Two parallel chords are drawn in a circle of diameter 20 cm. The length of one chord is 16 cm and the distance between the two chords is 12 cm. The length of the other chord is:

Options

A. a)12 cm
B. b)18 cm
C. c)16 cm
D. d)20 cm

Answer

To solve the problem step-by-step, we can follow these steps:

### Step 1: Understand the Circle and Chords
- We have a circle with a diameter of 20 cm.
- The radius of the circle is half of the diameter:
\[ \text{Radius} = \frac{20}{2} = 10 \text{ cm} \]

### Step 2: Identify the Chords and Their Lengths
- One chord (let's call it CD) has a length of 16 cm.
- The distance between the two parallel chords is given as 12 cm.

### Step 3: Calculate the Half-Length of the First Chord
- Since the chord CD is 16 cm long, its half-length is:
\[ \text{Half-length of CD} = \frac{16}{2} = 8 \text{ cm} \]

### Step 4: Draw a Perpendicular from the Center to the Chord
- Draw a perpendicular line from the center of the circle (let's call it O) to the chord CD. This line will bisect the chord.
- The distance from the center O to the chord CD is denoted as OP.

### Step 5: Use the Pythagorean Theorem
- In the right triangle OAP (where A is one endpoint of the chord CD), we can apply the Pythagorean theorem:
\[ OA^2 = OP^2 + AP^2 \]
where:
- \( OA = 10 \) cm (radius),
- \( AP = 8 \) cm (half-length of chord CD),
- \( OP \) is the distance from the center to the chord.

### Step 6: Calculate OP
- Plugging in the values:
\[ 10^2 = OP^2 + 8^2 \]
\[ 100 = OP^2 + 64 \]
\[ OP^2 = 100 - 64 = 36 \]
\[ OP = 6 \text{ cm} \]

### Step 7: Determine the Distance to the Second Chord
- The distance from the center O to the second chord (let's call it EF) will be:
\[ OQ = OP + 12 = 6 + 12 = 18 \text{ cm} \]

### Step 8: Use the Pythagorean Theorem for the Second Chord
- For the second chord EF, we again use the Pythagorean theorem:
\[ OE^2 = OQ^2 + EP^2 \]
where \( OE = 10 \) cm (radius) and \( OQ = 18 \) cm.
- Let \( EP \) be half the length of the second chord EF:
\[ 10^2 = 18^2 + EP^2 \]
\[ 100 = 324 + EP^2 \]
\[ EP^2 = 100 - 324 = -224 \]
Since this gives a negative value, we realize that the distance OQ cannot exceed the radius.

### Step 9: Correct the Calculation
- The distance from the center to the second chord should be calculated as:
\[ OQ = OP - 12 = 6 - 12 = -6 \text{ cm} \]
This indicates that the second chord is actually closer to the center than the first chord.

### Step 10: Calculate the Length of the Second Chord
- Since the distance is negative, we can calculate the length of the second chord using the correct distance:
\[ OE^2 = OP^2 + EP^2 \]
where \( OP = 6 \) cm (distance to the first chord) and \( EP \) is unknown.
- The calculation will yield:
\[ 10^2 = 6^2 + EP^2 \]
\[ 100 = 36 + EP^2 \]
\[ EP^2 = 100 - 36 = 64 \]
\[ EP = 8 \text{ cm} \]

### Step 11: Find the Length of the Second Chord
- The total length of the second chord EF is:
\[ \text{Length of EF} = 2 \times EP = 2 \times 8 = 16 \text{ cm} \]

### Final Answer
The length of the other chord is **16 cm**.