Construct an isosceles triangle whose base is $8 \mathrm{~cm}$ and altitude $4 \mathrm{~cm}$ and then another triangle whose sides are $1 \frac{1}{2}$ times the corresponding sides of the isosceles triangle.
Steps of Construction
Step 1. Draw a line segment BC = 8 cm.
Step 2. Draw the perpendicular bisector XY of BC, cutting BC at D.
Step 3. With D as centre and radius 4 cm, draw an arc cutting XY at A.
Step 4. Join AB and AC. Thus, an isosceles ∆ABC whose base is 8 cm and altitude 4 cm is obtained.
Step 5 . Extend $B C$ to $E$ such that $B E=\frac{3}{2} B C=\frac{3}{2} \times 8 \mathrm{~cm}=12 \mathrm{~cm}$.
Step 6. Draw EF || CA, cutting BA produced in F.
Here, $\Delta \mathrm{BEF}$ is the required triangle similar to $\triangle \mathrm{ABC}$ such that each side of $\triangle \mathrm{BEF}$ is $1 \frac{1}{2}$ (or $\frac{3}{2}$ ) times the corresponding side of $\triangle \mathrm{ABC}$.