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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
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