by Tiwari Academy
NCERT Class 9 Maths Ganita Manjari Chapter 5 Solutions – I’m Up and Down, and Round and Round – for Session 2026-27 School test and exams. Chapter 5 of the NCERT Ganita Manjari textbook for Class 9 Mathematics, titled “I’m Up and Down, and Round and Round,” is a beautifully written introduction to the geometry of circles โ one of the most ancient and universally recognised shapes in human history. From raindrops forming ripples on water to the cross-section of a plant stem and the apparent shape of the sun and moon, circles appear everywhere in nature and mathematicians across civilisations have been drawn to their perfect symmetry.
Class 9 Ganita Manjari Chapter 5 Quick Links:
Chapter 5 Ganita Manjari takes Grade 9 students on a structured journey through the formal geometry of circles, beginning with precise definitions and building steadily toward elegant theorems about chords, arcs, angles, and cyclic quadrilaterals. Students will encounter 12 theorems – each explained with a “Given, To Show, Why is this true?” structure that develops genuine mathematical reasoning rather than rote proof memorisation. The chapter connects geometric intuition with rigorous justification, using activities, congruence arguments, and the BaudhฤyanaโPythagoras theorem throughout. Whether you are a student preparing for your Class 9 exams, a teacher planning a theorem-based lesson or a parent trying to understand what your child is learning, this page covers everything you need for Chapter 5 of Ganita Manjari โ clearly structured and fully aligned with the 2026-27 NCERT syllabus.
What is Class 9 Maths Ganita Manjari Chapter 5 About?
Chapter 5, “I’m Up and Down, and Round and Round”, is the circles chapter of Ganita Manjari. Unlike the old NCERT textbook which presented circle theorems in a more formulaic way, this chapter builds each result from first principles using triangle congruence, symmetry arguments, and activity-based discovery. The chapter is structured across eight sections and contains 12 theorems, 6 exercise sets and 26 end-of-chapter questions – including several starred (*) advanced problems.
The key sections are as follows:
- Section 5.1: Definitions: Introduces circle, centre, radius, chord, diameter and the concept of locus in formal mathematical language.
- Section 5.2: Symmetries of a Circle: Explores complete rotational symmetry and reflection symmetry across diameters – the properties that make circles geometrically unique.
- Section 5.3: How Many Circles?: Establishes that infinitely many circles pass through two points, and proves Theorem 1 – that a unique circle passes through any three non-collinear points (circumcircle). Discusses circumcentre position for acute, obtuse, and right-angled triangles.
- Section 5.4: Chords and the Angles They Subtend: Proves Theorems 2 and 3 – equal chords subtend equal angles at the centre and vice versa โ using SSS and SAS congruence.
- Section 5.5: Midpoints and Perpendicular Bisectors of Chords: Proves Theorems 4 and 5 – the line joining the centre to the midpoint of a chord is perpendicular to it and the perpendicular from the centre bisects the chord.
- Section 5.6: Distance of Chords from the Centre: Proves Theorems 6, 7 and 8 – equal chords are equidistant from the centre, equidistant chords are equal and longer chords are closer to the centre.
- Section 5.7: Angles Subtended by an Arc: Introduces major and minor arcs, and proves Theorem 9 – the central angle is double the inscribed angle. Derives the corollary that the angle in a semicircle is always 90ยฐ.
- Section 5.8: Concyclicity of Points: Proves Theorems 10, 11, and 12 – conditions for four points to be concyclic, opposite angles of a cyclic quadrilateral sum to 180ยฐ and the converse.
NCERT Class 9 Maths Ganita Manjari Chapter 5 Solutions
Class 9 Maths Ganita Manjari Chapter 5 Exercise Set 5.1 Solutions
Exercise Set 5.1
1. Draw ฮABC with AB = 5 cm, โ A = 70ยฐ and โ B = 60ยฐ. Draw the circumcircle of ฮABC. Is the centre inside or outside the triangle?
Answer:
Steps of Construction:
1. Draw the line segment AB = 5 cm.
2. At point A, construct โ CAB = 70ยฐ.
3. At point B, construct โ CBA = 60ยฐ.
4. Let the two rays meet at C. Then ฮABC is formed.
5. Draw the perpendicular bisector of AB.
6. Draw the perpendicular bisectors of BC and AC.
7. Let the three perpendicular bisectors meet at O. This point O is the circumcentre.
8. With centre O and radius OA, draw a circle passing through A, B and C. This is the circumcircle of ฮABC.
Now:
โ C = 180ยฐ โ (70ยฐ + 60ยฐ) = 50ยฐ
Since all the angles of the triangle are less than 90ยฐ, the triangle is acute-angled.
In an acute-angled triangle, the circumcentre lies inside the triangle.
Therefore, the centre is inside the triangle.
2. Draw ฮABC with AB = 5 cm, โ A = 100ยฐ, AC = 4 cm. Draw the circumcircle of ฮABC. Is the centre inside or outside the triangle?
Answer:
Steps of Construction:
1. Draw the line segment AB = 5 cm.
2. At point A, construct โ BAC = 100ยฐ.
3. On the ray AX, mark point C such that AC = 4 cm.
4. Join C to B. Then ฮABC is formed.
5. Draw the perpendicular bisector of AB.
6. Draw the perpendicular bisector of AC and BC.
7. Let these bisectors meet at O. This point O is the circumcentre.
8. With centre O and radius OA, draw the circle through A, B and C. This is the circumcircle.
Reasoning:
Since โ A = 100ยฐ, the triangle is obtuse-angled.
In an obtuse-angled triangle, the circumcentre lies outside the triangle.
Therefore, the centre is outside the triangle.
3. Draw ฮABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ฮABC. Let the circumcentre be O. Measure OA, OB, OC.
Answer:
Steps of Construction:
1. Draw AB = 6 cm.
2. With centre A and radius 7 cm, draw an arc.
3. With centre B and radius 7 cm, draw another arc cutting the first arc at C.
4. Join AC and BC. Then ฮABC is formed.
5. Draw the perpendicular bisector of AB.
6. Draw the perpendicular bisectors of AC and BC).
7. Let they intersect at O. This point O is the circumcentre.
8. With centre O and radius OA, draw the circumcircle.
Observation:
Since O is the circumcentre, it is equidistant from all three vertices.
Therefore, OA = OB = OC
Approximate measurement:
For the triangle with sides 7 cm, 7 cm, 6 cm,
OA = OB = OC โ 4 cm
So, the three measurements are equal, each approximately 4 cm.
4. What is the least possible radius of a circle through two points A and B?
Answer:
For all circles passing through two fixed points A and B, the centre lies on the perpendicular bisector of AB.
The smallest circle occurs when the centre is the midpoint of AB.
In that case, AB becomes the diameter of the circle.
Therefore, the least possible radius is:
Radius = AB/2
Hence, the least possible radius of a circle through A and B is half of AB.
Class 9 Maths Ganita Manjari Chapter 5 Exercise Set 5.2 Solutions
Exercise Set 5.2
1. Show that the triangle formed by a chord and the centre of the circle is isosceles.
Answer:
Let AB be a chord of a circle with centre O.
Join OA and OB.
In triangle OAB
OA = OB [Radii of the same circle]
Therefore, triangle OAB is an isosceles triangle.
Hence proved.
2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Answer:
Let AB and PQ be two equal chords of the same circle with centre O.
Then triangles formed are โณOAB and โณOPQ.
Now, in โณOAB and โณOPQ
OA = OP [Radii of the same circle]
OB = OQ [Radii of the same circle]
AB = PQ [Given]
So, by SSS congruence criterion
โณOAB โ
โณOPQ.
Hence, if two such isosceles triangles have equal base length, they are congruent to each other.
Hence proved.
Quick Reference – All 12 Theorems in Class 9 Ganita Manjari Chapter 5
| Theorem | Statement |
|---|---|
| Theorem 1 | There is a unique circle passing through three non-collinear points. |
| Theorem 2 | Equal chords of a circle subtend equal angles at the centre. |
| Theorem 3 | Chords subtending equal angles at the centre are equal. |
| Theorem 4 | The line joining the centre to the midpoint of a chord is perpendicular to the chord. |
| Theorem 5 | The perpendicular from the centre to a chord bisects the chord. |
| Theorem 6 | Chords of equal length are equidistant from the centre. |
| Theorem 7 | Chords equidistant from the centre have equal length. |
| Theorem 8 | The longer of two chords is closer to the centre. |
| Theorem 9 | The angle subtended by an arc at the centre is double the angle at any point on the remaining circle. |
| Corollary | The angle subtended by a diameter at any point on the circle is 90ยฐ. |
| Theorem 10 | If a segment subtends equal angles at two points on the same side, all four points are concyclic. |
| Theorem 11 | Opposite angles of a cyclic quadrilateral sum to 180ยฐ. |
| Theorem 12 | If opposite angles of a quadrilateral sum to 180ยฐ, it is cyclic. |
Important Theorems Explained Simply
- Theorem 9 (Central Angle = Double Inscribed Angle) is the single most important result in this chapter. It states that if you pick any arc of a circle, the angle it creates at the centre is exactly double the angle it creates at any point on the remaining circle. The proof uses the exterior angle theorem and the isosceles triangle property (since all radii are equal). The famous corollary – that any angle in a semicircle is 90ยฐ – follows directly from this, because a diameter subtends a 180ยฐ angle at the centre, making the inscribed angle exactly 90ยฐ.
- Theorem 11 (Cyclic Quadrilateral) states that opposite angles of any cyclic quadrilateral always add up to 180ยฐ. The proof uses Theorem 9 twice – once for each arc โ and the fact that together both arcs make a complete 360ยฐ rotation at the centre, so their half-angles sum to 180ยฐ.
- Theorem 1 (Unique Circumcircle) establishes that three non-collinear points always determine exactly one circle. The proof is elegant: the centre must be equidistant from all three points, so it must lie on the perpendicular bisector of each pair. Two non-parallel lines meet at exactly one point, giving a unique centre.
Key Definitions in Ganita Manjari Chapter 5 – Circles
- Circle: The set of all points on a plane that are equidistant from a given point. That given point is the centre and the fixed distance is the radius.
- Locus: The set of all points satisfying a given condition. A circle is the locus of points equidistant from a fixed centre point.
- Chord: A line segment whose both endpoints lie on the circle.
- Diameter: A chord that passes through the centre of the circle. It is the longest possible chord and equals twice the radius.
- Arc: A connected portion of a circle defined by two endpoints. The larger portion is the major arc and the smaller is the minor arc.
- Circumcircle: The unique circle passing through all three vertices of a triangle. Its centre is called the circumcentre, which lies at the intersection of the perpendicular bisectors of the triangle’s sides.
- Concyclic Points: Points that all lie on the same circle.
- Cyclic Quadrilateral: A quadrilateral whose four vertices all lie on a single circle. Its opposite angles always add up to 180ยฐ.
Chapter 5 Ganita Manjari – Exercise-Wise Overview
- Exercise Set 5.1: Drawing circumcircles of triangles with given dimensions; determining whether circumcentre lies inside or outside the triangle; finding minimum radius through two points.
- Exercise Set 5.2: Proving that the triangle formed by a chord and the centre is isosceles; proving congruence of two such triangles with equal base lengths.
- Exercise Set 5.3: Proving the converse of Theorem 4 (perpendicular from centre bisects chord); inscribed isosceles triangle altitude through centre; distance between midpoints of parallel chords.
- Exercise Set 5.4: Using the BaudhฤyanaโPythagoras theorem to prove Theorem 6; showing AB = GF when equidistant chords are given.
- Exercise Set 5.5: Finding chord length given radius and perpendicular distance; deriving the general chord length formula 2โ(rยฒ โ dยฒ).
- Exercise Set 5.6: Finding chord length from central angle and radius; angles in the same segment; finding unknown angles in circle diagrams.
- End-of-Chapter Exercises: 26 questions covering all chapter theorems, cyclic quadrilateral angle problems, chord-distance calculations and advanced starred (*) proofs on rectangles inscribed in circles, intersecting chords and shortest chord through a point.
Starred (*) Advanced Questions – For High Achievers
Chapter 5 contains several starred questions in the end-of-chapter exercises that go well beyond standard exam expectation:
- Q12: When two chords of equal length intersect, prove that corresponding sub-segments are equal.
- Q14: Prove that a rectangle is the only parallelogram that can be inscribed in a circle.
- Q15: Prove that if a rectangle is inscribed in a circle, its diagonals intersect at the centre.
- Q16: Determine the shape formed by the midpoints of all chords of a fixed length in a given circle.
- Q19: A regular hexagon is inscribed in a circle of radius r – find the side length and the distance of each side from the centre.
- Q23: Show that the shortest chord through any interior point of a circle is the one perpendicular to the line joining that point to the centre.
- Q25: Two chords are drawn perpendicular to a diameter – prove that the segment joining the midpoints of two related chords is also perpendicular to the diameter.
These questions are excellent for NTSE, Mathematical Olympiad and Class 10 advanced preparation.
Frequently Asked Questions (FAQs) – Class 9 Maths Ganita Manjari Chapter 5
Is Class 9 Maths Ganita Manjari Chapter 5 easy or difficult?
It sits in the moderate-to-difficult range. The early sections on definitions, symmetry and basic chord properties are quite approachable. The difficulty increases significantly from Section 5.7 onward, where Theorem 9 on arc angles requires careful multi-step reasoning.
The cyclic quadrilateral theorems in Section 5.8 are conceptually demanding but become manageable once Theorem 9 is fully understood. Students who are comfortable with triangle congruence from earlier chapters will find the proof structure of this chapter much easier to follow.
How many theorems are in Class 9 Ganita Manjari Circles Chapter 5 and do I need to memorise all of them?
There are 12 theorems plus one important corollary (angle in a semicircle is 90ยฐ). You do not need to memorise the full word-for-word proofs, but you must understand the logic of each proof well enough to reproduce it in your own words.
More importantly, you must remember the statement of every theorem clearly โ because exam questions will ask you to apply them to find unknown angles, prove chord relationships or identify cyclic quadrilaterals.
The Chapter Summary on the last page of Chapter 5 lists all key results and is the best revision tool in the book.
Is Class 9 Ganita Manjari Chapter 5 I’m Up and Down Round and Round difficult for average students?
It is genuinely more demanding than earlier geometry chapters because it requires students to think in terms of formal mathematical proof, not just calculation. The challenge is two-fold โ understanding why each theorem is true, and then being able to apply the correct theorem to an unseen problem.
Students who struggle with abstract reasoning often find the arc angle theorems in Section 5.7 particularly confusing. However, the Ganita Manjari textbook uses a “Given, To Show, Why is this true?” format throughout that helps structure thinking. Regular drawing of diagrams alongside reading โ not just reading alone โ makes a significant difference for average learners.
How much time does Class 9 Maths Ganita Manjari Chapter 5 Circles take to complete?
For thorough preparation covering all six exercise sets and the end-of-chapter exercises, plan for 12 to 15 days. The chapter is longer and proof-heavier than Chapter 4. Sections 5.3, 5.7 and 5.8 each deserve at least two days โ one for understanding the theorems and one for attempting exercises.
Daily sessions of 35โ45 minutes with diagram practice work far better than long occasional sittings, because spatial reasoning for circle geometry builds gradually through repeated drawing. If your child has a test approaching, prioritise Theorems 1, 9, 11 and 12 along with their direct applications in Exercise Sets 5.1 and 5.6 โ these cover the majority of exam-likely questions.
How to solve Class 9 Maths Ganita Manjari Chapter 5 in one day?
One day works only for revision, not first-time learning. Start with the Chapter Summary โ read all theorem statements carefully. Then work through one solved example or activity from Sections 5.3, 5.6, and 5.7 as these carry the highest exam weight.
Attempt the non-starred end-of-chapter questions focusing on Q1, Q2, Q3, Q7, Q8, and Q9 โ these cover the most frequently tested skills. Skip starred questions for now. For a student encountering this chapter for the first time, a realistic minimum is five to six focused days with daily revision of theorem statements.
Can the old NCERT Maths textbook help my child study Ganita Manjari Class 9 Chapter 5 Circles?
Partially. The old NCERT Class 9 Maths book also covers circles and includes theorems on chord properties, equal chords and cyclic quadrilaterals. For those overlapping topics, the old textbook can serve as a supplementary practice resource.
However, Ganita Manjari Chapter 5 approaches proofs differently โ using the “Why is this true?” reasoning format, activity-based discovery and integration of the BaudhฤyanaโPythagoras theorem as a primary tool.
The concyclicity discussion in Section 5.8 and several starred end-of-chapter problems go beyond the scope of the old textbook. For these sections, the Ganita Manjari book itself is the only reliable resource aligned with the 2026-27 curriculum.
What is the best classroom approach to teach theorems in Ganita Manjari Class 9 Chapter 5?
The most effective approach follows the chapter’s own “Given, To Show, Why is this true?” structure explicitly in every lesson. Before stating a theorem, run the corresponding activity โ fold a paper circle for chord properties, measure inscribed angles for Theorem 9, or use tracing paper rotation for the equidistant chord result. This ensures students experience the result before they see its proof.
For Theorem 9 specifically, which has two cases depending on where the diameter extension falls, walk through both figures (Figs. 5.21 and 5.22) in the same lesson rather than treating the second case as optional. Students who see only one case frequently make errors in exams when the second configuration appears.
Which topics in Ganita Manjari Grade 9 Circles Chapter 5 need the most classroom time?
Three sections consistently need more time than a surface reading suggests. Section 5.7 on Theorem 9 requires at least two full periods โ one to establish the central angle equals double inscribed angle result and a separate period for the semicircle corollary and its applications in Exercise Set 5.6. Section 5.8 on cyclic quadrilaterals needs careful treatment of Theorems 11 and 12 together with their converses, as students frequently confuse which direction the logic runs.
Section 5.3 on circumcircles and the position of the circumcentre for different triangle types is often rushed but provides essential geometric intuition that supports everything that follows. The three construction exercises in Exercise Set 5.1 are worth doing as in-class activities, not homework, as they make the circumcentre concept concrete before the formal theorem is introduced.
How should teachers connect Class 9 Ganita Manjari Maths Chapter 5 to Class 10 topics and higher mathematics?
A brief forward-connection conversation at the chapter’s end significantly motivates students who wonder why they are learning circle proofs. The most direct link is to Class 10 geometry, where tangents to circles, the tangent-chord angle theorem and the alternate segment theorem all build on Theorems 9 and 11 from this chapter.
The cyclic quadrilateral property (opposite angles summing to 180ยฐ) reappears in trigonometry and complex number geometry in Classes 11 and 12. For students interested in competitive mathematics, Theorems 10 and 12 โ the concyclicity conditions โ are foundational for NTSE geometry sections and Olympiad circle problems. Starred questions Q14 and Q15, which prove that only rectangles can be inscribed in circles and that their diagonals meet at the centre, introduce the kind of elegant constraint-based reasoning that characterises higher mathematical thinking.
Framing Chapter 5 as the beginning of circle geometry rather than its entirety gives students the right perspective and genuine motivation to master it fully.
