Question
Construct a parallelogram $ABCD,$ in which diagonal $AC = 3.8\ cm,$ diagonal $BD = 4.6\ cm$ and the angle between $AC$ and $BD$ is $60^\circ .$
✓
Answer
We know that the diagonals of a parallelogram bisect each other.
Steps of construction:
Step 1: Draw $AC = 3.8\ cm.$
Step 2: Bisect $AC$ at $O.$
Step 3: Make $\angle\text{COX}=60^\circ$ Produce $XO$ to $Y.$
Step 4:
$\text{OB}=\frac{1}{2}(4.6)\text{cm}$
$\text{OB}=2.3\text{cm}$ and $\text{OD}=\frac{1}{2}(4.6)\text{cm}$
$\text{OD}=2.3\text{cm}$
Step 5: Join $AB, BC, CD$ and $AD.$
Thus, $ABCD$ is the required parallelogram.
Steps of construction:
Step 1: Draw $AC = 3.8\ cm.$
Step 2: Bisect $AC$ at $O.$
Step 3: Make $\angle\text{COX}=60^\circ$ Produce $XO$ to $Y.$
Step 4:
$\text{OB}=\frac{1}{2}(4.6)\text{cm}$
$\text{OB}=2.3\text{cm}$ and $\text{OD}=\frac{1}{2}(4.6)\text{cm}$
$\text{OD}=2.3\text{cm}$
Step 5: Join $AB, BC, CD$ and $AD.$
Thus, $ABCD$ is the required parallelogram.
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