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# 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.