10.212. The vertices of a rectangle inscribed in a circle divide it into four arcs. Find the distances from the midpoint of one of the larger arcs to the vertices of the rectangle, if its sides are 24 and 7 cm.
## Solution.

Since $AC$ is the diameter of the circle (Fig. 10.22), then $R=0.5 \sqrt{24^{2}+7^{2}}=12.5$ cm. In $\triangle B O F$, we have $O F=\sqrt{O B^{2}-B F^{2}}=\sqrt{12.5^{2}-12^{2}}=3.5$ cm; hence, $M F=12.5-3.5=9$ cm, $M K=12.5+3.5=16$ cm. From $\triangle M B F$ and $\triangle M A K$, we find the required distances: $M B=\sqrt{M F^{2}+B F^{2}}=\sqrt{9^{2}+12^{2}}=15 \text{ cm}$, $M A=\sqrt{M K^{2}+K A^{2}}=\sqrt{16^{2}+12^{2}}=20 \text{ cm}$.

Answer: 15 and $20 \text{ cm}$.