Question
In a triangle $$ABC,$$ let $$\angle C = \frac{\pi }{2}.$$ If $$r$$ is the inradius and $$R$$ is the circumradius of the triangle $$ABC,$$ then $$2 (r+ R)$$ equals
A.
$$b + c$$
B.
$$a + b$$
C.
$$a + b + c$$
D.
$$c + a$$
Answer :
$$a + b$$
Solution :
We know by sinc rule $$\frac{c}{{\sin C}} = 2R$$
$$\eqalign{
& \Rightarrow \,\,c = 2R\sin C \cr
& \Rightarrow \,\,c = 2R\,\,\,\,\,\,\,\left( {\because \,\,\angle C = {{90}^ \circ }} \right) \cr
& {\text{Also }}\tan \frac{C}{2} = \frac{r}{{s - c}} \cr
& \Rightarrow \,\,\tan \frac{\pi }{4} = \frac{r}{{s - c}}\,\,\left( {\because \,\,\angle C = {{90}^ \circ }} \right) \cr
& \Rightarrow \,\,r = s - c = \frac{{a + b - c}}{2} \cr
& \Rightarrow \,\,2r + c = a + b \cr
& \Rightarrow \,\,2r + 2R = a + b\,\left( {{\text{using }}c = 2R} \right) \cr} $$