A segment as what is left when the triangle is taken away

What to assume they know

• A sector's area as its share of the full turn — the sector's area, and the two properties the unitary argument needed
• Sector versus segment, and which one major and minor refer to — segment, sector, chord, and the major/minor convention set on p. 154
• Area of a triangle as half the base times the height
• Sine and cosine of an acute angle as ratios of sides in a right triangle, and the exact values at 30°, 45° and 60°, from Chapter 8
• RHS congruence, and that congruent triangles have equal corresponding parts
• Comfort with surds: leaving √3 in an answer and factorising a common number out

What they should be able to do

• Explain why the sector argument cannot be re-run for a segment
• Decompose a sector into a segment and a triangle, and justify that the two pieces cover it exactly once
• Compute a segment's area as the difference of a sector's area and a triangle's
• Justify that the perpendicular dropped from the centre to a chord bisects both the chord and the angle, using the congruence the chapter names
• Use the half-angle and the radius to obtain the chord and the height of the triangle
• Compute a major segment, and say why it is not the sector minus the triangle
• Check a segment result at the straight-angle case, where the triangle vanishes
• Apply the same subtraction repeatedly in a composite figure, and carry the result into a cost

Where it usually goes wrong

• "A segment is a sector, so it must scale with the angle too." It does not, and the 271-to-40 collapse above is the fastest way to feel it. The wedge scales; the triangle stubbornly does not, and at 60° and 120° on the same circle the triangle is the very same size.
• "The subtraction is an approximation, so the answer is approximate." The two pieces meet along the chord and cover the wedge exactly once. The only approximation anywhere in these questions is the value chosen for π, and for √3 when one is offered.
• "The triangle to subtract has the chord as one side and the arc as another." The triangle's three sides are two radii and the chord. It is bounded by straight lines only; if a curve has crept into it, the wrong region is being subtracted.
• "Major segment equals major sector minus the triangle." It is the major sector plus the triangle. The triangle sits on the major segment's side of the chord, not the minor one, and question 6 is built to punish the slip.
• "Use the full angle in the sine." The chapter's route halves the angle first, because the perpendicular from the centre splits the isosceles triangle into two right ones. Feeding 120° into a ratio table that stops at 90° is where students stall.
• "M is the mid-point because it looks like it." It is the mid-point because the two halves are congruent by RHS, and the same congruence is what halves the angle. Both facts come from one argument, and both are used.
• "Convert the surd to a decimal as soon as it appears." Example 2 carries √3 right to the end and factorises, and no value for √3 is offered there. Where the exercise wants a decimal it says which value to use — 1.73 in questions 6 and 7, and 1.7 in question 13, which are not the same number.
• "The six designs are sectors." They are the pieces left over between the straight sides and the rim, and the cost of confusing the two is far worse than it sounds. Six wedges of 60° are the whole cover — 2464 cm² against 464.8 cm² of design, over five times as much, and a bill of ₹862.40 against ₹162.68. A student who makes this substitution has not overshot slightly; they have quoted for covering the entire table.

Questions to check understanding

• Given radius and angle, find the minor segment
• Given radius and angle, find the major segment — testing the plus rather than the minus
• Given the radius and a chord that opens 90° at O, find the segment with no trigonometry at all
• A question naming a segment and a sector of the same circle in its two parts, as question 4 does
• Composite figures where one design is a segment and the answer is a count times an area, optionally carried into a cost
• Leave the answer in surd form when no value for √3 is supplied, and in decimals when one is
• Reasoning: why a segment's area cannot be found by scaling from the whole disc

Examples worth working on the board

Inputs only. Values marked verified are worked out here on data printed inside pp. 154–160.

• The failure of proportionality, demonstrated before it is claimed. Take radius 21 cm, the chapter's own radius in Example 2 and in Exercise 11.1 question 5, with π = 22/7. Verified: at 120° the wedge is 462 cm², the triangle is 441√3/4 ≈ 190.95 cm², and the segment is about 271.05 cm². At 60° the wedge is 231 cm² — exactly half, as the previous topic promised — but the triangle is again 441√3/4 ≈ 190.95 cm², so the segment collapses to about 40.05 cm². Halving the angle cut the segment by a factor of nearly seven. Open the explanation on this pair of numbers. It is the entire justification for changing method, and it also shows why: the triangle refused to halve.
• Fig. 11.4 (§11.1, p. 155, claimed by name on p. 156; read from the printed page). A circle with centre O labelled at the top of a drawn triangle; the two radii run down to A on the left and B on the right, one marked r, with θ set in the angle at O. The chord AB closes the triangle, and the thin region between that chord and the arc below it is shaded, with P on that arc and Q on the far arc. This is the decomposition picture: the triangle is drawn but unshaded, the segment is shaded, and together they are the wedge. Everything in section 3 can be said against this one figure.
• Example 2 inputs (§11.1, pp. 157–158, with Fig. 11.6 and Fig. 11.7). Radius 21 cm, angle at the centre 120°, π taken as 22/7, and the region wanted is the one the chord cuts off. Fig. 11.6 draws the circle with the chord high across it, both radii annotated 21 cm, 120° set at O, the label Y on the arc above the chord, and the region between chord and that arc shaded. Fig. 11.7 draws the triangle alone, apex O at the bottom, A and B at the ends of the chord above, M marked on AB with a right-angle mark, the two halves of the apex angle each marked 60°, and both slanting sides annotated 21 cm.
• The wedge. Verified: 120/360 reduces to a third, and a third of (22/7) × 441 = 1386 is 462 cm².
• The triangle, the chapter's way (p. 157). Drop a segment from O onto the chord and call its foot M. The two small triangles share that segment, have equal hypotenuses because both are radii, and each has a right angle at M — so RHS congruence pairs them, and M is the middle of the chord while the apex angle is split in half. That is two conclusions from one congruence, and both are needed. Verified: with the half-angle at 60°, the height is 21 cos 60° = 21/2 cm, half the chord is 21 sin 60° = 21√3/2 cm, the whole chord is 21√3 cm, and the triangle measures ½ × 21√3 × 21/2 = 441√3/4 cm².
• The segment. Verified: 462 − 441√3/4, which factorises as (21/4)(88 − 21√3) cm² — the form the chapter prints — because 462 is exactly 21 × 22. Taking √3 ≈ 1.732 gives about 271.05 cm², but note that the chapter supplies no value for √3 in this example and leaves the surd standing; the decimal above is added here. Say so rather than quietly evaluating.
• The general shape of the triangle half. Verified by an added algebra from the chapter's own steps: the chord is 2r sin(θ/2) and the height is r cos(θ/2), so the triangle measures r² sin(θ/2) cos(θ/2). Every argument in that expression is acute whenever θ is less than a straight angle, which is why the chapter's route through the half-angle works at 120° even though Chapter 8 tabulates nothing beyond 90°. This general form is added here; the chapter computes the triangle case by case and states no formula for it anywhere in pp. 154–160.
• The straight-angle check. Verified: put θ at 180°. The chord becomes a diameter, the "triangle" flattens onto it and measures nothing, and the formula returns half the disc — which is what the segment visibly is. A check that costs one line and catches a sign error or a swapped ratio instantly.
• The major segment (the note on p. 156). Verified by an added algebra: the major segment is the whole disc minus the minor segment, which rearranges to the major sector plus the triangle. Check it on Exercise 11.1 question 6 — radius 15 cm, chord opening 60°, π = 3.14 and √3 = 1.73. Verified: the minor wedge is 117.75 cm², the triangle is 97.3125 cm², so the minor segment is about 20.44 cm²; the whole disc is 706.5 cm², so the major segment is about 686.06 cm²; and the major wedge of 300° is 588.75 cm², which with the triangle added back gives the same 686.06. Both routes.
• Exercise 11.1, the other segment questions (p. 158). Q4 (i) — radius 10 cm, with a chord that opens a right angle out at O, π = 3.14. Q5 (iii) — radius 21 cm, 60°, π = 22/7, no value given for √3. Q7 — radius 12 cm, chord opening 120°, π = 3.14 and √3 = 1.73. Verified: Q4 (i) gives 78.5 − 50 = 28.5 cm², and it is the one case in the exercise whose triangle is right-angled at the centre, which is what makes its area a bare half-base-times-height. It is not the only question that dodges trigonometry, though: 5 (iii), 6 and 13 all sit at 60°, where the triangle is equilateral on a side equal to the radius and its area comes straight off the standard formula. Q5 (iii) gives 231 − 441√3/4; Q7 gives 150.72 − 62.28 = 88.44 cm².
• Exercise 11.1 question 13 and Fig. 11.11 (p. 159). A round table cover of radius 28 cm carries six equal designs; the cost is ₹0.35 per cm² and √3 is to be taken as 1.7. Read from the printed page and confirmed on the printed page: the figure is a circle with a six-sided figure inscribed, its six corners on the circle and its interior tinted plain, and the six regions between its sides and the circle filled with a repeating ornament. The figure carries no lettering and no measurements. So the designs are the six segments cut off by the six equal chords, each opening 60° at the centre, and each triangle is therefore equilateral with side equal to the radius. Verified: one wedge is 1232/3 ≈ 410.67 cm², one triangle is 333.2 cm², one design is about 77.47 cm², six come to 464.8 cm², and the bill is ₹162.68. The chapter names neither the six-sided figure nor the 60°; that reading is added here, and section 11 should show the six equal chords being counted round the circle before any number is used.

Figures to have open

• Fig. 11.4 redrawn (p. 155) as the decomposition figure: wedge, chord, triangle and segment, with the triangle able to slide out of the wedge. This is the asset the whole topic rests on, and the printed figure is static.
• A two-panel comparison of the same circle at 120° and 60°, with wedge, triangle and segment areas printed under each. Not in the book; it carries the thesis.
• Fig. 11.6 and Fig. 11.7 redrawn together (pp. 157) so the triangle can be lifted out of the circle and set beside it, keeping O at the apex as the book draws it. The right-angle mark at M and the two halves of the apex angle must survive the redraw — they are the content of the congruence step.
• A live plot of wedge area and triangle area against the angle from 0° to 180°, one a straight line and one not. Not in the book; nothing like it is printed, and it is the clearest statement of why the method had to change.
• Fig. 11.11's table cover (p. 159) reduced to a plain circle with six equal chords and one design tinted. The printed ornament is decorative and carries no data.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 11 "Areas Related to Circles", §11.1, p. 156 — the opening of the page, which sets the segment against Fig. 11.4 and states the difference, and the note beneath it giving the major-segment relation
• Example 2, pp. 157–158, with Fig. 11.6 and Fig. 11.7, including the RHS congruence step and the two ratios at 60°
• Exercise 11.1, pp. 158–159 — questions 4 (i), 5 (iii), 6, 7 and 13, together with the instruction at the head of the exercise fixing π as 22/7 by default
• §11.2 Summary, p. 160, item 3, which records the subtraction and nothing else — no formula for the triangle is given there or anywhere in the chapter
• Chapter 8, for the trigonometric ratios used without re-deriving; Example 2 draws on cosine and sine at 60° and on nothing else, while the wider 30°/45°/60° table is needed for the exercise questions rather than for the example. This is the only place the topic reaches outside pp. 154–160, and it is a deliberate backward reference