I'm getting 10-15 tonnes of topsoil delivered, what size tarpaulin do I need? I want to order 10-15 tonnes of top soil.
It'll be delivered by lorry and "tipped" onto the ground.
This will make roughly pyramid shape, or perhaps a truncated square pyramid. a cone shape, or a flat topped cone (conical frustrum).

Soil repose values: https://structx.com/Soil_Properties_005.html
Topsoil = 35 to 45 repose depending on wetness. (It's not going to be "dry" so 40 feels a good average.)
Standard density for topsoil is around 100 lb/ft3 (1600 kg/m3). ref This other site gives a value of 1250 kg/m3 ref2 but perhaps thats because its "loose" and not defined as 'topsoil'.
The soil should not go straight on the ground, it needs to go on a plastic sheet to cover the ground.
Given this, is it possible the estimate the meters squared area of the base of the shape; to work out how much tarpaulin / plastic sheet is needed?
I don't mind a margin of error up to 1 meter in either direction, if that helps you estimate. :-)
 A: As mentioned in the comments, the pile will have a cone shape. This is of course assuming that the unloading is done appropriately. The volume of this cone is given by
$$V=\frac13\pi r^2h,$$
where $h$ is the height of the cone and $r$ is the base radius. With the angle of repose $\theta_{\mathrm{soil}}$ we can relate $r$ and $h$ by
$$ h = r \tan(\theta_{\mathrm{soil}}). $$
Using further that the volume $V$ taken by a mass $M$ of soil is
$$ V = \frac{M}{\rho_{\mathrm{soil}}}, $$
where $\rho_{\mathrm{soil}}$ is the bulk density of soil. So, putting everything together, we arrive at
$$r = \sqrt[3]{ \frac{3M}{\pi\rho_{\mathrm{soil}}\tan(\theta_{\mathrm{soil}})} }.$$

Using the values in your case:
Plugging in that $M$ is between $10000 \mathrm{kg}$ and $15000 \mathrm{kg}$ and common values for the bulk density and angle of response are $\rho_{\mathrm{soil}}$ between $1250\mathrm{kg}/m^3$ and $1600 \mathrm{kg}/m^3$ and $\theta_{\mathrm{soil}}$ between $35°$ and $45°$ (see comments; assuming smaller angles and densities is more conservative), we arrive that the cone will have a base circle radius in the range of
$$r \approx 1.81\mathrm{m}~...~2.54 \mathrm{m}. $$
This range is obtained by using the extremal values of the mentioned ranges.
So, if you want to be on the (very) safe side, you could use a square-shape tarp that has a side length of $5\mathrm{m}$ for the pile plus $1\mathrm{m}$ margin ($0.5\mathrm{m}$ on each side), i.e., a total side length of $6\mathrm{m}$ and an area of $6\mathrm{m}\cdot6\mathrm{m}=36\mathrm{m}^2$, as displayed in the sketch below.

