Horses grazing in a circle. Question:
Diagram:

Note that

*

*The circle with center $C$ is touching the arc of semi-circle $AB$ also; I couldn't draw it.

*The figure wasn't drawn on cartesian planes; so, though it may seem $AB=10$, and same for others, it is not so. No measurements are given in org. diagram.

Given: $O$ is the center of $AB$; $P$ and $R$ are the center of $AO$ and $BO$.
The horses are tied to $P$, $R$ and $C$. It is identifiable which horse grazes till what extent.

Find the percentage of area of the semi-circle with diameter $AB$ not grazed by any horses to the nearest integer.

My attempt:
Let $AB=100cm$, then $AO=BO=50cm$, and $AP=PO=OR=BR=25cm$.
$\text{Area of circle } AO = \text{Area of circle } BO = \pi(25\text{cm})^2$.
But I seem to have zero idea on finding the area grazed by the horse tied to point $C$. Any hints/guide is appreciated.
 A: 
Draw the tangent (passing though $O$) to the 2 equal circles with centres in $P,R$: this line'll pass through the centre of the smallest circle in $C$ because of symmetrical reasons...then it's easy to notice that $COR$ is a rectangular triangle, and thus $CR\sin \angle ORC = CO,CR\cos \angle ORC = OR$. Now the reasoning should be easier to be understood...
By letting $AP = PO = OR = RB = \frac{AB}{4} = b$,$CD = a$ and $x=\angle CRO$, you can write the system below and find $a$ in terms of $b$:
\begin{cases}
a+(a+b)\sin x  =2b\\
(a+b)\cos x = b
\end{cases}
$$(a+b)^2\sin^2 x + (a+b)^2\cos^2 x=(2b-a)^2+b^2\\ (a+b)^2=4b^2-4ab+a^2+b^2\\a^2+b^2+2ab=4b^2-4ab+a^2+b^2 \\ 6ab = 4b^2 \Rightarrow b=0 \lor a=b\frac23$$
Finally the requested percentage is easy to be found and equals $$100-100\frac{\frac{b^2\pi}{2}+\frac{b^2\pi}{2}+\left(b\frac23\right)^2\pi}{\frac{(2b)^2\pi}{2}}\text{%}=\\100-100\frac{\frac12+\frac12+\frac49}{2}\text{%}=\\100-100\frac{13}{18}\text{%}=100-\frac{50\cdot 13}{9}\text{%}=100-72,\overline{2}\text{%}=27,\overline{7}\text{%}$$
