10.204. From a point on the circumference, two chords of lengths 10 and 12 cm are drawn. Find the radius of the circle if the distance from the midpoint of the shorter chord to the longer chord is 4 cm.
Solution.

From point $B$ of the circle (Fig. 10.14), chords $AB=10$ cm and $BC=12$ cm are drawn, $M$ is the midpoint of $AB$, $MN \perp BC$, $MN=4$ cm. Drop a perpendicular $AK$ from point $A$ to segment $BC$. In $\triangle ABK - MN$ is the midline. Then $AK=2MN=8$ cm. From $\triangle MNB \quad (\angle MNB=90^{\circ})$: $BN=\sqrt{BM^2-MN^2}=3$ cm. Then $BK=2BN=6$ cm. Since $BC=12$ cm,

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0719.jpg?height=468&width=678&top_left_y=111&top_left_x=97)

Fig. 10.15

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0719.jpg?height=500&width=444&top_left_y=79&top_left_x=822)

Fig. 10.16

then $K$ is the midpoint of $BC$, and therefore, $AC=AB=10$ cm, segment $AK$ lies on the diameter $AF$ of the given circle. Let $AF=2R$, where $R$ is the radius of the given circle. Then $KF=AF-AK=2R-8$. $K$ is the point of intersection of the diameter $AF$ and the chord $BC$. By the property of the chord $AK \cdot KF=CK \cdot KB; 8(2R-8)=6 \cdot 6; R=6.25$.

Answer: 6.25 cm.