Q. 124.2( 605 Votes )

A triangle ABC is

Answer :

To Find: AB and AC

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 ∆ABC,

From the theorem which states that the lengths of two tangents drawn from an external point to a circle are equal.
So, 

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

By heron's formula,

where, a, b and c are sides of triangle and s is semi perimeter.

2S = AB + BC + CA

     = x + 8 + 14 + 6 + x

     = 28 + 2x


S



=√(14 + x)(x)(8)(6)



Also, area of triangle =   
And from the given figure it is clear that
Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

Area of ΔOBC =

Area of ΔOCA =

Area of ΔOAB =

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

⇒ 

⇒ 

⇒ 


⇒ 3x = 14 + x

⇒ 2x = 14

⇒ x = 7


Therefore, x = 7

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

CA = 6 + x = 6 + 7 = 13 cm
Hence, AB = 15 cm and AC = 13 cm

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