NCERT Solutions for Class 9 Maths Chapter 5 I M Up and Down and Round and Round provide clear, step-by-step answers to every exercise and in-text question from the chapter I M Up and Down and Round and Round of the NCERT textbook Ganita Manjari. Prepared by subject experts as per the latest NCERT (CBSE) syllabus for 2026-27, these NCERT Solutions for Class 9 Maths help you understand each concept, write exam-ready answers, and check your own solutions. You can read them online below or download the free Class 9 Maths Chapter 5 question-answer PDF.
NCERT Solutions for Class 9 Maths Chapter 5 I M Up and Down and Round and Round
- Class: Class 9
- Subject: Maths
- Chapter: Chapter 5 – I M Up and Down and Round and Round
- Textbook: Ganita Manjari (NCERT)
- Study material: NCERT Solutions – questions with answers, free PDF
These solutions answer all the exercise questions of Chapter 5 I M Up and Down and Round and Round — including the in-text questions, short-answer and long-answer questions, and activities — with complete explanations so you can follow the method, not just the final answer. Read the full solutions below.
NCERT Solutions Class 9 Maths Chapter 5 I M Up and Down and Round and Round View Download
NCERT Solutions for Class 9 Maths Chapter 5 PDF Download
You can read the NCERT Solutions for Class 9 Maths Chapter 5 online above, or download the complete question-answer PDF to study I M Up and Down and Round and Round offline at any time.
NCERT Solutions for Class 9 Maths Chapter 5 PDF Download Link – Click Here to Download Solutions PDF
Questions Covered in This Chapter
These NCERT Solutions answer all 64 questions of this chapter. The questions solved are:
- Can you recognise the origin of the shapes in Fig. 5.1?
- Activity: List some objects from nature that resemble a circle.
- Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?
- What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?
- What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?
- The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points? (Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)
- How many circles pass through two points on a plane?
- Are there circles of all possible radii passing through A and B? What is the radius of the smallest circle passing through A and B? What is the radius of the largest circle passing through A and B?
- As you move away from segment AB along its perpendicular bisector, do the radii of the circles containing A and B increase or decrease?
- As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or less curved?
- You are given two points A and B on a plane. How many squares can you draw on the same plane with A and B on the boundary? How many squares can you draw on the plane with A and B as the corners of the square?
- What if A, B and C lie on a straight line, i.e., are collinear? Can you explain why, in this case, there is no circle through A, B and C?
- Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
- Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
- Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.
- What is the least possible radius of a circle through two points A and B?
- A, B and C are three collinear points. Can you find a point P such that PA = PB = PC ? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?
- The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?
- Show that the triangle formed by a chord and the centre of the circle is isosceles.
- Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
- Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)
- An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.
- Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.
- Activity: Take a paper circle. Fold the circle from the boundary, inwards. Open the fold. The crease is now a chord (see Fig. 5.13 B). Now fold the paper again, so that the end points of the chord meet. Open the fold (see Fig. 5.13 C). Measure the lengths of the parts into which the chord is divided. Measure the angle between the creases. Measure the distance from the centre to the midpoint of the chord.
- Now draw another chord of the same length. How will you do this? We will let you figure this out yourself. Join the centre to the midpoint of the new chord and measure its length. Is it the same as distance from the centre to the first chord?
- Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.
- Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.
- Solve the previous question using the Baudhāyana–Pythagoras theorem.
- Activity: Draw a circle. Draw chords of various lengths. Drop a perpendicular to each chord from the centre. Record the length of the chord and its distance from the centre in a table. (Table 1: Length of Chord / Distance from Centre.) What do you observe?
- Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
- Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√(r² − d²).
- In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.
- Exercise: A circle with centre O is drawn, and A, B, C, D are points on the circle (see Fig. 5.19). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.
- Activity: Draw a circle and a chord AB. Fix an arc AKB formed by AB and a point K between A, B on the circle. Measure the angle subtended at the centre by arc AKB. Take three points P, Q, R on the circle outside arc AKB. Measure the angles subtended by arc AKB at points P, Q, R. What do you notice?
- In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?
- Let A and B be two points on a circle with centre O. (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?
- Find x in Fig. 5.26.
- Exercise: A cyclic quadrilateral has angles measuring ∠A = 80°, ∠B = 110°, ∠C = 100°, and ∠D = 70°. Can such a quadrilateral be drawn? Explain why or why not.
- In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?
- An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
- The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.
- A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?
- Prove that the perpendicular bisector of a chord passes through the centre of the circle.
- The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.
- ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?
- Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find the value of x and the measures of ∠P and ∠R.
- The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
- A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
- Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?
- When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.
- Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre. (Hint: Is it a circumcircle of a suitable triangle?)
- Show that rectangle is the only parallelogram that can be inscribed in a circle.
- Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.
- Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
- In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: “The centre of the circle lies on the angle bisector of ∠BAC”.
- Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.
- A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.
- A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.
- Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).
- “There is no chord of a circle that is longer than its diameter.” How do you justify this statement?
- Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.
- How would you use the following figure to justify the statement that the angle in a semicircle is 90°? (Fig. 5.30)
- In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. Prove that the segment MM' joining the midpoints of the chords CD and C' D' is perpendicular to AB.
- How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°? (Fig. 5.31)
Chapter at a Glance
- 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) .
- An arc subtends at the centre twice the angle it subtends at any point of the circle outside the arc. Hence the angle in a semicircle is 90°, and all angles in the same segment are equal.
- Four points are concyclic when a segment subtends equal angles at two points on the same side of it. In a cyclic quadrilateral each pair of opposite angles adds to 180°, and the converse is also true.
How to Download NCERT Solutions for Class 9 Maths Chapter 5 PDF
Follow these simple steps to get the I M Up and Down and Round and Round questions-and-answers PDF from Ganita Manjari.
- Search NCERT Solutions for Class 9 Maths Chapter 5 aglasem and open this page.
- Read the exercise questions with answers for I M Up and Down and Round and Round shown above.
- Click the Download PDF link to save the I M Up and Down and Round and Round solutions to your device.
NCERT Solutions for Class 9 Maths – All Chapters
There are more chapters to study besides I M Up and Down and Round and Round in Maths. Here are the NCERT Solutions for all chapters of Class 9 Maths.
- Chapter 1 Orienting Yourself the Use of Coordinates
- Chapter 2 Introduction to Linear Polynomials
- Chapter 3 The World of Numbers
- Chapter 4 Exploring Algebraic Identities
- Chapter 5 I M Up and Down and Round and Round
- Chapter 6 Measuring Space Perimeter and Area
- Chapter 7 The Mathematics of Maybe Introduction to Probability
- Chapter 8 Predicting What Comes Next Exploring Sequences and Progressions
NCERT Solutions for Class 9 – All Subjects
Just like Chapter 5 of Maths, you can get the exercise questions with answers for every other subject of Class 9. Here are the NCERT Solutions for all subjects of Class 9.
NCERT Solutions for Class 9 Maths Chapter 5 – An Overview
The key highlights of this study material are as follows.
| Aspects | Details |
|---|---|
| Class | Class 9 |
| Subject | Maths |
| Chapter Number | Chapter 5 |
| Chapter Name | I M Up and Down and Round and Round |
| Book Name | Ganita Manjari |
| Book By | NCERT (National Council of Educational Research and Training) |
| Educational Resource Here | NCERT Solutions of Class 9 Maths Chapter 5 for all exercises |
| More Questions Answers of This Subject | NCERT Solutions for Class 9 Maths |
| Download Book Chapter | NCERT Book Class 9 Maths |
| All Questions Answers For This Class | NCERT Solutions for Class 9 |
| Complete Solutions | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 5 I M Up and Down and Round and Round – FAQs
What are the NCERT Solutions for Class 9 Maths Chapter 5 I M Up and Down and Round and Round?
They are the complete, step-by-step answers to all the exercise and in-text questions of Chapter 5 I M Up and Down and Round and Round from the NCERT Class 9 Maths textbook Ganita Manjari, written by experts as per the latest NCERT syllabus.
How can I download the Class 9 Maths Chapter 5 solutions PDF for free?
Open this page on aglasem, read the I M Up and Down and Round and Round questions with answers, and click the “Download Solutions PDF” link. The Class 9 Maths Chapter 5 NCERT Solutions PDF is completely free to download.
Are these NCERT Solutions as per the latest 2026-27 syllabus?
Yes. The NCERT Solutions for Class 9 Maths Chapter 5 are based on the latest NCERT textbook Ganita Manjari and the current 2026-27 CBSE syllabus, so the questions and answers match what you study in class.
Where can I get NCERT Solutions for the other chapters of Class 9 Maths?
You can find the answers to every chapter on the NCERT Solutions for Class 9 Maths page, and solutions for every subject on the NCERT Solutions for Class 9 page.
How do NCERT Solutions help in exam preparation?
They show the correct method to solve each question, help you write answers the way they are expected in exams, let you check and correct your own work, and save revision time — which together improve your marks in Class 9 Maths.
If you have any queries on NCERT Solutions for Class 9 Maths Chapter 5 I M Up and Down and Round and Round, then please ask in the comments below.
