Question

A circle has two parallel chords of lengths 6 cm and 8 cm. If the chords are 1 cm apart and the center is on the same side of the chords, then diameter of the circle is of length

A. 5 cm
B. 6 cm
C. 8 cm
D. 10 cm

Answer 1

The diameter of the circle, determined by parallel chords of lengths 6 cm and 8 cm, spaced 1 cm apart, with the center on the same side, is approximately 10 cm (Option D).

- Chords AB and CD are parallel with lengths 8 cm and 6 cm, respectively.

- The chords are 1 cm apart.

- The center O is on the same side of the chords.

Concept Used:

- The perpendicular bisector of a chord passes through the center of a circle.

Calculation:

1. Let O be the center of the circle with radius r, and chords AB = 8 cm and CD = 6 cm parallel to each other.

2. Drop perpendiculars OM and ON from O to chords AB and CD, respectively. It is given that MN = 1 cm.

3. Using the Pythagorean theorem:

`\(\triangle OAM\): \[ OM^2 + AM^2 = OA^2 \] \[ O \cdot M^2 + 4^2 = r^2 \]` (Equation 1)

4. In
`\(\triangle OCN\)`:

`\[ ON^2 + CN^2 = OC^2 \] \[ (OM + 1)^2 + 3^2 = r^2 \]` (Equation 2)

5. Now, using equations (1) and (2):

`\[ O \cdot M^2 + 16 = (O \cdot M^2 + 2OM + 1) + 9 \] \[ O \cdot M^2 + 16 = O \cdot M^2 + 2OM + 10 \] \[ 2OM = 6 \] \[ OM = 3 \]`

6. Substitute OM = 3 into equation (1):

`\[ O \cdot 3^2 + 16 = r^2 \] \[ 9O + 16 = r^2 \]`

7. Substitute \( r^2 \) from equation (2):

`\[ 9O + 16 = (O \cdot 3 + 1)^2 + 9 \] \[ 9O + 16 = 9O^2 + 6O + 1 + 9 \] \[ 9O^2 + 6O - 16 = 0 \]`

8. Solve the quadratic equation for O:

`\[ O = (-b \pm √(b^2 - 4ac))/(2a) \] \[ O = (-6 \pm √(6^2 - 4(9)(-16)))/(2(9)) \] \[ O = (-6 \pm √(36 + 576))/(18) \] \[ O = (-6 \pm √(612))/(18) \]`

Since O is the radius of the circle, choose the positive root:

`\[ O = (-6 + √(612))/(18) \] \[ O \approx (24.72)/(18) \] \[ O \approx 1.373 \]`

9. Now, use
`\( r^2 = 9O + 16 \)` to find r:

`\[ r^2 = 9 \cdot 1.373 + 16 \] \[ r^2 \approx 29.357 \] \[ r \approx √(29.357) \] \[ r \approx 5.42 \]`

10. Finally, the diameter of the circle is 2r:

`\[ \text{Diameter} = 2 \cdot 5.42 \] \[ \text{Diameter} \approx 10.84 \]`

Therefore, the closest option among the given choices is D. Rs. 10 cm.