Concept Scroll LogoConceptScroll
Join or Sign In
MathematicsClass 9I’m Up and Down, and

I’m Up and Down, and | Class 9 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 3 min read

I’m Up and Down, and | Class 9 Mathematics Notes

I’m Up and Down, and – this guide gives you a concise, exam-ready overview of I’m Up and Down, and from Class 9 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

5.2 SYMMETRIES OF A CIRCLE

This section explores the symmetry properties of circles. Circles exhibit perfect rotational symmetry, meaning that rotating a circle about its centre by any angle results in the same shape. Reflection symmetry is also present: every diameter acts as a line of reflection symmetry. Folding a circular paper along a diameter causes the boundaries to overlap perfectly, illustrating this property. The section encourages students to think about symmetries in other polygons, such as squares, pentagons, and hexagons, and to consider the longest and shortest chords in a circle. It also introduces the concept of the locus of points equidistant from two given points, which is the perpendicular bisector of the segment joining those points.

📊 Diagram: No specific figure referenced in this section.

🧪 Activity: Activity: Fold a circular paper to find lines of reflection symmetry.

🔗 Connection: Understanding symmetries leads to exploring how many circles can pass through given points.

Frequently asked questions

1. Show that the triangle formed by a chord and the centre of the circle is isosceles.

Consider a circle with centre O and a chord AB. Join OA and OB. Since OA and OB are radii of the circle, OA = OB. Therefore, triangle OAB has two equal sides and is isosceles.

2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

Let two isosceles triangles OAB and OCD be formed by chords AB and CD respectively, with OA = OB and OC = OD (radii). If AB = CD (equal bases), then by the Side-Side-Side (SSS) congruence criterion, triangles OAB and OCD are congruent because OA = OC (radii), OB = OD (radii), and AB = CD (given).

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.)

Solution:

Given: A chord AB of a circle with centre C, and a perpendicular CM from C to AB.

To prove: CM bisects AB, i.e., AM = BM.

Proof: 1. Since CM is perpendicular to AB, ∠CMA = ∠CMB = 90°. 2. In triangles CMA and CMB:

  • CM = CM (common side)
  • CA = CB (radii of the circle)
  • ∠CMA = ∠CMB = 90°

3. By RHS congruence criterion, ΔCMA ≅ ΔCMB. 4. Therefore, corresponding parts are equal, so AM = BM.

Hence, the perpendicular from the centre of a circle to a chord bisects the chord.

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.

Solution:

Given: Triangle ABC is isosceles with AB = AC, inscribed in a circle with centre O.

To prove: The altitude from A to BC passes through the centre O.

Proof: 1. Since AB = AC, triangle ABC is isosceles with vertex A. 2. The altitude from A to BC is also the median and angle bisector of angle A. 3. Let the altitude from A meet BC at D. 4. Since ABC is inscribed in the circle, points A, B, C lie on the circle. 5. The perpendicular bisector of chord BC passes through the centre O. 6. Sin

Ready to ace this chapter?

Get the full I’m Up and Down, and chapter — interactive notes, diagrams, worked solutions, polls and a free practice quiz — in the ConceptScroll app.

Open in ConceptScroll →

Study smarter with ConceptScroll

Daily NCERT-aligned reels, AI doubt solving and chapter quizzes — all free.

Start learning free
#cbse notes#class 9#mathematics#ncert

Continue reading

← All articles