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 $area (Δ QRS) = area (Δ STU)$ so that (A) has larger area than (B) and the difference in areas is equal to the area of the small triangle $Δ UVW$ .

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

Packing (A)

Triangle and parallelogram overlaid.

The triangle has side length

4 + 2 \sqrt{3}

so that area is

\frac{1}{2} (4 + 2 \sqrt{3})^{2} \sin 60^{\circ}

, or (equivalently, using half base times height)

\frac{1}{2} (4 + 2 \sqrt{3}) (3 + 2 \sqrt{3})

, which is 24.12 square units to 2 decimal places.

Packing (B)

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

Packing (C)

Skewed rectangle.

\begin{matrix} {AE}^{2} & = 16 + 4 - 16 \cos 120^{\circ} = 28 \end{matrix}

so that

AE = 2 \sqrt{7}

. Using the Sine Rule for

Δ ACE

\begin{matrix} \frac{\sin θ}{CE} & = \frac{\sin 120^{\circ}}{AE}, \end{matrix}

hence

\sin θ = \sqrt{3} / 2 \sqrt{7}

. Thus

\cos θ = 5 / 2 \sqrt{7}

, and the area is 30.40 square units to 2 decimal places.

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