We begin by drawing in three radii:  one to \(\displaystyle \small A\), one to \(\displaystyle \small B\), and one perpendicular to \(\displaystyle \small \overline{AB}\) with endpoint \(\displaystyle \small C\) on our circle.

We must also recall that our central angle \(\displaystyle \small \angle AOB\) has a measure equal to its intercepted arc.  Therefore, \(\displaystyle \small m\angle AOB=60^\circ\).  Our perpendicular radius actually divides \(\displaystyle \small \triangle AOB\) into two congruent triangles.  Therefore, it also bisects our central angle, meaning that \(\displaystyle \small m\angle AOC=m\angle COB=30^\circ\)

Therefore, each of these triangles is a 30-60-90 triangle, meaning that each half of our chord is simply half the length of the hypotenuse (our radius which is 6).  Therefore, each half is 3, and the entire chord is 6 feet.

Draw a segment perpendicular to the chord from the center, and this line will bisect the chord. Setting up the Pythagorean Theorem with the radius as the hypotenuse and the distance as one of the legs, we solve for the other leg.

\(\displaystyle 7^2-5^2=24\)  \(\displaystyle \rightarrow \sqrt{24}=4.8989\)

Since this leg is half of the chord, the total chord length is 2 times that, or 9.798.

\(\displaystyle 2(4.8989)=9.798\)

Draw a segment perpendicular to the chord from the center, and this line will bisect the chord. Setting up the Pythagorean Theorem with the radius as the hypotenuse and the distance as one of the legs, we solve for the other leg.

\(\displaystyle 10^2-6^2=64 \rightarrow \sqrt{64}=8\)

Since this leg is half of the chord, the total chord length is 2 times that, or 16.

\(\displaystyle 2(8)=16\)

Draw a segment perpendicular to the chord from the center, and this line will bisect the chord. Setting up the Pythagorean Theorem with the radius as the hypotenuse and the distance as one of the legs, we solve for the other leg.

\(\displaystyle 6^2-4.5^2=15.75 \rightarrow \sqrt{15.75}=3.9686\)

Since this leg is half of the chord, the total chord length is 2 times that, or 7.937.

\(\displaystyle 2(3.9686)=7.937\)

Draw a segment perpendicular to the chord from the center, and this line will bisect the chord. Setting up the Pythagorean Theorem with the radius as the hypotenuse and the distance as one of the legs, we solve for the other leg.

\(\displaystyle 3.5^2-3^2=3.25 \rightarrow \sqrt{3.25}=1.8027\)

Since this leg is half of the chord, the total chord length is 2 times that, or 3.606.

\(\displaystyle 2(1.8027)=3.606\)

Draw a segment perpendicular to the chord from the center, and this line will bisect the chord. Setting up the Pythagorean Theorem with the radius as the hypotenuse and the distance as one of the legs, we solve for the other leg.

\(\displaystyle 5^2-4^2=9 \rightarrow \sqrt{9}=3\)

Since this leg is half of the chord, the total chord length is 2 times that, or 6.

\(\displaystyle 2(3)=6\)

Draw a segment perpendicular to the chord from the center, and this line will bisect the chord. Setting up the Pythagorean Theorem with the radius as the hypotenuse and the distance as one of the legs, we solve for the other leg.

\(\displaystyle 12^2-10^2=44 \rightarrow \sqrt{44}=6.6332\)

Since this leg is half of the chord, the total chord length is 2 times that, or 13.266.

\(\displaystyle 2(6.6332)=13.266\)

Draw a segment perpendicular to the chord from the center, and this line will bisect the chord. Setting up the Pythagorean Theorem with the radius as the hypotenuse and the distance as one of the legs, we solve for the other leg.

\(\displaystyle 3^2-2^2=5 \rightarrow \sqrt{5}=2.2361\)

Since this leg is half of the chord, the total chord length is 2 times that, or 4.472.

\(\displaystyle 2(2.2361)=4.472\)

Draw a segment perpendicular to the chord from the center, and this line will bisect the chord. Setting up the Pythagorean Theorem with the radius as the hypotenuse and the distance as one of the legs, we solve for the other leg.

\(\displaystyle 4^2-1^2=15 \rightarrow \sqrt{15}=3.873\)

Since this leg is half of the chord, the total chord length is 2 times that, or 7.746.

\(\displaystyle 2(3.873)=7.746\)

Draw a segment perpendicular to the chord from the center, and this line will bisect the chord. Setting up the Pythagorean Theorem with the radius as the hypotenuse and the distance as one of the legs, we solve for the other leg.

\(\displaystyle 12^2-11^2=23 \rightarrow \sqrt{23}=4.7958\)

Since this leg is half of the chord, the total chord length is 2 times that, or 9.592.

\(\displaystyle 2(4.7958)=9.592\)