Frustum of a cone

A reservoir contains 28.54 m3 of water when completely full. The diameter of the upper base is 3.5 m while at the lower base is 2.5 m. Determine the height if the reservoir is in the form of a frustum of a right circular cone.

Result

h =  4.001 m

Solution:

V=28.54 m3 D1=3.5 m D2=2.5 m  r1=D1/2=3.5/2=74=1.75 m r2=D2/2=2.5/2=54=1.25 m  V=13 π h(r12+r1 r2+r22)  h=3 Vπ (r12+r1 r2+r22)=3 28.543.1416 (1.752+1.75 1.25+1.252)4.0005=4.001  m V = 28.54 \ m^3 \ \\ D_{ 1 } = 3.5 \ m \ \\ D_{ 2 } = 2.5 \ m \ \\ \ \\ r_{ 1 } = D_{ 1 }/2 = 3.5/2 = \dfrac{ 7 }{ 4 } = 1.75 \ m \ \\ r_{ 2 } = D_{ 2 }/2 = 2.5/2 = \dfrac{ 5 }{ 4 } = 1.25 \ m \ \\ \ \\ V = \dfrac{ 1 }{ 3 } \cdot \ \pi \cdot \ h ( r_{ 1 }^2 + r_{ 1 } \cdot \ r_{ 2 } + r_{ 2 }^2) \ \\ \ \\ h = \dfrac{ 3 \cdot \ V }{ \pi \cdot \ ( r_{ 1 }^2 + r_{ 1 } \cdot \ r_{ 2 } + r_{ 2 }^2) } = \dfrac{ 3 \cdot \ 28.54 }{ 3.1416 \cdot \ ( 1.75^2 + 1.75 \cdot \ 1.25 + 1.25^2) } \doteq 4.0005 = 4.001 \ \text { m }







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