Advanced geometry problem

Since we know that orange shape is making one fifth of the area of a cricle, we can calculate its radius: $\begin{array}{l} {S_{EOP}} = \frac{1}{5} \times {S_C}\\ {S_C} = 5 \times 5.655 = 28.274{\rm{ }}c{m^2}\\ {S_C} = \pi \times r_1^2 \Rightarrow {r_1} = \sqrt {\frac{{{S_C}}}{\pi }} = \sqrt {\frac{{28.274}}{\pi }} \buildrel\textstyle.\over= 3{\rm{ }}c{m^2} \end{array}$ We can notice that: ${r_1} = b$ ${c_a} = {r_1} - \left| {PR} \right| = 3 - 1.4 = 1.6{\rm{ }}cm$ By using Pythagorean theorem, we can calculate a height from the point C : ${v_c} = \sqrt {{b^2} - {c_a}^2} = \sqrt {{3^2} - {{1.6}^2}} = 2.538{\rm{ }}cm$ By using Euclid's formula, we can calculate 2nd part of the side c : $v_c^2 = {c_a} \times {c_b} \Rightarrow {c_b} = \frac{{v_c^2}}{{{c_a}}} = \frac{{{{2.538}^2}}}{{1.6}} = 4.026{\rm{ }}cm$ Via trigonometric formula for sin , we can calculate angle beta : $\sin \left( \beta \right) = \frac{b}{c} \Rightarrow \beta = \arcsin \left( {\frac{b}{c}} \right) = \arcsin \left( {\frac{3}{{4.026 + 1.6}}} \right) = 32.2$ We can tell, that: $\beta = \delta$ Where delta is XBY angle and also for the radius of a 2nd circle: ${r_2} = {c_b}$ Our last step involves a formula for a circle segment: $S = \frac{1}{2} \times r_2^2\left( {\frac{{\pi \times \delta }}{{180}} - \sin \left( \delta \right)} \right) = \frac{1}{2} \times {4.026^2}\left( {\frac{{\pi \times 32.2}}{{180}} - \sin \left( {32.2} \right)} \right) = 0.236{\rm{ }}c{m^2}$