Question

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.

Fig. 10.14

Answer 1

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 = 6 cm (Tangents on the circle from point C)
BE = BD = 8 cm (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
2s = x + 8 + 14 + 6 + x
2s = 28 + 2x
s = 14 + x
Area of ΔABC = \(\sqrt {s(s-a)(s-b)(s-c)}\)
= \(\sqrt {(14+x)(((14+x)-14){(14+x)-(6+x)}{(14+x)-(8+x)}}\)
= \(\sqrt {(14+x)(x)(8)(6)}\)
= \(4\sqrt {3(14x+x^2)}\)
Area of ∆OBC = \(\frac 12\) x OD x BC = \(\frac 12\) x 4 x 14 = 28

Area of ΔOCA = \(\frac 12\) x OF x AC = \(\frac 12\) x 4 x (6+x) = 12+2x

Area of ΔOAB = \(\frac 12\) x OE x AB = \(\frac 12\) x 4 x (8+x) = 16+2x
Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB
\(4\sqrt {3(14x+x^2)}\) = \(28+12+2x+16+2x\)
⇒ \(4\sqrt {3(14x+x^2)}\)= \(56+4x\)
⇒ \(\sqrt {3(14x+x^2)}\) = \(14+4x\)
⇒ \(3(14x+x^2)\) = \((14+4x)^2\)
⇒ \(42x+3x^2\) = \(196 +x^2+28x\)
⇒ \(2x^2+14x-196\) = \(0\)
⇒ \(x^2+7x-98\) = \(0\)
⇒ \(x^2+14x-7x-98\) = \(0\)
⇒ \(x(x+14) -7(X+14)\) =\(0\)
⇒ \((x+14)(x-7)\) = \(0\)
Either \(x+14\) =\(0\) or \(x - 7\)= \(0\)
Therefore, \(x\) = \(- 14\) 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