help

Not as  hard as it seems

Call the lower left vertex of the  triangle, B

Let the center of the bottom left circle  be A

Let the radius of this circle intersect the base of the triangle at C

So.....triangle  ABC  is a 30 -60 -90   right triangle

And   BC  =  radius  of the circle * sqrt (3)  = 1 * √ (3)   =  √ (3)

So....using symmetry...the side of the  triangle  =

2BC + 8 r  =

2sqrt (3)  +  8 r   =

2sqrt (3)  +  8  =

√12 + 8

So....the area of the  triangle =

(1/2) ( √12 + 8)^2 * (√3/2) =

(√3/4)  ( 12 + 16√12 + 64)  =

(√3/4) ( 76 + 16*2√3)   =

√3 ( 19 + 8√3)  =

19√3 + 24  units^2  ≈  56.9  units^2

All you need is a bottom row:

Circle diameter   >  D = 1

Radius                >  R = 0.5

Triangle side       >  a = 4D + 2[R / tan(30°)]     a = 5.732    a/2 = 2.866

Triangle height    >  H = sqrt[ a² - (a/2)²]      H = 4.964

Triangle area       >  A = H * a/2              A = 14.227 u²