Question
Construct a quadrilateral ABCD, where$\angle\text{A}=65^\circ,\text{B}=105^\circ,\text{C}=75^\circ,$ BC = 5.7 and CD = 6.8cm.

Answer


We know that the sum of all the angles in a quadrilateral is 360.
i.e.,$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$\Rightarrow\angle\text{D}=115^\circ$
Steps of construction:
Step I: Draw BC = 5.7cm.
Step II: Construct $\angle\text{XBC}=105^\circ$ at B and $\angle\text{BCY}=105^\circ$ at C.
Step III: With C as the centre and radius 6.8cm, cut off CD = 6.8cm.
Step IV: At D, draw $\angle\text{CDZ}=115^\circ$ such that it meets BY at A.
The quadrilateral so obtained is the required quadrilateral.

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