The Atomium is modelled on an iron atom that has been magnified 165 billion times - Leaving Cert Mathematics - Question a - 2016

The diameter of each sphere is given as 18 metres. The radius $r$ can be calculated using the formula:
$r = \frac{diameter}{2} = \frac{18}{2} = 9 \text{ m}.$

The formula for the volume $V$ of a sphere is:
$V = \frac{4}{3} \pi r^3.$
Substituting $r = 9$ m:
$V = \frac{4}{3} \pi (9)^3 = \frac{4}{3} \pi (729) \approx 3053.63 \text{ m}^3.$
Thus, the volume of each sphere is approximately $3053.63 \text{ m}^3$.

The surface area $A$ of a single sphere is given by:
$A = 4\pi r^2.$
Using $r = 9$ m:
$A = 4 \pi (9)^2 = 4 \pi (81) \approx 1017.88 \text{ m}^2.$
For 9 spheres, the total surface area is:
$A_{total} = 9 \times 1017.88 \approx 9160.88 \text{ m}^2.$
Rounding to the nearest m² gives us $9161 \text{ m}^2$.

The formula for the curved surface area (CSA) of a cylinder is:
$CSA = 2\pi r h.$
For one pipe with radius $1.65$ m and height $23$ m:
$CSA = 2\pi (1.65) (23) \approx 239.85 \text{ m}^2.$
For 8 pipes, the total curved surface area is:
$CSA_{total} = 8 \times 239.85 \approx 1918.82 \text{ m}^2.$
Thus, rounding to the nearest m² gives $1919 \text{ m}^2$.

The total curved surface area of the 12 pipes is given as $3170 \text{ m}^2$. The curved surface area of one pipe with a radius of $1.45$ m is calculated as:
$CSA = 2\pi (1.45) h.$
Letting $h$ be the length of one pipe, we have:
$3170 = 12 \times (2\pi (1.45) h).$
Solving for $h$:
$h = \frac{3170}{12 \times 2\pi (1.45)} \approx 29 \text{ m}.$

The total surface area to be resurfaced is:
$A_{total} = 1919 + 9161 = 11080 \text{ m}^2.$
The cost is €670 per m², so:
$Total \; cost = 11080 \times 670 = €7,409,600.$
Thus, the approximate cost is €7,409,600.