On a circle of radius 7 units, PQ and QR are chords of length 7 units each. What is the length of the chord PR in units?

Correct — C. A chord equal in length to the radius forms an equilateral triangle with the two radii drawn to its ends.

OP = OQ = PQ = 7, so ∠POQ = 60°
OQ = OR = QR = 7, so ∠QOR = 60°
∠POR = 60° + 60° = 120°

Chord length = 2R sin(half the central angle).
PR = 2 × 7 × sin 60°
= 14 × √3⁄2 = 7√3

Option (c) prints 21⁄√3, and 21⁄√3 = 21√3⁄3 = 7√3 → option (c)

Why the others are wrong

• (a)
  Option (a) prints 28⁄√3 ≈ 16.17 units, longer than the diameter of 14. No chord can exceed the diameter, so this one goes on sight.
• (b)
  Option (b) prints 14⁄√3 ≈ 8.08 units. That length belongs to a central angle of about 70°, not the 120° that two 60° arcs add up to.
• (d)
  Option (d) prints 7⁄√3 ≈ 4.04 units, shorter than PQ and QR themselves. PR joins the far ends of two 7-unit chords meeting at 120°, so it must come out longer than 7.

Two facts carry this question.

First, a chord equal to the radius subtends 60° at the centre, because that chord and the two radii to its ends make an equilateral triangle.

Second, chord = 2R sin(θ⁄2) for a central angle θ. With θ = 120° that gives PR = 14 × sin 60° = 7√3.

The cosine rule on triangle POR reaches the same place: PR² = 7² + 7² − 2 × 7 × 7 × cos 120° = 49 + 49 + 49 = 147, and √147 = 7√3.

Key facts

• A chord whose length equals the radius subtends 60° at the centre of the circle.
• Chord length = 2R sin(θ⁄2), where θ is the central angle the chord subtends.
• Here θ = 120°, so PR = 2 × 7 × sin 60° = 7√3 ≈ 12.12 units.
• 21⁄√3 rationalises to 7√3, which is why the printed option looks unfamiliar at first.

Common traps

• Adding the chords and answering 14 units, which would need P, Q and R to be collinear.
• Stopping at one 60° angle and answering 7 units, which is just the chord PQ again.
• Reaching 7√3 and failing to recognise it printed as 21⁄√3.

No directly related past PYQ was found.