The arc's length by the same share argument

What to assume they know

• A sector's area as its share of the full turn — the unitary derivation for a sector's area, and the two properties it rests on
• Sector versus segment, and which one major and minor refer to — sector, segment, arc, chord, and the major/minor convention
• Circumference of a circle as 2πr, with π ≈ 22/7 and π ≈ 3.14
• That an arc is part of the circle's curve, not part of the filled region
• Units: that a length is reported in cm and an area in cm², and that the two are never interchangeable

What they should be able to do

• State the two properties that let the share argument be re-used, and check each against arc length rather than assuming it
• Derive the arc-length rule from the circumference by the unitary route
• Compute an arc's length from a radius and an angle in degrees
• Show that the arc-to-circumference ratio, the sector-to-disc ratio and the angle-to-360 ratio are one and the same number
• Distinguish an arc from the boundary of the sector it belongs to, and compute each
• Use powers of r to decide whether a given expression can be a length or an area
• Recover an unknown angle or radius from a given arc length
• Assemble a total wire length from several circular and straight pieces

Where it usually goes wrong

• "Arc length and sector area are different formulas to memorise." They are one derivation applied to two anchor values. Memorise the fraction θ/360 and the two anchors, and there is nothing left to remember.
• "The arc is the boundary of the sector." It is one of three edges. The other two are radii, and a perimeter question wants all three.
• "θ/360 × 2πr and θ/360 × πr² look alike, so either will do." One is reported in cm and the other in cm². Reading the units off the answer catches this particular confusion — but do not sell it as the whole trick to question 14, because it only gets a student halfway there. Two of that question's wrong options are lengths and fall to the units check; the third is an area, and the only thing separating it from the right answer is the number underneath.
• "An arc is a curved region." An arc has no width and no area. The region between the arc and the chord is the segment, and it is a different object with a different topic.
• "Doubling the radius doubles both." It doubles the arc and quadruples the area. The two rules disagree about radius even though they agree about angle, and that is the sharpest way to feel that they are not the same formula.
• "Five diameters means five extra wires and five sectors." Five diameters cut ten wedges, because each diameter is two radii. Miscounting here is the intended difficulty of question 9.
• "For the brooch, use the diameter as the radius." 35 mm is across, not out from the centre. Every number in that question moves if the halving is missed.
• "Half the arc times the radius is a formula I was taught, so it is examinable." It is a consequence, not a printed result. Derive it and label it as a consequence; the marks are awarded for the printed rule.

Questions to check understanding

• Given radius and angle, find the arc's length
• Given the arc's length and the radius, recover the angle; or given arc and angle, recover the radius
• Find the whole way round a sector, arc and radii together
• Given a circumference, recover the radius before answering anything else
• Assemble a total length from curved and straight pieces, as in the brooch
• Multiple choice distinguishing a length expression from an area expression
• Reasoning: why the same fraction appears in both results, argued from the unitary route rather than asserted

Examples worth working on the board

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

• Where the arc rule sits on the page (§11.1, p. 155). The chapter raises the arc as a follow-on question, runs the same unitary route with 2πr in place of πr², and sets the result in bold at the foot of the page. The whole derivation is four lines long, and its brevity is the argument: nothing new had to be introduced.
• A caution about the figure beside it (read from p. 155 and confirmed on the printed page). The figure printed alongside the arc paragraph is Fig. 11.4, and the arc paragraph itself carries no explicit pointer to it. Fig. 11.4 shows a circle with centre O, the two radii drawn to A and B with r and θ marked, the chord AB drawn across, and the sliver between the chord and the near arc shaded — P sits on that near arc, Q on the far one. It is claimed by name on the following page, where it illustrates the segment. So the one shaded thing next to the arc rule is not the arc. Redraw it for this topic with the arc APB inked heavily and the chord and shading removed; otherwise students copy the shading and answer segment questions with the arc formula.
• The derivation, as lines to show (p. 155). Whole turn ↔ 2πr. One degree ↔ 2πr/360. θ degrees ↔ (θ/360) × 2πr. Set it beside the area derivation from the previous topic, line for line, and change only the anchor value. The parallel is the lesson.
• Exercise 11.1 question 5 (p. 158) — radius 21 cm, an arc opening 60° at the centre, with three parts: the arc's length, the wedge's area, and the area of the piece the chord cuts off. π is 22/7 by the instruction at the head of the exercise. Verified: the arc is 22 cm and the wedge is 231 cm². Use this one question as the spine of sections 6, 9 and 10, because it is the only place in the chapter where a student is made to produce a length and an area from the same two inputs. Its third part belongs to A segment as what is left when the triangle is taken away.
• The three ratios, on those numbers. Verified: the circumference is 132 cm and the arc is 22 cm, a sixth of it; the disc is 1386 cm² and the wedge is 231 cm², a sixth of it; and 60/360 is a sixth. Show the three fractions at once. This is the cheapest possible demonstration that the two rules are one idea.
• The boundary of that same wedge. Verified: going the whole way round it is 22 + 21 + 21 = 64 cm, not 22 cm. Nothing in Exercise 11.1 asks for this, but board papers do, and a student who has only ever seen the arc rule reaches for 22.
• The identity between the two rules. Verified by an added algebra: half the arc times the radius, ½ × (θ/360)(2πr) × r, collapses to (θ/360)πr², which is the area rule exactly. Check it on question 5's numbers: ½ × 22 × 21 = 231. This is worth a section on its own, because it says a thin wedge behaves like a triangle whose base is the arc and whose height is the radius — and that is the intuition a student will need again in any later course. The identity is added here; the chapter states the two rules and never multiplies them together.
• Exercise 11.1 question 9 and Fig. 11.9 (p. 159). A brooch made from silver wire: the circle has diameter 35 mm, and further wire forms 5 diameters, which between them cut the disc into 10 equal wedges. The question wants the total wire and the area of one wedge. Fig. 11.9 is a decorative drawing — a double-ringed circle with ten spokes and small ornaments in each wedge; the drawing carries no lettering and no measurements, so every number must come from the words. Verified: the radius is 17.5 mm; the rim takes 110 mm of wire, the five diameters take 175 mm, so 285 mm in all; each wedge spans 36° and measures 96.25 mm². The interest of the question is that it mixes a curved length with straight lengths in one total, and that the 10 wedges are counted from the 5 diameters rather than given.
• Exercise 11.1 question 14 (p. 159). Of the four offered expressions, two carry R to the first power — those are lengths, and one of them is precisely this topic's rule with a 360 under it. The other two are both areas, so the choice between them turns on the divisor alone. Verified by an added algebra: the option reading p/360 × 2πR is the arc length, not the area; the option reading p/180 × πR² is an area but comes to exactly twice the right one; the option reading p/720 × 2πR² simplifies to (p/360)πR² and is the answer. Work this question here as well as in the area topic, from the opposite side.
• Question 2's hidden use (p. 158). A circumference of 22 cm is given and a quadrant's area wanted. Verified: the radius comes out at 3.5 cm. The circumference is being used backwards — this topic's anchor value read as an equation to solve rather than a formula to evaluate.

Figures to have open

• A clean redraw of the p. 155 arc figure with the arc APB inked heavily, the radii thin, and no chord and no shading — deliberately unlike the printed Fig. 11.4, for the reason given above.
• A wheel of radius r unrolling along a line to leave a track of length 2πr, with a partial roll of θ degrees leaving the arc. Not in the book; this is the asset that makes the anchor value concrete.
• The two derivations side by side as a two-column table, differing only in the anchor. Not in the book.
• One circle of radius 21 cm carrying, simultaneously, the 60° wedge, its arc measured as 22 cm and its area as 231 cm², with the sixth-of-the-whole fraction shown for each. Not in the book; assembled from Exercise 11.1 question 5.
• Fig. 11.9's brooch (p. 159) reduced to a plain disc with five diameters drawn through it, ten wedges numbered. The printed drawing is ornamental and carries no data; a schematic serves better and avoids reproducing the art.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 11 "Areas Related to Circles", §11.1, p. 155 — the closing half-page, from the question about the arc to the bold statement of the result, with Fig. 11.4 printed beside it
• p. 156, where Fig. 11.4 is claimed by name for the segment — the reason this topic must not inherit its shading
• Exercise 11.1, pp. 158–159 — questions 2, 5 (i), 5 (ii), 9 and 14
• §11.2 Summary, p. 160, item 1, which states the arc result first, ahead of the area it was derived alongside