ABCD is a trapezium in which AB is parallel to CD. The sides AD and BC when extended, intersect at point E. If AB = 2 cm, CD = 1 cm, and perimeter of ABCD is 6 cm, then the perimeter, in cm, of $$\triangle AEB$$ is

The simplest way to visualise this would be with symmetry. 

We can take AD to be x and CB to be 3-x; since the perimeter of ABCD is 6, and AB and CD are given as 2 and 1 cm, respectively, that leaves AD+BC as three only. 

We can see that triangles AEB and DEC are similar, with the lengths of AB being double of CD, essentially making D the mid-point of AE and C the mid-point of EB. 

Through this, we get the length of DE and CE to be x and 3-x. 
Give us the AE and BE lengths as 2x and 6-2x, respectively. 

 Giving the perimeter of AEB as 2x+6-2x+2 = 8 cm

Therefore, Option A is the correct answer.