# 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

Correct result:

S =  831 mm2

#### Solution:

$a=25 \ \text{mm} \ \\ b=29 \ \text{mm} \ \\ c=36 \ \text{mm} \ \\ \ \\ c_{1}=\sqrt{ a^2+b^2 }=\sqrt{ 25^2+29^2 } \doteq \sqrt{ 1466 } \ \text{mm} \doteq 38.2884 \ \text{mm} \ \\ c_{1} \ne c \ \\ \ \\ s=\dfrac{ a+b+c }{ 2 }=\dfrac{ 25+29+36 }{ 2 }=45 \ \text{mm} \ \\ A=\sqrt{ s \cdot \ (s-a) \cdot \ (s-b) \cdot \ (s-c) }=\sqrt{ 45 \cdot \ (45-25) \cdot \ (45-29) \cdot \ (45-36) }=360 \ \text{mm}^2 \ \\ \ \\ R=\dfrac{ a \cdot \ b \cdot \ c }{ 4 \cdot \ A }=\dfrac{ 25 \cdot \ 29 \cdot \ 36 }{ 4 \cdot \ 360 }=\dfrac{ 145 }{ 8 }=18.125 \ \text{mm} \ \\ \ \\ r=\dfrac{ A }{ s }=\dfrac{ 360 }{ 45 }=8 \ \text{mm} \ \\ \ \\ S_{1}=\pi \cdot \ R^2=3.1416 \cdot \ 18.125^2 \doteq 1032.0623 \ \text{mm}^2 \ \\ S_{2}=\pi \cdot \ r^2=3.1416 \cdot \ 8^2 \doteq 201.0619 \ \text{mm}^2 \ \\ \ \\ S=S_{1}-S_{2}=1032.0623-201.0619=831 \ \text{mm}^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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