Question

If $A,{A_1},{A_2}$ and ${A_3}$ are the areas of the incircle and excircles, then $\dfrac{1}{{\sqrt {{A_1}} }} + \dfrac{1}{{\sqrt {{A_2}} }} + \dfrac{1}{{\sqrt {{A_3}} }}$ is equal to
A) $\dfrac{1}{{\sqrt A }}$
B) $\dfrac{2}{{\sqrt A }}$
C) $\dfrac{3}{{\sqrt A }}$
D) $\dfrac{4}{{\sqrt A }}$

Answer 1

Hint: First of all we have to know what an incircle is and an excircle of a triangle. So, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. And, an excircle or inscribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extension of the other two. In order to solve the problem, we have to assume the radii of the circles and simplify the given condition. Then, we have to use the formula that relates the radii of incircle to the radii of excircles, i.e.,
$\dfrac{1}{r} = \dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{r_3}}}$
Where, $r = $radius of incircle
${r_1},{r_2},{r_3} = $ radii of excircles
In order to find the required result.

Complete step by step answer:
Given, $A,{A_1},{A_2}$ and ${A_3}$ are the incircle and excircles.
Let $r$ be the radius for area $A$.
${r_1}$ be the radius for area ${A_1}$.
${r_2}$ be the radius for area ${A_2}$.
${r_3}$ be the radius for area ${A_3}$.
Now, we have, $\dfrac{1}{{\sqrt {{A_1}} }} + \dfrac{1}{{\sqrt {{A_2}} }} + \dfrac{1}{{\sqrt {{A_3}} }}$
We know, formula of area of a circle is $\pi {r^2}$.
Therefore, $\dfrac{1}{{\sqrt {{A_1}} }} + \dfrac{1}{{\sqrt {{A_2}} }} + \dfrac{1}{{\sqrt {{A_3}} }}$
$ = \dfrac{1}{{\sqrt {\pi r_1^2} }} + \dfrac{1}{{\sqrt {\pi r_2^2} }} + \dfrac{1}{{\sqrt {\pi r_3^2} }}$
Taking, $\dfrac{1}{{\sqrt \pi }}$ common, we get,
$ = \dfrac{1}{{\sqrt \pi }}\left( {\dfrac{1}{{\sqrt {r_1^2} }} + \dfrac{1}{{\sqrt {r_2^2} }} + \dfrac{1}{{\sqrt {r_3^2} }}} \right)$
Simplifying the equation, we get,
$ = \dfrac{1}{{\sqrt \pi }}\left( {\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{r_3}}}} \right)$
We know, the relation between radii of incircles and excircles is,
$\dfrac{1}{r} = \dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{r_3}}}$
Now, using this relation, we get,
$\dfrac{1}{{\sqrt {{A_1}} }} + \dfrac{1}{{\sqrt {{A_2}} }} + \dfrac{1}{{\sqrt {{A_3}} }} = \dfrac{1}{{\sqrt \pi }}\left( {\dfrac{1}{r}} \right)$
$ = \dfrac{1}{{\sqrt \pi }}\left( {\dfrac{1}{{\sqrt {{r^2}} }}} \right)$
$ = \dfrac{1}{{\sqrt {\pi {r^2}} }}$
Here, $A = \pi {r^2}$.
So, substituting the equality, we get,
 $\dfrac{1}{{\sqrt {{A_1}} }} + \dfrac{1}{{\sqrt {{A_2}} }} + \dfrac{1}{{\sqrt {{A_3}} }} = \dfrac{1}{{\sqrt A }}$
Therefore, the correct option is (A).

Note:
In geometry, the incircle and excircles are really useful in order to know the length of sides and area of the triangle if only the radii of the excircles and incircles are known. Many other properties of the triangle can be deduced if we have the information about the excircle and incircle.