In a triangle ABC,let angle C = frac{pi}{2}. | vedclass

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(B) In a right-angled triangle $	riangle ABC$ with $\angle C = \frac{\pi}{2}$,the sides are $a$ (opposite to $A$),$b$ (opposite to $B$),and $c$ (hypo

Vedclass · Accepted Answer

$\frac{a+b}{2}$

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(B) In a right-angled triangle $	riangle ABC$ with $\angle C = \frac{\pi}{2}$,the sides are $a$ (opposite to $A$),$b$ (opposite to $B$),and $c$ (hypotenuse opposite to $C$).From the properties of the incircle,the distance from vertex $C$ to the points of tangency on sides $AC$ and $BC$ is $r$.Thus,the sides can be expressed as $b = x+r$ and $a = y+r$,where $x$ and $y$ are the lengths of the tangents from vertices $A$ and $B$ to the incircle respectively.The hypotenuse $c = x+y$.Since $c$ is the diameter of the circumcircle,$c = 2R$,so $x+y = 2R$.Now,adding the expressions for $a$ and $b$:$a+b = (y+r) + (x+r) = (x+y) + 2r = 2R + 2r$.Dividing by $2$,we get:$R+r = \frac{a+b}{2}$.

In a $\triangle ABC$,let $\angle C = \frac{\pi}{2}$. If $r$ and $R$ are the inradius and circumradius of $\triangle ABC$ respectively,then $R+r=$

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