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 Fig. 10.14). 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


2S = AB + BC + CA


= x + 8 + 14 + 6 + x


= 28 + 2x


S =






Area of ΔOBC =


Area of ΔOCA =


Area of ΔOAB =


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






= 42x+3x2 = 196+x2+28x


= 2x2+14x-196 = 0


= x2 +7x-98 = 0


= x2 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 = - 14 and 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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