Length and distance from the centre are the same fact twice (Theorems 6–8)

What to assume they know

• The centre-to-midpoint line is perpendicular, and the converse (Theorems 4–5) — the perpendicular from the centre bisects a chord, and the converse; what "distance from the centre to a chord" means
• Chords of equal length cut off equal central angles, and the converse (Theorems 2–3) — the SSS argument on the chord-plus-centre triangle
• Total rotational symmetry, and why every diameter is an axis of reflection — rotational symmetry about the centre
• The Pythagorean relation among the sides of a right triangle — the chapter calls it the Baudhāyana–Pythagoras theorem
• SSS and RHS congruence; that congruent triangles have equal altitudes
• Comparing two positive quantities through their squares

What they should be able to do

• State what a chord's distance from the centre means, and say why the perpendicular is the only distance worth measuring
• Carry out the two-fold paper activity and identify the crease that gives the chord and the crease that gives the perpendicular
• Explain why a demonstration on many drawn chords does not settle a general claim, citing the chapter's own statement of that principle
• Prove that chords of equal length lie equally far from the centre, both by congruent whole triangles and by RHS on the half-triangles
• Prove the converse, and say where the chapter locates that proof
• Derive r² = d² + (half-chord)² and use it in either direction
• Prove that of two unequal chords the longer lies nearer the centre, and identify the step where squares are compared
• State the two extreme cases: the chord through the centre, and the chord pushed out to a single point
• Explain why doubling a chord's distance from the centre neither halves nor doubles its length

Where it usually goes wrong

• "A chord's distance from the centre means its distance to one of the ends." It means the perpendicular distance, and the chapter says so explicitly when it states Theorem 8. The distance to an end is always the radius and tells you nothing.
• "Longer chord, farther from the centre." Exactly backwards, and it is the most common error on this topic. Theorem 8 and Table 1 both exist to break it. The intuition comes from thinking of the chord as being "further round" the circle.
• "Chord length is proportional to distance from the centre, or inversely proportional to it." Neither. The relation is through squares. Starred Q3 is built to catch this.
• "The demonstration with tracing paper settles Theorem 6." The chapter itself says it does not. Reproducing the demonstration and stopping there teaches students that seeing several cases is proof, which is precisely what this page is written to prevent.
• "The two proofs of Theorem 6 are the same proof." They are not: one compares the whole triangles and uses the equality of corresponding altitudes; the other splits each chord at its midpoint and uses RHS on the halves. The second needs Theorem 5 first, and saying so shows the results stacking.
• "Every chord has a well-defined distance and a well-defined length, so both can be chosen freely." On a fixed circle, choosing one fixes the other. Two free numbers is one too many.
• "A chord of length zero is not a thing." The chapter's own Comment treats the collapse to a point as the end of the range. Present it as a limiting case and say so, rather than letting students think a point is a chord.
• "Same side and opposite side are interchangeable for parallel chords." End-of-Chapter Q18 says same side and Exercise Set 5.3 Q3 says opposite; the arithmetic differs. Read the words.

Questions to check understanding

• Given a radius and a chord's distance from the centre, find the chord; and the reverse
• Given a chord and its distance, find the radius
• Prove that equal chords are equally far from the centre, and prove the converse
• Prove that of two unequal chords the longer is nearer the centre
• Justify the chord-length formula 2√(r² − d²)
• Parallel chords of given lengths, same side or opposite sides: find the radius or the separation
• Explain why no chord can be longer than a diameter
• Find the shortest chord through a given interior point, with a reason
• Identify the locus of the midpoints of all chords of one fixed length
• Disprove a stated proportionality between distance and length by producing a counterexample

Examples worth working on the board

The chapter prints no answers, so every value marked verified is worked out here on the chapter's own stated inputs.

• Fig. 5.13 A, B, C (p. 102). Three photographs of a white paper disc against a dark ground. A: the plain disc, with a small dot marking the centre. B: the disc after one inward fold has been opened, the crease visible as a chord. C: after a second fold that brought the chord's ends together, so two creases now cross. The activity's instructions are printed alongside, and the measurements it asks for are the two pieces of the chord, the angle between the creases, and the distance from the centre to the crossing point.
• Fig. 5.14 (p. 103, read as the printed page). Theorem 6's figure. Circle with centre C. Chord BA runs across the top with E marked on it; chord GF runs down the left with H marked on it; the segments CB, CA, CE, CG, CF and CH are all drawn. There is a further labelled point D on the circle, above the chord BA, with no segment drawn to it and no role in the argument — read off the printed page, and worth telling the teacher, who will otherwise redraw D and then have to explain it. Given: AB = FG, with E and H the midpoints. To show: CE = CH.
• The two arguments for Theorem 6, as data. First: CA = CF and CB = CG, all radii, plus the given AB = FG — three side pairs, so SSS makes triangles CAB and CFG congruent, and CE and CH are corresponding altitudes, hence equal. Second: E and H bisect the chords by Theorem 5, so AE = FH; the angles at E and H are right angles; and CA = CF as radii — so RHS makes triangles CEA and CHF congruent and again CE = CH. Two routes to the same conclusion, and the chapter gives both deliberately.
• Fig. 5.15 and Exercise Set 5.4 (p. 104). Q1 asks for Theorem 6 by way of the Baudhāyana–Pythagoras theorem. Q2 supplies Fig. 5.15 — the same arrangement as Fig. 5.14, with B, E, A across the top, G on the left, H and F below, right-angle squares at both E and H, and the centre C — both marks matter, since Q2 gives both perpendicularities — and asks, given the two perpendiculars equal, for the two chords to be equal. Q3 asks for the same by Baudhāyana–Pythagoras. The chapter then states Theorem 7 and says Q2 established it. Verified: the shortest route is RHS on triangles CEA and CHF, giving AE = FH and hence AB = GF.
• Table 1 (p. 104). Printed as a two-row table with three empty columns; the row labels are "Length of Chord" and "Distance from Centre". The student draws chords of various lengths, drops the perpendicular from the centre to each, and fills the table. Nothing is filled in. Verified, as a filling the explanation can offer for a circle of radius 5: lengths 10, 8, 6, 4 against distances 0, 3, 4, √21 ≈ 4.58 — and the pattern is visible in one glance, which is the point of the activity.
• Fig. 5.16 (p. 105). Theorem 8's figure. Circle with centre C; chord AB with A at the top and B on the left, F its midpoint; chord DE on the right with G its midpoint; the perpendiculars CF and CG drawn. Given: AB > DE. To show: CF < CG.
• The Theorem 8 argument, as data. AC = CD, both radii. By Baudhāyana–Pythagoras, AC² = CF² + AF² and CD² = CG² + GD². Since the two hypotenuses are the same radius, CF² + AF² = CG² + GD². AB > DE gives AF > GD, because F and G bisect the two chords. So AF² > GD², so CF² < CG², so CF < CG. The step to slow down on is the last one: comparing squares tells you about the quantities themselves only because both are positive lengths.
• The Comment (§5.6.1, p. 105). Two extremes, both stated by the chapter: the chord that contains the centre sits at distance zero, so no chord beats the diameter for length; and pushing a chord away from the centre eventually collapses it to a single point of length zero at distance equal to the radius. Between those the whole family lives.
• Exercise Set 5.5 (pp. 105–106). Q1 inputs: radius 7 cm, perpendicular distance 6 cm; find the chord. Verified: 2√(49 − 36) = 2√13 ≈ 7.21 cm. Q2 asks the student to justify the general formula, printed on the page as 2√(r² − d²). Starred Q3 inputs: chord AB is twice as far from the centre as chord CD; may we conclude CD = 2 AB? Verified: no. On a circle of radius 5 take CD at distance 1 and AB at distance 2: CD = 2√24 ≈ 9.80 and AB = 2√21 ≈ 9.17, a ratio of about 1.07. Worse, take CD at distance 3 and AB would have to be 6 units from the centre, which exceeds the radius, so no such chord exists at all. Both counterexamples are not in the book; the chapter asks for reasons and prints none.
• End-of-chapter items this topic owns (pp. 114–115). Q1: chord 5 cm from the centre, radius 13 cm — verified: 24 cm. Q3: diameter 26 cm, chord 24 cm — verified: distance 5 cm. Q4: radius 15 cm, distance 9 cm — verified: chord 24 cm. Q9: chord 16 cm at distance 6 cm — verified: radius 10 cm. Q18: parallel chords 10 cm and 24 cm on the same side, 7 cm apart — verified: radius 13 cm. Starred Q16: the midpoints of all chords of one fixed length — verified: they form a circle concentric with the given one, of radius √(r² − ℓ²/4). Starred Q22: no chord beats the diameter — this is the Comment, set as a question. Starred Q23: through an interior point A, the shortest chord is the one at right angles to OA — verified, and it is Theorem 8 doing the work: that chord is the one whose distance from the centre is as large as OA, and no chord through A can be farther.

Figures to have open

• Fig. 5.13 A–C, or a redrawn fold sequence. The three-panel progression is the chapter's own photography (p. 102) and it is what makes "distance from the centre" a thing you can point at rather than define.
• Fig. 5.14 redrawn. Two equal chords, all four radii, both midpoints, both perpendiculars. Whether or not the stray point D is kept, the redraw must not give it a segment it does not have in the book.
• Fig. 5.16 redrawn: two unequal chords with their perpendiculars, and the two right triangles that share the radius picked out. This is the argument of Theorem 8 and the shared hypotenuse is the thing to make visible.
• A filled Table 1 for one stated radius, shown step by step so the two rows move in opposite directions. Standard schematic built on the chapter's own empty table (p. 104).
• A single annotated right triangle carrying r, d and the half-chord, reused across sections 9, 10 and 11. Standard schematic; the formula is printed at Exercise Set 5.5 Q2, p. 105, and no figure accompanies it.
• The starred-Q3 counterexample panel: two chords at distances 1 and 2, with their measured lengths. Standard schematic; not in the book.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 5, "I’m Up and Down, and Round and Round", §5.6, printed heading "Distance of Chords from the Centre", pp. 102–104, and §5.6.1, printed as a question about which of two unequal chords lies farther out, pp. 104–106.
• Figs. 5.13 A, B and C, p. 102. Fig. 5.14, p. 103. Fig. 5.15, p. 104. Fig. 5.16, p. 105.
• Theorem 6 with its two arguments, pp. 103–104. Theorem 7 as stated, p. 104. Theorem 8 with its argument, p. 105. The Comment on the two extreme chords, p. 105.
• Activity and Table 1, p. 104. Exercise Set 5.4, Q1–Q3, p. 104. Exercise Set 5.5, Q1–Q2 on p. 105 and starred Q3 on p. 106.
• End-of-Chapter Exercises Q1, Q3, Q4, Q9, starred Q16, Q18, starred Q22 and starred Q23, pp. 114–115.
• Chapter Summary, p. 117 — the bullet pairing equal chords with equal distances and its converse, and the bullet on the longer of two unequal chords.