Draw an isosceles triangle $ABC$ in which $AB = AC = 6 \, cm$ and $BC = 5 \, cm$. Construct a triangle $PQR$ similar to $ABC$ in which $PQ = 8 \, cm$. Also,justify the construction.

(N/A) Let $\triangle PQR$ and $\triangle ABC$ be similar triangles. The scale factor between the corresponding sides is $\frac{PQ}{AB} = \frac{8}{6} = \frac{4}{3}$.
Steps of construction:
$1.$ Draw a line segment $BC = 5 \, cm$.
$2.$ Taking $B$ and $C$ as centers,draw two arcs of radius $6 \, cm$ intersecting each other at $A$.
$3.$ Join $BA$ and $CA$. Thus,$\triangle ABC$ is the required isosceles triangle.
$4.$ From $B$,draw any ray $BX$ making an acute $\angle CBX$.
$5.$ Locate four points $B_1, B_2, B_3,$ and $B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$6.$ Join $B_3C$ and from $B_4$,draw a line $B_4R \parallel B_3C$ intersecting the extended line segment $BC$ at $R$.
$7.$ From point $R$,draw $RP \parallel CA$ meeting the extended line segment $BA$ at $P$.
Then,$\triangle PBR$ is the required triangle.
Justification:
Since $B_4R \parallel B_3C$ (by construction),
$\therefore \frac{BC}{CR} = \frac{3}{1}$.
Now,$\frac{BR}{BC} = \frac{BC + CR}{BC} = 1 + \frac{CR}{BC} = 1 + \frac{1}{3} = \frac{4}{3}$.
Also,$RP \parallel CA$,therefore $\triangle ABC \sim \triangle PBR$.
And $\frac{PB}{AB} = \frac{RP}{CA} = \frac{BR}{BC} = \frac{4}{3}$.
Hence,the new triangle is similar to the given triangle with sides $\frac{4}{3}$ times the corresponding sides of the isosceles $\triangle ABC$.