The cone: Surface area and volume
We have a tank of conical shape (as an inverted cone, with the apex below and the basis up) to fill it with rainwater. The entire tank measures $5$ metres high. Define the radius of the cone so, when the water rises the half of its height, it contains $1000$ L of water.
First, we have to find the dimensions of the part that contains water, which is the half a cone. So, on one hand, $h' =\dfrac{5}{2}=2,5$.
On the other hand, using the information of the volume: $$V_{water}=1000L\cdot\dfrac{1\ m^3}{1000L}=1\ m^3$$ $$V_{water}=\dfrac{1}{3}\cdot h_{water}\cdot\pi\cdot(r_{water})^2$$ $$r_{water}=\sqrt{\dfrac{3\cdot V_{water}}{\pi\cdot h_{water}}}=0,618 \ m$$
Finally, we use the proportions (because there is a direct relation between $h$ and $r$) $$\dfrac{r_{water}}{r_{tank}}=\dfrac{h_{water}}{h_{tank}} $$ $$r_{tank}=r_{water}\cdot\dfrac{h_{tank}}{h_{water}}=1,236 \ m^3$$
$$r_{tank}=1,236 \ m^3$$