282. Given a circle of radius $R$, $O$-its center, $AB$-its diameter, point $M$-on the radius $OA$, such that $\frac{|AM|}{|MO|}=k$. Through $M$ a random chord $CD$ is drawn. What is the maximum value of the area of the quadrilateral $ACBD$?
282. Since $\frac{S_{A C D}}{S_{O C D}}=\frac{|A M|}{|M O|}=k, \frac{S_{B C D}}{S_{O C D}}=\frac{|B M|}{|O M|}=k$. 2, then $S_{A C B D}=S_{A C D}+S_{B C D}=2(k+1) S_{O C D}$. Therefore, the area of $A C B D$ will be the largest when the area of triangle $C O D$ is the largest. But triangle $C O D$ is isosceles, with the lateral side equal to $R$; hence, its area is maximized when the sine of the angle at the vertex $O$ reaches its maximum. Let this angle be $\varphi$. Clearly, $\varphi_{0} \leqslant \varphi \leqslant 90^{\circ}$, then the value of $\varphi_{0}$.

Answer: if $k \leqslant \sqrt{2}-1$, then $S_{\max }=(k+1) R^{2}$;

$$
\text { if } k>\sqrt{2}-1, \text { then } S_{\max }=\frac{2 \sqrt{k(k+2)}}{k+1} R^{2}
$$