Two parallel chords of length 5 units and 8 units are on opposite sides of the center of a circle of radius 7 units. What is the distance between the chords? (round your answer to two decimal places)

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SSC CHSL Exam 2024 Tier-I Official Paper (Held On: 04 Jul, 2024 Shift 4)
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  1. 12.28 units
  2. 12.82 units
  3. 11.28 units
  4. 11.82 units

Answer (Detailed Solution Below)

Option 1 : 12.28 units
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SSC CHSL Exam 2023 Tier-I Official Paper (Held On: 02 Aug 2023 Shift 1)
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100 Questions 200 Marks 60 Mins

Detailed Solution

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Given:

Two parallel chords of length 5 units and 8 units are on opposite sides of the center of a circle of radius 7 units.

We need to find the distance between the chords.

Formula used:

In a right triangle:

OM = √(r² - AM²), where OM is the perpendicular distance from the center to the chord.

Calculation:

F1 SSC Priya 19 10 24 D5

For the chord of length 5 units:

⇒ AM = 1/2 × 5 = 2.5 units

⇒ OM² = r² - AM²

⇒ OM = √(7² - 2.5²)

⇒ OM = √(49 - 6.25)

⇒ OM = √42.75 = 6.54 units

For the chord of length 8 units:

⇒ CN = 1/2 × 8 = 4 units

⇒ ON² = r² - CN²

⇒ ON = √(7² - 4²)

⇒ ON = √(49 - 16)

⇒ ON = √33 = 5.74 units

Therefore, the distance between the chords:

⇒ Distance = OM + ON

⇒ Distance = 6.54 + 5.74 = 12.28 units

∴ The distance between the chords is 12.28 units.

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