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In Figure 4, a $\vartriangle ABC$ is drawn to circumscribe a circle of radius $3\ cm$, such that the segments BD and DC are respectively of lengths $6\ cm$ and $9\ cm$. If the area of $\vartriangle ABC$ is $54\ cm^{2}$, then find the lengths of sides AB and AC.
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Given:$\vartriangle ABC$ is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths $6\ cm$ and $9\ cm$. Area of $\vartriangle ABC$ is $54\ cm^{2}$

To do: 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 F and E respectively and let the 

length of line segment AF be x.

Now, it can be observed that:

$BF = BD = 6 cm \ \ \ \ \ \ ( tangents\ from\ point\ B)$

$CE = CD = 9 cm \ \ \ \ \ ( tangents\ from\ point\ C)$

$AE = AF = x \ \ \ \ \ \ \ ( tangents\ from\ point\ A)$

$AB=AF+FB=x+6$

$BC = BD + DC = 6 + 9 = 15$

$CA = CE + EA = 9 + x$

$2s = AB + BC + CA = x + 6 + 15 + 9 + x = 30 + 2x=2( 15+x)$

$s = 15 + x$

$s – a = 15 + x – 15 = x$

$s – b = 15 + x – (x + 9) = 6$

$s – c = 15 + x – (6 + x) = 9$

Area of $\vartriangle ABC$,

$\sqrt{s\left( s-a\right)\left( s-b\right)\left( s-c\right)}$

$\Rightarrow 54=\sqrt{\left( 15+x\right) .x\times 6\times 9}$

$\Rightarrow 54=3\sqrt{6\left( x^{2} +15x\right)}$

$\Rightarrow \sqrt{6\left( x^{2} +15x\right)}$=$\frac{54}{3}$=18

$\Rightarrow 6( x^{2}+15x)=324$

$\Rightarrow x^{2}+15x=54$

$\Rightarrow x^{2}$+15x-54=0$

$\Rightarrow x^{2}+18x-3x-54=0$

$\Rightarrow x(x+18)-3(x+18)=0$

$\Rightarrow ( x-3)( x+18)=0$

If $x-3=0$ 

$\Rightarrow x=3$

if $x+18=0$

$\Rightarrow x=-18$

$\because$ length can't be negative , therefore we reject the value $x=-18$.

$\therefore x=3$

$\therefore AB=x+6=3+6=9\ cm$

And $AC=x+9=3+9=12\ cm$

Therefore, The length of the sides $AB=9\ cm$ and $AC=12\ cm$.
 

Updated on: 10-Oct-2022

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