NCERT Solutions for Class 9 Maths Chapter 5 I M Up and Down and Round and Round – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.
Chapter 5 builds the circle up from one idea — every point of a circle is the same distance from its centre — and squeezes twelve theorems out of it. Chords, the angles they subtend, their distance from the centre, arcs and cyclic quadrilaterals all follow from that single definition plus the congruence rules you already know.
NCERT Solutions for Class 9 Maths Chapter 5 I M Up and Down and Round and Round PDF
All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.
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NCERT Solutions for Class 9 Maths Chapter 5 I M Up and Down and Round and Round – Questions and Answers
All 64 questions of the chapter (textbook pages 92 – 117) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
In-text Questions — Pages 92–93Chapter opening
All three pictures in Fig. 5.1 come from nature, and in each of them the shape is a circle.
- Left: raindrops striking still water. Each drop sends out a ripple that spreads outward at the same speed in every direction, so every point of the ripple stays at the same distance from the point where the drop fell — a circle.
- Middle: the cross-section of a tree trunk (a plant stem). The growth rings are nearly circular, because the stem thickens outward by roughly the same amount all round.
- Right: the head (inflorescence) of a sunflower. The florets are packed around a single centre, and the outer boundary is a circle.
Any object that spreads out equally in all directions from a centre will look circular. A few from around us:
- Ripples made by a stone dropped into a pond.
- The full Moon, and the Sun during a total solar eclipse.
- The annual rings on the cut end of a log; the cross-section of a bamboo or a sugarcane stem.
- The face of a sunflower; the flat top of a lotus seed-pod.
- Cross-sections of fruits — orange, lemon, tomato.
- A raindrop resting on a leaf; a soap bubble seen from above.
- The pupil and iris of an eye; the outline of a bird’s nest.
Think and Reflect — Page 93Before Section 5.1 Definitions
Amina almost certainly told her: fold the paper in half twice; the point where the two creases cross is the centre.
- Fold the circular paper so that the boundary falls exactly on itself. Open it. The crease is a line of reflection symmetry of the circle, so it is a diameter — and every diameter passes through the centre.
- Fold again in a different direction so the boundary again falls on itself. This crease is a second diameter.
- Two different diameters both pass through the centre, and two distinct lines meet at only one point. So their point of intersection is the centre.
Think and Reflect — Page 94Section 5.2 Symmetries of a Circle
A regular polygon with n sides has rotational symmetry of order n and exactly n lines of reflection symmetry.
Regular pentagon (n = 5): rotations of 72°, 144°, 216°, 288°, 360° — order 5; 5 lines of symmetry (each joins a vertex to the midpoint of the opposite side).
Regular hexagon (n = 6): rotations of 60°, 120°, 180°, 240°, 300°, 360° — order 6; 6 lines of symmetry (3 long diagonals, 3 through midpoints of opposite sides).
The longest chord is the diameter.
Is there a smallest chord? No. A chord drawn closer and closer to the edge of the circle becomes shorter and shorter, but its length never reaches 0 (then it would be just a point, not a chord). So there is no shortest chord.
The locus of points equidistant from two given points A and B is the perpendicular bisector of AB.
Part 1 (given in the hint): every point P with PA = PB lies on the perpendicular bisector of AB.
Part 2 (to show): every point of the perpendicular bisector is equidistant from A and B. Let Q be any point on the perpendicular bisector, meeting AB at its midpoint M. In ΔQMA and ΔQMB: QM common, ∠QMA = ∠QMB = 90°, AM = BM. By SAS, ΔQMA ≅ ΔQMB, so QA = QB.
So the locus is exactly the perpendicular bisector of AB.
Think and Reflect — Page 95Section 5.3 How Many Circles?
Infinitely many. Every point of the perpendicular bisector of AB can be the centre of a circle through A and B, and that line has infinitely many points.
No — not all radii are possible. Only radii of at least half of AB occur.
Smallest radius = c/2 = ½ AB, obtained when h = 0, i.e. when the centre is the midpoint of AB. For that circle AB is a diameter.
Largest radius: there is none. As h increases, r increases without bound, so the radii can be made as large as we like but never reach a maximum.
The radii increase.
h ↑ ⇒ h² ↑ ⇒ h² + (AB/2)² ↑ ⇒ r ↑
Less curved — the arc through A and B gets flatter and flatter, because the radius keeps increasing (Q3) and a bigger circle bends more gently.
(a) A and B anywhere on the boundary: infinitely many squares.
For example, place the segment AB along one side of a square: every side length greater than or equal to AB works. So there are infinitely many such squares (and many more tilted ones).
(b) A and B as corners (vertices) of the square: exactly 3 squares.
Case 2 — AB is a diagonal: the other two vertices are fixed, one on each side of AB → 1 square
Total = 2 + 1 = 3 squares
In-text Questions — Page 96Section 5.3 How Many Circles?
There is no circle at all through three collinear points.
Given: A, B, C on one line, all different. To show: no point O has OA = OB = OC.
OA = OB ⇒ O lies on the perpendicular bisector of AB
OB = OC ⇒ O lies on the perpendicular bisector of BC
But AB and BC lie along the same line ℓ.
So both perpendicular bisectors are perpendicular to ℓ ⇒ they are parallel to each other.
They pass through different points (the midpoints of AB and of BC are different), so they are distinct parallel lines — they never meet.
Hence no such point O exists. No circle passes through three collinear points.
Exercise Set 5.1 — Page 98Section 5.3 How Many Circles?
The centre lies inside the triangle.
Construction:
- Draw AB = 5 cm.
- At A draw a ray making 70° with AB; at B draw a ray making 60° with BA. They meet at C.
- Construct the perpendicular bisectors of AB and of BC. They meet at O.
- With centre O and radius OA, draw the circle. It passes through B and C as well.
Angles are 70°, 60°, 50° — all acute, so ΔABC is an acute-angled triangle.
For an acute-angled triangle the circumcentre lies inside the triangle.
The centre lies outside the triangle — on the far side of BC from A.
Construction: draw AB = 5 cm; at A draw a ray making 100° with AB and cut off AC = 4 cm on it; join BC. Then draw the perpendicular bisectors of AB and AC; they meet at O, outside the triangle. Draw the circle with centre O and radius OA.
For an obtuse-angled triangle the circumcentre lies outside the triangle,
beyond the side opposite the obtuse angle — here beyond BC.
OA = OB = OC ≈ 3.9 cm. All three must be equal — they are radii of the same circle.
Construction: draw AB = 6 cm; with centre A and radius 7 cm, and centre B and radius 7 cm, draw arcs cutting at C. Construct the perpendicular bisectors of AB and BC; they meet at O. Draw the circle with centre O, radius OA.
The least possible radius is ½ AB — half the distance between the two points.
No chord can exceed the diameter, so AB ≤ 2r, i.e. r ≥ AB/2.
The value r = AB/2 is actually reached: take the midpoint M of AB as centre.
Then MA = MB = AB/2, and the circle with centre M, radius AB/2, passes through both.
Least radius = AB/2, and for that circle AB is a diameter.
Think, Draw and Infer — Page 98Section 5.3 How Many Circles?
- No such point P exists. PA = PB puts P on the perpendicular bisector of AB, and PB = PC puts P on the perpendicular bisector of BC — but these two lines never meet (next answer).
- The perpendicular bisectors of AB and BC are parallel. Both are perpendicular to the same line ℓ that contains A, B and C, so they are parallel. They pass through different points (the midpoints of AB and BC), so they are two distinct parallel lines.
- No circle passes through three collinear points — its centre would have to be such a point P.
- No line cuts a circle in three distinct points — those three points would be collinear points on one circle, which is impossible.
Yes — infinitely many. Rotate ΔABC about the circumcentre O through any angle. The new triangle is congruent to ΔABC, and its vertices stay on the same circle. There are infinitely many angles to choose, so there are infinitely many such triangles.
Exercise Set 5.2 — Page 100Section 5.4 Chords and the Angles They Subtend
Given: a circle with centre O and a chord AB. Join OA and OB. To show: ΔOAB is isosceles.
OB = radius of the circle
∴ OA = OB
A triangle with two equal sides is isosceles, so ΔOAB is isosceles with base AB.
Consequence used again and again in this chapter: the base angles are equal, ∠OAB = ∠OBA.
Given: chords AB and DE of the same circle, centre C, with AB = DE. To show: ΔCAB ≅ ΔCDE.
CD = CE = r (radii)
⇒ CA = CD and CB = CE
AB = DE (given — the two bases are equal)
By the SSS congruence rule, ΔCAB ≅ ΔCDE
Exercise Set 5.3 — Page 101Section 5.5 Midpoints and Perpendicular Bisectors of Chords
Given: circle with centre C, chord AB, and CM ⊥ AB with M on AB. To show: AM = BM.
∠CMA = ∠CMB = 90° (given — CM is perpendicular to AB)
CA = CB = r (radii — these are the hypotenuses)
CM = CM (common side)
By the RHS congruence rule, ΔCMA ≅ ΔCMB
∴ AM = BM (corresponding sides) — CM bisects the chord AB
Given: ΔABC inscribed in a circle with centre O, and AB = AC. Let AD be the altitude from A to BC. To show: O lies on AD.
Step 1 — AD is the perpendicular bisector of BC. In ΔABD and ΔACD: AB = AC (given), AD common, ∠ADB = ∠ADC = 90°. So ΔABD ≅ ΔACD (RHS) and BD = DC.
Step 2 — O lies on the perpendicular bisector of BC. OB = OC (radii), so O is equidistant from B and C, and every such point lies on the perpendicular bisector of BC.
O lies on the perpendicular bisector of BC (Step 2)
∴ O lies on AD — the altitude from A passes through the centre
Distance = 7 cm.
Drop perpendiculars from the centre O to each chord. By Theorem 5 each perpendicular bisects its chord, so its foot is the midpoint of that chord.
d₁² = r² − 3² = 5² − 3² = 25 − 9 = 16 ⇒ d₁ = 4 cm
Chord PQ = 8 cm ⇒ half-chord = 4 cm
d₂² = r² − 4² = 5² − 4² = 25 − 16 = 9 ⇒ d₂ = 3 cm
The chords are on opposite sides of O, and both perpendiculars lie along the same line through O.
Distance between midpoints = d₁ + d₂ = 4 + 3 = 7 cm
In-text Questions — Page 102Section 5.6 Distance of Chords from the Centre
What you will find:
- The two parts of the chord are equal — the second crease bisects the chord.
- The angle between the creases is 90°.
- The second crease passes through the centre, so the distance from the centre to the midpoint of the chord is the distance of the chord from the centre (the perpendicular distance). Its value depends on your circle and chord.
Crease 2 passes through the centre O, meets AB at M
AM = MB, ∠OMA = ∠OMB = 90°, and OM = distance from O to the chord
How to draw a second chord of the same length: open a compass to the length of the first chord AB. Put the compass point anywhere on the circle, at a point P, and cut the circle at Q. Then PQ = AB.
Yes — the distance is the same. The two equal chords are equidistant from the centre.
In ΔOMA and ΔONP: OA = OP = r, AM = ½AB = ½PQ = PN, ∠OMA = ∠ONP = 90°
By RHS congruence, ΔOMA ≅ ΔONP ⇒ OM = ON
Exercise Set 5.4 — Page 104Section 5.6 Distance of Chords from the Centre
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.
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)
Given: CE ⊥ AB, CH ⊥ GF, CE = CH. To show: AB = GF.
∠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
Given: CE ⊥ AB, CH ⊥ GF, CE = CH, with C the centre and radius r. To show: AB = GF.
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
In-text Questions — Page 104Section 5.6.1 Which of the two unequal chords is farther from the centre?
| Length of Chord | Distance from Centre |
|---|---|
Observation: the longer the chord, the smaller its distance from the centre.
Here is the table filled in for a circle of radius 5 cm (do the same for your own circle and compare).
| Length of chord (cm) | 10 | 9 | 8 | 6 | 4 | 2 |
|---|---|---|---|---|---|---|
| Distance from centre (cm) | 0 | 2.18 | 3 | 4 | 4.58 | 4.90 |
d = √(r² − (½ chord)²) = √(25 − (½ chord)²)
e.g. chord 8 ⇒ d = √(25 − 16) = 3; chord 6 ⇒ d = √(25 − 9) = 4
Exercise Set 5.5 — Pages 105–106Section 5.6.1 Which of the two unequal chords is farther from the centre?
Chord = 2√13 cm ≈ 7.2 cm.
By Theorem 5, M is the midpoint of AB, and ΔOMA is right-angled at M.
AM² = OA² − OM² = 7² − 6² = 49 − 36 = 13
AM = √13 cm
AB = 2 × AM = 2√13 cm ≈ 7.21 cm
Given: circle with centre O, radius r; chord AB whose perpendicular distance from O is d. To show: AB = 2√(r² − d²).
By Theorem 5 (the perpendicular from the centre bisects the chord), AM = MB = ½ AB.
In the right triangle OMA, by Baudhāyana–Pythagoras:
OA² = OM² + AM²
r² = d² + AM²
AM² = r² − d²
AM = √(r² − d²)
AB = 2 AM = 2√(r² − d²)
No, we cannot. Distance and chord length are not proportional: they are linked by (½ chord)² + d² = r². We can only say (Theorem 8) that AB, being farther from the centre, is the shorter chord. A counterexample:
AB = 2√(r² − 4d²) and CD = 2√(r² − d²)
Take r = 10 cm and d = 3 cm:
CD = 2√(100 − 9) = 2√91 ≈ 19.08 cm
AB = 2√(100 − 36) = 2√64 = 16 cm
CD / AB ≈ 1.19, not 2.
In-text Questions — Page 107Section 5.7 Angles Subtended by an Arc
Arc AKB is a minor arc (about 100° at O); arc CLD is a major arc (about 204° at O). For arc CLD you turn from OC to OD through L — the long way round — so its angle is the reflex angle COD.
Measuring the two angles in Fig. 5.19 with a protractor:
| Arc | Angle swept at O | Less / greater than 180° | Type |
|---|---|---|---|
| arc AKB | about 100° | less than 180° | minor arc |
| arc CLD | about 204° (reflex) | greater than 180° | major arc |
Two things emerge, and together they are Theorem 9.
- The three angles ∠APB, ∠AQB and ∠ARB come out equal to one another — it makes no difference where on the circle (outside arc AKB) the point is taken.
- Each of them is exactly half the angle that arc AKB subtends at the centre.
angle at the centre, ∠AOB = 110°
∠APB = ∠AQB = ∠ARB = 55° = ½ × 110°
Exercise Set 5.6 — Pages 110–111Section 5.7.1 Angle subtended by an arc at a point on the circle outside the arc
AB = 12 cm — the chord equals the radius.
∠OAB = ∠OBA (base angles of an isosceles triangle)
∠OAB + ∠OBA + 60° = 180°
2 ∠OAB = 120° ⇒ ∠OAB = ∠OBA = 60°
All three angles are 60°, so ΔOAB is equilateral
∴ AB = OA = OB = 12 cm
(i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?
(ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?
(iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?
(i) No. X and Y on the circle on the same side of AB lie outside the same arc AB, so by Theorem 9 each of ∠AXB and ∠AYB is half the angle that this arc subtends at O. Hence ∠AXB = ∠AYB always.
(ii) Not always. If AB is a diameter, take X on one side of AB and Y on the other: ∠AXB = ∠AYB = 90° (angle in a semicircle), yet X and Y are on opposite sides.
(iii) Yes, if X and Y are on the same side of AB. A, B, X are not collinear, so one circle passes through them (Theorem 1); by Theorem 10, Y, which sees AB under the same angle from the same side, lies on it too. (If X and Y are on opposite sides of AB, this need not happen.)
x = 80°.
In Fig. 5.26 the four points A, D, C, B lie on a circle, so ADCB is a cyclic quadrilateral. The angle 100° is marked at D and the angle x is marked at B, and D and B are opposite vertices.
∠ADC + ∠ABC = 180°
100° + x = 180°
x = 180° − 100° = 80°
In-text Questions — Page 113Section 5.8 Concyclicity of Points
Yes, such a quadrilateral can be drawn. Both tests are satisfied.
80° + 110° + 100° + 70° = 360° ✓
Test 2 — opposite angles of a cyclic quadrilateral:
∠A + ∠C = 80° + 100° = 180° ✓
∠B + ∠D = 110° + 70° = 180° ✓
Both pairs of opposite angles add to 180°, so by Theorem 12 the four vertices must lie on a circle. Such a cyclic quadrilateral exists.
End-of-Chapter Exercises — Pages 114–116Whole chapter
Chord = 24 cm.
Half-chord = 12 cm
Chord = 2 × 12 = 24 cm
35°, at any point of the circle lying outside that arc.
70° = 2 × angle at the point
Angle at the point = 70° ÷ 2 = 35°
Distance = 5 cm.
Half-chord = 24 ÷ 2 = 12 cm (Theorem 5: the perpendicular from the centre bisects the chord)
d² = r² − (half-chord)² = 13² − 12² = 169 − 144 = 25
d = 5 cm
Chord = 24 cm.
= 2√(15² − 9²)
= 2√(225 − 81)
= 2√144 = 2 × 12 = 24 cm
Given: a circle with centre O and a chord AB. To show: the perpendicular bisector of AB passes through O.
So O is equidistant from A and B.
Every point equidistant from A and B lies on the perpendicular bisector of AB.
∴ O lies on the perpendicular bisector of AB.
The locus step, written out. Let M be the midpoint of AB and join OM. In ΔOMA and ΔOMB: OA = OB (radii), AM = BM (M is the midpoint), OM common. By SSS, ΔOMA ≅ ΔOMB, so ∠OMA = ∠OMB. These are angles on a straight line, so ∠OMA + ∠OMB = 180°, giving ∠OMA = ∠OMB = 90°. Hence OM is perpendicular to AB and bisects it — OM is the perpendicular bisector, and it passes through O.
∠ACB = 90° — the angle in a semicircle is a right angle.
Take the arc from A to B not containing C. The angle it sweeps at O is the straight angle:
∠AOB = 180°
C lies on the circle, outside that arc, so by Theorem 9
∠ACB = ½ ∠AOB = ½ × 180° = 90°
∠C = 105° and ∠D = 70°.
∠A + ∠C = 180° ⇒ 75° + ∠C = 180° ⇒ ∠C = 105°
∠B + ∠D = 180° ⇒ 110° + ∠D = 180° ⇒ ∠D = 70°
x = 38, ∠P = 86° and ∠R = 94°.
∠P + ∠R = 180°
(2x + 10) + (3x − 20) = 180
5x − 10 = 180
5x = 190
x = 38
∠P = 2(38) + 10 = 76 + 10 = 86°
∠R = 3(38) − 20 = 114 − 20 = 94°
Radius = 10 cm.
r² = d² + (half-chord)² = 6² + 8² = 36 + 64 = 100
r = 10 cm
Area = 60 square units.
Let the quadrilateral be ABCD with AB = BC = 5 and CD = DA = 12 — the two equal pairs adjacent, so it is a kite. Join the diagonal BD.
ABCD is cyclic, so ∠A + ∠C = 180° (Theorem 11)
⇒ 2∠A = 180° ⇒ ∠A = ∠C = 90°
So ΔABD and ΔCBD are right-angled at A and at C.
BD² = AB² + AD² = 5² + 12² = 25 + 144 = 169 ⇒ BD = 13
Area = area ΔABD + area ΔCBD
= ½ × 5 × 12 + ½ × 5 × 12
= 30 + 30 = 60 square units
Best method: draw one diagonal and look at the two triangles it makes.
A diagonal, say AC, cuts the cyclic quadrilateral ABCD into ΔABC and ΔACD. Both triangles are inscribed in the same circle, so the circumcentre of the quadrilateral is the circumcentre of each of them. And we already know where a triangle’s circumcentre sits:
| What you see in ΔABC and ΔACD | Centre of the circle |
|---|---|
| at least one of them is acute-angled | inside ABCD (inside that triangle) |
| a right angle at B (and so also at D) | midpoint of the diagonal AC — inside ABCD (AC is a diameter) |
| a right angle at A or at C in one of them | midpoint of a side — on the boundary (that side is a diameter) |
| both are obtuse-angled | outside ABCD |
Given: equal chords AB and CD of a circle with centre O intersect at P; M and N are their midpoints. To show: AP = CP and PB = PD (naming the ends suitably).
In ΔOMP and ΔONP: ∠OMP = ∠ONP = 90°, OP common, OM = ON
⇒ ΔOMP ≅ ΔONP (RHS) ⇒ PM = PN
Also AM = ½AB = ½CD = CN
So AP = AM − PM = CN − PN = CP
and PB = AB − AP = CD − CP = PD
The circle is forced: its radius must be 3√2 ≈ 4.24 cm.
r = √18 = 3√2 ≈ 4.24 cm
Construction (direct).
- Draw the chord AB = 6 cm.
- Construct the perpendicular bisector of AB; let it cut AB at M.
- Mark O on that bisector with OM = 3 cm.
- With centre O and radius OA (≈ 4.2 cm), draw the circle. AB is then a chord of length 6 cm at distance 3 cm from O.
Answer to the hint: yes. Draw ΔABC with AB = BC = 6 cm and ∠B = 90°. Its circumcircle has the hypotenuse AC as a diameter, so its centre O is the midpoint of AC, and O is 3 cm (half of BC) from AB.
Given: a parallelogram ABCD inscribed in a circle. To show: ABCD is a rectangle.
∠A = ∠C and ∠B = ∠D
ABCD is cyclic ⇒ opposite angles are supplementary (Theorem 11):
∠A + ∠C = 180°
Substituting ∠C = ∠A:
2∠A = 180° ⇒ ∠A = 90°, and hence ∠C = 90°
Similarly 2∠B = 180° ⇒ ∠B = ∠D = 90°
All four angles are right angles ⇒ ABCD is a rectangle.
Given: rectangle ABCD inscribed in a circle. To show: the diagonals meet at the centre.
Proof — using the properties of a rectangle. The diagonals of a rectangle are equal and bisect each other. Let them meet at P. Then
⇒ PA = PB = PC = PD
So P is equidistant from all four vertices. Only one circle passes through A, B, C (Theorem 1), and its centre is the only point equidistant from them. Hence P = O, the centre.
Another circle, with the same centre — a concentric circle of radius √(r² − ℓ²/4), where r is the radius of the given circle and ℓ the fixed chord length.
By Theorem 4, OM ⊥ AB, so OM is the distance from O to the chord.
OM² = OA² − AM² = r² − (ℓ/2)²
OM = √(r² − ℓ²/4), the same value for every such chord
So every midpoint is at a fixed distance from O ⇒ the midpoints lie on a circle with centre O and that radius.
Given: circle with centre O; chords AB and AC with AB = AC. To show: AO bisects ∠BAC.
AB = AC (given)
OB = OC = r (radii)
OA = OA (common)
By SSS congruence, ΔOAB ≅ ΔOAC
∴ ∠OAB = ∠OAC (corresponding angles)
That is, the ray AO divides ∠BAC into two equal parts — AO is the bisector of ∠BAC, so O lies on it.
Radius = 13 cm.
Both chords are on the same side, and the longer chord is the nearer one (Theorem 8),
so the distance to the 10 cm chord is d + 7.
For the 24 cm chord: r² = d² + 12² = d² + 144
For the 10 cm chord: r² = (d + 7)² + 5² = d² + 14d + 49 + 25
Equating: d² + 144 = d² + 14d + 74
144 − 74 = 14d
70 = 14d ⇒ d = 5 cm
r² = 5² + 12² = 25 + 144 = 169 ⇒ r = 13 cm
Side = r, and the distance of each side from the centre = (√3/2) r.
central angle = 360° ÷ 6 = 60°
Side: in the triangle formed by one side and the centre, the two radii are equal and the angle between them is 60°.
So the base angles are (180° − 60°) ÷ 2 = 60° each — the triangle is equilateral.
∴ side = r
Distance from the centre (the apothem): the side is a chord of length r, so
d = √(r² − (r/2)²) = √(r² − r²/4) = √(3r²/4) = (√3/2) r ≈ 0.866 r
∠MOP = ∠MNP — they are equal, being angles in the same segment standing on the chord MP.
Consider the chord MP. Both N and O lie on the same arc determined by MP
(the arc from M to P that runs through N and then O).
By Theorem 9, each of ∠MNP and ∠MOP is half the angle that the other arc MP subtends at the centre.
∴ ∠MOP = ∠MNP
The exterior angle at a vertex equals the interior angle at the opposite vertex.
∠ADC + ∠CDE = 180° (linear pair — angles on the straight line ADE)
∠ADC + ∠ABC = 180° (Theorem 11 — opposite angles of a cyclic 4-gon)
Comparing the two: ∠CDE = ∠ABC ∎
Justification 1 — the chord formula.
chord = 2√(r² − d²), with 0 ≤ d < r
Since d² ≥ 0, we get r² − d² ≤ r², so √(r² − d²) ≤ r
∴ chord ≤ 2r = diameter
Equality holds only when d = 0, i.e. when the chord passes through the centre — the chord is the diameter itself.
Given: a point A inside a circle with centre O (A ≠ O). To show: among all chords through A, the shortest is the one perpendicular to OA.
Step 1. A chord at distance d from O has length 2√(r² − d²) (Exercise 5.5, Q2), which is smaller when d is larger. So we need the chord through A that is farthest from O.
Step 2. Let PQ be any chord through A, and M the foot of the perpendicular from O to PQ.
If PQ ⊥ OA, then M = A and OM = OA.
So the chord through A perpendicular to OA is the farthest from O, and hence the shortest.
In Fig. 5.30 the tick marks show that OA and the two halves of the diameter are equal — all three are radii. Call the ends of the diameter P (angle a) and Q (angle b). OA splits ΔPAQ into two isosceles triangles.
In ΔOQA: OQ = OA ⇒ ∠OAQ = ∠OQA = b
The angle at A in the big triangle is ∠PAQ = ∠OAP + ∠OAQ = a + b
Angle sum of ΔPAQ:
a + b + (a + b) = 180°
2(a + b) = 180°
a + b = 90°
∴ ∠PAQ = a + b = 90°
Given: a circle with diameter AB; chords CC′ ⊥ AB and DD′ ⊥ AB. M is the midpoint of CD and M′ the midpoint of C′D′. To show: MM′ ⊥ AB.
Step 1. AB is perpendicular to the chord CC′, so it bisects it (Theorem 5). Thus AB is the perpendicular bisector of CC′: C′ is the mirror image of C in the line AB. In the same way, D′ is the mirror image of D.
Step 2. Reflection in AB maps the segment CD onto the segment C′D′, so it maps the midpoint M of CD onto the midpoint M′ of C′D′. The segment joining a point to its mirror image is perpendicular to the mirror line. Hence MM′ ⊥ AB.
In Fig. 5.31, OA = OB = OC = OD (radii, shown by the tick marks), so the triangles OAB, OBC, OCD and ODA are isosceles. Their equal base angles are:
Isosceles triangle OBC (OB = OC) ⇒ ∠OCB = ∠OBC = q
Isosceles triangle OCD (OC = OD) ⇒ ∠ODC = ∠OCD = u
Isosceles triangle ODA (OD = OA) ⇒ ∠OAD = ∠ODA = v
Angle sum of ABCD: 2(p + q + u + v) = 360° ⇒ p + q + u + v = 180°
∴ ∠A + ∠C = (p + v) + (q + u) = 180°
and ∠B + ∠D = (p + q) + (u + v) = 180°
Chapter 5 of 8 – Maths
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- All chapters: NCERT Solutions Class 9 Maths
Also for This Chapter
NCERT Solutions Class 9 Maths Chapter 5 I M Up and Down and Round and Round – Overview
| Class | Class 9 |
| Subject | Maths |
| Chapter | 5 – I M Up and Down and Round and Round |
| Textbook pages | 92 – 117 |
| Questions solved | 64 |
| Session | 2027 |
| Format | Full answers on this page + PDF |
NCERT Solutions Class 9 Maths Chapter 5 – FAQs
Where can I download NCERT Solutions Class 9 Maths Chapter 5 I M Up and Down and Round and Round PDF?
Read every answer on this page, or download the free PDF from the table at the top.
Is this for the 2027 session?
Yes, it follows the latest NCERT Class 9 Maths book for 2027.
How many questions are solved in NCERT Class 9 Maths Chapter 5?
All 64 questions of the chapter (textbook pages 92 – 117) – the in-text questions, activities and the end-of-chapter exercises.
How should I use these solutions?
Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.