Question
Construct a $\triangle\text{ABC}$ in which $BC = 3.6\ cm, AB = 5\ cm$ and $AC = 5.4\ cm.$ Draw the perpendicular bisector of the side $BC.$
✓
Answer
Steps for construction:
Step I: Draw a line segment $(AB)$ of length $5\ cm.$
Step II: Draw an arc of radius $5.4\ cm$ from the centre $(A).$
Step III: With $B$ as the centre, draw another arc of radius $3.6\ cm,$ cutting the previous arc at $C.$
Step IV: Join $AC$ and $BC.$
Step V: Taking $B$ as the centre and the radius more than half of $BC,$ draw two arcs on both the sides of $BC.$
Step VI: Similarly, taking $C$ as the centre and the same radius, draw arcs on both the sides of $BC,$ cutting the previous arcs at $P$ and $Q.$ Then, $PQ$ is the required perpendicuar bisector of $BC,$ meeting $BC$ at $D.$
Step I: Draw a line segment $(AB)$ of length $5\ cm.$
Step II: Draw an arc of radius $5.4\ cm$ from the centre $(A).$
Step III: With $B$ as the centre, draw another arc of radius $3.6\ cm,$ cutting the previous arc at $C.$
Step IV: Join $AC$ and $BC.$
Step V: Taking $B$ as the centre and the radius more than half of $BC,$ draw two arcs on both the sides of $BC.$
Step VI: Similarly, taking $C$ as the centre and the same radius, draw arcs on both the sides of $BC,$ cutting the previous arcs at $P$ and $Q.$ Then, $PQ$ is the required perpendicuar bisector of $BC,$ meeting $BC$ at $D.$
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