Class 9 Maths Part 1 Chapter 5 – I’m Up and Down, and Round and Round from Ganita Manjari with easy Notes and complete Book Solutions. Get clear explanations, important concepts, questions and answers, and revision material for Chapter 5 to help Class 9 students learn, practise and prepare effectively for examinations.
Chapter 5 – I’m Up and Down, and Round and Round
Class 9 Mathematics – Easy Notes, Concepts & Book Solutions
📘 Easy Notes & Concepts
⭕ Circle
A circle is the set of all points in a plane which are at the same distance from a fixed point called the centre.
📏 Radius
The distance from the centre of a circle to any point on the circle is called its radius.
📏 Diameter
A diameter is a chord which passes through the centre.
Diameter = 2 × Radius
📐 Chord
A chord is a line segment joining any two points on a circle.
🔄 Symmetry
Every diameter of a circle is a line of reflection symmetry. A circle also has rotational symmetry through every angle.
⭕ Circumcircle
Three non-collinear points determine one unique circle. This circle is called their circumcircle.
📏 Equal Chords
Equal chords of the same circle are equally distant from the centre.
📐 Angle in a Circle
The angle at the centre is twice the angle subtended by the same arc at a point on the circle.
🔲 Cyclic Quadrilateral
A quadrilateral whose four vertices lie on a circle is called a cyclic quadrilateral.
Opposite angles = 180°
Important formula:
If a chord has length L, the radius is r and the perpendicular distance of the chord from the centre is d:
L = 2√(r² − d²)
If a chord has length L, the radius is r and the perpendicular distance of the chord from the centre is d:
L = 2√(r² − d²)
💡 Remember:
Radius → centre to circle
Diameter → chord through centre
Chord → joins two points of a circle
Longer chord → nearer to centre
Radius → centre to circle
Diameter → chord through centre
Chord → joins two points of a circle
Longer chord → nearer to centre
5.1 Definitions – Think and Reflect
Think and Reflect: 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?
Draw any two different chords of the circle.Construct the perpendicular bisector of each chord.The two perpendicular bisectors meet at one point.That point is the centre of the circle.
Why? The centre is equally far from the two endpoints of every chord, so it lies on the perpendicular bisector of the chord.
Why? The centre is equally far from the two endpoints of every chord, so it lies on the perpendicular bisector of the chord.
5.2 Symmetries of a Circle – Think and Reflect
1. 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?
Square:It looks the same after rotations of 90°, 180°, 270° and 360°.It has 4 lines of reflection symmetry.
Regular pentagon:Rotational symmetry has order 5.It has 5 lines of reflection symmetry.
Regular hexagon:Rotational symmetry has order 6.It has 6 lines of reflection symmetry.
Regular pentagon:Rotational symmetry has order 5.It has 5 lines of reflection symmetry.
Regular hexagon:Rotational symmetry has order 6.It has 6 lines of reflection symmetry.
2. What is the length of the longest chord in a circle of radius
5 units? Is there a smallest chord?
The longest chord is the diameter.Diameter = 2 × radius= 2 × 5= 10 units.
There is no smallest positive chord because chords can be made smaller and smaller.
There is no smallest positive chord because chords can be made smaller and smaller.
3. 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?
The locus is the perpendicular bisector of the line segment
joining the two given points.Every point on this line is equally far from the two points.
5.3 How Many Circles? – Think and Reflect
1. How many circles pass through two points on a plane?
Infinitely many circles.Their centres can be any points on the perpendicular bisector of the
line segment joining the two points.
2. 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?
The smallest possible radius is:
AB ÷ 2
This happens when AB is the diameter.There is no largest radius. The radius can become larger and larger.
AB ÷ 2
This happens when AB is the diameter.There is no largest radius. The radius can become larger and larger.
3. As you move away from segment AB along its perpendicular bisector,
do the radii of the circles containing A and B increase or decrease?
They increase.The centre moves farther away from A and B, so the radius increases.
4. As you go along the perpendicular bisector, will the circle drawn
from that point through A and B appear more curved or less curved?
It appears less curved as the radius becomes larger.A very large circle looks almost like a straight line over a small part.
5. 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?
If A and B only have to be on the boundary, there can be
infinitely many squares.If A and B are two vertices of the square, there are
4 possible squares.
Exercise Set 5.1
1. 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 AB = 5 cm.At A construct an angle of 70°.At B construct an angle of 60°.The two rays meet at C.Now construct the perpendicular bisectors of two sides.Their intersection is the circumcentre.Since all three angles are less than 90°, the triangle is acute.Therefore the circumcentre lies inside the triangle.
2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the
circumcircle of ΔABC. Is the centre inside or outside the triangle?
Construct AB = 5 cm.At A construct 100°.On this ray mark AC = 4 cm.Construct perpendicular bisectors.Their intersection is the circumcentre.Since the triangle has an obtuse angle of 100°, the circumcentre
lies outside the triangle.
3. 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.
Construct the triangle using the three given sides.Draw perpendicular bisectors.Their intersection is O.Since OA, OB and OC are radii of the same circle:
OA = OB = OC
So all three measurements should be equal.
OA = OB = OC
So all three measurements should be equal.
4. What is the least possible radius of a circle through two points
A and B?
The least possible radius is:
AB ÷ 2
This happens when AB is the diameter.
AB ÷ 2
This happens when AB is the diameter.
5.4 Chords and the Angles They Subtend – Exercise Set 5.2
1. Show that the triangle formed by a chord and the centre of the
circle is isosceles.
Let AB be a chord and O be the centre.Join OA and OB.OA and OB are radii.Therefore:OA = OBA triangle having two equal sides is an isosceles triangle.Hence ΔAOB is isosceles.
2. Show that if two such isosceles triangles have equal base length,
they are congruent to each other.
Both triangles have equal radii.Their bases are also equal.Therefore all three corresponding sides are equal.Hence the triangles are congruent by SSS.
5.5 Midpoints and Perpendicular Bisectors of Chords – Exercise Set 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.)
Join CA and CB.CA = CB because both are radii.CM is common.∠CMA = ∠CMB = 90°.Therefore ΔCMA and ΔCMB are congruent by RHS.Hence:AM = BMTherefore CM bisects chord AB.
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.
Since AB = AC, the triangle is isosceles.The altitude from A also bisects BC.Therefore its midpoint is equally distant from B and C.The perpendicular bisector of BC passes through the centre of the
circle.Hence the altitude from A passes through the centre.
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.
For the 6 cm chord:Half chord = 3 cm.Distance from centre:√(5² − 3²)= √(25 − 9)= √16= 4 cm.
For the 8 cm chord:Half chord = 4 cm.Distance from centre:√(5² − 4²)= √(25 − 16)= √9= 3 cm.
The chords are on opposite sides.Therefore distance between their midpoints:4 + 3 = 7 cm
For the 8 cm chord:Half chord = 4 cm.Distance from centre:√(5² − 4²)= √(25 − 16)= √9= 3 cm.
The chords are on opposite sides.Therefore distance between their midpoints:4 + 3 = 7 cm
5.6 Distance of Chords from the Centre – Exercise Set 5.4
1. Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must
be true.
For equal chords, their half-chords are equal.Let the common half-chord be x and radius be r.Using Pythagoras:r² = x² + d²Since r and x are the same for both chords, d must also be the same.Therefore equal chords are equally distant from the centre.
2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is
perpendicular to GH, and CE = CH, show that AB = GF.
The perpendicular from the centre to a chord bisects the chord.Since:CE = CHthe corresponding half-chords are equal.Therefore:AB = GF
3. Solve the previous question using the Baudhāyana–Pythagoras theorem.
Let half of AB be x and half of GF be y.For the first chord:r² = CE² + x²For the second chord:r² = CH² + y²Given CE = CH.Therefore:x² = y²Since lengths are positive:x = yHence:AB = GF
5.6.1 Which of the two unequal chords is farther from the centre? – Exercise Set 5.5
1. Find the length of the chord of a circle where the radius is
7 cm and perpendicular distance is 6 cm.
Half chord:√(7² − 6²)= √(49 − 36)= √13 cm.Therefore chord:2√13 cm≈ 7.21 cm
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√(r²−d²).
The perpendicular from the centre bisects the chord.Let half the chord be x.Using Pythagoras:r² = d² + x²Therefore:x² = r² − d²x = √(r² − d²)The full chord is twice x.Therefore:Chord = 2√(r² − d²)
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 = 2 AB? Give reasons for your answer.
No.Chord length is not directly proportional to its distance from the
centre.For a fixed circle:Chord = 2√(r² − d²)So doubling d does not mean doubling the chord length.
5.7 Angles Subtended by an Arc
Theorem:The angle subtended by an arc at the centre of a circle is twice the
angle subtended by the same arc at a point on the circle.
Angle at centre = 2 × Angle at circumference
What is the angle subtended by a diameter at any point on the circle?
A diameter subtends 180° at the centre.Therefore:Angle at circle = 180° ÷ 2= 90°.Hence an angle in a semicircle is always 90°.
Exercise Set 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?
Join O to the midpoint M of AB.∠AOM = 30°.OA = 12 cm.Half chord:12 × sin 30°= 12 × 1/2= 6 cm.Therefore:AB = 12 cm.
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?
(i) No. Angles standing on the same chord in the same segment
are equal.
(ii) No. Equal angles alone do not prove that the points lie on the same side.
(iii) No. The result about equal angles applies when the points are in the required cyclic configuration. If X and Y are arbitrary points outside the circle, equality alone is not sufficient.
(ii) No. Equal angles alone do not prove that the points lie on the same side.
(iii) No. The result about equal angles applies when the points are in the required cyclic configuration. If X and Y are arbitrary points outside the circle, equality alone is not sufficient.
3. Find x in Fig. 5.26.
Angles standing on the same chord are equal.Therefore:x = 100°
5.8 Concyclicity of Points
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.
For a cyclic quadrilateral, opposite angles add to 180°.Check:∠A + ∠C= 80° + 100°= 180°Also:∠B + ∠D= 110° + 70°= 180°Therefore yes, such a cyclic quadrilateral can be drawn.
⭐ Important Results for Revision
Result 1
The perpendicular from the centre to a chord bisects the chord.
The perpendicular from the centre to a chord bisects the chord.
Result 2
Equal chords are equally distant from the centre.
Equal chords are equally distant from the centre.
Result 3
The longer chord is nearer to the centre.
The longer chord is nearer to the centre.
Result 4
Angle at centre = 2 × angle at circumference.
Angle at centre = 2 × angle at circumference.
Result 5
An angle in a semicircle is 90°.
An angle in a semicircle is 90°.
Result 6
Opposite angles of a cyclic quadrilateral add to 180°.
Opposite angles of a cyclic quadrilateral add to 180°.
📐 Picture – Chord and Centre
The perpendicular from the centre to a chord bisects the chord.
📏 Diameter – Longest Chord
AB passes through the centre, so AB is the diameter. The diameter is the longest chord of a circle.
📝 One-Page Revision
- A circle has a fixed centre and fixed radius.
- Diameter = 2 × radius.
- A diameter is the longest chord.
- Three non-collinear points determine one unique circle.
- The perpendicular from the centre to a chord bisects it.
- Equal chords are equally distant from the centre.
- The longer chord is closer to the centre.
- Central angle = 2 × angle at circumference.
- An angle in a semicircle is 90°.
- Opposite angles of a cyclic quadrilateral add to 180°.
prepare more with the given Worksheet of chapter 5, which will help you to improve your knowledge .
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