Question

Construct $△ABC$ in which $BC=6.3cm$ $\angle B={75}^{\circ }$ and $AB+AC=9cm$.

Answer 1

Steps of construction:

(1) Draw a base $BC$ of length $6.3cm$

(2) Draw a ray makes an angle ${75}^{\circ }$ at the centre $B$.

(3) With $B$ as center and radius equal to $9cm$, draw an arc to intersect ray $BX$ at $D$.

(4) Join $DC$.

(5) Draw the perpendicular bisector of $CD$, which intersects $BD$ at the point $A$.

(6) Join $CA$.

(7) Measure $CA$ and $DA$, both lengths are equal.

i.e., $AB+AC=AB+AD=BD=9cm$

i.e., In $\Delta ABC,AB+AC=9cm,BC=6.3cm,\angle B={75}^{\circ }$

Hence, required $\Delta ABC$ is formed.