Question

In a $ \triangle \, ABC, \, let \, \angle C = \pi / 2 $. If r is th e in rad iu s 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 1

Answer: (A)

Here, $ R^2 = MC^2 = \frac{1}{4} (a^2 + b^2)$ [by distance from origin]
= $ \frac{1}{4} c^2 $ $ $ [by Pythagoras theorem]
$\Rightarrow R = \frac{ c}{2}$
Next, r = (s- c) tan (C/2) = (s- c) tan $\pi$ /4 = s - c
$\therefore $ 2(r + R ) = 2 r + 2 R = 2 s - 2 c + c
$ $ = a + b + c - c = a + b.