Construct a triangle 𝐴𝐵𝐶 with 𝐵𝐶 = 7 𝑐𝑚, ∠𝐵 = 60° and 𝐴𝐵 = 6 𝑐𝑚. Construct another triangle whose sides are $\left(\frac{3}{4}\right)$ times of the corresponding sides of ∆𝐴𝐵𝐶.

We need to construct a triangle whose sides are ( times of the corresponding $\left(\frac{3}{4}\right)$ sides of ∆𝐴𝐵𝐶.

The following are the steps to be followed: 

a) Draw a line segment of 𝐵𝐶 = 7 𝑐𝑚

 b) Taking 𝐵 as centre, draw ∠𝐵 = 60°. 

c) Again taking 𝐵 as centre and radius 𝐴𝐵 = 6 𝑐𝑚, draw an arc and mark point A. 

d) Now, join 𝐴𝐶 to get ∆𝐴𝐵𝐶. 

e) Extend 𝐵 downwards making an acute angle with 𝐵𝐶. f) Along that line mark four points, namely$B_1,B_2,B_3,and B_4$such that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$

g) Join $B_4$ and C. 

h) Now, draw a line from $B_3$ such that it meets 𝐵𝐶 at 𝐶' at ensuring that . $B_{3}C\text{'}\parallel B_{4}C$

 i) Draw 𝐴'𝐶' ∥ 𝐴𝐶. 

The final figure would look like as shown in the following figure.