Question
Construct a quadrilateral ABCD when BC = 5.5cm, CD = 4.1cm, $\angle\text{A}=70^\circ,\text{AB}=110^\circ$ and $\angle\text{D}=85^\circ.$

Answer


We know that the sum of all the angles in a quadrilateral is 360.
i.e.,$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$\Rightarrow\angle\text{C}=95^\circ$
Steps of construction:
Step I: Draw BC = 5.5cm.
Step II: Construct $\angle\text{XBC}=110^\circ$ at A and $\angle\text{BCY}=95^\circ$
Step III: With C as the centre and radius 4.1cm, cut off CD = 4.1cm.
Step IV: At D, draw $\angle\text{CDZ}=85^\circ$ such that it meets BY at A.
The quadrilateral so obtained is the required quadrilateral.

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