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 D1=3 ft D2=5 ft   r=D1/2=3/2=32=1.5 ft R=D2/2=5/2=52=2.5 ft  l2=h2+(Rr)2 l=h2+(Rr)2=2.52+(2.51.5)22.6926 ft  S3=π l (r+R)=3.1416 2.6926 (1.5+2.5)33.83633.836 ft2h=2.5 \ \text{ft} \ \\ D_{1}=3 \ \text{ft} \ \\ D_{2}=5 \ \text{ft} \ \\ \ \\ \ \\ r=D_{1}/2=3/2=\dfrac{ 3 }{ 2 }=1.5 \ \text{ft} \ \\ R=D_{2}/2=5/2=\dfrac{ 5 }{ 2 }=2.5 \ \text{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 \ \text{ft} \ \\ \ \\ S_{3}=\pi \cdot \ l \cdot \ (r+R)=3.1416 \cdot \ 2.6926 \cdot \ (1.5+2.5) \doteq 33.836 \doteq 33.836 \ \text{ft}^2
S1=π r2=3.1416 1.527.0686 ft2 S2=π R2=3.1416 2.5219.635 ft2  S=S1+S2+S3=7.0686+19.635+33.83660.539560.54 ft2S_{1}=\pi \cdot \ r^2=3.1416 \cdot \ 1.5^2 \doteq 7.0686 \ \text{ft}^2 \ \\ S_{2}=\pi \cdot \ R^2=3.1416 \cdot \ 2.5^2 \doteq 19.635 \ \text{ft}^2 \ \\ \ \\ S=S_{1}+S_{2}+S_{3}=7.0686+19.635+33.836 \doteq 60.5395 \doteq 60.54 \ \text{ft}^2
V=13 π h (r2+r R+R2)=13 3.1416 2.5 (1.52+1.5 2.5+2.52)32.070432.07 ft3V=\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 \doteq 32.07 \ \text{ft}^3



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