Give the justification of the construction also:
Draw a right triangle in which the sides (other than hypotenuse) are of lengths $4 \,cm$ and $3 \,cm$. Then construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle.

(N/A) It is given that sides other than hypotenuse are of lengths $4 \,cm$ and $3 \,cm$. Clearly,these will be perpendicular to each other.
The required triangle can be drawn as follows:
$1.$ Draw a line segment $AB = 4 \,cm$. Draw a ray $AX$ making $90^{\circ}$ with it at $A$.
$2.$ Draw an arc of $3 \,cm$ radius while taking $A$ as its centre to intersect the ray at $C$. Join $BC$. $\triangle ABC$ is the required triangle.
$3.$ Draw a ray $AY$ making an acute angle with $AB$,opposite to vertex $C$.
$4.$ Locate $5$ points $A_1, A_2, A_3, A_4, A_5$ on $AY$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5$.
$5.$ Join $A_3B$. Draw a line through $A_5$ parallel to $A_3B$ intersecting the extended line segment $AB$ at $B'$.
$6.$ Through $B'$,draw a line parallel to $BC$ intersecting the extended line segment $AC$ at $C'$. $\triangle AB'C'$ is the required triangle.
Justification:
The construction can be justified by proving that $AB' = \frac{5}{3} AB, B'C' = \frac{5}{3} BC, AC' = \frac{5}{3} AC$.
In $\triangle ABC$ and $\triangle AB'C'$,
$\angle ABC = \angle AB'C'$ (Corresponding angles)
$\angle BAC = \angle B'AC'$ (Common)
$\therefore \triangle ABC \sim \triangle AB'C'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'}$ .......$(1)$
In $\triangle AA_3B$ and $\triangle AA_5B'$,
$\angle A_3AB = \angle A_5AB'$ (Common)
$\angle AA_3B = \angle AA_5B'$ (Corresponding angles)
$\therefore \triangle AA_3B \sim \triangle AA_5B'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{AA_3}{AA_5} = \frac{3}{5}$ .........$(2)$
On comparing equations $(1)$ and $(2)$,we obtain
$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{3}{5}$
$AB' = \frac{5}{3} AB, B'C' = \frac{5}{3} BC, AC' = \frac{5}{3} AC$
This justifies the construction.