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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. Find the sides AB and AC.

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13 cm and 15 cm

15 cm and 13 cm

16 cm and 12 cm

None of these

answer is C.

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Detailed Solution

Suppose there is a triangle ABC. Here O is the circle inscribed in it.
Question Image

Now, we know that length of the two tangent drawn from the same point in a circle are equal. Thus,

\begin{matrix} CF = CD = 6 cm \\ BE = BD = 8 cm \\ AE = AF = x (say) \end{matrix}

Now, from the figure we have,

\begin{matrix} AB + AE + EB = x + 8 \\ BC = BD + DC = 8 + 6 = 14 \\ CA = CF + FA = 6 + x \end{matrix}

Semi perimeter of the triangle ABC,

s = \frac{AB + BC + CA}{2}

s = \frac{(x + 8) + 14 + (6 + x)}{2}

s = 14 + x

From heron’s formula, area of triangle ABC,

\begin{matrix} A = \sqrt{s (s - AB) (s - BC) (s - CA)} \\ A = \sqrt{(14 + x) (14 + x - 14) (14 + x - x - 6) (14 + x - x - 8)} \\ A = \sqrt{(14 + x) (x) (8) (6)} \\ A = \sqrt{(14 + x) 48 x} \end{matrix}

Now, we have,
Area of triangle ABC = 2

\times

(area of triangle AOF+ area of triangle COD+ area of triangle DOB)

\begin{matrix} A = 2 \times [(\frac{1}{2} \times OF \times AF) + (\frac{1}{2} \times CD \times OD) + (\frac{1}{2} \times DB \times OD)] \\ A = 2 \times \frac{1}{2} (4 x + 24 + 32) \end{matrix}

A = 56 + 4 x

Now equate the area as,

\sqrt{(14 + x) 48 x} = 56 + 4 x

Square both side of the equation,

48 x (14 + x) = (56 + 4 x)^{2}

\begin{matrix} \Rightarrow 48 x = \frac{[4 (14 + x)]^{2}}{14 + x} \\ \Rightarrow 48 x = 16 (14 + x) \\ \Rightarrow 48 x = 224 + 16 x \\ \Rightarrow 32 x = 224 \\ \Rightarrow x = 7 cm \end{matrix}

Therefore,

AB = x + 8

AB = 7 + 8

AB = 15 cm

Also,

CA = 6 + x

CA = 6 + 7

CA = 13 cm

So, the correct option is 3.

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