If $d_{1}, d_{2}, d_{3}$ are the diameters of the three ascribed circles of a triangle, then $d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1}$ is equal to

$4 s^{2}$

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$△^{2}$

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$4 △^{2}$

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$2 △^{2}$

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Solution

The correct option is A
$4 s^{2}$

Explanation for the correct option:
Step 1. Find the value of $d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1}$ :
As we know, $d_{1} = 2 r_{1}, d_{2} = 2 r_{2}, d_{3} = 2 r_{3}$
Also we have, $r_{1} = \frac{∆}{(s - a)}$
$r_{2} = \frac{∆}{(s - b)}$
$r_{3} = \frac{∆}{(s - c)}$
Step 2. Put the values of $r_{1}, r_{2}, r_{3}$ in given expression, we get
$\begin{array}{rcl} ∴ d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1} & = & 4 (r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}) \\ = & 4 ∆^{2} [\frac{1}{(s - a) (s - b)} + \frac{1}{(s - b) (s - c)} + \frac{1}{(s - c) (s - a)}] \\ = & 4 ∆^{2} [\frac{(s - c + s - a + s - b)}{(s - a) (s - b) (s - c)}] \\ = & 4 ∆^{2} [\frac{3 s - (a + b + c)}{(s - a) (s - b) (s - c)}] \\ = & 4 ∆^{2} [\frac{(3 s - 2 s)}{(s - a) (s - b) (s - c)}] \\ = & 4 ∆^{2} [\frac{s}{(s - a) (s - b) (s - c)}] \\ = & 4 ∆^{2} [\frac{s^{2}}{s (s - a) (s - b) (s - c)}] \\ = & 4 ∆^{2} (\frac{s^{2}}{∆^{2}}) \\ = & 4 s^{2} \end{array}$
Hence, Option ‘A’ is Correct.

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If d1,d2,d3 are the diameters of the three ascribed circles of a triangle, then d1d2+d2d3+d3d1 is equal to

If $d_{1}, d_{2}, d_{3}$ are the diameters of the three ascribed circles of a triangle, then $d_{1} d_{2} + d_{2} d_{3} + d_{3} d_{1}$ is equal to