NCERT Solutions for Class 9th Maths Chapter 5 Exercise Set 5.4 — Distance of Chords from the Centre

Book page 104 Updated on2026-09-08

Q1.

Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.

Answer

Theorem 6: chords of a circle having the same length are at the same distance from the centre.

Given: circle with centre C and radius r; chords AB and FG with AB = FG; E and H are the feet of the perpendiculars from C to AB and FG. To show: CE = CH.

By Theorem 5 the perpendicular from the centre bisects the chord, so
AE = ½ AB and FH = ½ FG
AB = FG (given) ⇒ AE = FH

ΔCEA is right-angled at E, so by Baudhāyana–Pythagoras:
CE² = CA² − AE² = r² − AE²
Likewise CH² = CF² − FH² = r² − FH²

Since AE = FH, the two right-hand sides are equal:
CE² = CH² ⇒ CE = CH (lengths are positive)

Why this proof is worth having: the congruence proof in the text uses SSS or RHS; this one turns the whole theorem into a single equation. Written as a formula it says d = √(r² − (ℓ/2)²): the distance depends on the chord length ℓ and on nothing else. Equal ℓ forces equal d — and, read backwards, it also gives Theorem 7 and Theorem 8 immediately.

Q2.

Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.

Answer

(In Fig. 5.15, C is the centre; AB and GF are chords, E lies on AB and H lies on GF. The condition is that CH is perpendicular to GF — the chord — as the figure shows.)

Given: CE ⊥ AB, CH ⊥ GF, CE = CH. To show: AB = GF.

In ΔCEA and ΔCHF:
∠CEA = ∠CHF = 90° (given)
CA = CF = r (radii — the hypotenuses)
CE = CH (given)
By the RHS congruence rule, ΔCEA ≅ ΔCHF
⇒ AE = FH (corresponding sides)

By Theorem 5, E and H are the midpoints of AB and GF, so
AB = 2 AE and GF = 2 FH
∴ AB = 2 AE = 2 FH = GF

Why the midpoint step is essential: the congruence only gives us equal half-chords. It is Theorem 5 — the perpendicular from the centre bisects the chord — that turns AE = FH into AB = GF. This exercise proves Theorem 7, the converse of Theorem 6.

Note: the printed question says "CH is perpendicular to GH". Since H lies on the chord GF, the segment intended is GF; Fig. 5.15 marks the right angle at H on the chord GF.

Q3.

Solve the previous question using the Baudhāyana–Pythagoras theorem.

Answer

Given: CE ⊥ AB, CH ⊥ GF, CE = CH, with C the centre and radius r. To show: AB = GF.

ΔCEA is right-angled at E: AE² = CA² − CE² = r² − CE²
ΔCHF is right-angled at H: FH² = CF² − CH² = r² − CH²

Given CE = CH, so CE² = CH², hence
AE² = FH² ⇒ AE = FH

By Theorem 5, E and H bisect the chords:
AB = 2 AE and GF = 2 FH
∴ AB = GF

Tip: the whole of Sections 5.6 and 5.6.1 sits inside one identity — (½ chord)² + d² = r². Fix r. Then d decides the chord and the chord decides d. Equal chords ⇒ equal d (Theorem 6); equal d ⇒ equal chords (Theorem 7); bigger chord ⇒ smaller d (Theorem 8).

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