Question
Construct an isosceles triangle whose base is 8cm and altitude 4cm and then another triangle whose sides are $\frac{3}{2}$ times the corresponding sides of the isosceles triangle.
✓
Answer
Given that Construct an isosceles triangle ABC in which AB = BC = 6cm and altitude = 4cm then another triangle similar to it whose sides are $\frac{3}{2}$ of the corresponding sides of $\triangle\text{ABC}.$ We follow the following steps to construct the given 
Step of construction:
Step I: First of all we draw a line segment $A B=6 cm$.
Step II: With $B$ as centre and radius $=B C=6 cm$, draw an arc.
Step III: From point $A$ and $B$ construct altitutde $C D=4 cm$, which cut the line $B S$ at point $C$.
Step IV: Join $A C$ to obtain $\triangle A B C$. Step $V$ : Below $A B$, makes an acute angle $\angle B A X=60^{\circ}$.
Step VI: Along $A X$, mark off five points $A_1$, $A_2$ and $A_3$ such that $A A_1=A_1 A_2=A_2 A_3$.
Step VII: Join $A_2 B$.
Step VIII: Since we have to construct a triangle $\triangle A Q R$ each of whose sides is $\left(1.5\right.$ times $\left.=\frac{3}{2}\right)$ of the corresponding sides of $\triangle A B C$. So, we draw a line $A_3 Q$ on $A X$ from point $A_3$ which is $A_3 Q \| A_2 B$, and meeting $A B$ at $Q$.
Step IX: From $Q$ point draw $Q R \| B C$, and meeting $A C$ at $R$. Thus, $\triangle AQR$ is the required triangle, each of whose sides is $\left(\frac{3}{2}\right)$ of the corresponding sides of $\triangle ABC$.
Step of construction:Step I: First of all we draw a line segment $A B=6 cm$.
Step II: With $B$ as centre and radius $=B C=6 cm$, draw an arc.
Step III: From point $A$ and $B$ construct altitutde $C D=4 cm$, which cut the line $B S$ at point $C$.
Step IV: Join $A C$ to obtain $\triangle A B C$. Step $V$ : Below $A B$, makes an acute angle $\angle B A X=60^{\circ}$.
Step VI: Along $A X$, mark off five points $A_1$, $A_2$ and $A_3$ such that $A A_1=A_1 A_2=A_2 A_3$.
Step VII: Join $A_2 B$.
Step VIII: Since we have to construct a triangle $\triangle A Q R$ each of whose sides is $\left(1.5\right.$ times $\left.=\frac{3}{2}\right)$ of the corresponding sides of $\triangle A B C$. So, we draw a line $A_3 Q$ on $A X$ from point $A_3$ which is $A_3 Q \| A_2 B$, and meeting $A B$ at $Q$.
Step IX: From $Q$ point draw $Q R \| B C$, and meeting $A C$ at $R$. Thus, $\triangle AQR$ is the required triangle, each of whose sides is $\left(\frac{3}{2}\right)$ of the corresponding sides of $\triangle ABC$.
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