Question

Construct a triangle $PQR$ in which $\angle Q={60}^o$ and $\angle R={45}^o$ and $PQ+QR+PR=11cm$.

Answer 1

Step 1: Draw a line segment $AB$ of $11cm$

$XY+YZ+ZX=11cm$

Step 2: Using a protractor mark ${30}^{\circ }$ at $A$, i.e $\angle PAB$ and $\angle QBA$ as ${90}^{\circ }$

Step 3: Bisect ${30}^{\circ }$ at $A$, i.e $\angle PAB$ and $\angle QBA$ at point X.

Step 4: Join $XZ$ and make $XZ$ perpendicular to $AB$.

Step 5: Draw perpendicular bisector of $AX$ i.e $ST$.

Step 6: Mark the point $Y$ where $ST$ and $AB$ intersect. Join $XY$

$\therefore$ We get $△XYZ$