To compare the volumes of the sphere, cylinder, and cone, we first need to write out their volume formulas based on a shared radius \( r \) and the height of the cylinder and cone as \( h = 2r \).
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Volume of the Sphere (V_s): \[ V_s = \frac{4}{3} \pi r^3 \]
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Volume of the Cylinder (V_c): \[ V_c = \pi r^2 h = \pi r^2 (2r) = 2 \pi r^3 \]
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Volume of the Cone (V_co): \[ V_co = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3 \]
Next, let's express the total volume of the cylinder and cone: \[ V_c + V_co = 2 \pi r^3 + \frac{2}{3} \pi r^3 \]
To combine these, we need a common denominator: \[ V_c + V_co = 2 \pi r^3 + \left(\frac{2}{3} \pi r^3\right) = \frac{6}{3} \pi r^3 + \frac{2}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]
Now, we compare \( V_s \) to \( V_c + V_co \): \[ V_s = \frac{4}{3} \pi r^3 \] \[ V_c + V_co = \frac{8}{3} \pi r^3 \]
From this, it becomes clear that: \[ V_c + V_co = 2 V_s \]
Thus, rearranging gives us: \[ V_s = \frac{1}{2} (V_c + V_co) \]
This shows that: \[ V_s < V_c + V_co \]
Now, we can analyze the options:
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sphere = cylinder + cone ⇒ \( \frac{4}{3} \pi r^3 \neq 2 \pi r^3 + \frac{2}{3} \pi r^3 \)
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cylinder = sphere – cone ⇒ \( 2 \pi r^3 \neq \left(\frac{4}{3} \pi r^3 - \frac{2}{3} \pi r^3\right) \)
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cone = cylinder – sphere ⇒ \( \frac{2}{3} \pi r^3 \neq \left(2 \pi r^3 - \frac{4}{3} \pi r^3\right) \)
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cone = sphere – cylinder ⇒ \( \frac{2}{3} \pi r^3 \neq \left(\frac{4}{3} \pi r^3 - 2 \pi r^3\right) \)
None of the options provided correctly express the relationship we found.
Therefore, the correct conclusion based on the volume comparison is: $V_s$ is less than the sum of the volumes of the cylinder and the cone. However, this is not directly reflected in the options given.