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AB and CD are two parallel chords drawn on two opposite sides of their parallel diameter such that AB = 6 cm, CD = 8 cm. If the radius of the circle is 5 cm, the distance between the chords, in cm, is
  • a)
    2
  • b)
    7
  • c)
    6
  • d)
    3
Correct answer is option 'B'. Can you explain this answer?
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Given information:
- AB = 6 cm
- CD = 8 cm
- Radius of the circle = 5 cm

Calculating the distance between the chords:
- Let's consider the center of the circle as point O.
- Join OA, OB, OC, and OD.
- Since AB and CD are parallel chords, the distance between them is equal to the perpendicular distance between the chords from the center of the circle O.

Calculating the perpendicular distance from O to AB:
- In triangle OAB,
- OA = OB = radius = 5 cm (given)
- AB = 6 cm (given)
- Using Pythagorean theorem:
- (OA)^2 + (AB/2)^2 = (OB)^2
- (5)^2 + (3)^2 = (OB)^2
- 25 + 9 = (OB)^2
- OB = √34 cm
- The perpendicular distance from O to AB = OB = √34 cm

Calculating the perpendicular distance from O to CD:
- In triangle OCD,
- OC = OD = radius = 5 cm (given)
- CD = 8 cm (given)
- Using Pythagorean theorem:
- (OC)^2 + (CD/2)^2 = (OD)^2
- (5)^2 + (4)^2 = (OD)^2
- 25 + 16 = (OD)^2
- OD = √41 cm
- The perpendicular distance from O to CD = OD = √41 cm

Calculating the distance between the chords:
- Distance between the chords = √41 - √34
- ≈ 7 cm
Therefore, the distance between the chords is 7 cm, which corresponds to option B.
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