If $d_{1}, d_{2}, d_{3}$ are the diameters of the three escribed circles of a triangle, then $d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1}$ is equal to $\frac{a △^{2}}{r^{2}}$ , then $a$ is:

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Solution

From above,
$d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1} = 4 (r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1})$
$= [\frac{△}{s - a} . \frac{△}{s - b} + \frac{△}{s - b} \frac{△}{s - c} + \frac{△}{s - c} \frac{△}{s - a}]$
$= \frac{4 △^{2} [s - c + s - a + s - b]}{(s - a) (s - b) (s - c)}$
$= \frac{4 △^{2} (3 s - 2 s)}{(s - a) (s - b) (s - c)}$
Multiply Numerator and denominator by $s$ we get
$= \frac{4 △^{2} s^{2}}{s (s - a) (s - b) (s - c)}$
$= \frac{4 △^{2} s^{2}}{△^{2}}$
$= 4 s^{2}$
$= {(2 s)}^{2}$
$= {(a + b + c)}^{2}$
Also, $s = \frac{△}{r}$
$\Rightarrow 4 s^{2} = \frac{4 △^{2}}{r^{2}}$
$\frac{4 △^{2}}{r^{2}}$ = $\frac{a △^{2}}{r^{2}}$
$a = 4$

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If d1,d2,d3 are the diameters of the three escribed circles of a triangle, then d1d2+d2d3+d3d1 is equal to a△2r2, then a is:

If $d_{1}, d_{2}, d_{3}$ are the diameters of the three escribed circles of a triangle, then $d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1}$ is equal to $\frac{a △^{2}}{r^{2}}$ , then $a$ is: