Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1\frac { 1 }{ 2 }\) times the corresponding sides of the isosceles triangle.

Steps to construct figure by given data are as follows:
Step 1: Construct an isosceles triangle ABC with BC=8cm and altitude AD=4cm.

Step 2:Draw a ray BX, making an acute angle with BC.

Step 3: Locate 3 points on BX, such that BP=PQ=QR.

Step 4: Join QC.

Step 5: Through R, draw a line RC parallel to OC, meeting produced line BC at C’.

Step 6: Through C, draw a line CA parallel to CA, meeting the produced line BA at A’.



Thus, \(\triangle \) A’BC’ is the required isosceles triangle

JUSTIFICATION:
In \(\triangle \) ABC and \(\triangle \) A’BC’, we have

\(\triangle ACB=\triangle A’C’B \)(corresponding angles)

\(∠B=∠ B \)(common)

\(∴ ΔABC∼ΔA’BC’ \) (By AA similarity)

\(∴ \frac{AB}{A’B}=\frac{AC}{A’C’}=\frac{BC}{BC’} \)

But, \(\frac{BC}{BC’}=\frac{BQ}{BR}=\frac{2}3 ∴ \frac{BC}{BC’}=\frac{2}3 \)

\(⇒ \frac{A’B}{AB}=\frac{A’C’}{AC}=\frac{B’C}{BC}=\frac{3}2.\)

Hence, we get the required triangle whose sides are \(\frac{3}{2} \) , i.e., \(1\frac{1}{2}\) times of the corresponding sides of the isosceles \(\therefore \) ABC.