Class 9 Mathematics Chapter 11: Areas Related to Circles – Previous Year Questions (2020–2025)

Studying Chapter 11 theory once is enough; solving past papers is what converts that knowledge into exam points. Here's why: (1) Pattern Recognition — CBSE examiners repeat certain question types (e.g., 'find area of shaded region' or 'sector + triangle combinations'). Spotting these patterns saves time in the exam hall. (2) Formula Confidence — Writing Area = θ/360° × πr² repeatedly embeds the sector formula into muscle memory; theory revision alone doesn't lock it in. (3) Common Pitfalls — Past papers expose typical errors: forgetting to subtract the triangle area when finding segment area, mixing degrees and radians, or misidentifying which part is shaded. (4) Time Management — Exam questions often combine circles with rectangles or triangles. Solving 5-mark composites from past papers teaches you the 8–10 minute pacing needed on test day. (5) Confidence in Exam Pressure — Facing a 'find area of four sectors at corners of a square' question is far less daunting if you've solved it twice before. This page bundles 13 questions carefully selected from CBSE-standard papers, ordered by difficulty and weight, so you build momentum from easy 1-mark recall to harder 5-mark spatial reasoning.

One-mark questions test quick formula recall and definition clarity. These five are representative of question types seen consistently: **Q1:** If a sector of a circle with radius 6 cm has a central angle of 60°, what is the arc length of the sector? **Answer:** Arc length = θ/360° × 2πr = 60°/360° × 2π(6) = 1/6 × 12π = 2π cm ≈ 6.28 cm. **Q2:** Define a segment of a circle. **Answer:** A segment is the region between a chord and the arc subtended by that chord. It can be minor (smaller, on one side of the chord) or major (larger, on the other side). **Q3:** A circle has radius 7 cm. Find the area of a sector with central angle 90°. **Answer:** Area = θ/360° × πr² = 90°/360° × π(7)² = 1/4 × 49π = 12.25π cm² ≈ 38.5 cm². **Q4:** The area of a sector is 77 cm² and the radius is 7 cm. Find the central angle (in degrees). **Answer:** 77 = θ/360° × π(7)² ⇒ 77 = θ/360° × 154 ⇒ θ = (77 × 360°)/154 = 180°. **Q5:** A semicircle has radius 5 cm. What is its perimeter (including the diameter)? **Answer:** Perimeter = πr + 2r = π(5) + 2(5) = 5π + 10 ≈ 25.7 cm.

Three-mark questions require two or three calculation steps, often combining circle formulas with coordinate geometry or area subtraction. **Q1:** A circle has radius 14 cm. Two sectors have central angles 60° and 120°. Find the ratio of their areas. **Solution:** Area of sector 1 = 60°/360° × π(14)² = 1/6 × 196π. Area of sector 2 = 120°/360° × π(14)² = 1/3 × 196π. Ratio = (1/6 × 196π) : (1/3 × 196π) = 1/6 : 1/3 = 1 : 2. **Q2:** Find the area of a minor segment of a circle with radius 10 cm and central angle 90°. **Solution:** Area of sector = 90°/360° × π(10)² = 1/4 × 100π = 25π cm². Area of triangle (isoceles right angle at center) = 1/2 × 10 × 10 = 50 cm². Area of segment = 25π − 50 ≈ 78.5 − 50 = 28.5 cm². **Q3:** A square has side 20 cm. Four equal circles are inscribed at its four corners, each touching two adjacent sides. Find the total area of the four circles. **Solution:** Each circle has radius r = 10 cm (half the side, as it touches two sides). But only a quarter of each circle lies inside the square (the rest is outside). Total area inside square = 4 × (1/4 × π × 10²) = π × 100 = 100π ≈ 314.2 cm². **Q4:** A chord of length 12 cm is at a distance of 8 cm from the center of a circle. Find the radius and the area of the minor segment. **Solution:** Using perpendicular from center to chord bisects it: half-chord = 6 cm. By Pythagoras: r² = 8² + 6² = 64 + 36 = 100 ⇒ r = 10 cm. Central angle θ: sin(θ/2) = 6/10 = 0.6 ⇒ θ/2 ≈ 36.87° ⇒ θ ≈ 73.74° (or cos(θ/2) = 8/10 = 0.8 ⇒ θ/2 ≈ 36.87°). Area of sector ≈ 73.74°/360° × π(10)² ≈ 64.3 cm². Area of triangle = 1/2 × 12 × 8 = 48 cm². Area of segment ≈ 64.3 − 48 = 16.3 cm². **Q5:** A circular park has radius 35 m. A 4 m wide path runs along the inner boundary. Find the area of the path. **Solution:** Outer radius = 35 m, inner radius = 35 − 4 = 31 m. Area of path = π(35)² − π(31)² = π(1225 − 961) = 264π ≈ 829.4 m².

Five-mark questions demand multi-step reasoning, often combining circles with rectangles, squares, or multiple sectors. Full working is shown. **Q1:** A rectangular playground measures 120 m × 90 m. At one corner, a circular fountain of radius 20 m is to be built. A 5 m wide path is to be laid outside the circular region (but inside the rectangle). Find the area available for planting grass (i.e., total playground area minus fountain and path). **Solution:** Total playground area = 120 × 90 = 10,800 m². The fountain is a quarter-circle (at a corner) of radius 20 m. Fountain area = 1/4 × π(20)² = 100π ≈ 314.2 m². The path is a 5 m wide ring outside the fountain but still a quarter-circle, with outer radius 25 m and inner radius 20 m. Path area = 1/4 × [π(25)² − π(20)²] = 1/4 × π(625 − 400) = 1/4 × 225π = 56.25π ≈ 176.7 m². Remaining area for grass = 10,800 − 314.2 − 176.7 ≈ 10,309.1 m². **Q2:** A square park ABCD has side 56 m. Semicircular flower beds are made on each side of the square (outside it). Find the total area of the four semicircles. **Solution:** Each side = 56 m is the diameter of each semicircle. Radius of each semicircle r = 28 m. Area of one semicircle = 1/2 × π(28)² = 1/2 × 784π = 392π m². Total area of four semicircles = 4 × 392π = 1568π ≈ 4,926.4 m². **Q3:** A circular cardboard sheet of radius 21 cm has two concentric circles drawn on it such that they divide the sheet into three regions of equal area. Find the radii of the two concentric circles. **Solution:** Total area of sheet = π(21)² = 441π cm². Each region has area 441π/3 = 147π cm². Region 1 (innermost circle) has area = πr₁² = 147π ⇒ r₁² = 147 ⇒ r₁ = √147 = 7√3 ≈ 12.1 cm. Region 2 (annulus between r₁ and r₂) has area = π(r₂² − r₁²) = 147π ⇒ r₂² − 147 = 147 ⇒ r₂² = 294 ⇒ r₂ = √294 = 7√6 ≈ 17.1 cm. (Region 3 is the outer ring from r₂ to 21.)

CBSE has signalled subtle shifts in how Chapter 11 questions are framed in recent papers. Awareness of these patterns will sharpen your exam strategy. (1) More Composite Figures — Instead of isolated sectors or segments, examiners are increasingly embedding circles within polygons or vice versa (e.g., a semicircle on a side of a triangle, or a circular park inside a square field). Expect 3- and 5-mark questions to require identifying which shapes to combine and subtract. (2) Real-World Contexts — Questions now often include 'practical' setups: garden paths, fountains at corners, racing tracks, or wheels on vehicles. The math is the same, but the story-telling adds a distraction layer; re-read the problem carefully to extract which region you actually need. (3) Emphasis on Segment Area (Not Just Sectors) — Older papers favored sector-only questions; newer papers push minor and major segment definitions and calculations. Ensure you can quickly recall: Segment Area = Sector Area − Triangle Area (for minor segments with central angle < 180°). (4) Angle Formats — A few recent papers have given central angles in radians (e.g., π/3) alongside degrees; brush up on radian–degree conversion: θ(radians) = θ(degrees) × π/180. (5) 'Shaded Region' Ambiguity — Diagrams are now less clear about which region is shaded; the text description matters. Always identify: Is it inside or outside the circle? Is it a sector, segment, or a combination with a polygon? This cognitive step is now more heavily tested.

On exam day, efficient problem-solving saves minutes and reduces careless errors. Follow this checklist: (1) Read the Entire Question First — Don't start calculating until you've identified all given data (radius, angle, dimensions of combined shapes) and what you need to find. For composite figures, sketch a rough diagram in the margin. (2) Identify the Shape — Is it a sector (pie slice), a segment (between chord and arc), or a composite? For composites, decide which areas to add and which to subtract. (3) Choose the Right Formula — Sector area: θ/360° × πr². Segment area: Sector area − Triangle area. Arc length: θ/360° × 2πr. Perimeter of sector: 2r + arc length. Write the formula clearly before substituting numbers. (4) Substitute Values Carefully — Confirm the angle is in degrees (or convert from radians). Confirm the radius is the same across all shapes. Use π ≈ 22/7 or 3.14 (or leave as π if the answer asks for exact form). (5) Check Units — If radius is in cm, area is in cm²; if in m, then m². A common slip is forgetting to square the radius. (6) Estimate Your Answer — Before writing the final line, do a quick sanity check: Is a sector of a circle with r = 10 cm and angle 90° roughly a quarter of 300 cm² (the full circle)? That's about 75 cm², so your answer should be in that ballpark. (7) For 5-Mark Composites — Break into sub-questions: What's the area of the circle part? What's the area of the polygon? Do I add or subtract? Write each step on a separate line so the marker can award partial credit even if your final answer is wrong. (8) Leave π in the Answer If Possible — Unless the question says 'use π = 22/7' or 'find the approximate area,' write your answer in terms of π (e.g., 25π cm²) to show exactness.

Use this checklist in your final 2–3 days of revision to ensure you're exam-ready for Chapter 11: ✓ Can you write the sector area formula and the segment area formula without looking? ✓ Do you know the difference between a minor and major segment? ✓ Have you solved at least one 5-mark composite figure question where the circle is inside a polygon (not the other way around)? ✓ Can you convert a central angle from radians to degrees and vice versa? ✓ Have you worked a problem where you had to find the radius given the sector area and angle? ✓ Do you understand why the triangle area must be subtracted from the sector area to get the segment area? ✓ Have you timed yourself solving a full 5-mark question; can you do it in under 10 minutes? ✓ Have you re-read the 13 previous year questions above and checked your answers? If any box is unticked, spend 30 minutes on that skill before exam day. Chapter 11 is scoring territory — the formulas are fixed, the logic is clear, and past papers reveal what examiners want. Master this chapter, and you'll walk into your final exam with real confidence.