Areas Related to Circles
Segments and real problems
Find the area of a segment and solve real-life area problems.
There is no new formula for a segment. A sector is a segment with a triangle added on, so:
segment = sector – triangle
Here, the triangle is formed by the two radii and the chord. Its area depends on the angle at the centre:
| Angle | The triangle | Its area |
|---|---|---|
| 90° | half a square | ½ × r × r |
| 60° | equilateral (two sides are radii, so the base angles are equal, and with 60° at the top all three angles are 60°) | (√3/4) r² |
| 120° | split it in half with a line from O at right angles to the chord | (√3/4) r² |
The 120° case, step by step: drop OM at right angles to the chord AB. The two halves are congruent (RHS), so OM cuts AB in half and splits the 120° into 60° + 60°. Then OM = r cos 60° = r/2, and AM = r sin 60° = (√3/2) r, so AB = √3 r. Area = ½ × AB × OM = (√3/4) r².
The same idea, half the chord with sin and the height with cos, works for any angle. Don't learn it as 'only 60°, 90° or 120°'.
Major segment = the whole circle – the minor segment.
Keep the arithmetic small. With π = 3.14 and √3 = 1.73, take r² out of both terms first: segment = r²(θπ/360 – triangle part).
Real things made of sectors and segments:
- an umbrella with 8 equal ribs makes 8 sectors of 45°
- a brooch with 5 diameters makes 10 sectors of 36° (wire needed = circumference + 5 diameters)
- a lighthouse lights up a sector of the sea
- the designs on a round table cover are segments around a regular hexagon
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