I’m Up and Down, and Round and Round Class 9 (Ganita Manjari I)
Master I’m Up and Down, and Round and Round Class 9 (Ganita Manjari I) with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
NCERT Solutions
I’m Up and Down, and Round and Round Class 9 (Ganita Manjari I) – NCERT Solutions
Each question below opens its complete step-by-step Teachoo solution.
Exercise Set 5.1
4 questionsEx 5.1, 1
Draw △ABC with AB=5" " cm,∠" " A=70^∘ and ∠B=60^∘. Draw the circumcircle of △ABC. Is the centre inside or outside the triangle?
Since side AB is given, we follow these steps
Draw side AB
Make ∠" " 𝐀=〖𝟕𝟎〗^∘ and ∠𝐁=〖𝟔𝟎〗^∘
To make circumcircle, we construct perpendicular bisectors of two sides
Where the bisectors meet is center of circle, let’s mark it as O
Make a circle, taking center as O, radius as OA (or OB, OC since OA = OB = OC)
Ex 5.1, 2
Draw △ABC with AB=5" " cm,∠" " A=100^∘,AC=4" " cm. Draw the circumcircle of △ABC. Is the centre inside or outside the triangle?
We follow these steps
Draw side AB
Make ∠" " 𝐀=〖𝟏𝟎𝟎〗^∘
In this ray, we mark point C such that AC = 4 cm
To make circumcircle, we construct perpendicular bisectors of two sides
Where the bisectors meet is center of circle, let’s mark it as O
Make a circle, taking center as O, radius as OA (or OB, OC since OA = OB = OC)
Ex 5.1, 3
Draw △ABC, with AB=6" " cm,BC=7" " cm and CA=7" " cm. Draw the circumcircle of △ABC. Let the circumcentre be O . Measure OA, OB,OC.
We follow these steps
Draw side AB
Make AC = 7 cm from point A, BC = 7 cm from point B
Where they intersect is point C
To make circumcircle, we construct perpendicular bisectors of two sides
Where the bisectors meet is center of circle, let’s mark it as O
Make a circle, taking center as O, radius as OA (or OB, OC since OA = OB = OC)
Let’s do it step by step
Steps of construction – For making Triangle
Ex 5.1, 4
What is the least possible radius of a circle through two points A and 𝐵?
View solutionExercise Set 5.2
2 questionsEx 5.2, 1
Show that the triangle formed by a chord and the centre of the circle is isosceles.
Let’s make chord AB on the circle
Triangle formed by chord and center of circle is ∆ AOB
Ex 5.2, 2
Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Let’s consider same circle and ∆ AOB from previous question
Exercise Set 5.3
3 questionsEx 5.3, 1
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.)
Let’s do the proof of the converse
Ex 5.3, 2
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.
Given: ∆ ABC inscribed in circle
With AB = AC
Ex 5.3, 3
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.
Let’s draw the figure
Exercise Set 5.4
3 questionsEx 5.4, 1
Use the Baudhāyana–Pythagoras theorem to show why Theorem
6 must be true.
Theorem 6 is
Chords of a circle having the same length are all at the same distance from the centre of the circle
Ex 5.4, 2
Consider Fig. 5.15. If CE is perpendicular to AB,CH is perpendicular to GH, and CE =CH, show that AB=GF.
Given: : 𝐶𝐸⊥𝐴𝐵 and 𝐶𝐻⊥𝐺𝐹.
And, distances are equal i.e. CE = CH
Ex 5.4, 3
Solve the previous question using the Baudhāyana-Pythagoras theorem.
Given: : 𝐶𝐸⊥𝐴𝐵 and 𝐶𝐻⊥𝐺𝐹.
And, distances are equal i.e. CE = CH
Exercise Set 5.5
3 questionsEx 5.5, 1
Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
Let’s draw the diagram
Here, we have circle with center O
And, Radius = 7 cm
Let AB be the chord
And, OM be perpendicular distance to AB from O
∴ OM = 6 cm
Ex 5.5, 2
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√(𝑟^2−𝑑^2 ).
This is the same as previous question, but without the values
Ex 5.5, 3
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=2AB? Give reasons for your answer.
We draw a diagram
Let Radius = r
Exercise Set 5.6
3 questionsEx 5.6, 1
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’s draw a figure
Ex 5.6, 2
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?
Ex 5.6, 3
(Method 1) Find 𝑥 in Fig. 5.26.
Here, we will use the property
Opposite angles of Cyclic Quadrilateral are supplementary
Theorems (and their Proofs)
14 questionsTheorem 1
There is a unique circle passing through three non-collinear points
View solutionTheorem 2
Equal chords of a circle subtend equal angles at the centre of the circle
View solutionTheorem 3
Chords of a circle that subtend equal angles at the centre are equal.
View solutionTheorem 4
The line joining the centre of a circle and the midpoint of
a chord of the circle is perpendicular to the chord
Theorem 5
The perpendicular from the centre of a circle to a chord of the circle bisects the chord
View solutionTheorem 6
Chords of a circle having the same length are all at the same distance from the centre of the circle
View solutionTheorem 7
Chords of a circle that are equidistant from the centre have equal length
View solutionTheorem 8
Let AB and DE be two chords of a circle with centre C. Suppose AB > DE. Then the distance from C to AB is less than the distance from C to DE
View solutionTheorem 9
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.
View solutionAngle subtended by Diameter is 90°
Theorem
Angle subtended by a diameter/semicircle on any point of circle is 90°
Angles in the same segment are equal
Theorem
Angles in the same segment of a circle are equal.
Theorem 10
If a line segment AB joining two points A, B subtends equal angles at two other points C, D that lie on the same side of AB, then the four points lie on a circle.
View solutionTheorem 11
The sum of two opposite angles of a cyclic quadrilateral is 180°.
View solutionTheorem 12
If two opposite angles of a quadrilateral add up to 180°, then the vertices of the quadrilateral lie on a circle, i.e., they are concyclic.
View solutionEnd-of-Chapter Exercises
26 questionsQuestion 1
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?
Let’s draw the diagram
Here, we have circle with center O
And, Radius = 13 cm
Let AB be the chord
And, OM be perpendicular distance to AB from O
∴ OM = 5 cm
Question 2
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?
Drawing figure
Let arc AB subtend ∠ AOB = 70° at the center
Question 3
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.
Let’s draw the diagram
Here, we have circle with center O
Since Diamter is 26 cm
Radius = 26/2 cm = 13 cm
Question 4
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?
Let’s draw the diagram
Here, we have circle with center O
And, Radius = 15 cm
Let AB be the chord
And, OM be perpendicular distance to AB from O
∴ OM = 9 cm
Question 5
Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Given: AB is a chord
And PM is perpendicular bisector of AB
∴ PM ⊥ AB
& M is mid-point of AB, i.e. AM = BM
Question 6
The diameter of a circle is AB . Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.
We know that
Angle in a semicircle is always a right angle
Question 7
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 ?
In a cyclic quadrilateral
Opposite angles are supplementary
Question 8
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.
In a cyclic quadrilateral
Opposite angles are supplementary
Question 9
The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
Let’s draw the diagram
Here, we have circle with center O
Let Radius = r cm
Question 10
A cyclic quadrilateral has sides 5, 5, 12, 12 units Find its area.
Depending on the order of the sides, this shape is either
a Rectangle (5, 12, 5, 12)
or a Kite (5, 5, 12, 12)
Question 11
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?
Our property is
Centre of any circumcircle (the circumcentre) is always located where the perpendicular bisectors of the shape's sides intersect
Question 12
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.
Given: Let AB & CD be the two equal chords
intersecting at point X.
∴ AB = CD
Question 13
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?)
Let’s draw a rough diagram
Here, radius of circle and center is not given
Question 14
Show that rectangle is the only parallelogram that can be inscribed in a circle.
Given: Let ABCD be a cyclic parallelogram
Question 15
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.
Given: A circle with center O
ABCD is a rectangle inscribed in a circle
with diagonals AC & BD
Question 16
Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
Let’s do this step-by-step
Step 1 of 6
The Single Chord
Let's start with an outer circle of radius
. We draw a single chord of a fixed length
.
The green dot marks the exact midpoint of this chord.
Previous
Next StepStep 2 of 6
2. Multiple Chords
Now, imagine drawing many different chords around the circle, all with the exact same length
.
We mark the midpoint of each one. Notice a pattern starting to emerge?
Previous
Next StepStep 3 of 6
3. The Locus (The Answer)
If we draw every possible chord of length L, their midpoints form a continuous path.
Question 17
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 ".
Given: A circle with center O. Two equal chords, AB and AC
i.e. AB = AC.
Question 18
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.
Let’s draw the figure
Question 19
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.
We have a regular hexagon inscribed in a circle with center O and radius r.
Question 20
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’s draw the figure
Question 21
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).
Let’s draw the figure
Question 22
There is no chord of a circle that is longer than its diameter." How do you justify this statement?
Let’s do this step-by-step
Step 1 of 5
The Diameter
Let's start with a circle with center
and a radius of
.
The diameter is a special chord that passes straight through the center. Its total length is exactly
.
Previous
Next StepStep 2 of 5
2. A Random Chord
Now, let's draw any other random chord, AB, that does not pass through the center.
We want to prove that this red chord AB must be strictly shorter than the blue diameter.
Previous
Next StepStep 3 of 5
3. Forming a Triangle
Draw a line from the center
to point
, and another from
to
.
These are both radii of the circle, so their lengths are both exactly
.
Notice that we have now formed a triangle:
.
Previous
Next StepStep 4 of 5
4. The Triangle Inequality
A fundamental rule of geometry is the Triangle Inequality Theorem: The shortest distance between two points is a straight line.
This means walking from A to B directly (the chord) is always shorter than taking the detour from A to O , and then O to B .
Previous
Next StepStep 5 of 5
5. The Final Justification
Let's write this out mathematically:
Distance
Distance
Question 23
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.
View solutionQuestion 24
How would you use the following figure to justify the statement that the angle in a semicircle is 90^∘?
Let’s label the points at the end as M & N
We need to prove
∠ MAN = 90°
Question 25
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.
We use the concept of reflectional symmetry across the diameter to prove this.
Question 26
How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180^∘?
Here,
Each small triangle (eg: ∆ OBC) is an isosceles triangle
And, angles opposite to equal sides are equal
So, ∠ OCB = ∠ OBC = q
Similarly, we can find all angles
And, at the end we can use Angle sum property of Quadrilateral to prove opposite angles are supplementary
Why Learn This With Teachoo?
I’m Up and Down, and Round and Round is Chapter 5 of NCERT Class 9 Ganita Manjari Part 1. It is the book’s circle-geometry chapter, covering circle definitions, symmetry, chords, arcs, perpendicular bisectors, distances from the centre, concyclic points and cyclic quadrilaterals. Students learn not only the results but also how to prove major circle theorems. Teachoo provides concept-wise explanations and solutions for Exercise Sets 5.1 to 5.6 and the end-of-chapter exercises.
Basic circle ideas and symmetry
A circle is the set of all points in a plane at a fixed distance from a fixed point. The fixed point is the centre and the fixed distance is the radius. A chord joins two points on the circle; a diameter is a chord through the centre and has length twice the radius. An arc is part of the circumference, and a sector and segment describe different enclosed regions.
Circles possess rotational and reflection symmetry. Every diameter is an axis of reflection, and rotation about the centre through any angle maps the circle onto itself. Students also investigate how many circles can pass through one, two or three given points. Through three non-collinear points, exactly one circle can be drawn; its centre is the intersection of perpendicular bisectors of two joining segments.
Chords and their relation to the centre
Equal chords of the same circle subtend equal angles at the centre, and the converse is also true. A perpendicular from the centre to a chord bisects the chord. Conversely, the line from the centre to the midpoint of a chord is perpendicular to that chord.
Equal chords are equidistant from the centre, while chords closer to the centre are longer. The diameter is therefore the longest chord. For unequal chords, comparing their distances from the centre allows their lengths to be ordered.
Arcs, concyclicity and cyclic quadrilaterals
The angle subtended by an arc depends on where the vertex lies. The angle at the centre is twice the angle subtended by the same arc at a point on the remaining circle. Angles in the same segment are equal, and an angle in a semicircle is a right angle.
Points lying on the same circle are concyclic. A cyclic quadrilateral has all four vertices on one circle. Its opposite angles are supplementary, and an exterior angle equals the opposite interior angle. These properties allow unknown angles to be found and concyclicity to be established.
Topics covered on Teachoo
Teachoo includes:
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circle definitions and terminology;
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symmetries of a circle;
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number of circles through given points;
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Exercise Set 5.1;
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Think, Draw and Infer on page 98;
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angles subtended by chords;
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Exercise Set 5.2;
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midpoints and perpendicular bisectors of chords;
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Exercise Set 5.3;
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distances of equal and unequal chords from the centre;
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Exercise Sets 5.4 and 5.5;
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angles subtended by arcs;
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Exercise Set 5.6;
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concyclic points and cyclic quadrilaterals;
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theorems with proofs; and
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end-of-chapter exercise solutions.
Learning outcomes
Students should be able to use correct circle terminology, construct the centre from chords and apply chord, arc and cyclic-quadrilateral theorems. They should determine unknown angles, compare chord lengths through centre distances and prove points concyclic. They should also write a proof as a chain of given facts, established theorems and justified conclusions.
Why is this chapter important?
Circle geometry supports constructions, coordinate geometry, trigonometry and design. It develops proof more deeply than formula-based chapters. Many results are connected: perpendicular bisectors locate the centre, central angles relate to chords, and arc-angle theorems lead to cyclic quadrilateral properties.
How Teachoo helps you study
Teachoo organises the chapter by theorem and exercise set. Draw a clean diagram and label the centre, radii, chords and arcs. Join the centre to chord endpoints when equal radii may create isosceles triangles. Mark the exact arc associated with an angle before applying the central-angle theorem.
For proofs, write the reason beside each claim. Attempt the proof independently, then compare its logical sequence with Teachoo’s solution. A diagram can suggest a result, but only the stated facts and theorems establish it.
Common mistakes to avoid
Do not assume the centre lies at the visual middle of a sketch. Chord theorems apply within the same circle or equal circles under stated conditions. The angle at the centre and angle at the circumference must subtend the same arc. Opposite angles are supplementary only after the quadrilateral is known to be cyclic.
Deeper reasoning and concept connections
A student has understood I’m Up and Down, and Round and Round (Ganita Manjari Part 1) only when the idea can be moved between words, diagrams, examples and mathematical notation. Start with a concrete example, identify what changes and what remains fixed, represent the relationship clearly and then state the rule. This movement between representations is important because school and competency questions often present a familiar idea in an unfamiliar form.
The chapter should also be connected to earlier and later mathematics. Definitions supply the language, worked examples reveal the method, and mixed questions test whether the method can be selected without a hint. Instead of memorising the appearance of a solved question, ask what information triggered the method, which condition made it valid and how the answer could be checked. That makes learning transferable to later chapters rather than limited to one exercise.
How to solve unfamiliar and competency-based questions
Read the complete problem before calculating. Underline the quantities, conditions and command word—find, compare, construct, justify, estimate or prove. Rephrase the task in one sentence and choose a representation such as a table, labelled figure, number line, expression or graph. Solve in small steps, keeping units and labels visible.
For an application question, the final line must answer the situation, not only display a number. For an assertion–reason question, test the assertion and reason separately before deciding whether one explains the other. For an MCQ, eliminate options using definitions, signs, size estimates or boundary cases before performing long calculations. If the answer is visual, check it against the stated scale or construction conditions rather than the appearance of the drawing.
What complete mastery looks like
For I’m Up and Down, and Round and Round (Ganita Manjari Part 1), a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.
Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.
Additional frequently asked questions
What should a student know before starting I’m Up and Down, and Round and Round (Ganita Manjari Part 1)?
Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.
How can a student check an answer in I’m Up and Down, and Round and Round (Ganita Manjari Part 1)?
Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.
How many questions are enough for strong preparation?
There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.
How should Teachoo solutions be used without becoming dependent on them?
Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.
Frequently asked questions
How many circles pass through three points?
Exactly one circle passes through three non-collinear points. No circle passes through three distinct collinear points.
What happens when a perpendicular is drawn from the centre to a chord?
It bisects the chord.
What is the key property of a cyclic quadrilateral?
Its opposite interior angles add to 180°.
Redraw the figure, identify the relevant arc or chord and justify every step. That is the reliable method for circle proofs.