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A water tank has a cylindrical shape with a diameter of 2 m and a perpendicular height of 3 m. Since it was already old and leaking, it is to be replaced by another tank of the same capacity but in the form of a frustum of a cone. The diameters of the ends of the frustum are designed to be 1 m and 2 m respectively. What must be its height?
Solution:

The volume of the shapes must be equal thus, $$V_{cy} = V_{fcone}$$

Volume of the cylinder: $$V_{cy} = \pi r^2h$$

$$V_{cy} = \pi (1)^2 (3)$$

$$V_{cy} = 3\pi$$

Volume of the frustum of a cone: $$V_{fc} = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1r_2)$$

$$V_{fc} = \frac{1}{3} \pi h ((0.5)^2 + (1)^2 +(0.5)(1))$$

$$V_{fc} = \frac{7}{12}\pi h$$

$$3\pi = \frac{7}{12}\pi h$$

$$h = 5.14 ~ m$$