Question

Q23. ​In a triangle ABC, let angle C=π/2.If r is the inradius and R is the circumradius of the triangle, then 2(r+R) is equal to,
(A). a+b-c (B). b+c (C). c+a (D). a+b+c

Answer 1

Dear Student,
$$\begin{gathered} Given,InatriangleABC,\angle C={90}^{\circ } \\ Weknowthat \\ R=\frac{c}{2\sin C} \\ \Rightarrow R=\frac{c}{2\sin {90}^{\circ }} \\ \Rightarrow R=\frac{c}{2} \\ \Rightarrow 2R=c\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right) \\ And, \\ r=\left(s-c\right)\tan \frac{C}{2} \\ \Rightarrow r=\left(s-c\right)\tan \frac{{90}^{\circ }}{2} \\ \Rightarrow r=\left(s-c\right)\tan {45}^{\circ } \\ \Rightarrow r=\left(s-c\right) \\ \Rightarrow 2r=2s-2c\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \left(2\right) \\ Addingequation\left(1\right)and\left(2\right),wehave \\ 2R+2r=c+2s-2c \\ \Rightarrow 2R+2r=c+\left(a+b+c\right)-2c \\ \Rightarrow 2R+2r=a+b \\ \Rightarrow 2\left(R+r\right)=a+b \\ Regards. \end{gathered}$$