In any triangle ABC, r_1 r_2 + r_2 r_3 + r_3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We know that $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, and $r_3 = \frac{\Delta}{s-c}$.Then, $r_1 r_2 + r_2 r_3 + r_3 r_1 = \frac{\De

Vedclass · Accepted Answer

$\frac{\Delta^2}{r^2}$

Vedclass · Answer

(A) We know that $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, and $r_3 = \frac{\Delta}{s-c}$.Then, $r_1 r_2 + r_2 r_3 + r_3 r_1 = \frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}$.Taking $\frac{\Delta^2}{(s-a)(s-b)(s-c)}$ as a common factor, we get:$\frac{\Delta^2}{(s-a)(s-b)(s-c)} [(s-c) + (s-a) + (s-b)]$.Since $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$, the expression becomes:$\frac{\Delta^2}{\Delta^2/s} [3s - (a+b+c)]$.Using $a+b+c = 2s$, we have:$s [3s - 2s] = s^2$.Since $r = \frac{\Delta}{s}$, we have $s = \frac{\Delta}{r}$, so $s^2 = \frac{\Delta^2}{r^2}$.

In any $\triangle ABC$, $r_1 r_2 + r_2 r_3 + r_3 r_1$ is equal to

Similar Questions

In a $\Delta ABC$,side $b$ is equal to

In a $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides of the triangle are in

The sum of the radii of the inscribed and circumscribed circles for an $n$-sided regular polygon of side $a$ is:

In $\triangle ABC$,if $r=3$ and $R=5$,then $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = $

If the angles of a triangle are in the ratio $1:2:3$,then their corresponding sides are in the ratio

Vedclass Test Series

Exam Paper Generator

Online Exam Module