Perimeter puzzles: composite curved boundaries reduce to arcs you already know

What to assume they know

• Arc length as the central angle's share of the circumference — arc length as θ/360 of the circumference
• That a triangle with three equal sides has three 60° angles
• Angles about a point on one side of a line summing appropriately, and adding two adjacent angles
• The relation between a semicircle's diameter and its arc length
• That lengths along a straight line add

What they should be able to do

• Decompose a composite curved boundary into arcs, stating each arc's radius and its fraction of a full circle
• Recognise when the geometry of a figure fixes an angle that was not given, and say which property supplied it
• Work Example 1: two equal circles each through the other's centre, and find the outer boundary in terms of the radius
• Work Example 2: compare one semicircle against three semicircles standing on parts of the same segment, and prove the answer independent of how the segment is divided
• Explain why the second result survives any number of sub-semicircles, not just three
• Read each of the nine shapes of Fig. 6.14 correctly, identifying every arc's radius and share
• Compute the total petal boundary in the square and hexagon flower figures
• Explain why arcs centred at the midpoints of a square's sides meet at the square's centre

Where it usually goes wrong

• "Curved boundaries need a new formula." They need the same formula used more than once. Every one of these figures is a list of arcs, and the only skill is writing the list down honestly.
• "The dotted arcs in Fig. 6.12 must be included somehow." The question says to leave them out (p. 127), and they are dotted for that reason. They are still doing work in the solution, though — they are how the 120° gets counted. Ignored in the total, essential in the argument.
• "Where did the 60° come from? Nobody gave an angle." Three radii of equal length made a triangle equilateral. This is the single move the example is built to teach: a length condition delivering an angle.
• "The bumpy route must be longer — look at all that extra wiggling." The wiggling is up and down, and a semicircle's length is fixed by its diameter alone. Get the class to vote before the algebra; the vote is the lesson.
• "It works out equal because the three are equal." They are not required to be. The chapter's own figure draws them unequal, and the algebra never assumes otherwise. Any split of PQ gives the same total.
• "The petals' arcs are semicircles." In Fig. 6.15A they are quarter circles of radius 7 cm, not semicircles of any radius; in Fig. 6.15B they are sixths of a circle of radius 42 cm. Getting the share wrong is the whole of the error in these two questions.
• "14 cm in Fig. 6.14(v) is the side of the whole square." It is one cell of the three-by-three grid, so the outer dashed square is 42 cm across. See Notes — this reading is added here and it is the one judgement in this brief a reviewer should check.

Questions to check understanding

• Find the perimeter of a composite figure made of straight pieces and arcs, given its dimensions
• Identify the share each arc in a given figure takes — 90°, 180° or 270° — and state its radius
• Find the total boundary of a petal figure built from arcs centred at named points
• Prove that two routes between the same pair of points have equal length, where one is a single semicircle and the other several
• Deduce an angle in a figure from equal lengths, and use it to find an arc
• Multiple-choice with three options, in the chapter's own Example 2 style, where the correct answer is "equal"
• Extension: state and prove the general version of Example 2 for n semicircles

Examples worth working on the board

Inputs, not answers, except where the chapter itself prints the result. Values marked Verified are worked out here; the chapter prints no answer key.

• Example 1 (pp. 127–128), Fig. 6.12. Two circles of equal radius r, centred A and B, each passing through the other's centre, meeting at C above and D below. The outer boundary — the two long arcs — is drawn in red; the two inner arcs, which lie inside the other circle, are dotted. The question asks for the total length of the outer boundary, in terms of r, and says the dotted parts are to be ignored.
  • The chapter's argument: AB, AC and BC are all radii, so triangle ABC is equilateral and the angle at A is 60°; the same below gives 60° again; so the dotted arc of each circle subtends 120° and is one third of its circle.
  • Printed conclusion: the red boundary totals 8πr/3. Verified: each red arc is 240°, so two-thirds of 2πr, and two of them give 8πr/3.
• Example 2 (p. 128), Fig. 6.13. Points P and Q with two routes between them. The first route is a single semicircle a arching above PQ. The second is three semicircles: b below the line on the left, c above the line in the middle, d below the line on the right. The reader is asked which route is longer, with three printed options — the first, the second, or neither — and told to answer before reading on.
  • The chapter's argument in its own notation: with radii a′, b′, c′, d′, the first route has length πa′ and the second π(b′ + c′ + d′). Since PQ = 2a′ and also 2b′ + 2c′ + 2d′, the radii add, so the lengths are equal.
  • Printed conclusion: equal. Verified, and worth saying out loud: the equality holds for any number of sub-semicircles and any split, because π factors out of a sum of diameters.
• A concrete instance. PQ = 12 cm. Big semicircle arc = 6π. Three small ones of diameters 5, 4 and 3 cm: arcs 2.5π + 2π + 1.5π = 6π. Verified equal. Then redo with 8, 3, 1 and get 6π again. Two instances make the independence visible before the algebra explains it.
• Fig. 6.14 (p. 129), the nine shapes of Exercise Set 6.1 Q5. The question restricts every arc in the nine figures to one of three shares — a quarter, a half, or 270° — whichever fits, and π is 22/7 unless stated. Added readings from the printed page, with every printed dimension:
  • (i) A rectangle with a semicircular cap on each short end. The vertical dashed line inside is labelled 60 m and the horizontal one 80 m. So the straight sides are 80 m each and the caps are semicircles of diameter 60 m.
  • (ii) A semicircular arch: outer semicircle on a base of 12 cm, inner semicircle of 8 cm cut out of it, and two straight feet of 2 cm each closing the figure at the bottom.
  • (iii) A square of side 10 cm with an outward semicircle on each of the four sides.
  • (iv) An equilateral triangle of side 12 cm with an outward semicircle on each of the three sides.
  • (v) A dashed square ruled into a three-by-three grid, one cell measuring 14 cm. Each side carries an outward semicircle of diameter 14 cm on its middle third, and each corner is replaced by a quarter arc of radius 14 cm centred at the nearest inner grid point.
  • (vi) A base line of 28 cm with tick marks at each quarter, so four 7 cm pieces. One large semicircular arc spans the whole 28 cm above the line, and four semicircles of diameter 7 cm alternate below and above the line along it.
  • (vii) A right-angled triangle with legs 8 cm and 6 cm and hypotenuse 10 cm, drawn dashed with the right angle marked, and a semicircle standing outward on each of the three sides.
  • (viii) A base of three 4 cm pieces, so 12 cm; one large semicircle over the whole 12 cm and three semicircles of diameter 4 cm sitting under it.
  • (ix) A base of two 10 cm pieces, carrying three arcs, not two. Over the whole 20 cm base sits one large semicircle; inside it, a semicircle stands on the left 10 cm and a third hangs below the right 10 cm. The two printed dots on the base are the centres of those two small ones. The S-curve is only the inner boundary — the outer edge of the region is the big semicircle, and reading the shape as an S alone loses it.
  • Verified perimeters, for anyone checking arithmetic added here: (i) 160 + 60π; (ii) 10π + 4; (iii) 20π; (iv) 18π; (v) 8 × 22 = 176 cm exactly, since with π = 22/7 both a 14 cm semicircle and a 14 cm quarter circle measure 22 cm; (vi) 14π + 14π = 28π; (vii) 12π; (viii) 12π; (ix) 10π + 5π + 5π = 20π, which is 440/7 ≈ 62.86 cm.
  • (viii) is Example 2 with numbers in it. Three small semicircles under one big one, and the totals are 6π and 6π. Put them side by side; it is the cheapest reinforcement in the chapter.
  • (ix) makes the same point a second time, and more sharply. It has the same structure as (viii) and as Fig. 6.13 with one small semicircle flipped below the line, so the single outer arc measures 10π and the two inner arcs measure 5π + 5π — equal again. Flipping an arc below the base changes the shape entirely and changes the boundary length not at all, which is worth a beat.
• Exercise Set 6.1 Q7 (p. 130), the two flowers.
  • Fig. 6.15A. A dashed square with 14 cm marked on a side, and four petals drawn inside it whose arcs are centred at the midpoints of the sides. The four petals meet at the square's centre and each points at a corner. Verified: each arc is a quarter circle of radius 7 cm; each petal has two such arcs; four petals give eight arcs, 8 × 11 = 88 cm with π as 22/7.
  • Fig. 6.15B. A dashed regular hexagon with 42 cm marked on a side, and six petals whose arcs are centred at the hexagon's vertices. The petals meet at the centre and each reaches a vertex. Verified: each arc is a sixth of a circle of radius 42 cm, so 44 cm with π as 22/7; twelve arcs give 528 cm.
  • Why the arcs meet where they do, which neither caption says: an arc centred at the midpoint of a square's side with radius half the side passes through the two nearer corners and through the square's centre; in the hexagon, an arc centred at a vertex with radius equal to the side passes through both neighbouring vertices and through the centre, because a regular hexagon's circumradius equals its side.

Figures to have open

• Fig. 6.12 (p. 128), redrawn: two equal circles through each other's centres, with the outer boundary in one colour and the inner arcs dotted, and the equilateral triangle appearing on cue.
• Fig. 6.13 (p. 128), redrawn, with the three small semicircles drawn visibly unequal so nobody suspects the answer depends on symmetry. Then a second version with a different split.
• All nine shapes of Fig. 6.14 (p. 129) redrawn as clean line figures with their printed dimensions. Every dimension in the printed figure sits inside the artwork, so these must be rebuilt from the readings above rather than lifted.
• Fig. 6.15A and Fig. 6.15B (p. 130) redrawn, each with the construction circles shown faintly so that a petal is visibly the overlap of two of them. This is the step that makes the quarter and the sixth obvious rather than asserted.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics (NCF-SE 2023), Chapter 6, §6.5 "Problems, Puzzles, and Paradoxes on Perimeter", opening on p. 127 and running through Example 1 and Example 2 to the end of p. 128.
• Figures 6.12 and 6.13, both on p. 128, with their labels.
• Exercise Set 6.1 in full (pp. 129–130): Q5 with Fig. 6.14 (i) to (ix) on p. 129, and Q7 with Fig. 6.15A and Fig. 6.15B on p. 130. Q1 to Q4, Q6 and Q8 belong to the two earlier topics of this chapter.
• The instruction at the head of Exercise Set 6.1 (p. 129) fixing π as 22/7 unless a question overrides it.
• Related but later: end-of-chapter Q22 (p. 152, Fig. 6.50) is the same four-petal construction on a square of side 2 units and asks for area as well as perimeter, and Q25 (p. 153, Fig. 6.53) is Example 1's configuration again, asked for area. Both are handled in Slicing a disc into sectors to see where πr² comes from.
• Arc length itself is §6.4 (pp. 125–126), handled in Arc length as the central angle's share of the circumference.