\left.\begin{array}{l}\text { The Law of Sines } \\ \text { [Midline of a Triangle] }\end{array}\right]

Two chords, equal to 10 and 12, are drawn from one point on a circle. Find the radius of the circle if the distance from the midpoint of the smaller chord to the larger one is 4.
Connect the midpoints of the given chords.

## Solution

Let $M$ be the midpoint of the chord $A B=10$, and $M P$ be the perpendicular dropped from point $M$ to the larger chord $A C=12$. Let $K$ be the midpoint of the larger chord. Then,

$$
A P=\sqrt{A M^{2}-M P^{2}}=\sqrt{5^{2}-4^{2}}=3, \quad P K=A K-A P=3
$$

Therefore, triangle $A M K$ is isosceles.

Since $M K$ is the midline of triangle $A B C$,

$$
B C=2 M K=10, \quad \sin \angle B A C=\frac{M P}{A M}=\frac{4}{5}
$$

If $R$ is the desired radius, then

$$
R=\frac{B C}{2 \sin \angle B A C}=\frac{10}{2 \cdot \frac{4}{5}}=\frac{25}{4}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-11.jpg?height=463&width=458&top_left_y=1524&top_left_x=800)

## Answer

$\frac{25}{4}$.