2) Construct a triangle ABC in which AB = 5 cm, $\angle B=60°$ and altitudes CD = 3 cm. Construct a $△AQR$ similar to $△ABC$ such that each side of $△PQR$ is 1.5 times that of corresponding side of $△ABC.$

Steps of construction will be: -

1) Construct a line segment AB of length 5 cm.

2) From B, construct an angle of 60° using compass. Name it as ∠ABX.

3) Construct a line EF parallel to line segment AB at a height of 3 cm such that it intersect  at point C. Join AC.

4) Construct CD perpendicular to AB. CD is the altitude of the triangle ABC of height 3 cm.

5) Construct  making an acute angle with AB, on the side opposite to vertex C and mark 3 points (Corresponding to the greater of 2 or 3) X₁, X₂ and X₃ on AL such that AX₁ =  X₁X₂ = X₂X₃.

6) Join X₂ B (Corresponding to the denominator of the given ratio) and construct a line passing through X₃(Corresponding to the numerator of the given ratio) and parallel to X₂B. Let this line intersect the extended line AB at R.

7) Now, construct a line passing through R and parallel to BC, which intersects the extended line AC at point Q.

8) Thus, AQR is the required triangle which is similar to the given triangle ACB and with sides  of the sides of the given triangle.