Question

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

Answer 1

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

Complete step by step answer:
According to the problem, we need to construct a $\mathrm{\Delta }ABC$ in which $AB=4cm$ , $\mathrm{\angle }B={60}^{\circ }$ and altitude $CL=3cm$ . We then need to construct a $\mathrm{\Delta }ADE$ similar to $\mathrm{\Delta }ABC$ such that each side of $\mathrm{\Delta }ADE$ is ${\displaystyle \frac{3}{2}}$ times that of the corresponding side of $\mathrm{\Delta }ABC$ .
Let us first construct $\mathrm{\Delta }ABC$ . 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 $\mathrm{\angle }ABQ={60}^{\circ }$ 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 $\mathrm{\angle }BAP={90}^{\circ }$ of 3 cm to intersect the arc using a protractor.

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

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

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

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

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

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 $\mathrm{\Delta }ADE$ .