Question

Draw a triangle $ABC$ with side $BC=7cm,\angle B={45}^o,\angle A={105}^o.$ Then construct a triangle whose side are $\frac{4}{3}$ times the corresponding sides of $△ABC.$ Write the construction.

Answer 1

In order to construct $△ABC$, we follow the following steps :

$STEPI$ Draw $BC=7cm$.

$STEPII$ At $B$ construct $\angle CBX={45}^o$ and at $C$ construct
$\angle BCY={180}^o-({45}^o-{105}^o)={30}^o$

Suppose $BX$ and $CY$ intersect at $A.△ABC$ so obtained is the given triangle.

To construct a triangle similar to $△ABC$, we follow the following steps.

$STEPI$ Construct an acute angle $\angle CBZ$ at $B$ on opposite side of vertex $A$ of $△ABC$.

$STEPII$ Mark-off four (greater $4$ of and $3$ in $\frac{4}{3}$) points $B_1,B_2,B_3,B_4$ on $BZ$ such that
$B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$.

$STEPIII$ Join $B_3$ (the third point) to $C$ and draw a line through $B_{4}C$ parallel to $B_{3}C$, intersecting the extended line segment $BC$ at $C$.

$STEPIV$ Draw a line through $C$ parallel to $CA$ intersecting the extended line segment $BA$ at $A$.

Triangle $ABC$ so obtained is the required triangle such that

$\frac{AB}{AB}=\frac{BC}{BC}=\frac{AC}{AC}=\frac{4}{3}$