Question
Construct a parallelogram $\text{ABCD}$ in which $\text{AB} = 4\ cm, \text{BC} = 5\ cm$ and $\angle\text{B}=60^\circ.$
✓
Answer
The opposite sides of a parallelogram are equal.
So, $\text{AB = DC} = 4\ cm$
$\text{BC = AD} = 5\ cm$
$\angle\text{A}+\angle\text{B}=180^\circ [$sum of co$-$interior angles$]$
$\angle\text{A}=120^\circ$
Steps of Construction:
$1.$ Draw $\text{AB} = 4\ cm$
$2.$ Draw $\text{BX}$ such that $\angle\text{ABX}=60^\circ$
$3.$ Mark a point $C$ such that $\text{BC} = 5\ cm$
$4.$ Draw a ray $\text{AY}$ such that $\angle\text{YAB}=120^\circ$
$5.$ Mark a point $D$ such that $\text{AD} = 5\ cm$
$6.$ Join $C$ and $D$.
Hence, $\text{ABCD}$ is the required parallelogram.
So, $\text{AB = DC} = 4\ cm$
$\text{BC = AD} = 5\ cm$
$\angle\text{A}+\angle\text{B}=180^\circ [$sum of co$-$interior angles$]$
$\angle\text{A}=120^\circ$
Steps of Construction:
$1.$ Draw $\text{AB} = 4\ cm$
$2.$ Draw $\text{BX}$ such that $\angle\text{ABX}=60^\circ$
$3.$ Mark a point $C$ such that $\text{BC} = 5\ cm$
$4.$ Draw a ray $\text{AY}$ such that $\angle\text{YAB}=120^\circ$
$5.$ Mark a point $D$ such that $\text{AD} = 5\ cm$
$6.$ Join $C$ and $D$.
Hence, $\text{ABCD}$ is the required parallelogram.
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