Congratulations to Nicola Spittal, S4, Madras College, St Andrew's for your excellent solution to this problem.

We have four packings of six circles, each of radius 1 unit that touch their neighbours and the sides of the box namely: (A) triangular; (B) parallelogram; (C) rectangle (unusual packing) and (D) rectangle (usual packing).

By superimposing the triangular packing (A) on the parallelogram packing (B) as shown below we can see that area(ΔQRS)=area(ΔSTU) so that (A) has larger area than (B) and the difference in areas is equal to the area of the small triangle ΔUVW.

We use many 30-60-90 triangles with sides in the ratio 1:2:3. The areas of the packings are as follows.

Packing (A)

Triangle and parallelogram overlaid.

The triangle has side length 4+23 so that area is 1 2 (4+23)2 sin 60 , or (equivalently, using half base times height) 1 2 (4+23)(3+23), which is 24.12 square units to 2 decimal places.

Packing (B)

The base length is given by PW=tan 60 +4+tan 30 =3+4+ 1 3 =6.3094. The height is given by h=2+2cos 30 =2+3=3.7321, so that the area is 23.55 square units to 2 decimal places.

Packing (C)

Skewed rectangle.

The line joining AE is parallel to the bottom of the rectangle. The base of the rectangle has length AA'+ADcosθ+DD' which is 2+6cosθ. Similarly, the height of the rectangle is 2+6sinθ. It remains to find θ. As ΔCDE is equilateral, ACE= 120 . Using the Cosine Rule
AE2 =16+4-16cos 120 =28

so that AE=27. Using the Sine Rule for ΔACE,
sinθ CE = sin 120 AE ,

hence sinθ=3/27. Thus cosθ=5/27, and the area is 30.40 square units to 2 decimal places.

Packing (D) : as the rectangle is 6×4, the area is 24 square units.