To analyze the relationship between the volumes of the sphere, cylinder, and cone, let's first establish their volume formulas and the given dimensions.
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Sphere: The volume \( V \) of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \]
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Cylinder: The volume \( V \) of a cylinder is given by: \[ V = \pi r^2 h \] Since the height \( h \) of the cylinder is twice the radius \( r \), we have: \[ h = 2r \implies V = \pi r^2 (2r) = 2\pi r^3 \]
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Cone: The volume \( V \) of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \] With the same height \( h = 2r \): \[ V = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3 \]
Now, let's summarize the volumes:
- Volume of sphere: \( V_s = \frac{4}{3} \pi r^3 \)
- Volume of cylinder: \( V_c = 2 \pi r^3 \)
- Volume of cone: \( V_{cone} = \frac{2}{3} \pi r^3 \)
Next, we will compare these volumes to see which statement is true:
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Check \( V_c = V_s - V_{cone} \): \[ V_s - V_{cone} = \frac{4}{3} \pi r^3 - \frac{2}{3} \pi r^3 = \frac{2}{3} \pi r^3 = V_{cone} \] This statement does not hold true.
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Check \( V_s = V_c + V_{cone} \): \[ V_c + V_{cone} = 2 \pi r^3 + \frac{2}{3} \pi r^3 = \left(2 + \frac{2}{3}\right) \pi r^3 = \frac{6}{3} \pi r^3 + \frac{2}{3} \pi r^3 = \frac{8}{3} \pi r^3 \] This statement does not hold true either.
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Check \( V_{cone} = V_c - V_s \): \[ V_c - V_s = 2 \pi r^3 - \frac{4}{3} \pi r^3 = \left(2 - \frac{4}{3}\right) \pi r^3 = \frac{6}{3} \pi r^3 - \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 = V_{cone} \] This statement does hold true.
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Check \( V_{cone} = V_s - V_c \): \[ V_s - V_c = \frac{4}{3} \pi r^3 - 2 \pi r^3 = \frac{4}{3} \pi r^3 - \frac{6}{3} \pi r^3 = -\frac{2}{3} \pi r^3 \] This statement does not hold true.
The only statement that holds true is:
cone = cylinder – sphere.
2 answers