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In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ∆ABC is 84 cm^{2}.

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#### Solution

Here, D, E and F are the points of contact of the circle with the sides BC, AB and AC, respectively.

OD = OE = OF = 4 cm (Radii of the circle)

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴ BD = BE = 8 cm

CD = CF = 6 cm

AE = AF = *x* cm (say)

So, BC = BD + CD = 8 cm + 6 cm = 14 cm

AB = AE + BE = *x* cm + 8 cm = (*x* + 8) cm

AC = AF + FC = *x* cm + 6 cm = (*x* + 6) cm

Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OD ⊥ BC, OE ⊥ AB and OF ⊥ AC

Now,

ar(∆OBC) + ar(∆OAB) + ar(∆OCA) = ar(∆ABC)

\[\therefore \frac{1}{2} \times BC \times OD + \frac{1}{2} \times AB \times OE + \frac{1}{2} \times AC \times OF = 84 {cm}^2 \]

\[ \Rightarrow \frac{1}{2} \times 14 \times 4 + \frac{1}{2} \times \left( x + 8 \right) \times 4 + \frac{1}{2} \times \left( x + 6 \right) \times 4 = 84\]

\[ \Rightarrow 28 + 2x + 16 + 2x + 12 = 84\]

\[ \Rightarrow 4x + 56 = 84\]

\[\Rightarrow 4x = 84 - 56 = 28\]

\[ \Rightarrow x = 7\]

∴ AB = (*x* + 8) cm = (7 + 8) cm = 15 cm

AC = (*x* + 6) cm = (7 + 6) cm = 13 cm

Hence, the lengths of sides AB and AC are 15 cm and 13 cm, respectively.

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