Question
Construct a triangle whose sides are $3.6\ cm, 3.0\ cm$ and $4.8\ cm$. Bisect the smallest angle and measure each part.
✓
Answer
To construct a triangle $A B C$ in which $A B=3.6 \ cm, A C=3.0 \ cm$ and $B C=4.8 \ cm$, use the following steps.
1. Draw a line segment $B C$ of length $4.8 cm .$
2. From $B$, point $A$ is at a distance of $3.6 \ cm$ . So, having $B$ as centre, draw an arc of radius $3.6 \ cm . $
3. From $C$, point $A$ is at a distance of $3 \ cm$ . So, having $C$ as centre, draw an arc of radius $3 \ cm$ which intersect previous arc at $A$.
4. Join $A B$ and $A C$. Thus, $\triangle A B C$ is the required triangle.
Here, angle $B$ is smallest, as $A C$ is the smallest side. To direct angle $B$, we use the following steps.
1. Taking $B$ as centre, we draw an are intersecting $A B$ and $B C$ at $D$ and $E$, respectively.
2. Taking $D$ and $E$ as centres we draw arcs intersecting at $P$.
3. Joining $B P$, we obtain angle bisector of $\angle B$
4. Here, $\angle ABC =39^{\circ}$
Thus, $\angle ABD =\angle= DBC =\frac{1}{2} \times 139^{\circ}=19.5^{\circ}$
1. Draw a line segment $B C$ of length $4.8 cm .$
2. From $B$, point $A$ is at a distance of $3.6 \ cm$ . So, having $B$ as centre, draw an arc of radius $3.6 \ cm . $
3. From $C$, point $A$ is at a distance of $3 \ cm$ . So, having $C$ as centre, draw an arc of radius $3 \ cm$ which intersect previous arc at $A$.
4. Join $A B$ and $A C$. Thus, $\triangle A B C$ is the required triangle.
Here, angle $B$ is smallest, as $A C$ is the smallest side. To direct angle $B$, we use the following steps.
1. Taking $B$ as centre, we draw an are intersecting $A B$ and $B C$ at $D$ and $E$, respectively.
2. Taking $D$ and $E$ as centres we draw arcs intersecting at $P$.
3. Joining $B P$, we obtain angle bisector of $\angle B$
4. Here, $\angle ABC =39^{\circ}$
Thus, $\angle ABD =\angle= DBC =\frac{1}{2} \times 139^{\circ}=19.5^{\circ}$
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