Question
Construct a parallelogram $ABCD$ in which $AB = 4.5\ cm, \angle A = 105^\circ $ and the distance between $AB$ and $CD$ is $3.2\ cm.$
✓
Answer
Steps of construction:
$1)$ Draw line $AB = 4.5\ cm.$
$2)$ At $B$, draw $BX$ perpendicular to $AB.$
$3)$ From $BX,$ cut $BR = 3.2\ cm =$ distance between $AB$ and $CD.$
$4)$ Through $R$, draw a line perpendicular to $BX$ to get $QR$ parallel to $AB.$
$5)$ With $A$ as centre, draw a ray $AP$ making an angle of $105$ with $AB$ and meeting $QR$ at $D.$
$6)$ With $B$ as centre, draw an arc with radius $= AD$ on $QR$ and mark it as $C.$
$7)$ Join $BC.$
$8) \text{ABCD}$ is the required parallelogram.
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