Numerical answer
In a circle of radius 13 cm, AB is a chord of length 24 cm. CD is a chord of length 10 cm parallel to AB, lying on the opposite side of the centre. What is the area (in cm²) of the trapezium ABDC?
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Answer: 289
The perpendicular distance of a chord of length L from the centre is √(r² − (L/2)²).
- AB = 24: distance = √(169 − 144) = 5 cm.
- CD = 10: distance = √(169 − 25) = 12 cm.
The chords are on opposite sides of the centre, so the distance between them (height of the trapezium) = 5 + 12 = 17 cm.
Area = ½ × (24 + 10) × 17 = 17 × 17 = 289 cm².
Trap: If the chords were on the same side, the height would be 12 − 5 = 7 cm and the area 119 cm² – read 'opposite side' carefully.
- AB = 24: distance = √(169 − 144) = 5 cm.
- CD = 10: distance = √(169 − 25) = 12 cm.
The chords are on opposite sides of the centre, so the distance between them (height of the trapezium) = 5 + 12 = 17 cm.
Area = ½ × (24 + 10) × 17 = 17 × 17 = 289 cm².
Trap: If the chords were on the same side, the height would be 12 − 5 = 7 cm and the area 119 cm² – read 'opposite side' carefully.
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