+2440 The general formula for volume for any pyramid or cone is \(V=\frac{1}{3}Bh\)
Let V = volume of the cone or pyramid
Let B = area of the base
Let h = the perpendicular height of the cone or pyramid
Since all these figures are cones, we know that the area of the base equals \(V=\frac{1}{3}\pi r^2h\). Ok, let's start solving for the individual volumes.
| Cone 1 | Cone 2 | Cone 3 |
| \(V_1=\frac{1}{3}\pi r^2h;\\ r=11,h=9 \) | \(V_2=\frac{1}{3}\pi r^2h;\\ r=6,h=14 \) | \(V_3=\frac{1}{3}\pi r^2h;\\ r=14, h=8\) |
| \(V_1=\frac{1}{3}\pi*11^2*9\) | \(V_2=\frac{1}{3}\pi*6^2*14\) | \(V_3=\frac{1}{3}\pi*14^2*8\) |
| \(V_1=3\pi*121\) | \(V_2=12\pi*14\) | \(V_3=\frac{1}{3}\pi*1568\) |
| \(V_1=363\pi\text{cm}^3\) | \(V_2=168\pi\text{cm}^3\) | \(V_3=\frac{1568\pi}{3}\text{cm}^3\approx 500\text{cm}^3\) |
I estimated for cone 3 because 3 does not divide evenly with 1568. This approximation is good enough for these purposes anyway because we only care about their order of greatness. The otder, therefore, from least to greatest is the following:
Cone 2, Cone 1, Cone 3 (option 2)
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