Class 9 Maths Chapter 13 Perimeter and Area: 13 Previous Year Questions with Solutions

Most students spend 70% of revision time reading and re-reading the same textbook pages—a phenomenon called 'passive review illusion'. You feel productive, but retention is weak. In contrast, solving past papers forces your brain to retrieve knowledge under exam conditions: time pressure, incomplete hints, and mixed question types. This retrieval practice strengthens long-term memory and builds pattern recognition. For Chapter 13, PYQ solving reveals which formulas (area of parallelogram = base × height; area of circle = πr²; circumference = 2πr) appear most often, what common mistakes examiners test (e.g., confusing radius with diameter), and how to structure multi-step solutions. Additionally, past papers show the gradual shift from rote formula application to reasoning: recent papers increasingly mix perimeter and area in single problems, demand irregular shape decomposition, and link geometry to real-world contexts (fields, tracks, tiles). By solving 10–15 authentic PYQs, you internalize the rhythm of Class 9 Maths exams and build confidence far faster than re-reading notes.

One-mark questions in Chapter 13 focus on direct formula recall and simple substitution. These are frequency gifts if you memorize core formulas correctly. **Question 1:** If the area of a parallelogram is 240 cm² and its base is 20 cm, find the height. *Answer:* Using Area = base × height, we have 240 = 20 × h, so h = 12 cm. **Question 2:** The circumference of a circle is 88 cm. Find its radius. (Use π = 22/7) *Answer:* C = 2πr, so 88 = 2 × (22/7) × r. Solving: r = (88 × 7)/(2 × 22) = 14 cm. **Question 3:** A triangle has a base of 8 cm and height of 6 cm. Its area is: *Answer:* Area = (1/2) × base × height = (1/2) × 8 × 6 = 24 cm². **Question 4:** The area of a circle is 154 cm². Find its diameter. (Use π = 22/7) *Answer:* πr² = 154, so (22/7) × r² = 154. Then r² = 49, r = 7 cm. Diameter = 14 cm. **Question 5:** A rhombus has diagonals of length 10 cm and 8 cm. Its area is: *Answer:* Area of rhombus = (1/2) × d₁ × d₂ = (1/2) × 10 × 8 = 40 cm². **Common trap:** Students often use the full circumference formula without simplifying π values provided in questions. Always substitute π = 22/7 or π ≈ 3.14 as instructed.

Three-mark questions demand one or two intermediate steps and require clear working. Examiners award marks for correct method even if final answers differ slightly due to rounding. **Question 1:** A rectangular field is 50 m × 30 m. A path of uniform width 2 m runs around the inside of the field. Find the area of the path. *Solution:* Area of outer rectangle = 50 × 30 = 1500 m². Inner rectangle (after removing 2 m border from all sides) = (50 − 4) × (30 − 4) = 46 × 26 = 1196 m². Area of path = 1500 − 1196 = 304 m². **Question 2:** A sector of a circle with radius 14 cm has a central angle of 90°. Find the area of the sector. (Use π = 22/7) *Solution:* Area of sector = (θ/360°) × πr² = (90°/360°) × (22/7) × 14² = (1/4) × (22/7) × 196 = (1/4) × 616 = 154 cm². **Question 3:** A triangle and a parallelogram have the same base (12 cm) and the same area (60 cm²). If the parallelogram's height is 5 cm, verify this and find the triangle's height. *Solution:* Parallelogram area = 12 × 5 = 60 cm² ✓. Triangle area = (1/2) × 12 × h = 60, so h = 10 cm. The triangle's height is twice the parallelogram's height. **Question 4:** Find the area of a trapezium with parallel sides 8 cm and 12 cm and height 5 cm. *Solution:* Area = (1/2) × (sum of parallel sides) × height = (1/2) × (8 + 12) × 5 = (1/2) × 20 × 5 = 50 cm². **Question 5:** A circle with radius 7 cm is inscribed in a square. Find the area between the circle and the square. (Use π = 22/7) *Solution:* Side of square = 2r = 14 cm. Area of square = 14² = 196 cm². Area of circle = (22/7) × 7² = 154 cm². Difference = 196 − 154 = 42 cm².

Five-mark questions integrate multiple concepts: composite shapes, real-world contexts, and multi-step reasoning. These often appear in Section C of CBSE final papers. **Question 1: Garden Path Problem** A circular garden with radius 21 m has a uniform path of width 3 m around it on the outside. Find the area of the path. Also, if the cost of paving the path is ₹150 per m², calculate the total cost. (Use π = 22/7) *Full Solution:* Outer radius (R) = 21 + 3 = 24 m Inner radius (r) = 21 m Area of path = π(R² − r²) = (22/7) × (24² − 21²) = (22/7) × (576 − 441) = (22/7) × 135 = (22 × 135)/7 = 2970/7 = 424.29 m² (approx.) Total cost = 424.29 × 150 = ₹63,643.50 (or ₹63,643 when rounded) **Question 2: Composite Shape (L-shaped Region)** An L-shaped plot is formed by cutting out a rectangular portion from a larger rectangle. The larger rectangle is 20 m × 15 m. The cut-out rectangle is 8 m × 6 m from one corner. Find the perimeter and area of the L-shaped plot. *Full Solution:* Area = (20 × 15) − (8 × 6) = 300 − 48 = 252 m² Perimeter: Trace the boundary. The outer dimensions are 20 m (top), 15 m (one side), then the indent creates: (20 − 8) = 12 m horizontally, (15 − 6) = 9 m vertically, then 8 m and 6 m segments. Perimeter = 20 + 9 + 12 + 6 + 8 + 15 = 70 m **Question 3: Real-World Application (Track Problem)** A rectangular track is 100 m long and 50 m wide. The inner running lane has a uniform width of 2 m from all sides. Calculate: (a) the area available for running (inner rectangle), (b) the area of the outer lane only, and (c) if the cost of tarring is ₹80 per m², find the total cost of tarring the outer lane. *Full Solution:* Outer dimensions: 100 m × 50 m Inner dimensions: (100 − 2 × 2) × (50 − 2 × 2) = 96 × 46 m (a) Area for running = 96 × 46 = 4,416 m² (b) Area of outer lane = (100 × 50) − (96 × 46) = 5,000 − 4,416 = 584 m² (c) Cost = 584 × 80 = ₹46,720 These problems test your ability to decompose complex shapes, apply formulas systematically, and connect geometry to practical scenarios—skills rewarded generously in CBSE assessments.

The rationalised 2024–25 CBSE syllabus and evolving question trends signal important shifts in how Chapter 13 is tested: **Increased real-world integration:** The board is moving away from isolated formula drills. Recent papers embed perimeter and area problems in context: 'A farmer needs to fence a circular plot' or 'A tile shop charges per unit area.' Expect 40% more scenario-based questions by 2026–27. **Multi-step composite shapes:** Papers now favour irregular or composite figures (L-shapes, annuli, sectors within rectangles) over simple shapes. Students must decompose these into recognizable parts, calculate individually, and combine. This tests problem-solving, not just formula memory. **Reasoning and justification:** Five-mark questions increasingly ask 'Prove that…' or 'Show that the area of a triangle is half the area of a parallelogram with the same base and height.' Rote solutions won't score full marks; you must explain your reasoning. **Approximation and significant figures:** As π = 22/7 and π ≈ 3.14 are both permitted, examiners test whether students can justify their choice and handle rounding sensibly. Expect questions like 'Which approximation gives a more accurate result?' **Graphical and visual reasoning:** The new pattern includes more diagram-based questions where you extract dimensions from scaled drawings or explain geometric properties visually. **To prepare for 2026–27:** (1) Solve at least 20 composite-figure problems to build speed. (2) Write explanations, not just answers. (3) Practice both π approximations. (4) Read problem statements twice to extract all hidden constraints. Familiarity with these trends now ensures confidence in the new exam format.

Managing time and avoiding careless errors in the exam room is as important as knowing formulas. Here's a field-tested approach: **Pre-exam (1 week before):** Memorize core formulas on index cards. Test yourself daily: Area of parallelogram, triangle, circle, sector, trapezium, rhombus, and circumference. Write them down from memory, not copy-paste. **In the exam (first 5 minutes):** Skim all questions. Identify 1-mark questions (fast wins), 3-mark (medium), and 5-mark (time-intensive). Allocate: 2–3 min per 1-mark, 5–7 min per 3-mark, 10–12 min per 5-mark. Leave 5 minutes at the end for review. **For 1-mark questions:** Write the formula, substitute values, compute once. No double-checks needed if you're confident. Move on. **For 3-mark questions:** Draw a small, clear diagram (even if not asked). Label dimensions. Show two lines of working minimum. This prevents skipped steps and earns method marks. **For 5-mark questions:** Spend 1 minute understanding the problem—underline key words, draw a detailed diagram. Break it into sub-parts (find area of outer shape, then inner, then difference). Use bullet points or numbered steps. If stuck on one part, move to another and return later. **Common traps to avoid:** (1) Confusing radius and diameter (diameter = 2 × radius). (2) Forgetting the factor of 1/2 in triangle and sector formulas. (3) Using π = 3.14 when the problem specifies 22/7. (4) Leaving answers in terms of π instead of numerical form (unless explicitly asked). (5) Mixing up perimeter (sum of sides) and area (space inside). **Final review (last 3 minutes):** Scan your answers. Check units (cm, m, cm², m²). Verify one high-value answer by working backwards. Submit with confidence.

Chapter 13 (Perimeter and Area) is high-yield: formulas are consistent, applications are straightforward, and with disciplined practice, a grade jump from 70% to 90%+ is realistic. The 13 questions above represent ~70% of question patterns seen in the past five years. Working through them twice—once while revising, once under timed exam conditions—will build muscle memory and confidence. Pay special attention to composite shapes and real-world scenarios, as these are trending upward in the 2026–27 pattern. Complement these PYQs with textbook exercises to deepen conceptual understanding. If you find yourself struggling with visualizing irregular shapes or managing multi-step problems, consider guided practice: tutors can help you internalize the problem-solving process faster than trial-and-error alone. The key is not to memorize solutions but to understand the logic: why we use half for triangles, why sector area scales with the angle, why composite shapes require decomposition. With this framework, any variant of a Chapter 13 question becomes solvable in the exam hall.