10.306. A line intersects a circle of radius $R$ at points $A$ and $B$ such that $\cup A B=45^{\circ}$, and the line perpendicular to the diameter $A M$ of the circle and passing through its center intersects it at point $K$. The line passing through point $B$ and perpendicular to $A M$ intersects it at point $C$. Find the area of trapezoid $O C B K$.
## Solution.

Given that $\angle B O C=45^{\circ}$, therefore $B C=O C=\frac{R}{\sqrt{2}}, A C=R-\frac{R}{\sqrt{2}}=$ $=\frac{R(\sqrt{2}-1)}{\sqrt{2}}$ (Fig. 10.95). Since $\triangle A B C \sim \triangle A K O$, then $\frac{K O}{B C}=\frac{A O}{A C} \Rightarrow$ $\Rightarrow K O=\frac{A O \cdot B C}{A C}=\frac{R}{\sqrt{2}-1}=R(\sqrt{2}+1)$. Then the area of the trapezoid will be: $S=\frac{1}{2}(K O+B C) O C=\frac{1}{2}\left(R(\sqrt{2}+1)+\frac{R}{\sqrt{2}}\right) \frac{R}{\sqrt{2}}=\frac{R^{2}(3+\sqrt{2})}{4}$. Answer: $\frac{R^{2}(3+\sqrt{2})}{4}$.