NCERT Solutions for Class 9th Maths Chapter 5 I’m Up and Down, and Round and Round
Updated on 2026-09-19
About this chapter
A circle is the locus of points at a fixed distance (the radius) from a fixed point (the centre). Every diameter is a line of reflection symmetry, and the circle has rotational symmetry about its centre through any angle. Infinitely many circles pass through two given points A and B; all their centres lie on the perpendicular bisector of AB. Through three non-collinear points there is exactly one circle — the circumcircle. Equal chords ⇔ equal angles at the centre; equal chords ⇔ equal distances from the centre. The longer of two unequal chords is the nearer one, and the diameter is the longest chord of all. The line from the centre to the midpoint of a chord is perpendicular to it, and the perpendicular from the centre bisects the chord. This gives the workhorse formula chord = 2√(r² − d²
- Chapter opening
- Definitions
- Symmetries of a Circle
- How Many Circles?
- Chords and the Angles They Subtend
- Midpoints and Perpendicular Bisectors of Chords
- Distance of Chords from the Centre
- Which of the two unequal chords is farther from the centre?
- Angles Subtended by an Arc
- Concyclicity of Points
Quick revision
| Result | What it says | Reason in one line | Where you use it |
|---|---|---|---|
| Theorem 1 | Exactly one circle passes through three non-collinear points | The perpendicular bisectors of AB and AC meet at exactly one point | Drawing a circumcircle |
| Theorems 2 & 3 | Equal chords ⇔ equal angles at the centre | SSS (and SAS) on the two radius–chord triangles | Comparing two chords |
| Theorems 4 & 5 | Centre → midpoint of a chord is ⊥ to it; ⊥ from the centre bisects the chord | ΔCMA ≅ ΔCMB, and ∠CMA + ∠CMB = 180° | Splitting a chord in half |
| Theorems 6 & 7 | Equal chords ⇔ equidistant from the centre | RHS on ΔCEA and ΔCHF (half-chords equal, radii equal) | Parallel-chord problems |
| Theorem 8 | Of two unequal chords the longer is nearer the centre | r² = d² + (half-chord)², so d shrinks as the chord grows | Ordering chords by distance |
| Chord formula | chord = 2√(r² − d²) | Baudhāyana–Pythagoras in the half-chord right triangle | Almost every numerical question |
| Theorem 9 | Angle at the centre = 2 × angle at a point of the circle outside the arc | Exterior-angle theorem applied to two isosceles triangles | Finding unknown angles |
| Corollary | The angle in a semicircle is 90° | Half of the straight angle 180° subtended by a diameter | Spotting right angles in a circle |
| Theorem 10 | Equal angles at two points on the same side of AB ⇒ all four points concyclic | D inside or outside the circle both give a contradiction | Proving points concyclic |
| Theorems 11 & 12 | Opposite angles of a cyclic 4-gon add to 180° (and the converse) | The two halves add to ½ of the complete angle 360° | Cyclic quadrilateral angle chases |
Exercises
- Chapter opening — In-text Questions Page 92–93
- Definitions — Think and Reflect Page 93
- Think and Reflect — Symmetries of a Circle Page 94
- Think and Reflect — How Many Circles? Page 95
- In-text Questions — How Many Circles? Page 96
- Exercise Set 5.1 — How Many Circles? Page 98
- Think, Draw and Infer — How Many Circles? Page 98
- Exercise Set 5.2 — Chords and the Angles They Subtend Page 100
- Exercise Set 5.3 — Midpoints and Perpendicular Bisectors of Chords Page 101
- In-text Questions — Distance of Chords from the Centre Page 102
- Exercise Set 5.4 — Distance of Chords from the Centre Page 104
- In-text Questions — Which of the two unequal chords is farther from the centre? Page 104
- Exercise Set 5.5 — Which of the two unequal chords is farther from the centre? Page 105–106
- In-text Questions — Angles Subtended by an Arc Page 107
- Exercise Set 5.6 — Angle subtended by an arc at a point on the circle outside the arc Page 110–111
- In-text Questions — Concyclicity of Points Page 113
- Whole chapter — End-of-Chapter Exercises Page 114–116