AB and CD are chords of a circle with centre O. AB = 48 cm and its distance from centre O is 10 cm. If the distance of CD from centre O is 24 cm, find the length of CD.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have two chords AB and CD in a circle with center O. The length of chord AB is given as 48 cm, and its distance from the center O is 10 cm. The distance of chord CD from the center O is 24 cm. We need to find the length of chord CD.

### Step 2: Find the half-length of chord AB
Since the distance from the center O to chord AB is 10 cm, we can denote the midpoint of AB as M. The property of a chord states that the line from the center to the chord is the perpendicular bisector. Therefore, we can find the half-length of AB.

\[ AM = MB = \frac{AB}{2} = \frac{48 \text{ cm}}{2} = 24 \text{ cm} \]

### Step 3: Apply the Pythagorean theorem to triangle OMB
In triangle OMB, we can apply the Pythagorean theorem. Here, OM is the distance from the center to the chord (10 cm), and MB is half the length of the chord (24 cm). We need to find the radius OB.

Using the Pythagorean theorem:

\[ OB^2 = OM^2 + MB^2 \]

Substituting the values:

\[ OB^2 = 10^2 + 24^2 \]
\[ OB^2 = 100 + 576 = 676 \]
\[ OB = \sqrt{676} = 26 \text{ cm} \]

### Step 4: Set up for chord CD
Let the length of chord CD be denoted as 2x. The midpoint of CD is N, and the distance from the center O to chord CD is 24 cm (ON).

### Step 5: Apply the Pythagorean theorem to triangle ODN
In triangle ODN, we apply the Pythagorean theorem again:

\[ OD^2 = ON^2 + DN^2 \]

Substituting the known values:

\[ OB^2 = ON^2 + DN^2 \]
\[ 26^2 = 24^2 + x^2 \]
\[ 676 = 576 + x^2 \]
\[ x^2 = 676 - 576 = 100 \]
\[ x = \sqrt{100} = 10 \text{ cm} \]

### Step 6: Find the length of chord CD
Since CD = 2x, we have:

\[ CD = 2 \times 10 = 20 \text{ cm} \]

### Final Answer
The length of chord CD is **20 cm**.

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