Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.

Steps of construction:

1. Draw a line segment AB=7 cm.

2. Draw a ray AX, making an acute ∠BAX.

3. Along AX, mark 3+5= 8 points A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈ such that AA₁=A₁A₂=A₂A₃=A₃A₄=A₄A₅=A₅A₆=A₆A₇⁼A₇A₈.

4. Join A₈B.

5. From A₃, draw A₃P ││ A₈B meeting AB at P. (By making an angle equal to ∠BA₈ at A₃).

Then P is the point on AB which divides it in the ratio 3:5. Thus, AP : PB=3:5

Justification:

Let

AA₁=A₁A₂=A₂A₃=A₃A₄=........=A₇A₈=x

In ∆ABA₈, A₃P││A₈B

Hence, AP:PB=3:5