Question

Construct an isosceles triangle whose base is of length $4$ cm and the vertex angle is equal to ${30}^{\circ }$. Draw a perpendicular from the vertex to the base.

Answer 1

$\because$ Vertex angle $={30}^{\circ }$
$\therefore$ Each base angle $=\frac{{180}^{\circ }-{30}^{\circ }}{A^2}={75}^{\circ }$
Steps of construction :
1. Draw a line segment $BC=4$ cm as base.
2. With $B$ as centre, draw an angle of ${75}^{\circ }$.
3. With $C$ as centre, also draw an angle of ${75}^{\circ }$ which intersects at $A$.
4. Draw a perpendicular from the vertex $A$ to the base $BC$.
Hence, $\Delta ABC$ is the required isosceles triangles whose base is $4$ cm and vertex angle = ${30}^{\circ }$