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Hee Chan and Jeang Bin sent us their working with correct areas for shapes A and C:
The longer part of the rectangle is 4 because each circle has radius 1 unit.
The right-angled triangle with side length marked x has angles 30^{\circ}, 60^{\circ} and 90^{\circ} so its hypotenuse must be 2, and side x is \sqrt{3} according to Pythagoras's law.
So the total length of one side of the triangle is 4+2\sqrt{3}.
The area of the equilateral triangle is
\frac{1}{2}(4+2\sqrt{3})^2 \times \sin 60^{\circ}which is 24.12 square units to two decimal places.
For C, for the bottom there are 3 circles and this means that the length of the bottom is 6 units. Also for the height there are 2 circles so the length is 4 units. The area is 4 \times 6 = 24 square units.
In a similar problem Nicola Spittal, S4, Madras College, St Andrew's sent in this excellent solution, demonstrating that A must have a larger than B and also calculating the correct areas for shapes A, B and C.
By superimposing the triangular packing (A) on the parallelogram packing (B) as shown below we can see that area(\Delta QRS) = area(\Delta STU) so that (A) has larger area than (B) and the difference in areas is equal to the area of the small triangle \Delta 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)
The triangle has side length 4+2\sqrt{3} so that area is {1\over 2}(4+2\sqrt{3})^2 \sin 60^\circ, or (equivalently, using half base times height) {1\over 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 + {1\over \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)
As the rectangle is 6\times 4, the area is 24 square units.
Here is one method of finding the area of the remaining shapes:
Packing (D)
h = \sqrt{3} + 1 (from diagram)
\cos 30^\circ = \frac{h}{a} \Rightarrow a = \frac{h}{\cos 30^\circ} = 2(\frac{\sqrt{3}}{3} + 1)
Splitting the hexagon in to two trapezia, the area is therefore:
A_D = 2\{\frac{1}{2}(a + 2a)h\} = \frac{3}{2}ah = \frac{3}{2}(\frac{\sqrt{3}}{3}+1)(\sqrt{3}+1) = 8\sqrt{3}+12 = 25.86 \textrm{ (4 s.f.)}
Packing (E)
Area composed of triangle, width 2h height \frac{a}{2}, and rectangle, width 2h height 2+\frac{a}{2}
\therefore \ A_E = \frac{1}{2}(2h)(\frac{a}{2}) + (2h)(2+\frac{a}{2}) = \frac{3}{2}ah + 4h = 8\sqrt{3} + 10 = 23.86 \textrm{ (4 s.f.)}
Packing (F)
A_F = 5(2h) = 10h = 10(\sqrt{3}+1) = 27.32 \textrm{ (4 s.f.)}
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