(C) In a right-angled triangle $\Delta ABC$ with $\angle C = 90^{\circ}$, the inradius $r$ is given by $r = \frac{a+b-c}{2}$.Since $\angle C = 90^{\ci

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(C) In a right-angled triangle $\Delta ABC$ with $\angle C = 90^{\circ}$, the inradius $r$ is given by $r = \frac{a+b-c}{2}$.Since $\angle C = 90^{\circ}$, the hypotenuse $c$ is the diameter of the circumcircle, so the circumradius $R = \frac{c}{2}$, which implies $2R = c$.Now, consider the expression $2(r+R) = 2r + 2R$.Substituting the values, we get $2(\frac{a+b-c}{2}) + c$.$= (a+b-c) + c = a+b$.Therefore, $2(r+R) = a+b$.

Class 11 Mathematics Trigonometrical Equations Circle connected with triangle

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In a $\Delta ABC$, if $\angle C = 90^{\circ}$, $r$ and $R$ are the inradius and circumradius of the $\Delta ABC$ respectively, then $2(r+R)$ is equal to

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$b+c$
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$c+a$
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$a+b$
D
$a+b+c$

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In a $\Delta ABC$, if $\angle C = 90^{\circ}$, $r$ and $R$ are the inradius and circumradius of the $\Delta ABC$ respectively, then $2(r+R)$ is equal to

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