Three circle touch each other externally. A triangle is formed when the centres of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm

$r_{2} = 5.5 c m$
$r_{3} = 3.5 c m$
$r_{1} = 3.5 c m$

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$r_{2} = 5.9 c m$
$r_{3} = 3.5 c m$
$r_{1} = 2.5 c m$

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$r_{2} = 5.9 c m$
$r_{3} = 1.5 c m$
$r_{1} = 2.5 c m$

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$r_{2} = 5.9 c m$
$r_{3} = 9.5 c m$
$r_{1} = 2.5 c m$

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Solution

The correct option is A
$r_{2} = 5.5 c m$
$r_{3} = 3.5 c m$
$r_{1} = 3.5 c m$

Three circle touches each other externally
$Δ$ ABC is formed by joining the centres, A, B and C of the circles.
AB = 6 cm, AC = 8 cm and BC = 9 cm

Let radii of the circles having centres A, B and C
be $r_{1}, r_{2}, r_{3}$ respectively
$r_{1} + r_{3} = 8, r_{3} + r_{2} = 9, r_{1} + r_{2} = 6$
Adding $r_{1} + r_{3} + r_{3} + r_{2} + r_{1} + r_{2} = 6 + 8 + 9 c m$ $2 (r_{2} + r_{2} + r_{3}) = 23$
$r_{1} + r_{2} + r_{3} = \frac{23}{2} = 11.5 c m$
$r_{1} + 9 = 11.5 c m$
$r_{1} = 11.5 - 9 = 2.5 c m$
$r_{2} + 6 = 11.5 \Rightarrow r_{2} = 11.5 - 6 = 5.5 c m$
$r_{3} = 11.5 \Rightarrow r_{3} = 11.5 - 8 = 3.5 c m$
$r_{1} = 2.5 c m r_{2} = 2.5 a n d r_{3} = 3.5 c m$

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