Question
Construct a quadrilateral $ABCD$, where $AB = 5.5\ cm, BC = 3.7\ cm,$$\angle\text{A}=60^\circ,$ $\angle\text{B}=105^\circ$ and $\angle\text{D}=90^\circ$
✓
Answer
We know that the sum of all the angles in a quadrilateral is $360$.
i.e., $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$\Rightarrow\angle\text{C}=105^\circ$
Steps of construction:
Step $I$: Draw $AB = 5.5\ cm$.
Step $II$: Construct $\angle\text{XAB}=60^\circ$ at $A$ and $\angle\text{ABY}=105^\circ.$
Step $III$: With $B$ as the centre and radius $3.7\ cm$, cut off $BC = 3.7\ cm$.
Step $IV$: At $C$, draw $\angle\text{BCZ}=105^\circ$ such that it meets $AX$ at $D$.
The quadrilateral so obtained is the required quadrilateral.
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