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In a triangle $\mathrm{ABC}$, let $\angle \mathrm{C}=\frac{\pi}{2}$. If $\mathrm{r}$ is the inradius and $\mathrm{R}$ is the circumradius of the the triangle $A B C$, then $2(r+R)$ equals
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Verified Answer
The correct answer is:
$a+b$
$a+b$
$$
2 r+2 R=c+\frac{2 a b}{(a+b+c)}=\frac{(a+b)^2+c(a+b)}{(a+b+c)}=a+b \quad\left(\text { since } c^2=a^2+b^2\right)
$$
2 r+2 R=c+\frac{2 a b}{(a+b+c)}=\frac{(a+b)^2+c(a+b)}{(a+b+c)}=a+b \quad\left(\text { since } c^2=a^2+b^2\right)
$$
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