Annulus from triangle

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

Correct answer:

S =  831.0003 mm²

Step-by-step explanation:

$$\begin{gathered} a=25\text{mm} \\ b=29\text{mm} \\ c=36\text{mm} \\ {c}_1=\sqrt{a^2+b^2}=\sqrt{25^2+29^2}=\sqrt{1466}\text{mm}\doteq 38.2884\text{mm} \\ {c}_1\ne c \\ s=\frac{a+b+c}{2}=\frac{25+29+36}{2}=45\text{mm} \\ A=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}=\sqrt{45\cdot (45-25)\cdot (45-29)\cdot (45-36)}=360{\text{mm}}^2 \\ R=\frac{a\cdot b\cdot c}{4\cdot A}=\frac{25\cdot 29\cdot 36}{4\cdot 360}=\frac{145}{8}=18.125\text{mm} \\ r=\frac{A}{s}=\frac{360}{45}=8\text{mm} \\ {S}_1=\pi \cdot R^2=3.1416\cdot 18.125^2\doteq 1032.0623{\text{mm}}^2 \\ {S}_2=\pi \cdot r^2=3.1416\cdot 8^2\doteq 201.0619{\text{mm}}^2 \\ S=S_1-S_2=1032.0623-201.0619=831.0003{\text{mm}}^2 \end{gathered}$$

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