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

Answer



We know that the sum of all the angles in a quadrilateral is $360$.
i.e., $\angle\text{P}+\angle\text{Q}+\angle\text{R}+\angle\text{S}+360^\circ$
$\Rightarrow\angle\text{Q}=75^\circ$
Steps of construction:
Step $I$: Draw $PQ = 3.5\ cm$.
Step $II$: Construct $\angle\text{XPQ}=105^\circ$ at $P$ and $\angle\text{PQY}=75^\circ$ at $Q$.
Step $III$: With $Q$ as the centre and radius $6.5\ cm$, cut off $QR = 6.5$
Step $IV$​​​​​​​: At $R$, draw $\angle\text{QRZ}=105^\circ$ such that it meets $PX$ at $S$.
The quadrilateral so obtained is the required quadrilateral.

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