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.

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,

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 ΔOBC =

Area of ΔOCA =

Area of ΔOAB =

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

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

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