Segment Area in a Fixed-Perimeter Glass Sector
A designer is producing a curved glass panel for a hotel lobby. One component is a circular sector of radius r cm and angle θ radians, with a straight chord joining the ends of the arc. Because of the cutting jig, the perimeter of the sector (two radii plus the arc) must be exactly 26 cm. The decorative feature is the segment - the region between the chord and the arc - which is to be etched at £0.85 per cm². The designer first tries θ = 1.25 rad, which gives a workable shape, then proposes doubling the angle to 2.5 rad, arguing that 'the etched region will be twice as big, so the etching budget must double'. The workshop foreman is not convinced and wants the two areas computed before any glass is ordered. All angles are in radians, and the sector angle must stay below π so the panel fits the frame.
1. For θ = 1.25 rad, find the radius, the arc length and the chord length.
Read the worked answer
Perimeter constraint: 2r + rθ = 26, so r(2 + θ) = 26.
With θ = 1.25: r = 26/3.25 = 8.00 cm.
Arc length = rθ = 8 × 1.25 = 10.0 cm (this is the boundary the etching follows).
Chord: the triangle formed by the two radii and the chord gives chord = 2r sin(θ/2) = 16 sin 0.625 = 16 × 0.58510 = 9.36 cm.
Check that the perimeter is consistent: 8 + 8 + 10 = 26 cm, as required.
2. Find the area of the etched segment and hence the etching cost at θ = 1.25 rad.
Read the worked answer
Segment area = sector area − triangle area.
Sector = ½r²θ = ½ × 64 × 1.25 = 40.00 cm².
Triangle = ½r² sin θ = ½ × 64 × sin 1.25 = 32 × 0.94899 = 30.37 cm².
Segment = 40.00 − 30.37 = 9.63 cm² (calculator value 9.6325).
Cost = 9.6325 × 0.85 = £8.19.
Note the structural point for the next part: the triangle term is not small, so the segment is only about a quarter of the sector. Any argument that ignores how the triangle term behaves will mislead.
3. Compute the segment area at θ = 2.5 rad and rule on the designer's claim that doubling θ doubles the etched area and the budget.
Read the worked answer
With θ = 2.5, the radius is forced down by the perimeter constraint: r = 26/(2 + 2.5) = 26/4.5 = 5.778 cm, r² = 33.383.
Sector = ½ × 33.383 × 2.5 = 41.73 cm².
Triangle = ½ × 33.383 × sin 2.5 = 16.6915 × 0.59847 = 9.99 cm².
Segment = 41.73 − 9.99 = 31.74 cm².
Ratio to the first design: 31.74/9.63 = 3.30, not 2. Cost would be 31.74 × 0.85 = £26.98, so the budget more than triples.
The designer's error is to treat the two variables as independent. The perimeter is fixed, so increasing θ shrinks r (from 8.00 to 5.78 cm) and drives sin θ down from 0.949 to 0.598. Both effects shrink the triangle term, and the triangle is subtracted, so the segment grows far faster than θ. The claim is wrong in magnitude and in the direction of the cost risk; it is not wrong in direction, since the segment does grow.
A useful bound for the report: as θ → π the triangle vanishes and the segment becomes a semicircle of radius 26/(2 + π) = 5.057 cm, area ½π(5.057)² ≈ 40.2 cm². So a doubling of the etched area is comfortably achievable within the frame limit, but the cost model must be recalculated, not scaled.