A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC.

#### Solution

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be *x.*

In ΔABC,

CF = CD = 6cm (Tangents on the circle from point C)

BE = BD = 8cm (Tangents on the circle from point B)

AE = AF = *x* (Tangents on the circle from point A)

AB = AE + EB = *x* + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 +* x*

2*s = *AB + BC + CA

= *x* + 8 + 14 + 6 +* x*

*=* 28 + 2*x*

*s* = 14 + *x*

`Area of triangleABC = sqrt(s(s-1)(s-b)(s-c))`

`=sqrt({14+x}{(14+x)-14}{(14+x)-(6+x)}{(14+x)-(8+x)})`

`=sqrt((14+x)(x)(8)(6))`

`=4sqrt(3(14x+x^2))`

`Area of triangleOBC = 1/2xxODxxBC = 1/2xx4xx14=28`

`Area of triangleOCA = 1/2xxOFxxAC = 1/2xx4xx(6+x)=12+2x`

`Area of triangleOAB = 1/2xxOExxAB = 1/2xx4xx(8+x)=16+2x`

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

`4sqrt(3(14x+x^2)) = 28+12+2x+16+2x`

`4sqrt(3(14x+x^2)) = 56+4x`

`=>sqrt(3(14x+x^2)) = 14+x`

`=>3(14x+x^2) = (14+x)^2`

`=>42x+3x^2=196+x^2+28x`

`=>2x^2+14x-196 = 0`

`=>x^2+7x-98 = 0`

`=>x^2+14x-7x-98=0`

`=>x(x+14)-7(x+14) =0`

=>(x+14)(x-7) =0

Either *x+*14 = 0 or *x* − 7 =0

Therefore*, x = −*14and 7

However, *x *= −14 is not possible as the length of the sides will be negative.

Therefore, *x* = 7

Hence, AB = *x* + 8 = 7 + 8 = 15 cm

CA = 6 +* x *= 6 + 7 = 13 cm