In a triangle ABC, let angle C = π / 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

Question

In a triangle ABC, let angle C = π / 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

Tardigrade Admin · Accepted Answer

Here, R2 = MC2 = (1/4) (a2 + b2) [by distance from origin] = (1/4) c2 [by Pythagoras theorem] ⇒ R = ( c/2) Next, r = (s- c) tan (C/2) = (s- c) tan π /4 = s - c ∴ 2(r + R ) = 2 r + 2 R = 2 s - 2 c + c = a + b + c - c = a + b.

Q. In a $△ A BC, l e t ∠ C = π /2$ . 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 + b

b + c

c + a

a + b + c

Here, $R^{2} = M C^{2} = \frac{1}{4} (a^{2} + b^{2})$ [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 $π$ /4 = s - c
$∴$ 2(r + R ) = 2 r + 2 R = 2 s - 2 c + c
= a + b + c - c = a + b.

Q. In a △ABC,let∠C=π/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