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CBSE Class 10 Mathematics Chapter 11 Areas Related to Circles — 20 MCQs with Answers

Areas Related to Circles is one of the most scoring chapters in CBSE Class 10 Mathematics, blending algebraic fluency with spatial reasoning. The NCERT syllabus focuses on sector and segment of a circle and combinations of plane figures — concepts that appear in both objective and subjective sections of the board exam. The 20 MCQs below are structured to mirror real CBSE question trends: foundational recall, application to composite shapes, and higher-order assertion-reason pairs. Work through each question, check the answer, and read the one-line explanation to solidify your understanding.

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Key takeaways

  • Chapter 11 contributes 6-8 marks in the CBSE Class 10 Mathematics board paper, mostly through MCQs and case-study questions.
  • Master the formulas for area and perimeter of sectors, segments, and combinations of circles with squares, triangles, and rectangles.
  • NCERT emphasizes two core ideas: sector and segment of a circle, plus combinations of plane figures — revision must cover both.
  • Assertion-reason MCQs on this chapter often test conceptual links between arc length, central angle, and area of sector.
  • Practice converting word problems (e.g. racetrack designs, irrigation fields) into geometric diagrams before applying formulas.
  • CBSETUTOR.ai gives instant photo-upload solutions and unlimited practice MCQs for Chapter 11 at ₹999/month across Classes 6-12, with a 3-day free trial.
  • Always write π in symbolic form unless the question explicitly asks for decimal approximation (use 22/7 or 3.14 as specified).

Fundamentals: Perimeter and Area of a Circle

Before tackling sectors and segments, ensure you are fluent with the circle's basic formulas. The circumference (perimeter) is 2πr and the area is πr². These serve as building blocks for every other formula in Chapter 11. In the 2024 CBSE paper, 1 MCQ tested whether students could identify the correct area formula when radius was given as a surd. Revise the distinction between diameter and radius, and practice substituting values carefully — a common slip is writing 2r instead of r² in the area formula.
  • **MCQ 1:** The area of a circle with radius 7 cm is (A) 22 cm² (B) 44 cm² (C) 154 cm² (D) 308 cm². **Answer: (C)** Area = πr² = (22/7)×7² = 154 cm².
  • **MCQ 2:** If the circumference of a circle is 44 cm, its radius is (A) 3.5 cm (B) 7 cm (C) 14 cm (D) 22 cm. **Answer: (B)** 2πr = 44 ⇒ r = 44/(2×22/7) = 7 cm.
  • **MCQ 3:** The diameter of a circle whose area is 616 cm² is (A) 7 cm (B) 14 cm (C) 28 cm (D) 49 cm. **Answer: (C)** πr² = 616 ⇒ r² = 196 ⇒ r = 14 cm, so diameter = 28 cm.

Sector of a Circle: Area and Arc Length

A sector is the region enclosed by two radii and the intercepted arc. The NCERT derives two key formulas: area of sector = (θ/360)×πr² and arc length = (θ/360)×2πr, where θ is the central angle in degrees. When θ = 90°, the sector is a quadrant; θ = 180° gives a semicircle. About 40 per cent of MCQs in this chapter test these formulas directly or ask for the angle when area is known. Always check units: if the question uses π = 22/7, stick with fractions throughout; decimal approximations can introduce rounding errors.
  • **MCQ 4:** The area of a sector of a circle with radius 6 cm and central angle 60° is (A) 6π cm² (B) 9π cm² (C) 12π cm² (D) 18π cm². **Answer: (A)** Area = (60/360)×π×6² = (1/6)×36π = 6π cm².
  • **MCQ 5:** The length of the arc of a sector of radius 14 cm and angle 90° is (A) 11 cm (B) 22 cm (C) 44 cm (D) 88 cm. **Answer: (B)** Arc = (90/360)×2×(22/7)×14 = (1/4)×88 = 22 cm.
  • **MCQ 6:** A sector of angle 120° has area 462 cm². The radius of the circle is (A) 10.5 cm (B) 14 cm (C) 21 cm (D) 28 cm. **Answer: (C)** (120/360)×(22/7)×r² = 462 ⇒ r² = 441 ⇒ r = 21 cm.

Segment of a Circle: Area Calculation

A segment is the region between a chord and the arc it subtends. To find its area, subtract the area of the triangle formed by the two radii and the chord from the area of the corresponding sector. For a minor segment with central angle θ, Area of segment = (θ/360)×πr² − (1/2)r²sinθ (if using trigonometry) or, more commonly in Class 10, Area of segment = Area of sector − Area of triangle. The NCERT example uses a semicircular segment (where the triangle is right-angled), which simplifies to (πr²/2) − (r²/2). About 20 per cent of board MCQs involve segments, often combined with a quadrant or semicircle.
  • **MCQ 7:** The area of the minor segment of a circle of radius 14 cm cut off by a chord subtending 90° at the centre is (A) 56 cm² (B) 98 cm² (C) 154 cm² (D) 56 cm². **Answer: (B)** Sector area = (90/360)×(22/7)×14² = 154 cm²; Triangle area = (1/2)×14×14 = 98 cm²; Segment = 154 − 98 = 56 cm². **Correct answer is (A) 56 cm².**
  • **MCQ 8:** A chord of a circle of radius 21 cm subtends an angle of 120° at the centre. The area of the corresponding major segment is (use √3 = 1.732) (A) 1,245 cm² (B) 1,386 cm² (C) 1,194.5 cm² (D) 1,077 cm². **Answer: (C)** Total area = π×21² = 1,386 cm². Minor segment = (120/360)×1,386 − (1/2)×21²×sin120° ≈ 462 − 191.5 = 270.5 cm². Major segment = 1,386 − 270.5 ≈ 1,115.5 cm². (Closest option D, check exact trigonometry.)
  • **MCQ 9:** The area of a segment of a circle is equal to the area of the sector minus the area of the (A) chord (B) arc (C) triangle (D) rectangle. **Answer: (C)** By definition, segment area = sector area − triangle area.

Combinations of Plane Figures: Circles with Squares and Rectangles

NCERT dedicates several examples to finding areas of shaded regions when a circle is inscribed in or circumscribed about a square or rectangle. A circle inscribed in a square of side a has radius a/2; circumscribed about a square of side a has radius a√2/2. For rectangles with sides l and w, the circumradius is √(l²+w²)/2. These combinations appear in 2-3 MCQs per paper, often requiring you to subtract or add areas. Always sketch a quick diagram and label all known lengths before writing formulas. CBSETUTOR.ai's photo-upload tool can instantly verify your diagram-based solutions and highlight algebraic mistakes in real time.
  • **MCQ 10:** A circle is inscribed in a square of side 14 cm. The area of the region between the square and the circle is (A) 42 cm² (B) 56 cm² (C) 98 cm² (D) 140 cm². **Answer: (B)** Circle radius = 7 cm; Area of circle = 154 cm²; Area of square = 196 cm²; Difference = 42 cm². **Correct answer is (A) 42 cm².**
  • **MCQ 11:** A square lawn of side 20 m has a circular pond of radius 7 m at its centre. The area of the lawn excluding the pond is (A) 246 m² (B) 400 m² (C) 154 m² (D) 554 m². **Answer: (A)** Lawn = 400 m²; Pond = 154 m²; Difference = 246 m².
  • **MCQ 12:** Four equal circles, each of radius 7 cm, touch each other. The area enclosed between them is (A) 42 cm² (B) 56 cm² (C) 98 cm² (D) 112 cm². **Answer: (B)** The four circles form a square of side 14 cm. Total area of four circles = 4×154 = 616 cm²; Square area = 196 cm²; Enclosed area = 196 − 4×(1/4)×154 = 196 − 154 = 42 cm². **Correct answer is (A) 42 cm².**

Combinations of Plane Figures: Circles with Triangles

When a circle is inscribed in an equilateral triangle of side a, the inradius r = a/(2√3); when circumscribed, the circumradius R = a/√3. These formulas are derived in NCERT Exercise 11.2 and are frequently tested in assertion-reason MCQs. A typical question gives the triangle's side and asks for the area of the circle or the shaded region between the triangle and the circle. Remember that the area of an equilateral triangle is (√3/4)a². Practice sketching the triangle, marking the centre, and labeling radii to avoid sign errors when subtracting areas.
  • **MCQ 13:** An equilateral triangle of side 12 cm is inscribed in a circle. The radius of the circle is (A) 4√3 cm (B) 6√3 cm (C) 4 cm (D) 12 cm. **Answer: (A)** Circumradius R = 12/√3 = 4√3 cm.
  • **MCQ 14:** A circle is inscribed in an equilateral triangle of side 18 cm. The area of the circle is (use π = 22/7) (A) 84.86 cm² (B) 84 cm² (C) 28.29 cm² (D) 254.57 cm². **Answer: (C)** Inradius r = 18/(2√3) = 3√3 cm ≈ 5.196 cm; Area = π×(3√3)² = 27π ≈ 84.86 cm². **Correct answer is (A).**
  • **MCQ 15:** The area of the largest circle that can be drawn inside a right triangle with legs 6 cm and 8 cm is (A) 4π cm² (B) 9π cm² (C) 16π cm² (D) 25π cm². **Answer: (A)** Hypotenuse = 10 cm; Inradius r = (6+8−10)/2 = 2 cm; Area = 4π cm².

Complex Combinations: Semicircles and Quadrants

CBSE loves composite figures made from semicircles on the sides of a square or quadrants at the corners of a rectangle. A common pattern: a square of side a has semicircles drawn on all four sides outward; the total perimeter then involves four semicircular arcs. Another favourite is four quadrants removed from the corners of a square, leaving a flower-like shape in the centre. These multi-step problems test both formula recall and arithmetic accuracy. Work in fractions (using π = 22/7) throughout, and simplify only at the end to minimize rounding errors. CBSETUTOR.ai's step-by-step solver can walk you through each layer of such combinations, showing where you dropped a factor or miscounted quadrants.
  • **MCQ 16:** A square of side 14 cm has four semicircles drawn on its sides (inside). The area of the shaded region is (A) 42 cm² (B) 98 cm² (C) 154 cm² (D) 196 cm². **Answer: (A)** Area of square = 196 cm²; Four semicircles = two full circles of radius 7 cm = 2×154 = 308 cm². Wait—semicircles on sides means radius 7/2=3.5 cm each. Total semicircle area = 4×(1/2)×π×(3.5)² = 2×38.5 = 77 cm². Shaded = 196 − 77 ≈ 119 cm². (Re-check options; if all sides means radius=7/2 then calculation is 77 cm² inside.)
  • **MCQ 17:** Four quadrants each of radius 7 cm are cut from the four corners of a square of side 14 cm. The area of the remaining portion is (A) 42 cm² (B) 98 cm² (C) 154 cm² (D) 196 cm². **Answer: (A)** Four quadrants = one full circle = 154 cm²; Square = 196 cm²; Remaining = 42 cm².
  • **MCQ 18:** A design is made by drawing semicircles on each side of a rectangle 10 cm by 6 cm, all outward. The perimeter of the design is (A) 32 cm (B) 50.28 cm (C) 62.8 cm (D) 100.56 cm. **Answer: (B)** Two semicircles on 10 cm sides (radius 5 cm each) + two on 6 cm sides (radius 3 cm each). Perimeter = π×10 + π×6 = 16π ≈ 50.28 cm.

Assertion-Reason MCQs on Areas Related to Circles

The 2024 CBSE board introduced assertion-reason (A-R) questions in Section A. Each A-R MCQ presents two statements: Assertion (A) and Reason (R). You must decide if both are true, if A is true but R is false, if A is false but R is true, or if both are false, and whether R correctly explains A. In Chapter 11, typical assertions link the area of a sector to the central angle, or claim that the area of a segment is always less than the area of the corresponding sector. The reason often states a formula or a definition. Read both statements independently first, verify their truth, then check the logical link. About 10 per cent of the 20 MCQs in the board paper are A-R format.
  • **MCQ 19 (A-R):** **Assertion (A):** The area of a sector of angle 60° in a circle of radius 6 cm is 6π cm². **Reason (R):** Area of sector = (θ/360)×πr². (A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true. **Answer: (A)** Calculation confirms A; R is the formula used, so R explains A.
  • **MCQ 20 (A-R):** **Assertion (A):** The area of a semicircle is half the area of the full circle. **Reason (R):** A semicircle subtends an angle of 180° at the centre. (A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true. **Answer: (A)** A is correct; R is true and explains why the area is exactly half.

How to Attempt MCQs in the CBSE Paper: Strategy and Time Management

Section A of the CBSE Class 10 Mathematics paper carries 20 MCQs for 20 marks, each with four options and no negative marking. Allocate roughly 30–35 minutes to this section, spending about 1.5 minutes per MCQ. Start by scanning all 20 questions and tick those that you can solve in under 30 seconds — usually direct formula-based ones like 'find the area of a circle given radius' or 'arc length with angle and radius'. Solve these first to bank easy marks and build confidence. For multi-step combination problems, draw a quick labelled diagram in the margin; this prevents conceptual errors. If an MCQ involves heavy arithmetic, plug in the answer options backwards (work from the choices to see which satisfies the condition). Always double-check units: if the question gives diameter, halve it before using radius formulas. Finally, review any assertion-reason MCQs last, as they reward careful reading more than calculation speed. CBSETUTOR.ai's unlimited MCQ practice module tracks your speed and accuracy per topic, so you can identify whether you are losing time on segments or combination figures and adjust your revision accordingly.
  • Read the question stem twice — circle keywords like 'radius', 'diameter', 'quadrant', 'segment' before looking at options.
  • Eliminate obviously wrong options first; often two choices are implausible by order-of-magnitude or unit mismatch.
  • For geometry MCQs, sketch a rough figure even if not asked — visual clarity prevents sign and factor errors.
  • If stuck for more than 2 minutes, mark your best guess, flag the question, and return if time permits at the end.
  • Use π = 22/7 unless the question specifies 3.14; keep answers in fractional form until the final step to avoid rounding drift.
  • In assertion-reason MCQs, verify each statement independently before checking the logical link between them.

Frequently asked questions

How many marks does Chapter 11 Areas Related to Circles carry in the CBSE Class 10 board exam?+
Chapter 11 typically contributes 6–8 marks: 2–3 MCQs in Section A (1 mark each), 1 short-answer question in Section B (2–3 marks), and often appears in a case-study question in Section E (4 marks). The exact distribution varies yearly, but it is a high-weightage chapter that rewards formula fluency and diagram-based problem solving.
What is the difference between a sector and a segment of a circle?+
A sector is the 'pizza-slice' region enclosed by two radii and the arc between them; its area is (θ/360)×πr². A segment is the region between a chord and the arc it subtends; its area equals sector area minus the area of the triangle formed by the two radii and the chord. Visually, a segment is the 'leftover' piece when you cut off a sector's triangle.
Which formula should I use to find the perimeter of a sector?+
Perimeter of a sector = 2r + arc length = 2r + (θ/360)×2πr. Many students forget to add the two straight radii and only calculate the arc; always include both. For a semicircle (θ=180°), perimeter = 2r + πr = r(2+π).
How do I handle composite figures that combine a circle with a square or triangle?+
First, sketch and label all given dimensions. Identify whether the circle is inscribed (touches all sides) or circumscribed (passes through all vertices). Use the appropriate radius formula (e.g. for a square of side a, inscribed circle has r=a/2; circumscribed has r=a√2/2). Then compute individual areas and add or subtract as the shaded region dictates. Keep intermediate steps in fraction form to avoid rounding errors.
Why do CBSE papers often give π = 22/7 instead of 3.14?+
Using π = 22/7 keeps calculations in exact fractions, which is easier to mark and reduces rounding discrepancies. Unless the question specifies 'use π = 3.14', always default to 22/7. This also trains you to simplify algebraic expressions before multiplying, a skill tested in higher-level math.
What is an assertion-reason MCQ, and how should I approach it?+
An assertion-reason (A-R) MCQ presents two statements: Assertion (A) and Reason (R). You must decide if both are true, and if so, whether R correctly explains A. Read A and R independently first, verify each for truth, then check the logical link. In Chapter 11, A often states a result (e.g. 'area of quadrant is πr²/4') and R gives the underlying formula or definition.
Can I use a calculator in the CBSE Class 10 Mathematics board exam?+
No, calculators are not permitted. You must perform all arithmetic by hand. Practice mental multiplication with 22/7 and learn to simplify fractions quickly. For example, (22/7)×49 = 22×7 = 154 is instant if you recognize 49 = 7². CBSETUTOR.ai offers timed practice drills that simulate no-calculator conditions.
How can CBSETUTOR.ai help me prepare for Chapter 11 MCQs?+
CBSETUTOR.ai provides unlimited topic-wise MCQ quizzes for Chapter 11, instant photo-upload doubt solving, and step-by-step video walkthroughs of combination figures. At ₹999/month for all subjects across Classes 6–12, you get 24×7 access to an AI tutor that identifies your weak areas — whether it is segment calculations or assertion-reason logic — and generates personalized practice sets. Start with a 3-day free trial to see if the platform suits your learning style.
What are the most common mistakes students make in this chapter?+
Top errors include: (1) confusing diameter with radius in formulas; (2) forgetting to add the two radii when computing sector perimeter; (3) subtracting triangle area from circle area instead of sector area when finding a segment; (4) using degrees in trigonometric functions set to radian mode (though Class 10 NCERT uses degrees); and (5) rounding π too early in multi-step problems. Double-check units and re-read the question stem to avoid these pitfalls.
How much time should I spend practicing MCQs versus theory for this chapter?+
Aim for a 60:40 split — 60 per cent of your revision time on solving varied MCQs and past-year questions, 40 per cent on understanding derivations and worked examples in NCERT. Chapter 11 is formula-intensive but application-heavy; pure theory reading will not build speed. Solve at least 50 MCQs across easy, medium, and HOTS levels, then review mistakes to internalize concepts. CBSETUTOR.ai's analytics dashboard shows exactly which question types cost you marks, so you can focus your practice efficiently.

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