In triangle ABC,frac{1}{r_1^2}+frac{1}{r_2^2} | vedclass

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

Question

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.Therefore,$\frac{1}{r_1}

Vedclass · Accepted Answer

$\frac{a^2+b^2+c^2}{\Delta^2}$

Vedclass · Answer

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$.Therefore,$\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,$\frac{1}{r_3} = \frac{s-c}{\Delta}$,and $\frac{1}{r} = \frac{s}{\Delta}$.Now,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} = \frac{(s-a)^2}{\Delta^2} + \frac{(s-b)^2}{\Delta^2} + \frac{(s-c)^2}{\Delta^2} + \frac{s^2}{\Delta^2}$.$= \frac{(s^2+a^2-2as) + (s^2+b^2-2sb) + (s^2+c^2-2sc) + s^2}{\Delta^2}$.$= \frac{4s^2 + a^2+b^2+c^2 - 2s(a+b+c)}{\Delta^2}$.Since $a+b+c = 2s$,we have $2s(a+b+c) = 2s(2s) = 4s^2$.$= \frac{4s^2 + a^2+b^2+c^2 - 4s^2}{\Delta^2} = \frac{a^2+b^2+c^2}{\Delta^2}$.

In $\triangle ABC$,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} =$

Similar Questions

The radius of the incircle of a triangle with sides $18 \, cm$,$24 \, cm$,and $30 \, cm$ is ..... $cm$.

The circumradius of a triangle whose sides are $10 \ units$,$8 \ units$,and $6 \ units$ is (in $units$)

In a $\Delta ABC$, if $\angle C = 90^{\circ}$, $r$ and $R$ are the inradius and circumradius of the $\Delta ABC$ respectively, then $2(r+R)$ is equal to

In a $\triangle ABC$,if $a=13, b=14, c=15$,then find the value of $r_1$.

$PQR$ is an isosceles triangle with $PQ=PR$. If the radius of the circumcircle of $\triangle PQR$ is equal to the length of $PQ$,then $\angle P=$ (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module