Circles
Chords of a circle
Grade 11
Question:
<p>If two parallel chords of a circle, having diameter 4 units, lie on the opposite sides of the centre and subtend angles \(\cos^{-1}(1/7)\) and \(\sec^{-1}(7)\) at the centre, respectively, then the distance between these chords is</p>
<p>\(\dfrac{16}{7}\)</p>
<p>\(\dfrac{8}{\sqrt{7}}\)</p>
<p>\(\dfrac{8}{7}\)</p>
<p>\(\dfrac{4}{\sqrt{7}}\)</p>
Step-by-Step Solution
Key Concept: For a chord subtending angle 2θ at the center of a circle with radius r, the perpendicular distance from center is r·cos(θ). Use this to find distances of both chords from center, then add them since chords are on opposite sides.
<p><strong>Step 1:</strong> Given diameter = 4, so radius r = 2.</p><p><strong>Step 2:</strong> First chord subtends angle α = cos⁻¹(1/7) at center. If this is the full angle, the perpendicular distance d₁ from center is:</p><p>d₁ = r·cos(α/2)</p><p>Using cos(α) = 1/7, we get cos(α/2) = √[(1 + cos α)/2] = √[(1 + 1/7)/2] = √[4/7] = 2/√7</p><p>So d₁ = 2 · (2/√7) = 4/√7</p><p><strong>Step 3:</strong> Second chord subtends angle β = sec⁻¹(7) at center. Note that sec⁻¹(7) = cos⁻¹(1/7), so β = α.</p><p>Therefore d₂ = 4/√7</p><p><strong>Step 4:</strong> Since chords lie on opposite sides of the center, the distance between them is:</p><p>Distance = d₁ + d₂ = 4/√7 + 4/√7 = 8/√7 = 8√7/7</p><p>∴ Answer: A</p>
Correct Answer: A