# Annulus from triangle

Calculate the content of the area bounded by a circle circumscribed and a circle inscribed by a triangle with sides a = 25mm, b = 29mm, c = 36mm

Result

S =  831 mm2

#### Solution:

$a=25 \ \text{mm} \ \\ b=29 \ \text{mm} \ \\ c=36 \ \text{mm} \ \\ \ \\ c_{1}=\sqrt{ a^2+b^2 }=\sqrt{ 25^2+29^2 } \doteq \sqrt{ 1466 } \ \text{mm} \doteq 38.2884 \ \text{mm} \ \\ c_{1} \ne c \ \\ \ \\ s=\dfrac{ a+b+c }{ 2 }=\dfrac{ 25+29+36 }{ 2 }=45 \ \text{mm} \ \\ A=\sqrt{ s \cdot \ (s-a) \cdot \ (s-b) \cdot \ (s-c) }=\sqrt{ 45 \cdot \ (45-25) \cdot \ (45-29) \cdot \ (45-36) }=360 \ \text{mm}^2 \ \\ \ \\ R=\dfrac{ a \cdot \ b \cdot \ c }{ 4 \cdot \ A }=\dfrac{ 25 \cdot \ 29 \cdot \ 36 }{ 4 \cdot \ 360 }=\dfrac{ 145 }{ 8 }=18.125 \ \text{mm} \ \\ \ \\ r=\dfrac{ A }{ s }=\dfrac{ 360 }{ 45 }=8 \ \text{mm} \ \\ \ \\ S_{1}=\pi \cdot \ R^2=3.1416 \cdot \ 18.125^2 \doteq 1032.0623 \ \text{mm}^2 \ \\ S_{2}=\pi \cdot \ r^2=3.1416 \cdot \ 8^2 \doteq 201.0619 \ \text{mm}^2 \ \\ \ \\ S=S_{1}-S_{2}=1032.0623-201.0619 \doteq 831.0003 \doteq 831 \ \text{mm}^2$

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