Chapter 9.4 — Exercise 9.3 — Segments of a Secant
Segments of a circle formed by a secant. This is Lesson 4 of 4 in Chapter 9: Tangents and Secants to a Circle.
What a Chord Leaves Behind
Every secant's chord splits a circle's interior into two separate regions. Exercise 9.3 leaves tangents behind entirely and asks a purely area-based question instead: exactly how much space does each of those two regions actually cover? Every problem in this exercise, whether it's a single segment or an entire shaded design built from several circular pieces, comes back to the same one calculation repeated as many times as the figure needs.
Minor Segment, Major Segment
A chord passing exactly through the centre is the one special case where both regions come out identical — that's a diameter, and each half is called a semicircle rather than a minor or major segment, since neither one is genuinely smaller than the other. Every other chord, positioned anywhere off-centre, always produces one genuinely smaller region and one genuinely larger one — the closer the chord sits to the edge of the circle, the smaller the minor segment shrinks relative to the major one.
Segment Area, Built From a Sector Minus a Triangle
A sector (the pie-slice bounded by two radii and the arc between them) is easy to find the area of directly. A segment is almost the same shape, except a straight chord replaces one of the two radii-bounded edges — so subtracting the triangle formed by the two radii and the chord from the full sector leaves exactly the segment behind. The major segment is simpler still, once the minor segment is already known: it's just whatever area of the circle is left over once the minor segment is removed, so there's rarely any need to compute a major-segment area from scratch using its own sector and triangle.
Area of minor segment = area of sector OAB − area of △OAB
Area of major segment = area of circle − area of minor segmentA Right-Angle Chord, Fully Worked
A circle of radius 10 cm has a chord subtending a right angle (90°) at the centre. With the sector angle a clean 90°, both the sector and the triangle areas come out simply.
area of △OAB = ½×OA×OB = ½×10×10 = 50 cm²
area of sector OAB = (90/360)×π×10² = ¼×314 = 78.5 cm²
area of minor segment = 78.5−50 = 28.5 cm²
area of major segment = πr² − 28.5 = 314−28.5 = 285.5 cm²A 90° sector angle is what makes this problem clean: the triangle OAB is automatically right-angled at O too, so its area comes straight from ½×base×height without needing any trigonometry at all. That's precisely why this particular chord angle is the one every textbook reaches for first — every other value of the central angle needs the auxiliary-perpendicular construction shown in the next problem below, since the two radii no longer meet at a convenient right angle on their own.
A 120° Chord, Needing Trigonometry
A circle of radius 12 cm has a chord subtending 120° at the centre. Without a right angle to lean on directly, finding the triangle's area first needs an auxiliary perpendicular from the centre to the chord, splitting the angle exactly in half.
OM ⊥ AB ⟹ ∠AOM = 60°
sin 60° = AM/OA ⟹ AM = 12×(√3/2) = 6√3 ⟹ AB = 2×AM = 12√3
cos 60° = OM/OA ⟹ OM = 12×½ = 6
area of △OAB = ½×AB×OM = ½×12√3×6 = 36√3 ≈ 62.352 cm² (using √3≈1.732)
area of sector OAB = (120/360)×π×12² = ⅓×452.16 = 150.72 cm²
area of minor segment = 150.72−62.352 = 88.368 cm²The perpendicular from the centre does two jobs simultaneously here: it bisects the central angle (making the two halves easy to handle with sine and cosine separately) and it bisects the chord itself, since a perpendicular from a circle's centre to any chord always passes through that chord's midpoint. Both consequences come from the exact same construction step, which is worth remembering as a single reusable move rather than two separate facts to look up independently — draw one perpendicular from the centre, and both the angle and the chord split themselves in half automatically.
Six More Area Problems, Compressed
| Setup | Result |
|---|---|
| Two 25 cm wiper blades, each sweeping 115°, area cleaned per sweep | 1255 cm² (two sectors added directly) |
| Square of side 10 cm, four semicircles (radius 5) drawn on each side, shaded region between them | 57 cm² |
| Square of side 7 cm, two semicircles (APD, BPC) drawn inward, shaded region | 10.5 cm² |
| Quadrant of radius 3.5 cm, OD=2 cm, shaded region between sector and triangle | 6.125 cm² |
| Two concentric arcs, radii 21 cm and 7 cm, sector angle 30°, area between them | 102.67 cm² |
| Two overlapping quadrants of radius 10 cm each, common "lens" area | 57 cm² |
Every one of these six problems reduces to the identical two ingredients used throughout this exercise — a circular sector's area and a triangle's (or square's) area — just added, subtracted, or doubled in whatever combination the shaded region actually needs. The wiper-blade problem simply doubles one sector; the two square-and-semicircle problems subtract circular pieces from a straight-edged shape; the quadrant and concentric-arc problems subtract one curved region from another. None of the six needs a new formula beyond the sector-minus-triangle idea already established above.
The two problems sharing the identical 57 cm² answer are worth a direct comparison, since they reach that number by genuinely different routes. The four-semicircle square problem subtracts two full semicircle areas (equal to one full circle) from the square directly. The overlapping-quadrants problem instead adds two separate segment areas together — the "lens" shape common to both quadrants turns out to be built from two identical segments back to back, rather than from any square-minus-circle subtraction at all. Matching final answers here is a coincidence of the specific numbers chosen, not a sign the two problems are secretly the same calculation in disguise.
It's worth noticing, too, which of these six problems is the odd one out: the wiper-blade problem is the only genuine real-world word problem in the set, while the other five are all abstract shaded-region designs built from squares, quadrants, and circles with no everyday framing attached. Recognising a sector hiding inside a real-world description — a wiper's sweep, a searchlight's beam, a sprinkler's reach — is really the same skill as spotting one inside a purely geometric figure, just with an extra translation step at the start.
The Chapter, Complete
Tangents and Secants to a Circle moved through three genuinely different ideas: the introduction established the tangent-radius right angle and the resulting tangent-length formula, Exercise 9.1 and Exercise 9.2 built out everything a single tangent and a pair of tangents can prove, and this exercise turned to the areas a secant's chord carves out instead — a genuinely different kind of question from every proof earlier in the chapter, but built from tools (Pythagoras, basic trigonometry, sector and triangle areas) already familiar from earlier chapters. The next chapter, Mensuration, extends this same area-and-volume thinking from flat circles into three-dimensional solids.