# Truncated cone

A truncated cone has a bases radiuses 40 cm and 10 cm and a height of 25 cm. Calculate its surface area and volume.

Result

S =  11474.863 cm2
V =  54977.871 cm3

#### Solution:

$r_{1}=10 \ \text{cm} \ \\ r_{2}=40 \ \text{cm} \ \\ v=25 \ \text{cm} \ \\ \ \\ S_{1}=\pi \cdot \ r_{1}^2=3.1416 \cdot \ 10^2 \doteq 314.1593 \ \text{cm}^2 \ \\ S_{2}=\pi \cdot \ r_{2}^2=3.1416 \cdot \ 40^2 \doteq 5026.5482 \ \text{cm}^2 \ \\ \ \\ x=|r_{1}-r_{2}|=|10-40|=30 \ \text{cm} \ \\ \ \\ s=\sqrt{ v^2+x^2 }=\sqrt{ 25^2+30^2 } \doteq 5 \ \sqrt{ 61 } \ \text{cm}^2 \doteq 39.0512 \ \text{cm}^2 \ \\ \ \\ S_{3}=\pi \cdot \ (r_{1}+r_{2}) \cdot \ s=3.1416 \cdot \ (10+40) \cdot \ 39.0512 \doteq 6134.1558 \ \text{cm}^2 \ \\ S=S_{1} + S_{2} + S_{3}=314.1593 + 5026.5482 + 6134.1558 \doteq 11474.8633 \doteq 11474.863 \ \text{cm}^2$
$S_{4}=r_{1}^2+r_{1} \cdot \ r_{2}+r_{2}^2=10^2+10 \cdot \ 40+40^2=2100 \ \text{cm}^2 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ \pi \cdot \ v \cdot \ S_{4}=\dfrac{ 1 }{ 3 } \cdot \ 3.1416 \cdot \ 25 \cdot \ 2100 \doteq 54977.8714 \doteq 54977.871 \ \text{cm}^3$

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