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Class 9th Mathematics Chapter 11 Circles – PYQ

Previous year questions (PYQ) with answers for Class 9th Mathematics Chapter 11 Circles, based on the NCERT CBSE syllabus.

Topics covered in Circles

Frequently asked questions

Where can I find previous year questions (PYQ) for Class 9th Mathematics Chapter 11 Circles?

This page lists previous year questions with answers for Circles. Topics covered: Chord and arc, Angle subtended by chord, Perpendicular from centre, Cyclic quadrilateral, Equal chords.

Are Circles PYQs free?

Yes, Notebook gives free previous year questions for Class 9 Mathematics.

How to prepare Circles for Class 9 Mathematics exams?

Read the Circles notes first, revise the key topics (Chord and arc, Angle subtended by chord, Perpendicular from centre, Cyclic quadrilateral, Equal chords), then practise the important questions and previous year questions (PYQ) for this chapter.

Where can I find Class 9th Mathematics Chapter 11 Circles pyq for quick revision?

You are on the right page: this pyq page for Circles covers Class 9 Mathematics for the CBSE exam 2026-27. Use the tabs on this chapter for notes, important questions and PYQ.

Is Class 9th Mathematics Chapter 11 Circles important for the CBSE exam 2026-27?

Circles is part of the Class 9 Mathematics NCERT CBSE syllabus. Notebook gives notes, important questions and PYQs for it, free.

Also see: Class 9th Mathematics Chapter 11 Circles Notes · Class 9th Mathematics Chapter 11 Circles Important Questions

Circles

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1 Mark CBSE 2023
What is the total surface area of a cube with side 4 cm?
Ans: TSA of cube = 6a² = 6 × 16 = 96 cm².
3 Marks CBSE 2022
A cone has base radius 6 cm and slant height 10 cm. Find its curved surface area and volume.
Ans: CSA = πrl = π×6×10 = 60π ≈ 188.4 cm². Height h = √(l²−r²) = √(100−36) = 8 cm. Volume = ⅓πr²h = ⅓×π×36×8 = 96π ≈ 301.4 cm³.
5 Marks CBSE 2021
A hemispherical bowl has inner radius 5 cm. Find its volume and inner curved surface area. If it is made of metal 0.25 cm thick, find the outer TSA. (π=3.14)
Ans: Inner: CSA = 2πr² = 2×3.14×25 = 157 cm². Volume = (2/3)πr³ = (2/3)×3.14×125 = 261.67 cm³. Outer radius = 5.25 cm. Outer TSA = 2πR² + π(R²−r²) = π[2×5.25²+(5.25²−5²)] = π[55.125+2.5625] = 3.14×57.69 ≈ 181.2 cm².
Other chapters in Mathematics · Class 9 View all Mathematics chapters →
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