Worksheet on chord properties of a circle with five geometric problems involving chords, radii, and right triangles.

Let's solve each problem step by step.

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Problem 1: Find the length of the chord AB and CD.

#### (a) Chord AB
- The given triangle is a right triangle with:
- Hypotenuse \( OB = 10 \) (radius of the circle).
- One leg \( OM = 8 \) (distance from the center to the chord, which is perpendicular to the chord).
- The other leg \( MB \) is half the length of the chord \( AB \).

Using the Pythagorean theorem in \( \triangle OMB \):
\[
OB^2 = OM^2 + MB^2
\]
\[
10^2 = 8^2 + MB^2
\]
\[
100 = 64 + MB^2
\]
\[
MB^2 = 36
\]
\[
MB = 6
\]

Since \( MB \) is half the length of the chord \( AB \):
\[
AB = 2 \times MB = 2 \times 6 = 12
\]

#### (b) Chord CD
- The given triangle is a right triangle with:
- Hypotenuse \( OD = 13 \) (radius of the circle).
- One leg \( ON = 12 \) (distance from the center to the chord, which is perpendicular to the chord).
- The other leg \( ND \) is half the length of the chord \( CD \).

Using the Pythagorean theorem in \( \triangleOND \):
\[
OD^2 = ON^2 + ND^2
\]
\[
13^2 = 12^2 + ND^2
\]
\[
169 = 144 + ND^2
\]
\[
ND^2 = 25
\]
\[
ND = 5
\]

Since \( ND \) is half the length of the chord \( CD \):
\[
CD = 2 \times ND = 2 \times 5 = 10
\]

Answer for Problem 1:
\[
\boxed{12 \text{ and } 10}
\]

---

Problem 2: Solve for \( x \).

- The given circle has two chords intersecting inside the circle.
- The lengths of the segments are:
- One chord is divided into segments \( 4x - 3 \) and \( 6 \).
- The other chord is divided into segments \( 6 \) and \( 3x + 2 \).

Using the intersecting chords theorem:
\[
(4x - 3) \cdot 6 = 6 \cdot (3x + 2)
\]
\[
24x - 18 = 18x + 12
\]
\[
24x - 18x = 12 + 18
\]
\[
6x = 30
\]
\[
x = 5
\]

Answer for Problem 2:
\[
\boxed{5}
\]

---

Problem 3: Find the radius of the circle.

- The given right triangle has:
- One leg \( OF = 5 \) (distance from the center to the chord, which is perpendicular to the chord).
- The other leg \( FE = 24 \) (half the length of the chord).
- The hypotenuse \( OE \) is the radius of the circle.

Using the Pythagorean theorem in \( \triangle OFE \):
\[
OE^2 = OF^2 + FE^2
\]
\[
OE^2 = 5^2 + 24^2
\]
\[
OE^2 = 25 + 576
\]
\[
OE^2 = 601
\]
\[
OE = \sqrt{601}
\]

Answer for Problem 3:
\[
\boxed{\sqrt{601}}
\]

---

Problem 4: Find the length of PQ.

- The given circle has:
- Radius \( OR = 20 \) cm.
- Distance from the center to the chord \( RS \) is \( OP = 12 \) cm.
- \( PQ \) is the segment of the chord \( RS \) that is perpendicular to \( OP \).

Using the Pythagorean theorem in \( \triangle OPR \):
\[
OR^2 = OP^2 + PR^2
\]
\[
20^2 = 12^2 + PR^2
\]
\[
400 = 144 + PR^2
\]
\[
PR^2 = 256
\]
\[
PR = 16
\]

Since \( PR \) is half the length of the chord \( RS \):
\[
RS = 2 \times PR = 2 \times 16 = 32
\]

The length of \( PQ \) is the same as \( RS \) because \( PQ \) is the entire chord \( RS \):
\[
PQ = 32
\]

Answer for Problem 4:
\[
\boxed{32}
\]

---

Problem 5: Find the chord \( CB \).

- The given circle has:
- Diameter \( AB = 4 \) cm.
- \( O \) is the center of the circle.
- \( OC \) is perpendicular to \( AB \), so \( OC \) is the radius of the circle.

The radius \( OA = OB = \frac{AB}{2} = \frac{4}{2} = 2 \) cm.

Using the Pythagorean theorem in \( \triangle OCB \):
\[
OB^2 = OC^2 + CB^2
\]
\[
2^2 = 2^2 + CB^2
\]
\[
4 = 4 + CB^2
\]
\[
CB^2 = 0
\]
\[
CB = 0
\]

However, this result seems incorrect. Let's re-evaluate the problem. If \( C \) is on the circle and \( OC \) is perpendicular to \( AB \), then \( C \) must be at the same level as \( A \) or \( B \), making \( CB \) the diameter of the circle.

Thus, the correct interpretation is:
\[
CB = AB = 4
\]

Answer for Problem 5:
\[
\boxed{4}
\]

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Final Answers:

1. \( \boxed{12 \text{ and } 10} \)
2. \( \boxed{5} \)
3. \( \boxed{\sqrt{601}} \)
4. \( \boxed{32} \)
5. \( \boxed{4} \)