In a Delta ABC,if {r_1}

Question

(A) Given that in a $\Delta ABC$,${r_1}  \frac{1}{r_2} > \frac{1}{r_3}

Vedclass · Accepted Answer

$a < b < c$

Vedclass · Answer

(A) Given that in a $\Delta ABC$,${r_1} Taking the reciprocal of the inequality,we get:$\frac{1}{r_1} > \frac{1}{r_2} > \frac{1}{r_3}$.Using the formula for the exradii,$\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,and $\frac{1}{r_3} = \frac{s-c}{\Delta}$,where $s$ is the semi-perimeter and $\Delta$ is the area of the triangle:$\frac{s-a}{\Delta} > \frac{s-b}{\Delta} > \frac{s-c}{\Delta}$.Multiplying by $\Delta$ (since $\Delta > 0$):$s - a > s - b > s - c$.Subtracting $s$ from all parts:$-a > -b > -c$.Multiplying by $-1$ reverses the inequality signs:$a < b < c$.

In a $\Delta ABC$,if ${r_1} < {r_2} < {r_3}$,then:

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