check the picture below
so.. if you notice, the diameter is 20, the radius is half that, the radius and height are at a ratio of 5:6, if the larger cone is filled up to 9ft, that means, the conical gap above, is 3ft in height
so, simply getting the volume of the larger cone filled up, with a height of 12, and then subtracting that conical gap, you'd end up with what's leftover, whatever the volume is at 9ft
[tex]\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\\\\
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\textit{volume of larger cone}\qquad V_l=\cfrac{\pi 10^2\cdot 12}{3}\implies \boxed{V_l=400\pi }
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\textit{volume of smaller cone}\qquad V_s=\cfrac{\pi \left( \frac{5}{2} \right)^2\cdot 3}{3}\implies \boxed{V_s=\cfrac{25\pi }{4}}
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\textit{volume of liquid up to 9ft}\qquad V_l-V_s=\cfrac{1575\pi }{4}\implies 393.75\pi [/tex]
