Question

Construct a $△ABC$ in which $BC=4.5cm,\angle B={45}^{\circ }$ and $AB+AC=8cm$ Justify your construction.

Answer 1

Justification:-

After drawing $\Delta DBC$ with $BC=4.5$cm, $\angle B={45}^o$ and $DB=8$cm. The perpendicular bisector of DC is drawn and it intersects DB at A.

So, we get AC$=$AD

BD$=8$cm

$\Rightarrow BA+AD=8$cm

$\Rightarrow AB+AC=8$cm

Hence $\Delta ABC$ is the required triangle.