B2. The chord is 3 cm away from the center of a circle with a diameter of $6 \sqrt{2} \mathrm{~cm}$. Calculate the length of the chord, which is the intersection of the circle and the line. Draw a sketch.

(6 points)
B2. Let $S$ be the center of the circle, and $A$ and $B$ the endpoints of the chord $t$. The segments $S A$ and $S B$ are radii of the circle. Therefore, triangle $A B S$ is isosceles. The height from $S$ to the base of triangle $A B S$ bisects the chord $t$ and is 3 cm long (the distance from $S$ to the line). Using the Pythagorean theorem $\left(\frac{t}{2}\right)^{2}=r^{2}-v^{2}$ or $\left(\frac{t}{2}\right)^{2}=(3 \sqrt{2})^{2}-3^{2}$. We simplify the equation to $\left(\frac{t}{2}\right)^{2}=9$. The solution to the equation is $t=6$.

Drawing an appropriate sketch 1 point

Recording or considering the radius of the circle $r=3 \sqrt{2} \mathrm{~cm}$ 1 point

Recognizing or using that the height $v$ bisects the chord $t$ 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-13.jpg?height=71&width=1687&top_left_y=2626&top_left_x=230)

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-13.jpg?height=77&width=1685&top_left_y=2686&top_left_x=228)

![](https://cdn.mathpix.com/cropped/2024_06_07_d6b556bd14f8001ca6fag-14.jpg?height=474&width=617&top_left_y=337&top_left_x=228)