In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 6 cm and 9 cm respectively. If the area of △ABC = 54 cm² then find the lengths of sides of AB and AC.

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Solution

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
Area △ABC = Area △BOC + Area △AOB + Area △AOC
$\Rightarrow 54 = \frac{1}{2} \times B C \times O D + \frac{1}{2} \times A B \times O E + \frac{1}{2} \times A C \times O F$ [Radius $⊥$ Tangent]
$\Rightarrow 108 = 15 \times 3 + (6 + x) \times 3 + (9 + x) \times 3$
$\Rightarrow 36 = 15 + 6 + x + 9 + x$
$\Rightarrow 36 = 30 + 2 x$
$\Rightarrow 2 x = 6$
$\Rightarrow x = 3$ cm
$∴ A B = 6 + 3 = 9$ cm and $A C = 9 + 3 = 12$ cm

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