This question was uploaded on 03/12/21 on social media accounts.
Solution:
Let, side of the triangle be 'S'.
Area of the above figure is equal to the sum of areas of triangles ABO, BCO, CAO:
\[A = \frac{3S + 4S + 5S}2 = 6S\]
Area of the above figure is equal to:
\[A = \frac{h.S}2\]
From both equations, h = 12
\[\Rightarrow S = \frac2{\sqrt3}h = 8\sqrt3\]
\[\Rightarrow A = 6S = 48\sqrt3\]
