The inradius of Delta ABC is 3 units and the | vedclass

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(A) Let $r$ be the inradius and $r_a$ be the exradius opposite to vertex $A$. We are given $r = 3$ and $r_a = 4$.Using the formulas $r = \frac{\Delta}

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) Let $r$ be the inradius and $r_a$ be the exradius opposite to vertex $A$. We are given $r = 3$ and $r_a = 4$.Using the formulas $r = \frac{\Delta}{s}$ and $r_a = \frac{\Delta}{s-a}$,where $\Delta$ is the area of the triangle and $s$ is the semi-perimeter,we have:$\frac{\Delta}{s} = 3$ and $\frac{\Delta}{s-a} = 4$.Dividing the two equations,we get $\frac{s-a}{s} = \frac{3}{4}$,which implies $1 - \frac{a}{s} = \frac{3}{4}$,so $\frac{a}{s} = \frac{1}{4}$.Since $s = \frac{a+b+c}{2}$,we have $\frac{a}{s} = \frac{2a}{a+b+c} = \frac{1}{4}$,so $a+b+c = 8a$,or $b+c = 7a$.Using the area formula $\Delta = \frac{1}{2} a h_a$,where $h_a$ is the altitude from $A$,we have $r = \frac{\Delta}{s} = \frac{a h_a}{2s} = 3$,so $h_a = \frac{6s}{a}$.Since $\frac{s}{a} = 4$,we get $h_a = 6 	imes 4 = 24$ units.

The inradius of $\Delta ABC$ is $3$ units and the exradius opposite to vertex $A$ is $4$ units. Then,the length of the altitude from vertex $A$ is ............. $units$.

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