A triangle $A B C$ is drawn to circumscribe a circle of radius $4 c m$ such that the segments $B D$ and $D C$ into which $B C$ is divided by the point of contact $D$ are of lengths $8 c m$ and $6 c m$ respectively .Find the sides $A B$ and $A C$ .

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Solution

Firstly, consider that the given circle will touch the sides $A B$ and $A C$ of the triangle at point $E$ and $F$ respectively.

Let $A F = x$

In $Δ A B C$ ,

$C F = C D = 6$ (Tangents drawn from an exterior point to a circle are equal. Here,tangent is drawn from exterior point $C$ )

$B E = B D = 8$ (Tangents drawn from an exterior point to a circle are equal. Here,tangent is drawn from exterior point $B$ )

$A E = A F = x$ (Tangents drawn from an exterior point to a circle are equal. Here,tangent is drawn from exterior point $A$ )

Now, $A B = A E + E B = x + 8$

Also, $B C = B D + D C = 8 + 6 = 14 a n d C A = C F + F A = 6 + x$

Now, we get all the sides of a triangle so its area can be find out by using Heron's formula as

$2 s = A B + B C + C A$

$\Rightarrow 2 s = x + 8 + 14 + 6 + x$

$\Rightarrow 2 s = 28 + 2 x$

$\Rightarrow s = 14 + x$

Area of $Δ A B C = \sqrt{s (s - a) (s - b) (s - c)}$

$= \sqrt{(14 + x) (14 + x - 14) (14 + x - (6 - x)) (14 + x - (8 + x))}$

Area of $Δ A B C$ $= 4 \sqrt{3 (14 x + x^{2})}$

Area of traingle is also equal to $\frac{1}{2} \times b a s e \times h e i g h t$

Area of $Δ O B C = \frac{1}{2} \times O D \times B C = \frac{1}{2} \times 4 \times 14 = 28$

Area of $Δ O C A = \frac{1}{2} \times O F \times A C = \frac{1}{2} \times 4 \times (6 + x) = 2 (6 + x) = 12 + 2 x$

Area of $Δ O A B = \frac{1}{2} \times O E \times A B = \frac{1}{2} \times 4 \times (8 + x) = 2 (8 + x) = 16 + 2 x$

Area of $Δ A B C =$ Area of $O B C +$ Area of $Δ O C A +$ Area of $Δ O A B$

$4 \sqrt{3 (14 x + x^{2})} = 28 + 12 + 2 x + 16 + 2 x$

$\Rightarrow 4 \sqrt{3 (14 x + x^{2})} = 56 + 4 x$

$\Rightarrow \sqrt{3 (14 x + x^{2})} = 14 + x$

Squaring on both sides, we get

$3 (14 x + x^{2}) = (14 + x)^{2}$

$\Rightarrow 42 x + 3 x^{2} = 196 + x^{2} + 28 x$

$\Rightarrow 2 x^{2} + 14 x - 196 = 0$

$\Rightarrow x^{2} + 7 x - 98 = 0$

$\Rightarrow (x + 14) (x - 7) = 0$

$\Rightarrow (x + 14) = 0 o r (x - 7) = 0$

$\Rightarrow x = - 14 o r x = 7$

However, length of the side cannot be negative, $∴ x = 7$

Hence $A B = x + 8 = 7 + 8 = 15 c m$ and $A C = 6 + x = 6 + 7 = 13 c m$

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