Question

If a triangle of maximum area is inscribed within a circle of radius R, then

${\text{A}}{\text{. }}s = 2{R^2} \\\ {\text{B}}{\text{. }}\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{r_3}}} = \dfrac{{\sqrt 2 + 1}}{R} \\\ {\text{C}}{\text{. }}r = \left( {\sqrt 2 - 1} \right)R \\\ {\text{D}}{\text{. }}s = \left( {1 + \sqrt 2 } \right).2R \\\$

Answer 1

Hint: -First you have to draw a diagram of circle in which draw a right angled triangle assuming maximum area then apply the condition for finding maximum area and use properties of triangle to solve further.

Complete step-by-step solution -
We have
Let ABC be a right angled triangle inscribed in a circle of radius R
So BC = 2R =diameter
$\Delta = \dfrac{1}{2}.AB.AC\sin {90^0}$
It will be maximum if AB = AC
$A{B^2} + A{C^2} = B{C^2} \Rightarrow 2{\left( {AB} \right)^2} = 4{R^2}$
$\therefore s = \dfrac{1}{2}.AB.AC = \dfrac{1}{2}{\left( {AB} \right)^2} = {R^2}$
So the option is incorrect.
$2s = AB + BC + CA = 2R + R\sqrt 2 + R\sqrt 2 = 2R\left( {1 + \sqrt 2 } \right)$
$\therefore s = R\left( {1 + \sqrt 2 } \right)$
So option D is incorrect.
$r = \dfrac{S}{s} = \dfrac{{2{R^2}}}{{R\left( {\sqrt 2 + 1} \right)}} = 2\left( {\sqrt 2 - 1} \right)R$
So option C is also incorrect.
Hence we saw that option A,C, D are not correct.
Also we know that
$\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{r_3}}} = \dfrac{1}{r} = \dfrac{S}{s} = \dfrac{{R\left( {1 + \sqrt 2 } \right)}}{{{R^2}}}$(properties of solution of triangle)
$\therefore \dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{r_3}}} = \dfrac{1}{r} = \dfrac{{\sqrt 2 + 1}}{R}$
So option B is the correct option.

Note: -Whenever you get this type of question the key concept of solving is You have to remember relations like $r = \dfrac{S}{s}$and many relations like this to solve these types of questions. You have to understand the incenter circumcenter and inradius circumradius to solve this type of question.