# Cross-sections of a cone

Cone with base radius 16 cm and height 11 cm divide by parallel planes to base into three bodies. The planes divide the height of the cone into three equal parts.

Determine the volume ratio of the maximum and minimum of the resulting body.

Result

p =  19

#### Solution:

$r = 16 \ \\ v = 11 \ \\ \ \\ V_1 = \dfrac13 \pi (r/3)^2 (v/3) = 109.219 \ \\ V_2 = \dfrac13 \pi (2r/3)^2 (2v/3) = 873.751 \ \\ V_3 = \dfrac13 \pi (r)^2 v = 2948.908 \ \\ p = \dfrac{ V_3 - V_2 }{ V_1 } = \dfrac{ 2948.908 - 873.751 }{ 109.219 } = 19 \ \\ p = \dfrac{ V_3 - V_2 }{ V_1 } = \ \\ p = \dfrac{ \dfrac13 \pi r^2 v - \dfrac13 \pi (2r/3)^2 (2v/3)}{ \dfrac13 \pi (r/3)^2 (v/3) } \ \\ p = \dfrac{ r^2 v - (2r/3)^2 (2v/3)}{ (r/3)^2 (v/3) } \ \\ p = \dfrac{ r^2 - (2r/3)^2 (2/3)}{ (r/3)^2 \cdot (1/3) } \ \\ p = \dfrac{ 1 - (2/3)^2 (2/3)}{ (1/3)^2 /3 } \ \\ p = \dfrac{ 1 - (2/3)^2 (2/3)}{ (1/3)^2 /3 } \ \\ p = \dfrac{ 19/27 }{ 1/27 } \ \\ p = \dfrac{ 19/27 }{ 1/27 } \ \\ p = 19 = 19:1$

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