Question

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

Answer 1

The below-given steps will be followed to construct the required triangle.

Step I: Draw a line segment $AB$ of $11cm.$

(As $XY+YZ+ZX=11cm$)

Step II: Construct an angle, $\angle PAB,$ of ${30}^{\circ },$ at point $A$ and an angle, $\angle QBA,$ of ${90}^{\circ },$ at point B.

Step III: Bisect $\angle PAB$ and $\angle QBA.$ Let these bisectors intersect each other at point $X.$

Step IV: Draw perpendicular bisector $ST$ of $AX$ and $UV$ of $BX.$

Step V: Let $ST$ intersect $AB$ at $Y$ and $UV$ intersect $AB$ at $Z.$

Join $XY,XZ.$

$\Delta XYZ$ is the required triangle.