Need help w/ Finding X with semicircle inscribed in a right triangle

[Ottomatical]

I have no idea how to actually get the answer. I understand that the radius is used to create another equal but smaller triangle.

I have gotten to the stage where
c^2=a^2+b^2
(r+1,15)^2=r^2+(4-r)^2


But i have no clue if i am on the right track or not.
The final answer is 0.4

 

[MarkFL]

I would use similarity of the triangles formed by drawing radii from the center of the semi-circle to the tangent points to the large triangle to state (where $r$ is the radius of the semi-circle):

[MATH]\frac{r}{r+1.15}=\frac{3}{x+2r+1.15}[/MATH]
[MATH]\frac{r}{r+x}=\frac{4}{x+2r+1.15}[/MATH]
Can you proceed?

 

[MarkFL]

To follow up...

We know:

[MATH]x+2r+1.15=5[/MATH]
And so the first equation becomes:

[MATH]\frac{r}{r+1.15}=\frac{3}{5}[/MATH]
[MATH]5r=3r+3.45[/MATH]
[MATH]r=1.725[/MATH]
Using this, the second equation becomes:

[MATH]\frac{1.725}{1.725+x}=\frac{4}{x+2\cdot1.725+1.15}[/MATH]
Solving this for $x$, we obtain:

[MATH]x=\frac{207}{455}[/MATH]

 

[pka]

I have no idea how to actually get the answer. I understand that the radius is used to create another equal but smaller triangle.

View attachment 11671
The final answer is 0.4

I have a different way to solve this. (sorry I cannot reproduce the diagram.)
But label the $\displaystyle \Delta ABC$ starting at the leftmost vertex counterclockwise .
The centre of the semicircle $\displaystyle \mathcal{O}$ and $\displaystyle \overline{\mathcal{O}P}$ the radius to side of length 4.
Now we have similar triangles $\displaystyle \Delta ACB\approx \Delta \mathcal{O}PB$.
From the correspondence we get $\displaystyle \overline{\mathcal{O}P}\approx\overline{AC}~\&~\overline{PB}\approx\overline{CB}$
Thus $\displaystyle \frac{r}{3}=\frac{4-r}{4}$ or $\displaystyle r=\frac{12}{7}$ from which you can solve for $\displaystyle x=5-2\left(\frac{12}{7}\right)-1.15$.