Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are `3/4` of the corresponding sides of the ∆ABC.
A ΔA'BC' whose sides are `3/4`of the corresponding sides of ΔABC can be drawn as follows:
Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°
Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Locate 4 points (as 4 is greater in 3 and 4), B1, B2, B3, B4, on line segment BX.
Join B4C and draw a line through B3, parallel to B4C intersecting BC at C'
Draw a line through C' parallel to AC intersecting AB at A'. ΔA'BC' is the required triangle.