Question

Construct a triangle with perimeter $12cm$ and the ratio of their angles is $3:4:5$

Answer 1

Let the angles of the given triangle be $3x,4x$ and $5x$

Then, $3x+4x+5x={180}^{\circ }$

$\Rightarrow 12x={180}^{\circ }$

$\Rightarrow x=\frac{{180}^{\circ }}{12}$

$\therefore ,x={45}^{\circ }$ , ${60}^{\circ }$ and ${75}^{\circ }$

Steps of construction.

Let $\angle A={45}^{\circ },\angle B={60}^{\circ }$ and $\angle C={75}^{\circ }$

$\therefore$ Perimeter$=12$cm and base angles are ${60}^{\circ }$ and ${75}^{\circ }$

$1.$Draw a line segment $PQ=12$cm

$2.$Construct $\angle QPR={60}^{\circ }$ ad $\angle PQS={75}^{\circ }$

$3.$Draw the bisectors of $\angle QPR$ and $\angle PQS$ to meet at $A$

$4.$Draw the perpendicular bisectors of $PA$ and $QA$ to meet $PQ$ at $B$ and $C$ respectively.

$5.$Join $AB$ and $AC$

Result: $△ABC$ is the required triangle.