Construct a $ \Delta ABC $ in which $ AB=4cm $ , $ \angle B={{60}^{\circ }} $ and altitude $ CL=3cm $ . Construct a $ \Delta ADE $ similar to $ \Delta ABC $ such that each side of $ \Delta ADE $ is $ \dfrac{3}{2} $ times that of the corresponding side of $ \Delta ABC $ .

Hint:
 We start solving the problem by first drawing the line segment AB and then constructing the angle $ \angle ABQ={{60}^{\circ }} $ . We then draw an arc of length 3 cm and then constructing the angle $ \angle BAP={{90}^{\circ }} $ of 3 cm to intersect arc at P. We then constructing the angle $ \angle APC={{90}^{\circ }} $ to intersect BQ at C. We then connect the points A and C to complete the $ \Delta ABC $ . We then extend the side AB to D by using the fact that AD is $ \dfrac{3}{2} $ times of AB. We then construct the angle $ \angle ADS={{60}^{\circ }} $ using a protractor. We then extend the side AC to intersect DS at E which completes the $ \Delta ADE $ .

Complete step by step answer:
According to the problem, we need to construct a $ \Delta ABC $ in which $ AB=4cm $ , $ \angle B={{60}^{\circ }} $ and altitude $ CL=3cm $ . We then need to construct a $ \Delta ADE $ similar to $ \Delta ABC $ such that each side of $ \Delta ADE $ is $ \dfrac{3}{2} $ times that of the corresponding side of $ \Delta ABC $ .
Let us first construct $ \Delta ABC $ . The steps of construction are as shown below:
(i) Let us first draw the line segment AB of length 4 cm using a ruler.


(ii) Now, let us construct the angle $ \angle ABQ={{60}^{\circ }} $ using a protractor.

(c) Let us draw an arc of length 3 cm by taking A as a centre.

(d) Now, let us construct the angle $ \angle BAP={{90}^{\circ }} $ of 3 cm to intersect the arc using a protractor.

(e) Now, let us construct the angle $ \angle APC={{90}^{\circ }} $ to intersect the BQ at C using a protractor.

(f) Now, let us join the points A and C to complete $ \Delta ABC $ .

(g) Now, let us extend AB to D such that $ AD=\dfrac{3}{2}\times AB=\dfrac{3}{2}\times 4=6cm $ .

(h) Now, let us construct the angle $ \angle ADS={{60}^{\circ }} $ using a protractor.

(i) Now, let us extend AC to intersect DS at E to complete $ \Delta ADE $ .

Note:
 We can take the ratio of the sides of the triangles ABC and ADE to verify the similarity of both the triangles. We need to follow each step carefully in order to avoid confusion and mistakes in the diagram while constructing the triangles. Similarly, we can expect problems to draw the circum-center of the obtained triangle $ \Delta ADE $ .