Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to AABC in which PQ = 8 cm. Also justify the
construction.
Let $\triangle \mathrm{PQR}$ and $\triangle \mathrm{ABC}$ are similar triangles, then its scale factor between the corresponding sides is
$\frac{P Q}{A B}=\frac{8}{6}=\frac{4}{3}$
Steps of construction

 Draw a line segment BC = 5 cm.
 Construct OQ the perpendicular bisector of line segment BC meeting BC at P’.
 Taking B and C as centres draw two arcs of equal radius 6 cm intersecting each other at A
 Join BA and CA. So, ΔABC is the required isosceles triangle.
5. From $B$, draw any ray $B X$ making an acute $\angle C B X$
6. Locate four points $B_{1}, B_{2}, B_{3}$ and $B_{4}$ on $B X$ such that $B B_{1}=B_{1} B_{2}=B_{2} B_{3}=B_{3} B_{4}$
7. Join $B_{3} C$ and from $B_{4}$ draw a line $B_{4} R \ B_{3} C$ intersecting the extended line segment $B C$ at $R$.
8. From point R, draw RP\CA meeting BA produced at P
Then, $\triangle P B R$ is the required triangle.
Justification
$\because \quad \quad B_{4} R \ B_{3} C \quad$ (by construction)
$\therefore$ $\frac{B C}{C R}=\frac{3}{1}$
Now, $\frac{B R}{B C}=\frac{B C+C R}{B C}$
$=1+\frac{C R}{B C}=1+\frac{1}{3}=\frac{4}{3}$
Also, $R P \ C A$
$\therefore$ $\triangle A B C \sim \triangle P B R$
and $\frac{P B}{A B}=\frac{R P}{C A}=\frac{B R}{B C}=\frac{4}{3}$
Hence, the new triangle is similar to the given triangle whose sides are,$\frac{4}{3}$ times, of the
corresponding sides of the isosceles $\triangle A B C$.