# A concrete pedestal

A concrete pedestal has a shape of a right circular cone having a height of 2.5 feet. The diameter of the upper and lower bases are 3 feet and 5 feet, respectively. Determine the lateral surface area, total surface area, and the volume of the pedestal.

Result

S3 =  33.836 ft2
S =  60.54 ft2
V =  32.07 ft3

#### Solution:

$h = 2.5 \ ft \ \\ D_{ 1 } = 3 \ ft \ \\ D_{ 2 } = 5 \ ft \ \\ \ \\ \ \\ r = D_{ 1 }/2 = 3/2 = \dfrac{ 3 }{ 2 } = 1.5 \ ft \ \\ R = D_{ 2 }/2 = 5/2 = \dfrac{ 5 }{ 2 } = 2.5 \ ft \ \\ \ \\ l^2 = h^2 + (R-r)^2 \ \\ l = \sqrt{ h^2 + (R-r)^2 } = \sqrt{ 2.5^2 + (2.5-1.5)^2 } \doteq 2.6926 \ ft \ \\ \ \\ S_{ 3 } = \pi \cdot \ l \cdot \ (r+R) = 3.1416 \cdot \ 2.6926 \cdot \ (1.5+2.5) \doteq 33.836 = 33.836 \ ft^2$
$S_{ 1 } = \pi \cdot \ r^2 = 3.1416 \cdot \ 1.5^2 \doteq 7.0686 \ ft^2 \ \\ S_{ 2 } = \pi \cdot \ R^2 = 3.1416 \cdot \ 2.5^2 \doteq 19.635 \ ft^2 \ \\ \ \\ S = S_{ 1 }+S_{ 2 }+S_{ 3 } = 7.0686+19.635+33.836 \doteq 60.5395 = 60.54 \ ft^2$
$V = \dfrac{ 1 }{ 3 } \cdot \ \pi \cdot \ h \cdot \ (r^2 + r \cdot \ R + R^2) = \dfrac{ 1 }{ 3 } \cdot \ 3.1416 \cdot \ 2.5 \cdot \ (1.5^2 + 1.5 \cdot \ 2.5 + 2.5^2) \doteq 32.0704 = 32.07 \ ft^3$

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