The radius of the incircle of …

Ans. Let a circle with centre O be inscribed in triangle ABC.

OD=OE=OF=6cm   [radii of circle]

Let BD = 9 cm and CD = 12 cm

We know that, length of two tangents drawn from an external porint to a circle are equal.

BF = BD = 9 cm

CE = CD = 12 cm

Let AE =AF = x cm

CA = AE + CE  =  (x+12)  cm

AB = AF + FB = x + 9  cm

BC = BD + CD =  9+ 12 = 21 cm

Semi-perimeter of triangle ABC

$s = \frac{\left(x+21\right)+\left(x+9\right) +21}{2} = \frac{2x+42}{2} = x+21$

Area of triangle ABC = $\sqrt{\left(s\right)\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

=> $\sqrt{\left(x+21\right)\left(x+21-x-12\right)\left(x+21-x-9\right)\left(x+21-21\right)}$

=> $\sqrt{\left(x+21\right)\left(9\right)\left(12\right)\left(x\right)}$

=> $\sqrt{108x\left(x+21\right)} c{m}^2 .......\left(1\right)$

$Also, Area of ∆ABC = Area of (∆OBC) + Area of (∆OCA) + Area of (∆OAB)$

=> $\frac{1}{2}BC\times OD +\hspace{0.17em}\frac{1}{2}CA\times OE + \frac{1}{2}AB\times OF$

=> $\frac{1 }{2}\times 21\times 6 + \frac{1}{2}(x+12)\times 6 + \frac{1}{2}\left(x+9\right)\times 6$

=> $\frac{1}{2}\times 6 \left(21+ x+12 +x+9\right)$

=> $3 \times \left(2x +42\right)$

=> $6\left(x+21\right) c{m}^2 ...... \left(2\right)$

From (1) and (2)

$\sqrt{108 x(x+21)} = 6(x+21)$

$Squaring both sides, we get$

=> $108 x (x+21)=36(x+21{)}^2$

=> $\frac{108x}{36}= \frac{{\left(x+21\right)}^2}{\left(x+21\right)}$

=> $3x = x+21$

=> 2x = 21

=> $x = \frac{21}{2}$

=> $Side AC = 12\hspace{0.17em}+ \frac{21}{2} = \frac{45}{2}$cm

$Side AB = 9 + \frac{21}{2} = \frac{39}{2}cm$

Let a circle with centre O be inscribed in triangle ABC.

OD=OE=OF= 6cm

As tangents drawn from an external point to a circle are equal in length, so

BD=BE=9cm

CF=CE=12cm

Let AD=AF= x cm

AB=AD+DB= (x+9)cm

BC=BE+EC= 9+12=21cm

CA=CF+AF= (x+12)cm

Semi perimeter, s=

 $\frac{AB+BC+CA}{2}=\frac{2x+42}{2}= x+21Area of ABC= \sqrt{s(s-AB)(s-BC)(s-CA)}=\sqrt{(x+21)(x+21-x-9)(x+21-21)(x+21-x-12)}=\sqrt{(x+21)\left(9\right)\left(12\right)x}=\sqrt{108x(x+21)} c{m}^2 -\left(1\right)$

Also, area of triangle ABC= area (OBC)+area(OCA)+area(OAB)

= $\frac{1}{2}xBCxOE+\frac{1}{2}xACxOF+\frac{1}{2}xABxOD\frac{1}{2}x21x6+\frac{1}{2}x(x+12)x6+\frac{1}{2}x6x(x+9)= 6x+126= 6(x+21) -\left(2\right)$

Equating (1) and (2)

6(x+21)=$\sqrt{108x(x+21)}$

Squaring both sides,

36(x+21)²=108x(x+21)

x+21=3x

x=21/2

AB= 21/2+9 = 39/2 cm

AC= 21/2+12= 45/2 cm