A right - angled triangle ABC is right - angled at B. The point o contact of the incircle with the hypotenuse, divides it, in the ratio 3:2. If the perimeter of the triangle ABC is 36 cm, find its inradis.

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Solution

Given: AB + BC + CA = 36
Also, AD = AE (Tangent from same point are equal)
DC = CF
Let, BF = BE = y
$3 x + 3 x + 2 x + 2 x + y + y = 36$
$\Rightarrow 10 x + 2 y = 36 \Rightarrow 5 x + y = 18 \dots \dots (1)$
Also, $(3 x + y)^{2} + (2 x + y)^{2} = (5 x)^{2}$
$\Rightarrow 13 x^{2} + 2 y^{2} + 10 x y = 25 x^{2}$
$\Rightarrow 12 x^{2} = 2 y^{2} + 10 x y$
$\Rightarrow 12 x^{2} = 2 (18 - 5 x)^{2} + 10 x (18 - 5 x)$
$\Rightarrow 12 x^{2} = 2 (18 + 52) [18 - 5 x + 5 x]$
$\Rightarrow 12 x^{2} = 36 (18 - 5 x)$
$\Rightarrow x^{2} = 3 (18 - 52)$
$\Rightarrow x^{2} = 54 - 15 x$
$\Rightarrow x^{2} + 15 x - 54 = 0$
$\Rightarrow x^{2} + 18 x - 3 x - 54 = 0$
$\Rightarrow x (x + 18) - 3 (x + 18) = 0$
$\Rightarrow (x - 3) (x + 18) = 0$
$\Rightarrow x = 3$
$∴ X = 3$
$Y = 18 - 5 x = 3$
$∴$ In radius = 3 cm

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