Construct a triangle of sides 5 cm, 6cm and 7cm and then a triangle similar to it whose sides are \[\dfrac{7}{5}\] of the corresponding sides of the first triangle.

Hint: Here first we will construct the triangle with sides 5 cm, 6cm, and 7cm and then we will use the given fraction of the side to construct a similar triangle.

Complete step by step answer:
Let us first draw the triangle with sides 5 cm, 6cm and 7cm.
- First, we will draw the base AB of the triangle of length 5cm.
- Then A as the center we will mark an arc with the help of a compass at a distance of 6cm.
- Similarly, now with B as the center, we will mark an arc with the help of a compass at a distance of 7cm.
- Now finally we will mark the point so obtained as C and then join the points A and C and A and B.

Now we will construct a similar triangle of ABC whose sides are \[\dfrac{7}{5}\]of the corresponding sides of\[\Delta ABC\].
We will first construct a ray AY such that it makes an acute angle with line AB on the opposite side of vertex C.

Now we will mark seven points (as 7 is greater than 5 in the fraction\[\dfrac{7}{5}\]) at equal distances from each other on the ray AY i.e. \[A{A_1} = {A_1}{A_2} = {A_2}{A_3} = {A_3}{A_4} = {A_4}{A_5} = {A_5}{A_6} = {A_6}{A_7}\]

Now we will join \[{A_5}\]to \[B\]and then we will draw a line \[{A_7}\] (as the side of the similar triangle is \[\dfrac{7}{5}\]of the corresponding sides of\[\Delta ABC\]) to the extended line AB such that it intersect AB at \[{B^{'}}\]and is parallel to\[{A_5}B\].

Now we will finally draw a line from \[{B^{'}}\]to extended AC such that it intersects AC at \[{C^{'}}\]and the line \[{B^{'}}{C^{'}}\]is parallel to the line BC.

Now the new triangle so formed i.e. \[\Delta A{B^{'}}{C^{'}}\] is the similar triangle to \[\Delta ABC\].

Note:
- Students should use compass and scale for perfect measurement and arcs an accurate distance to get the desired triangles.
- Also, we have drawn three arcs on the ray AY because we had to construct a similar triangle such that its sides are \[\dfrac{7}{5}\]of the corresponding sides of the first triangle.