If the inradius is $2$ and the segments determined by the circumference inscribed on one side measure $3$ and $5$, find the area of triangle

For reference: Calculate the area of ​​a triangle if the inradius is $2$ and the segments determined by the circumference inscribed on one side measure $3$ and $5$.(Answer:$\frac{240}{11}$)

My progress

$r = 2\\\triangle CIF: CI^2 = 5^2+4^2 \implies CI = \sqrt29\\ \triangle FIB: BI^2=3^2+2^2 \implies BI = \sqrt13\\ S_{ABC} = p.r\\ p = \frac{10+6+2AD}{2}=8+AD\\ \therefore S_{ABC} = 16+2AD\\ S_{CIB}=\frac{8.2}{2}=8\\ S_{CDI} = \frac{5.2}{2} = 5$

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Solution 1:

Notice that if $AD = x$ then $AB = x+3$ and $AC = x+5$, so the half of perimeter is $s= x+8$ and then $S= s\cdot r = 2x+16$.

Use Heron's formula $$ (2x+16)^2 =S^2 = (x+8)\cdot 3\cdot 5\cdot x$$ So $4x+32 = 15x\implies x= {32\over 11}$ and thus $S= 240/11$.

Solution 2:

If $AD = x, AI = \sqrt{x^2 + 4}$. If the angle bisector of $\angle A$ meets $BC$ at $E$ then,

$ \displaystyle CE = \frac{8 (x+5)}{2x+8}$ and $ \displaystyle EF = 5 - \frac{4 (x+5)}{x+4} = \frac{x}{x+4}$

$ \displaystyle IE = \frac{8}{2x+8} \cdot AI = \frac{4}{x+4} \cdot \sqrt{x^2+4}$

Applying Pythagoras in $\triangle IEF$,

$ \displaystyle 2^2 + \left(\frac{x}{x+4}\right)^2 = \frac{16 (x^2 + 4)}{(x+4)^2} $

$ \displaystyle \implies 4 (x+4)^2 + x^2 = 16 x^2 + 64 \implies x = \frac{32}{11}$

So area of the the triangle is $ \displaystyle \left(5 + 3 + \frac{32}{11}\right) \cdot 2 = \frac{240}{11}$