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Let the long diagonal  of the rhombus    =  L

By the intersecting chord theorem  we have that

(3/2)r * (1/2)r    = [ ( 1/2) L ]^2

(3/4)r^2  = (1/4)L^2       multiply both sides by 4

(3)r^2  =  L^2       take the square root

(√3) r  = L

So.....the area of the rhombus  =   (1/2) product of the diagonals   =

(1/2) (r) (√3) r  =

[√3/2] r^2

Circle's radius      >  r = 2

Rhombus side      >  s = r

--II-- long diagonal > d₁ =?             d₁ = sqrt[s²- (r/2)²]*2      d₁ = 3.464

--II-- short diagonal > d₂ = r           

Rhombus  area       > A = ?             A = 1/2(d₁*r)  or  A = s * sqrt(3)  or A = (d₁*d₂)/2   

                                                                                      A = 3.464 u²