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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 figure). Find the sides $AB$ and $AC$.
"
Given:
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
To do:
We have to find the lengths of sides $AB$ and $AC$.
Solution:
Let the given circle touch the sides AB and AC of the triangle at points $E$ and $F$ respectively and let the
length of line segment $AF$ be $x$.
$AE = AF = x\ cm$
In $∆ABC$,
$a = 6 + 8$
$= 14\ cm$
$b = (x + 6)\ cm$
$c = (x + 8)\ cm$
Therefore,
$s=\frac{a+b+c}{2}$
$=\frac{14+x+6+x+8}{2}$
$=\frac{2 x+28}{2}$
$=(x+14) \mathrm{cm}$
$\operatorname{ar}(\Delta \mathrm{ABC})=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{(x+14) \times x \times 8 \times 6}$
$=\sqrt{48 x \times(x+14)} \mathrm{cm}^{2}$.........(i)
$\operatorname{ar}(\Delta \mathrm{ABC})=\operatorname{ar}(\Delta \mathrm{OBC})+\operatorname{ar}(\Delta \mathrm{OCA})+\operatorname{ar}(\Delta \mathrm{OAB})$
$=\frac{1}{2} \times 4 \times a+\frac{1}{2} \times 4 \times b+\frac{1}{2} \times 4 \times c$
$=2 a+2 b+2 c$
$=2(a+b+c)$
$=2 \times 2(x+14)$............(ii)
From (i) and (ii), we get,
$\sqrt{48 x(x+14)}=4(x+14)$
$48 x(x+14)=4^{2}(x+14)^{2}$
$48 x(x+14) =16(x+14)^{2}$
$3 x(x+14)=(x+14)^{2}$
$3 x =x+14$
$2 x=14$
$x=7$
$\mathrm{AB}=x+8$
$=7+8$
$=15 \mathrm{~cm}$
$\mathrm{AC} =x+6$
$=7+6$
$=13 \mathrm{~cm}$