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