Construct an isosceles triangle whose base is $8\, cm$ and altitude is $4\, cm$. Then,construct another triangle whose sides are $1 \frac{1}{2}$ times the corresponding sides of the isosceles triangle. Also,provide the justification for the construction.

(N/A) Let us assume that $\triangle ABC$ is an isosceles triangle having $CA = CB$,base $AB = 8\, cm$,and altitude $AD = 4\, cm$ (where $D$ is the midpoint of $AB$).
$A$ $\triangle AB'C'$ whose sides are $\frac{3}{2}$ times those of $\triangle ABC$ can be drawn as follows:
$1.$ Draw a line segment $AB = 8\, cm$. Construct the perpendicular bisector of $AB$ to find the midpoint $D$. Let the perpendicular line be $OO'$.
$2.$ Taking $D$ as the centre,draw an arc of radius $4\, cm$ on the perpendicular line to intersect at point $C$. Join $AC$ and $BC$ to form the isosceles $\triangle ABC$.
$3.$ Draw a ray $AX$ making an acute angle with $AB$ on the side opposite to vertex $C$.
$4.$ Locate $3$ points $A_1, A_2,$ and $A_3$ on $AX$ such that $AA_1 = A_1A_2 = A_2A_3$.
$5.$ Join $BA_2$. Draw a line through $A_3$ parallel to $BA_2$ to intersect the extended line segment $AB$ at $B'$.
$6.$ Draw a line through $B'$ parallel to $BC$ to intersect the extended line segment $AC$ at $C'$. $\triangle AB'C'$ is the required triangle.
Justification:
Since $A_2B \parallel A_3B'$,by Basic Proportionality Theorem in $\triangle AA_3B'$,we have $\frac{AB}{AB'} = \frac{AA_2}{AA_3} = \frac{2}{3}$. Thus,$AB' = \frac{3}{2}AB$.
Since $BC \parallel B'C'$,$\triangle ABC \sim \triangle AB'C'$. Therefore,$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{2}{3}$,which implies $B'C' = \frac{3}{2}BC$ and $AC' = \frac{3}{2}AC$. This justifies the construction.