Question
Construct a $\triangle\text{ABC}$ in which AB = 5cm, $\angle\text{B} = 60^\circ,$ altitude CD = 3cm. Construct a $\triangle\text{AQR}$ similar to $\triangle\text{ABC}$ such that side of $\triangle\text{AQR}$ is 1.5 times that of the corresponding sides of $\triangle\text{ABC}.$

Answer

Steps of construction:
i. Draw a line segment $A B=5 cm$.
ii. At $A$, draw a perpendicular and cut off $A E=3 cm$.
iii. From E, draw EF || AB.
iv. From $B$, draw a ray making an angle of 60 meeting $E F$ at $C$.
v. Join CA. Then $A B C$ is the triangle.
vi. From $A$, draw a ray $A X$ making an acute angle with $A B$ and cut off 3 equal parts making $A A_1=A_1 A_2=A_2 A_3$.
vii. Join $A_2$ and $B$.
iiii. From $A$, draw $A^{\prime} B^{\prime}$ parallel to $A_2 B$ and $B^{\prime} C^{\prime}$ parallel to $B C$.
Then $\triangle\text{C}'\text{AB}'$ is the required triangle.

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