Draw a triangle $ABC$ with side $BC = 6 \, cm$,$AB = 5 \, cm$ and $\angle ABC = 60^{\circ}$. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle $ABC$.

(N/A) $\triangle A'BC'$ whose sides are $\frac{3}{4}$ of the corresponding sides of $\triangle ABC$ can be drawn as follows:
$1.$ Draw a $\triangle ABC$ with side $BC = 6 \, cm$,$AB = 5 \, cm$ and $\angle ABC = 60^{\circ}$.
$2.$ Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.
$3.$ Locate $4$ points (as $4$ is the greater number in $\frac{3}{4}$),$B_1, B_2, B_3, B_4$,on line segment $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$4.$ Join $B_4C$ and draw a line through $B_3$ parallel to $B_4C$ intersecting $BC$ at $C'$.
$5.$ Draw a line through $C'$ parallel to $AC$ intersecting $AB$ at $A'$. $\triangle A'BC'$ is the required triangle.
Justification:
The construction can be justified by proving $A'B = \frac{3}{4} AB$,$BC' = \frac{3}{4} BC$,$A'C' = \frac{3}{4} AC$.
In $\triangle A'BC'$ and $\triangle ABC$:
$\angle A'C'B = \angle ACB$ (Corresponding angles)
$\angle A'BC' = \angle ABC$ (Common)
$\therefore \triangle A'BC' \sim \triangle ABC$ ($AA$ similarity criterion)
$\Rightarrow \frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$ ........$(1)$
In $\triangle BB_3C'$ and $\triangle BB_4C$:
$\angle B_3BC' = \angle B_4BC$ (Common)
$\angle BB_3C' = \angle BB_4C$ (Corresponding angles)
$\therefore \triangle BB_3C' \sim \triangle BB_4C$ ($AA$ similarity criterion)
$\Rightarrow \frac{BC'}{BC} = \frac{BB_3}{BB_4} = \frac{3}{4}$ ........$(2)$
From equations $(1)$ and $(2)$,we obtain:
$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{3}{4}$
$A'B = \frac{3}{4} AB, BC' = \frac{3}{4} BC, A'C' = \frac{3}{4} AC$.
This justifies the construction.