The perpendicular distances between the pairs of opposite sides of a parallelogram $A B C D$ are $3 c m$ and $4 c m$ and one of its angles measures $60^{o}$ . Using ruler and compasses only, construct $A B C D$ .

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Solution

$\Rightarrow$ Draw two parallel lines $L_{1}$ and $L_{2}$ which have distance between $3 c m .$
$\Rightarrow$ Let point $A$ on line $L_{2}$ .
$\Rightarrow$ Make $60^{o}$ angle at point $A$ and we get point $D$ on $L_{1} .$
$\Rightarrow$ Draw a arc of radius of $4 c m$ .
$\Rightarrow$ Let point $C$ and make $60^{o}$ angle at point $C$ which is touch the arc at point F$.
$\Rightarrow$ Join $D E$ and $D F$
$∴$ $A B C D$ is required parallelogram.

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The perpendicular distances between the pairs of opposite sides of a parallelogram ABCD are 3cm and 4cm and one of its angles measures 60o. Using ruler and compasses only, construct ABCD.

The perpendicular distances between the pairs of opposite sides of a parallelogram $A B C D$ are $3 c m$ and $4 c m$ and one of its angles measures $60^{o}$ . Using ruler and compasses only, construct $A B C D$ .