In a triangle ABC, let [ C = 2].

Question

In a triangle $ABC$, let $$\angle C = {\pi  \over 2}$$. If $r$ is the inradius and $R$ is the circumradius of the triangle $ABC$, then $2(r+R)$ equals :

Accepted Answer

We know by sin c rule

$${c \over {\sin C}} = 2R \Rightarrow c = 2R\sin C$$

$$ \Rightarrow c = 2R$$

$$\left( \, \right.$$as $$\,\,\,\angle C = {90^ \circ }$$ $$\left. \, \right)$$

Also $$\tan {C \over 2} = {r \over {s - c}}$$

$$ \Rightarrow \tan {\pi  \over 4} = {r \over {s - c}}$$

$$\left( \, \right.$$as $$\,\,\,\angle C = {90^ \circ }$$ $$\left. \, \right)$$

$$ \Rightarrow r = s - c = {{a + b - c} \over 2}$$

$$ \Rightarrow 2r + c = a + b$$

$$ \Rightarrow 2r + 2R = a + b$$

(using $c=2R$)

Solutions · 1