Circles 2

Calculate the area of the region between the circumscribed circle and the inscribed circle of a triangle with sides 29 cm, 16 cm, and 21 cm.

Final Answer:

S =  614.6 cm²

Step-by-step explanation:

$$\begin{gathered} a=29\text{cm} \\ b=16\text{cm} \\ c=21\text{cm} \\ s=\frac{a+b+c}{2}=\frac{29+16+21}{2}=33\text{cm} \\ {S}_1=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}=\sqrt{33\cdot (33-29)\cdot (33-16)\cdot (33-21)}=12\sqrt{187}{\text{cm}}^2\doteq 164.0975{\text{cm}}^2 \\ \rho =\frac{S_1}{s}=\frac{164.0975}{33}\doteq 4.9727\text{cm} \\ r=\frac{a\cdot b\cdot c}{4\cdot S_1}=\frac{29\cdot 16\cdot 21}{4\cdot 164.0975}\doteq 14.8448\text{cm} \\ S=\pi \cdot (r^2-{\rho }^2)=3.1416\cdot (14.8448^2-4.9727^2)=614.6{\text{cm}}^2 \end{gathered}$$

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