Question

In a triangle ABC, Let \(C = \frac {\pi} 2\), If r is the inradius and R is circumradius of the triangle ABC, then 2(r + R) equals
1. a + b
2. a + b + c
3. a + c
4. b + c

Answer 1

Correct Answer - Option 1 : a + b

Concept:

Sine law:

 \(\rm \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\), Where a,b,c are sides and R is circumradius.

Inradius:

r = (s - a) tan(\(\rm \frac A 2\)) = (s - b) tan(\(\rm \frac B 2\)) = (s - c) tan(\(\rm \frac C 2\)), Where a,b,c are sides, r is inradius, s is semiperimeter = \(\rm \frac{a+b+c}{2}\)

 

Calculation:

Here, in triangle ABC, Let \(C = \frac {\pi} 2\),

\(\rm \frac{c}{sinC}=2R\\ \Rightarrow 2R = \rm \frac{c}{sin(\pi/2)}=c\)

R = \(\rm \frac C 2\)

Now, inradius r = (s - c) tan(\(\rm \frac C 2\))

= (s - c) \(\rm tan (\frac \pi 4)\)

= (s - c)

So, 2(r + R) = 2 (s - c + \(\rm \frac C 2\)) = 2 (s - \(\rm \frac C 2\))

= a + b + c - c

= a + b

Hence, option (1) is correct.