A chord XY of a circle with centre O and radius 5.32cm has ∠XOY = 140°. What is the length of the chord to the nearest centimetre?

Analyze the question: We have a circle with centre O and radius 5.32 cm. The chord XY subtends an angle of 140° at the centre. We need the chord length to the nearest cm.

Concept: Triangle XOY is isosceles (OX = OY = r). The perpendicular from O to the chord bisects both the chord and the angle at O. So:

Chord length = 2r sin(θ/2)

Solution:

• θ/2 = 140° ÷ 2 = 70°
• Chord = 2 × 5.32 × sin 70°
• sin 70° ≈ 0.9397
• Chord = 10.64 × 0.9397 ≈ 9.99 cm

To the nearest centimetre, this is 10 cm.

Check using the cosine rule: XY² = r² + r² − 2r² cos 140° = 2r²(1 − cos 140°) = 2(28.30)(1 + 0.7660) = 56.60 × 1.7660 ≈ 99.96 XY ≈ 10.0 cm ✓

Answer: Option A, 10 cm.

Tip: For chord problems, drop a perpendicular from the centre to form a right-angled triangle, then use chord = 2r sin(θ/2). The chord can never exceed the diameter (10.64 cm here), which quickly rules out nothing too large, and since the angle 140° is close to 180°, the chord should be close to the diameter. That points straight to 10 cm.