Problems from 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.

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Development:

First, we have to find the dimensions of the part that contains water, which is the half a cone. So, on one hand, $h^{'} = \frac{5}{2} = 2, 5$ .

On the other hand, using the information of the volume: $V_{w a t e r} = 1000 L \cdot \frac{1 m^{3}}{1000 L} = 1 m^{3}$ $V_{w a t e r} = \frac{1}{3} \cdot h_{w a t e r} \cdot π \cdot (r_{w a t e r})^{2}$ $r_{w a t e r} = \sqrt{\frac{3 \cdot V_{w a t e r}}{π \cdot h_{w a t e r}}} = 0, 618 m$

Finally, we use the proportions (because there is a direct relation between $h$ and $r$ ) $\frac{r_{w a t e r}}{r_{t a n k}} = \frac{h_{w a t e r}}{h_{t a n k}}$ $r_{t a n k} = r_{w a t e r} \cdot \frac{h_{t a n k}}{h_{w a t e r}} = 1, 236 m^{3}$

Solution:

$r_{t a n k} = 1, 236 m^{3}$

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