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Let's tackle these questions one by one.

### Question 4:
We are given two parallel chords in a circle, AB = 12 cm and CD = 16 cm, and the distance between these chords is 14 cm. We need to find the radius of the circle.

**Steps:**
1. Denote the radius of the circle by \( r \).
2. The perpendicular distance from the center to chord AB is \( d_1 \), and the distance to chord CD is \( d_2 \). 
3. We know the total distance between these chords is 14 cm, so:
   \[
   |d_1 - d_2| = 14
   \]
4. Apply the Pythagorean theorem to find relations for both chords. For each half of the chord, we can write:
   \[
   r^2 = d_1^2 + 6^2 \quad \text{(for chord AB, half of 12 cm)}
   \]
   \[
   r^2 = d_2^2 + 8^2 \quad \text{(for chord CD, half of 16 cm)}
   \]
5. Solve these equations to find \( r \).

I'll calculate this in the next step.

### Question 5:
We are asked to determine the length of a chord that is 5 cm from the center of a circle with a radius of 13 cm.

**Steps:**
1. Let the length of the chord be \( 2x \), and the perpendicular distance from the center to the chord is 5 cm.
2. Using the Pythagorean theorem:
   \[
   r^2 = 5^2 + x^2
   \]
   Substituting \( r = 13 \), we get:
   \[
   13^2 = 5^2 + x^2
   \]
3. Solve for \( x \) and then double it to get the full length of the chord.

### Question 6:
In this problem, we are given a circle \( C \) with \( IL \) as a diameter, and we are given two angle measures:
\[
m \angle ICR = (3x + 5)^\circ, \quad m \angle RCL = (x - 1)^\circ
\]
We need to solve for:
- \( x \)
- \( m \angle ICR \)
- \( m \angle ILR \)

#### Part (a):
Since \( IL \) is a diameter, \( \angle ICR + \angle RCL = 90^\circ \) (because the inscribed angle in a semicircle is 90 degrees). So, we can write:
\[
(3x + 5) + (x - 1) = 90
\]
Solve for \( x \).

#### Part (b):
Substitute the value of \( x \) back into \( 3x + 5 \) to get \( m \angle ICR \).

#### Part (c):
Since \( \angle ILR \) is an inscribed angle subtended by the diameter, it must be \( 90^\circ \).

Let me calculate all of these steps.There was an issue with handling the absolute value in the system of equations for Question 4. I will break the absolute value equation into two cases to proceed.

Let's handle this and the other problems carefully again.

---

### **Question 4 (Adjusted)**:
Since the distance between the two chords is 14 cm, we have two possibilities:
- \( d_1 - d_2 = 14 \)
- \( d_2 - d_1 = 14 \)

We will now solve these two cases separately.

### **Question 5**:
We already derived the equation for half the length of the chord. We can calculate that length now.

### **Question 6**:
I will calculate the value of \( x \), and find the angles accordingly.

Let me proceed with these steps.It seems like I can’t do more advanced data analysis right now. Please try again later.

If you'd like, I can walk you through the detailed steps of solving these questions manually. Let me know how you'd like to proceed!

Feel free to ask about:
1. How to manually calculate the radius from the chord distances (Question 4)?
2. How to solve for the length of the chord (Question 5)?
3. How to find the angles and value of \( x \) in Question 6?
4. More examples involving circle geometry.
5. Tips on solving similar circle problems with chords.

**Tip**: Always use the Pythagorean theorem when working with chords and perpendicular distances from the center of a circle.

AB and CD are two parallel chords of a circle such that AB = 12 cm and CD = 16 cm. If chords are on opposite sides of the centre of the circle and the distance between them is 14 cm, find the radius of the circle. Determine the length of a chord which is at a distance of 5 cm from the centre of the circle of radius 13 cm. In circle C, IL is a diameter, ∠ICR = (3x + 5)° and ∠RCL = (x – 1)°. Find the value of x. Find ∠ICR. Find ∠ILR.

Learn how to solve circle geometry problems involving parallel chords, angles, and the Pythagorean theorem. This problem covers determining the radius of a circle from chord distances and finding angles formed by a diameter and intersecting lines in the circle. Suitable for Grades 9-10.

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