1.57. The vertices of a rectangle inscribed in a circle divide it into four arcs. Find the distance from the midpoint of one of the larger arcs to the vertices of the rectangle, if its sides are 24 and 7 cm.
1.57. Since $A C$ is the diameter of the circle (Fig. 1.44), then $R=0.5 \sqrt{24^{2}+7^{2}}=12.5 \quad$ (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: $\quad M B=\sqrt{M F^{2}+B F^{2}}=\sqrt{9^{2}+12^{2}}=15(\mathrm{~cm})$, $M A=\sqrt{M K^{2}+K A^{2}}=\sqrt{16^{2}+12^{2}}=20(\mathrm{~cm})$.

Answer: 15 and 20 cm.