# Rotation

The right triangle with legs 11 cm and 18 cm rotate around the longer leg.

Calculate the volume and surface area of the formed cone.

Result

S =  1109.124 cm2
V =  2280.796 cm3

#### Solution:

$a=11 \ \text{cm} \ \\ b=18 \ \text{cm} \ \\ b>a \ \\ \ \\ r=a=11 \ \text{cm} \ \\ s=\sqrt{ a^2+b^2 }=\sqrt{ 11^2+18^2 } \doteq \sqrt{ 445 } \ \text{cm} \doteq 21.095 \ \text{cm} \ \\ \ \\ S_{1}=\pi \cdot \ r^2=3.1416 \cdot \ 11^2 \doteq 380.1327 \ \text{cm}^2 \ \\ S_{2}=\pi \cdot \ r \cdot \ s=3.1416 \cdot \ 11 \cdot \ 21.095 \doteq 728.9917 \ \text{cm}^2 \ \\ S=S_{1} + S_{2}=380.1327 + 728.9917 \doteq 1109.1244 \doteq 1109.124 \ \text{cm}^2$
$V=\dfrac{ 1 }{ 3 } \cdot \ S_{1} \cdot \ b=\dfrac{ 1 }{ 3 } \cdot \ 380.1327 \cdot \ 18 \doteq 2280.7963 \doteq 2280.796 \ \text{cm}^3$

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