There are probably easier ways to do this...... but
We have three "outside" triangles and one inner one
Area of the the three "outside" triangles =
(1/2)(√3/2) [ 6*4 + 2 * 4 + 6 * 2) ] = 11√3 units^2
And we an construct an inner triangle with sides BF, BD and FD
Length of BF = √[ 6^2 + 2^2 – 2(6)(2)(1/2)] = √28
Length of BD = √[ 6^2 + 4^2 – 2(6)(4)(1/2)] = √28
Length of FD = √[ 2^2 + 4^2 – 2(2)(4)(1/2)] = √12
Deriving the semi-perimeter of this triangle → s = [2√28 + √12] / 2 =
√3 + 2√7 = √3 + √28 → s^2 = 31 + 2√84 = 31 + √336
And using Heron's formula ...
Area of inner triangle = √ [ (√3 + √28)(√3 + √28 – √28)^2 (√3 + √28 – √12) ] =
√ [ (√3 + √28)(√3 )^2 (√3 + √28 – √12) ] =
√3 * √ [ (√3 + √28) ((√3 + √28 – √12) ] =
√3 * √ [ (s) ((s – √12) ] =
√3 * √ [ s^2 – √12 s) ] =
√3 * √ [ 31 + √336 – √12 [ √3 + √28) ] =
√3 * √ [ 31 + √336 – √36 - √336) ] =
√3 * √ [ 31 – 6 ] =
√3 * √ [ 25 ] =
√3 * 5 =
5√3 units^2
So....the total area = [11√3 + 5√3] units^2 = 16√3 units^2