A sector's area is its angle's share of the whole

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What to assume they know

Slicing a disc into sectors to see where πr² comes from — area of a disc as πr²
Arc length as the central angle's share of the circumference — arc length as θ/360 of the circumference, and the rotation argument behind it
Reflection and rotation as motions that change no length and no area
That the angles round a point total 360°
Area of an equilateral triangle, needed for the segment questions

What they should be able to do

Define a sector, and distinguish it from a segment
Show that a half disc has area πr²/2 by reflection, and a quarter disc πr²/4 by a quarter-turn
Rewrite those two areas as πr² times 180/360 and 90/360, and explain what the rewriting is for
State the general sector-area formula for a central angle θ°
Justify the general formula from rotation-invariance and additivity of area, rather than by extending a pattern
Set the sector-area formula alongside the arc-length formula and identify what is shared and what differs
Compute a minor sector and the corresponding major sector, and check that the two total the disc
Compute a segment's area as a sector minus a triangle
Apply the formula to a minute hand and to a windscreen wiper
Compute the ratio of an inscribed regular polygon's area to the disc's

Where it usually goes wrong

"The sector formula is a separate thing to memorise." It is the disc's area times the angle's fraction. Written that way it is one formula and it never inverts by accident.
"A sector and a segment are the same thing." A sector is bounded by two radii and an arc; a segment by an arc and the chord across it. The chapter defines both, one paragraph apart (pp. 146–147), and the difference is exactly the triangle between the two radii and the chord.
"The formula was extrapolated from two examples." Two cases would be flimsy. The formula is forced by rotation-invariance and additivity, and the chapter itself names rotational symmetry as what its derivations used. Give the argument.
"Minor and major sectors need separate formulas." One formula, two angles that total 360°. Checking that the two areas total πr² is a free error-check on every such question, and Q4 is built to be checked that way.
"A segment's area is the sector's area." It is the sector minus the triangle. Q5 and Q7 both turn on that subtraction and it is the only place in the exercise set where the triangle formula is needed.
"The two wipers clean twice one blade's area, whatever the geometry." Only because the question says they do not overlap. Say why the phrase is there; overlapping sweeps would need the overlap subtracted.
"If the printed answer to a starred question looks strange, it must be right." Q7's printed expression comes out negative. An area cannot. Checking the sign of an answer before trusting it is a habit worth building on this exact question.

Questions to check understanding

Area of a sector from radius and angle, with π given
Radius or angle recovered from a given sector area
Area of a quadrant from a given circumference
Minor and major sector areas for one chord, with the total checked against the disc
Area of a segment as a sector minus a triangle
Area swept by a rotating arm — a clock hand or a wiper — in a stated time or through a stated angle
Ratio of an inscribed regular polygon's area to the disc's, for a triangle, square or hexagon
Explain why the sector formula and the arc formula share the same fraction

Examples worth working on the board

Inputs, not answers. Values marked Verified are worked out here; the chapter prints no answers and this volume has no appended answer key.

Fig. 6.38 (p. 146). A shaded sector standing alone: centre O at the left, the angle marked θ° at the centre, and two radii out to B above and A below. Mind which element is dashed: the two radii are dashed, while the arc joining B to A is drawn solid in magenta, and the sector between them is filled pink.
The definition as printed (p. 146). A sector is what an arc and the two radii through its endpoints fence off. The chapter then says the area will be found the same way the arc length was.
Fig. 6.39 (p. 147). A circle with a horizontal diameter AB through O and the radius marked r, the upper half filled pink and the lower half filled blue.
Fig. 6.40 (p. 147). A circle with two perpendicular diameters, B at the top, A right, C left, D below, and the four quadrants filled in four different colours.
The two derivations as printed (p. 147). By symmetry the half disc is 180/360 of the disc, so πr²/2, and the chapter notes reflection in a diameter as the reason. By symmetry the quarter disc is 90/360, so πr²/4, with the quarter-turn as the reason.
The general formula as printed (p. 147). For an arc AB subtending θ° at the centre O, the sector's area is πr² × θ/360. The chapter introduces it by saying that examining the two cases immediately yields the formula, and then remarks explicitly that rotational symmetry was what these derivations used.
The argument that makes it certain, which the chapter does not give. Cut 360° into n equal angles about the centre. Rotating the disc through one of those angles carries each sector exactly onto the next, so all n sectors have equal area; they do not overlap and they fill the disc, so each is πr²/n. Taking m of them side by side gives mπr²/n, which is πr² × θ/360 for every angle that is a whole-number fraction of a turn; and any other angle is squeezed between two such. Verified. This is the same construction as in Arc length as the central angle's share of the circumference, with area additivity in place of length additivity, and the two videos should present it identically so a student notices.
Side by side, for section 7. Arc = 2πr × θ/360; sector = πr² × θ/360. Same fraction, different whole. Verified: dividing the sector's area by the arc's length gives r/2 for every θ — which is Archimedes' half-perimeter-times-radius rule again, now for a piece of the disc rather than all of it. Worth one line; it ties this topic back to Slicing a disc into sectors to see where πr² comes from.
The segment definition (p. 147). The chapter closes §6.10.1 by defining a segment as the region between an arc and the chord across its ends. It gives no figure for a segment and no formula. The route: minor segment = minor sector − triangle formed by the chord and the two radii.
Exercise Set 6.3 Q1 (p. 148). Sector of radius 7 cm and angle 60°, π as 22/7. Verified: (60/360)(22/7)(49) = 77/3 ≈ 25.67 cm².
Exercise Set 6.3 Q2 (p. 148). A circle measures 44 cm round; find the area of one quadrant of it. Verified: radius 7 cm, quadrant area 38.5 cm².
Exercise Set 6.3 Q3 (p. 148). A clock's minute hand is 7 cm long; find the area it sweeps in 10 minutes. Verified: 10 minutes is a sixth of a turn, so 60°, and the answer is again 77/3 ≈ 25.67 cm² — the same number as Q1.
Exercise Set 6.3 Q4 (p. 148). In a circle of radius 10 cm, a chord makes 90° at the centre; both sectors are wanted — the minor one at 90° and the major one at 270°. The question fixes π as 3.14. Verified: 78.5 cm² and 235.5 cm², totalling 314 cm², which is πr². The check is the point of including both parts.
Exercise Set 6.3 Q5 (p. 148). In a circle of radius 15 cm, a chord makes 60° at the centre; both segment areas are wanted, with π as 3.14 and √3 as 1.73. Verified: the sector is 117.75 cm²; the triangle is equilateral of side 15, so (1.73/4)(225) = 97.3125 cm²; the minor segment is 20.4375 cm² and the major is 706.5 − 20.4375 = 686.0625 cm².
Exercise Set 6.3 Q6 (p. 148). A car has two wipers that do not overlap; each blade is 28 cm long and sweeps 120°. Find the total area cleaned in one sweep. Verified: each sweep is (120/360)(22/7)(784) = 2464/3 ≈ 821.33 cm², so the pair clean 4928/3 ≈ 1642.67 cm². The words "do not overlap" are doing real work: they are what licenses adding the two areas.
Exercise Set 6.3 Q7 (p. 148), starred. In a circle of radius r, a chord makes 60° at the centre; the smaller segment's area is asked for as a closed expression. The printed target expression is wrong. Verified: the correct value is r²(π/6 − √3/4). The printed version puts π outside a bracket containing both terms, which evaluates to a negative number and so cannot be an area. See Notes.
Exercise Set 6.3 Q8, Q9 and Q10 (p. 148), all starred. An equilateral triangle, a square and a regular hexagon inscribed in a circle of radius r; in each case the ratio of the polygon's area to the disc's is to be shown, and Q10 adds the observation that its answer is exactly twice Q8's and asks why. Verified: the triangle has side r√3 and area (3√3/4)r², giving 3√3/4π ≈ 0.413; the square has side r√2 and area 2r², giving 2/π ≈ 0.637; the hexagon has side r and area (3√3/2)r², giving 3√3/2π ≈ 0.827. The doubling has a one-picture reason, but take care which cut you use. Splitting the hexagon into its six equilateral triangles from the centre does not help: the inscribed triangle V₁V₃V₅ is not three of those six, because each of them lies only partly inside it — V₂ falls outside the chord V₁V₃ — so highlighting three alternate ones gives a pinwheel and not a triangle. Two cuts that do work, either alone: join the centre to V₁, V₃ and V₅, which divides the inscribed triangle into three triangles of area ½r²·sin 120° = (√3/4)r² each, exactly one of the six equilateral triangles apiece, so the inscribed triangle is three sixths of the hexagon; or note that the hexagon is the inscribed triangle plus the three corner triangles V₁V₂V₃, V₃V₄V₅, V₅V₆V₁, each again (√3/4)r², so the corners make up half the hexagon and the triangle is the other half. The chapter asks and does not answer.
End-of-chapter Q8 (p. 150). The same question as Q2 above with 66 cm round instead of 44. Verified: radius 10.5 cm, quadrant area 86.625 cm².
End-of-chapter Q21 (p. 152), Fig. 6.49. Inside a square, a quarter circle centred at one vertex passing through the two adjacent vertices, and two semicircles on two adjacent sides as diameters, creating two shaded regions labelled A and B; show they are equal in area. Worth including as the topic's hardest sector-and-segment bookkeeping.
End-of-chapter Q27 (p. 153), Fig. 6.55. Three overlapping figures — one quarter of a circle, one semicircle, one triangle — cut out two shaded pieces between them; A, O and C lie along a base line, B sits above O, and E, F, D are interior. Show the two pieces have equal area. The same skill again, with a harder figure.

Figures to have open

Fig. 6.38 (p. 146), redrawn: a single sector with θ° and the two radii marked.
Fig. 6.39 and Fig. 6.40 (p. 147), redrawn with the halves and quadrants in contrasting fills so the reflection and the quarter-turn can be shown moving. As in Arc length as the central angle's share of the circumference, the colour is the argument and not decoration.
A sector and a segment side by side, with the triangle between them shaded. The chapter defines the segment in words with no figure at all (p. 147), so this is added here and it is the figure the topic most needs.
A disc cut into n equal sectors with n adjustable, for section 6.
A regular hexagon inscribed in a circle with V₁V₃V₅ drawn in, cut so the doubling is visible: either the three centre-to-alternate-vertex triangles filling V₁V₃V₅, or the three corner triangles left outside it, each of which matches one of the hexagon's six equilateral triangles in area. Do not specify this as three alternate centre triangles highlighted — those form a pinwheel, not the inscribed triangle, and the figure cannot be drawn that way. Not in the book, and it is the answer to the question Q10 asks.
Fig. 6.49 (p. 152) and Fig. 6.55 (p. 153) redrawn if sections 8 to 10 are extended; both carry all their labels inside the artwork.

Where this sits in the book

NCERT Ganita Manjari, Class 9 Mathematics (NCF-SE 2023), Chapter 6, §6.10.1 "Area of Sector of a Circle" (pp. 146–147), including the printed bold heading that introduces the general formula, the remark on rotational symmetry, and the closing definition of a segment.
Figures 6.38 (p. 146), 6.39 and 6.40 (p. 147).
Exercise Set 6.3 in full (p. 148), Q1 to Q10, with the instruction at its head fixing π as 22/7 unless a question overrides it — Q4 and Q5 both do.
End-of-chapter exercises Q8 (p. 150), Q21 with Fig. 6.49 (p. 152) and Q27 with Fig. 6.55 (p. 153).
Chapter Summary (p. 154) carries the sector-area bullet with θ named as the central angle.
The parallel argument for arc length is §6.4 (pp. 125–126), handled in Arc length as the central angle's share of the circumference; the disc's own area is §6.10 (pp. 143–146), handled in Slicing a disc into sectors to see where πr² comes from.