+128406 Let the altitude of 1/3 be drawn to side A
Let the altitude of 1/4 be drawn to side B
Let the altitude of 1/6 be drawn to side C
Equating areas
(1/2) (A) (1/3) = (1/2)(B)(1/4) → B = (4/3)A
(1/2)(A) (1/3) = (1/2)(C)(1/6) → C = (2)A
The semi-perimeter = (A + (4/3)A + (2)A ) / 2 = (13/6) A
Using Heron's Formula
Area = sqrt [ (13/6)A * [ (13/6)A - A ] * [ (13/6)A - (4/3)A ] * [ (13/6)A - 2A ] ] =
sqrt [ A^4 (13/6) (7/6) (5/6) ( 1/6) ] =
A^2 sqrt [ 455 ] / 36
Equating areas
(1/2)A (1/3) = A^2 sqrt [ 455] / 36
(1/6)A = A^2 sqrt [ 455] / 36
(1/6) = A sqrt [ 455 ] / 36
6 /sqrt [ 455] = A
The area is (1/2) (6 / sqrt [455]) ( 1/3) = 1/ sqrt 455 units^2 ≈ .0468 units^2
