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