Question

Construct a quadrilateral $PQRS,$ where $PQ=3.5cm,QR=6.5cm,\angle P=\angle R={105}^{\circ }$ and $\angle S={75}^{\circ }.$

Answer 1

We know that the sum of all the angles in a quadrilateral is ${360}^{\circ }$.

i.e., $\angle P+\angle Q+\angle R+\angle S={360}^{\circ }$.

$\Rightarrow {105}^{\circ }+\angle Q+{105}^{\circ }+{75}^{\circ }={360}^{\circ }$

$\Rightarrow \angle Q={75}^{\circ }$.

Steps of construction:

Step I: Draw $PQ=3.5cm.$

Step II: Construct $\angle XPQ={105}^{\circ }$ at $P$ and $\angle PQY={75}^{\circ }$ at $Q.$

Step III: With $Q$ as the center and radius $6.5cm,$ cut off $QR=6.5cm$

Step IV: At $R,$ draw $\angle QRZ={105}^{\circ }$ such that it meets $PX$ at $S.$

The quadrilateral $PQRS$ so obtained is the required quadrilateral.