Problem:

A chord AB of a circle with centre O is 10 cm. If the chord is 12 cm away from the centre, then what is the radius of the circle?

Solution:

Step 1: Understand the problem

In this problem, we are given a circle with center O and a chord AB. The length of the chord AB is 10 cm, and it is 12 cm away from the center of the circle. We need to find the radius of the circle.

Step 2: Recall the properties of a circle

Before solving the problem, let's recall some properties of a circle:

- The radius of a circle is the distance from the center of the circle to any point on the circumference.
- The diameter of a circle is a chord that passes through the center of the circle.
- The perpendicular bisector of a chord passes through the center of the circle.

Step 3: Visualize the problem

Let's draw a diagram to visualize the problem. The circle will be represented by a circle with center O, and the chord AB will be drawn inside the circle. The chord AB will be 10 cm long, and it will be 12 cm away from the center O.

Step 4: Identify the key points

Based on the problem, we can identify the following key points:

- The chord AB is 10 cm long.
- The chord AB is 12 cm away from the center O.
- We need to find the radius of the circle.

Step 5: Solve the problem

To solve the problem, we can use the properties of a circle. Since the chord AB is 12 cm away from the center O, it means that the perpendicular bisector of the chord passes through the center O.

Let's draw the perpendicular bisector of AB, which will pass through the center O. Since the perpendicular bisector passes through the center O, it will bisect the chord AB into two equal parts.

Since the chord AB is 10 cm long and it is bisected, each half of the chord will be 5 cm long. Let's call the point where the perpendicular bisector intersects the chord AB as point M.

Using the Pythagorean theorem, we can find the length of OM, which is the radius of the circle. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

In this case, OM is the hypotenuse, and the lengths of the other two sides are 5 cm each. So, we have:

OM² = OA² + AM²

Since OA is the radius of the circle and AM is half of the chord AB, we can substitute the values:

OM² = r² + (5 cm)²

OM² = r² + 25 cm²

We are given that the chord AB is 12 cm away from the center O. This means that the length of OM is 12 cm. So, we can write:

12² = r² + 25 cm²

144 = r² + 25

r² = 144 - 25

r² = 119

Taking the square root of both sides, we have:

r = √119