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In the figure, a $ \triangle A B C $ is drawn to circumscribe a circle of radius $ 4 \mathrm{~cm} $ such that the segments $ B D $ and $ D C $ are of lengths $ 8 \mathrm{~cm} $ and $ 6 \mathrm{~cm} $ respectively. Find the lengths of sides $ A B $ and $ A C $, when area of $ \triangle A B C $ is $ 84 \mathrm{~cm}^{2} $.
"
Given:
In the figure, a \( \triangle A B C \) is drawn to circumscribe a circle of radius \( 4 \mathrm{~cm} \) such that the segments \( B D \) and \( D C \) are of lengths \( 8 \mathrm{~cm} \) and \( 6 \mathrm{~cm} \) respectively.
Area of \( \triangle A B C \) is \( 84 \mathrm{~cm}^{2} \).
To do: We have to find the lengths of sides \( A B \) and \( A C \).
Solution:
In the given figure,
A circle is inscribed in $\triangle ABC$ touching it at $D, E$ and $F$ respectively.
Radius of the circle $r = 4\ cm$
$OD\ \perp\ BC$
$OD = 4\ cm, BD = 8\ cm, DC = 6\ cm$
Join $OE$ and $OF$.
Area of $\Delta \mathrm{ABC}=84 \mathrm{~cm}^{2}$
$=\frac{1}{2}$ (Perimeter of $\Delta \mathrm{ABC}) \times r$
$\Rightarrow \frac{1}{2} r \times$ (Perimeter) $=84$
Therefore,
Perimeter $=\frac{84 \times 2}{r}$
$=\frac{84 \times 2}{4}$
$=42 \mathrm{~cm}$
$\mathrm{BD}=\mathrm{BF}=8 \mathrm{~cm}$ (Tangents to the cirlce are equal)
Similarly,
$\mathrm{CD}=\mathrm{CE}=6 \mathrm{~cm}$
$A E=A F$
$B D+C E+A F=\frac{1}{2}$ (Perimeter)
$=\frac{1}{2} \times 42$
$=21 \mathrm{~cm}$
$\Rightarrow 8+6+A F=21$
$\Rightarrow A F=21-8-6=7$
$A E=A F=7 \mathrm{~cm}$
$A B=A F+F B$
$=7+8$
$=15 \mathrm{~cm}$
$A C=A E+E C$
$=7+6$
$=13 \mathrm{~cm}$
The lengths of sides \( A B \) and \( A C \) are $15\ cm$ and $13\ cm$ respectively.