QUESTION IMAGE
Question
a conical tank is partially filled with water. the height of the tank is 11 m and the radius of the tank is 5 m. the height of the water is 6 m and the radius of the water is 3 m. how much more water, in cubic meters, could you fit in the tank? use 3.14 for pi and round to the nearest tenth. a. 56.5 cubic meters b. 287.8 cubic meters c. 231.3 cubic meters d. 344.3 cubic meters
Response
- First, recall the volume - formula for a cone:
- The volume of a cone is \(V=\frac{1}{3}\pi r^{2}h\).
- The volume of the entire conical tank with \(r = 5\) m and \(h = 11\) m is \(V_{total}=\frac{1}{3}\times3.14\times5^{2}\times11=\frac{1}{3}\times3.14\times25\times11=\frac{3.14\times275}{3}\approx287.83\) \(m^{3}\).
- The volume of the water - filled part of the cone with \(r = 3\) m and \(h = 6\) m is \(V_{water}=\frac{1}{3}\times3.14\times3^{2}\times6=\frac{1}{3}\times3.14\times9\times6 = 3.14\times18 = 56.52\) \(m^{3}\).
- Then, find the volume of the empty part of the cone:
- The volume of the empty part \(V_{empty}=V_{total}-V_{water}\).
- \(V_{empty}=\frac{1}{3}\times3.14\times5^{2}\times11-\frac{1}{3}\times3.14\times3^{2}\times6\).
- \(V_{empty}=\frac{3.14}{3}(25\times11 - 9\times6)=\frac{3.14}{3}(275 - 54)=\frac{3.14}{3}\times221\approx231.3\) \(m^{3}\).
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c. 231.3 cubic meters