Question

Construction of a triangle of sides 4$cm$, 5$cm$, and 6$cm$ and then a triangle similar to it whose sides are $\dfrac{2}{3}$ of the corresponding sides of the first triangle. The length of side $A'C'$ (in $cm$) is:

Answer 1

Hint: Here we will proceed by practical geometry, using a set of rulers, compasses and protractors, different shapes. Then by applying the conditions given in the question we will get our answer.

Complete step-by-step answer:
Steps of construction are:
1. Firstly draw a line segment $AB$ of length 4$cm$.

2. Now cut an arc of radius 5$cm$from point $A$an arc of 6$cm$from point B.

3. Name the point of intersection of arcs to be point $C$.

4. Now join point $AC$and $BC$. Thus $ABC$ is the required triangle.

5. Draw a line $AD$ which makes an acute angle with $AB$ and is opposite of vertex $C.$

6. Cut three equal parts of line $AD$ namely $A{A_1},A{A_2},A{A_3}$.

7. Now join ${A_3}$to $B.$ Draw a line ${A_2}B'$ parallel to ${A_3}B$.

8. And then draw a line $B'C'$ parallel to $BC$.

Hence,
$A'B'C'$is the required triangle.

Note: Whenever we come up with this type of problem, it is to be noted that in the construction of a triangle the three-sides and angle may or may not be equal in dimensions. Then by writing down the steps one by one we will get our required answer.