The radius of the circle is $R$. Find the chord drawn from the end of this diameter through the midpoint of the perpendicular radius.
Let $AB$ be the diameter, $OK$ be the perpendicular radius, and $AC$ be a chord passing through the midpoint $M$ of the radius $OK$. Then $AM^2 = R^2 + \left(\frac{R^2}{2}\right)^2 = \frac{5}{4} R^2$.

Since $\angle ACB = 90^\circ$, triangles $AOM$ and $ACB$ are similar. Therefore, $AC : AB = AO : AM$, from which $AC = \frac{4R\sqrt{5}}{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-15.jpg?height=471&width=486&top_left_y=479&top_left_x=795)

Answer

$\frac{4R\sqrt{5}}{5}$.