The length of a chord of a circle is calculated as follows:

 Chord length =  \(\displaystyle 2rsin(\frac{C}{2})\)

Chord length = \(\displaystyle 2\sqrt{r^2-d^2}\), where \(\displaystyle r\)  is the radius of the circle and \(\displaystyle d\)  is the perpendicular distance from the chord to the circle center.

Chord length =  \(\displaystyle 2\sqrt{r^2-d^2}=2\sqrt{5^2-4^2}\)

\(\displaystyle =2\sqrt{25-16}=2\sqrt{9}=2\times 3\)

 \(\displaystyle \Rightarrow\) Chord length = \(\displaystyle 6\ cm\)

The figure below shows \(\displaystyle \angle AOB\), which matches this description, along with its chord \(\displaystyle \overline{AB}\):

By way of the Isoscelese Triangle Theorem, \(\displaystyle \Delta AOB\) can be proved a 45-45-90 triangle with legs of length 18, so its hypotenuse - the desired chord length \(\displaystyle AB\) - is \(\displaystyle \sqrt{2}\) times this, or \(\displaystyle 18\sqrt{2}\).

The arc intercepted by a \(\displaystyle 60^{\circ}\) central angle is \(\displaystyle \frac{60}{360}= \frac{1}{6}\) of the circle, so the circumference of the circle is \(\displaystyle 7 \pi \cdot 6 = 42\pi\). The radius is the circumference divided by \(\displaystyle 2 \pi\), or \(\displaystyle 42 \pi \div 2 \pi = 21\). 

The figure below shows a \(\displaystyle 60^{\circ}\) central angle \(\displaystyle \angle AOB\), along with its chord \(\displaystyle \overline{AB}\):

By way of the Isoscelese Triangle Theorem, \(\displaystyle \Delta AOB\) can be proved equilateral, so \(\displaystyle AB = OA = 21\).

The arc intercepted by a \(\displaystyle 120^{\circ}\) central angle is \(\displaystyle \frac{120}{360}= \frac{1}{3}\) of the circle, so the circumference of the circle is \(\displaystyle \frac{4 \pi}{3}\cdot 3= 4\pi\). The radius is the circumference divided by \(\displaystyle 2 \pi\), or \(\displaystyle 4\pi \div 2 \pi = 2\). 

The figure below shows a \(\displaystyle 120^{\circ}\) central angle \(\displaystyle \angle AOB\), along with its chord \(\displaystyle \overline{AB}\)and triangle bisector \(\displaystyle \overline{OM}\). 

We will concentrate on \(\displaystyle \Delta AOM\), which is a 30-60-90 triangle. By the 30-60-90 Theorem, 

\(\displaystyle OM = \frac{1}{2} AO = \frac{1}{2} \cdot 2 = 1\)

and

\(\displaystyle AM = OM \cdot \sqrt{3}= 1\sqrt{3}= \sqrt{3}\)

\(\displaystyle M\) is the midpoint of \(\displaystyle \overline{AB}\), so

\(\displaystyle AB = 2 \cdot AM = 2 \sqrt{3}\)

The figure below shows \(\displaystyle \angle AOB\), which matches this description, along with its chord \(\displaystyle \overline{AB}\):

By way of the Isosceles Triangle Theorem, \(\displaystyle \Delta AOB\) can be proved equilateral, so \(\displaystyle AB = OA = 20\).

This answer is not among the choices given.

The figure below shows \(\displaystyle \angle AOB\), which matches this description, along with its chord \(\displaystyle \overline{AB}\) and triangle bisector \(\displaystyle \overline{OM}\). 

We will concentrate on \(\displaystyle \Delta AOM\), which is a 30-60-90 triangle. By the 30-60-90 Theorem, 

\(\displaystyle OM = \frac{1}{2} AO = \frac{1}{2} \cdot 12 = 6\)

and

\(\displaystyle AM = OM \cdot \sqrt{3}= 6\sqrt{3}\)

\(\displaystyle M\) is the midpoint of \(\displaystyle \overline{AB}\), so

\(\displaystyle AB = 2 \cdot AM = 2 \cdot 6 \sqrt{3}= 12 \sqrt{3}\)

If two chords of a circle intersect inside the circle, the product of the lengths of the parts of each chord is the same. In other words,

\(\displaystyle 40 t = 20 (t+12)\)

Solving for \(\displaystyle t\) - distribute:

\(\displaystyle 40 t = 20 (t+12)\)

\(\displaystyle 40 t = 20 t+ 20 \cdot 12\)

\(\displaystyle 40 t = 20 t+ 240\)

Subtract \(\displaystyle 20t\) from both sides:

\(\displaystyle 40 t- 20 t = 20 t+ 240 - 20 t\)

\(\displaystyle 20 t = 240\)

Divide both sides by 20:

\(\displaystyle 20 t \div 20 = 240 \div 20\)

\(\displaystyle t = 12\)

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle along the secant; in other words,

\(\displaystyle t^{2} = 12 \cdot (12+ 18) = 12 \cdot 30 = 360\)

Solving for \(\displaystyle t\):

\(\displaystyle t^{2} = 360\)

\(\displaystyle t = \sqrt{360}\)

Simplifying the radical using the Product of Radicals Principle, and noting that 36 is the greatest perfect square factor of 360:

\(\displaystyle t = \sqrt{36} \cdot \sqrt{10} = 6 \sqrt{10}\)