={90}^{\circ }$ from point $B$

Let the angle be $\angle MBA$

$4.$Bisect $\angle LAB$ and $\angle MBA$.Let these bisectors intersect at a point $X$

$5.$Make perpendicular bisector of $XA$.Let it intersect $AB$ at point $Y$

$6.$Make perpendicular bisector of $XB$.Let it intersect $AB$ at point $Z$

$7.$Join $XY$ and $XZ$

$\therefore △XYZ$ is the required triangle.