Class 9 Mathematics Chapter 6: Perimeter and Area – Important Questions with Answers

Chapter 6 accounts for approximately 8–12 marks in the CBSE Class 9 Mathematics final exam, distributed across multiple question types. The 2024-25 rationalized syllabus emphasizes conceptual understanding over rote learning, which means examiners test your ability to: 1. **Apply formulas contextually**: Rather than just memorizing P = 2(l + b), you must recognize when and how to use it in word problems. 2. **Convert units accurately**: Questions like 'Find area in m² when dimensions are in cm' are common. Mistakes in conversion (1 m² = 10,000 cm²) cost marks. 3. **Calculate irregular shapes**: Grid method (counting squares) or decomposing irregular polygons into known shapes (rectangles, triangles) is expected. 4. **Interpret diagrams**: Board exams present composite figures requiring you to break them into simpler components. 5. **Justify answers**: Even 1-mark MCQs often require reasoning in step-marking schemes. The pattern has shifted from pure calculation toward application. A typical board paper includes: 1 or 2 one-mark MCQs, 2–3 two-mark short answers, 1–2 three-mark questions, and 1 five-mark problem. This guide covers all these patterns with real board-style questions.

**Q1.** The perimeter of a square with side length 8 cm is: (A) 32 cm (B) 64 cm (C) 16 cm (D) 128 cm **Answer: (A) 32 cm** *Explanation:* Perimeter of square = 4 × side = 4 × 8 = 32 cm. --- **Q2.** If the area of a rectangle is 48 cm² and length is 12 cm, the breadth is: (A) 4 cm (B) 6 cm (C) 8 cm (D) 10 cm **Answer: (A) 4 cm** *Explanation:* Area = length × breadth → 48 = 12 × b → b = 48 ÷ 12 = 4 cm. --- **Q3.** 1 m² equals: (A) 100 cm² (B) 1,000 cm² (C) 10,000 cm² (D) 100,000 cm² **Answer: (C) 10,000 cm²** *Explanation:* 1 m = 100 cm, so 1 m² = 100 × 100 = 10,000 cm². --- **Q4.** The perimeter of an equilateral triangle with side 5 cm is: (A) 15 cm (B) 20 cm (C) 25 cm (D) 10 cm **Answer: (A) 15 cm** *Explanation:* Perimeter = 3 × side = 3 × 5 = 15 cm. --- **Q5.** A shape covers 12 complete squares and 8 half squares on a grid paper (1 square = 1 cm²). The area is: (A) 12 cm² (B) 16 cm² (C) 20 cm² (D) 8 cm² **Answer: (B) 16 cm²** *Explanation:* Area = 12 + (8 ÷ 2) = 12 + 4 = 16 cm².

**Q1.** A rectangular garden has length 25 m and breadth 15 m. Find its perimeter and area. **Answer:** Perimeter = 2(l + b) = 2(25 + 15) = 2 × 40 = **80 m** Area = l × b = 25 × 15 = **375 m²** --- **Q2.** Convert 5 m² into cm². **Answer:** 5 m² = 5 × 10,000 cm² = **50,000 cm²** *Note:* 1 m = 100 cm, so 1 m² = 100² = 10,000 cm². --- **Q3.** A square tile has a perimeter of 40 cm. What is its area? **Answer:** Side of square = Perimeter ÷ 4 = 40 ÷ 4 = 10 cm Area = side² = 10² = **100 cm²** --- **Q4.** Find the perimeter of a pentagon (five-sided polygon) with sides 4 cm, 5 cm, 6 cm, 5 cm, and 4 cm. **Answer:** Perimeter = sum of all sides = 4 + 5 + 6 + 5 + 4 = **24 cm** --- **Q5.** An L-shaped figure is made by joining a rectangle of dimensions 10 cm × 5 cm with another rectangle of dimensions 6 cm × 4 cm (sharing a common side). Find the area. **Answer:** Area = Area of rectangle 1 + Area of rectangle 2 = (10 × 5) + (6 × 4) = 50 + 24 = **74 cm²**

**Q1.** A rectangular field has length 60 m and breadth 40 m. A path of width 2 m runs all around it on the outside. Find the area of the path. **Answer:** Outer length = 60 + 2(2) = 60 + 4 = 64 m Outer breadth = 40 + 2(2) = 40 + 4 = 44 m Outer area = 64 × 44 = 2,816 m² Inner area (field) = 60 × 40 = 2,400 m² Area of path = 2,816 − 2,400 = **416 m²** --- **Q2.** A composite figure consists of a rectangle 12 cm × 8 cm with a triangle on top (base 12 cm, height 5 cm). Calculate total area. **Answer:** Area of rectangle = 12 × 8 = 96 cm² Area of triangle = ½ × base × height = ½ × 12 × 5 = 30 cm² Total area = 96 + 30 = **126 cm²** --- **Q3.** A shape on grid paper (1 small square = 1 cm²) has 18 complete squares and 6 half squares. Express the area in both cm² and mm². **Answer:** Area = 18 + (6 ÷ 2) = 18 + 3 = 21 cm² Convert to mm²: 1 cm = 10 mm, so 1 cm² = 100 mm² Area = 21 × 100 = **2,100 mm²** --- **Q4.** Two concentric squares have sides 15 cm and 10 cm respectively. Find the area between them (shaded region). **Answer:** Area of outer square = 15² = 225 cm² Area of inner square = 10² = 100 cm² Area of shaded region = 225 − 100 = **125 cm²**

**Q1.** A farmer wants to fence a rectangular plot measuring 80 m × 50 m. The cost of fencing is ₹150 per meter. Additionally, he wants to tile the plot at ₹25 per m². Calculate the total cost of fencing and tiling. **Solution:** **Step 1:** Find perimeter of the plot. Perimeter = 2(l + b) = 2(80 + 50) = 2 × 130 = 260 m **Step 2:** Calculate cost of fencing. Cost of fencing = 260 × 150 = ₹39,000 **Step 3:** Find area of the plot. Area = l × b = 80 × 50 = 4,000 m² **Step 4:** Calculate cost of tiling. Cost of tiling = 4,000 × 25 = ₹1,00,000 **Step 5:** Find total cost. Total cost = 39,000 + 1,00,000 = **₹1,39,000** --- **Q2.** A composite figure consists of a rectangle (12 cm × 8 cm) with two semicircles on opposite sides of the 8 cm width. Find the total perimeter of this figure. (Use π ≈ 3.14) **Solution:** **Step 1:** Identify the perimeter components. - Two lengths of rectangle = 12 + 12 = 24 cm - Two semicircles form one complete circle with diameter 8 cm (radius = 4 cm) **Step 2:** Calculate circumference of the circle. Circumference = 2πr = 2 × 3.14 × 4 = 25.12 cm **Step 3:** Total perimeter. Perimeter = 24 + 25.12 = **49.12 cm** --- **Q3.** An irregularly shaped field on a grid map (1 square = 5 m × 5 m) covers 24 complete grid squares and 10 half grid squares. The field requires irrigation at ₹80 per m². Find the total irrigation cost. **Solution:** **Step 1:** Calculate area in grid units. Area in grid squares = 24 + (10 ÷ 2) = 24 + 5 = 29 squares **Step 2:** Convert grid squares to m². Each grid square = 5 × 5 = 25 m² Total area = 29 × 25 = 725 m² **Step 3:** Calculate irrigation cost. Cost = 725 × 80 = **₹58,000**

**Case Study:** A school wants to install a playground and a garden in a rectangular plot of land measuring 60 m × 40 m. The playground (rectangular) is 35 m × 25 m. The remaining area will be used for a garden. A fence (₹100/m) will surround both the playground and the garden. A walking track of width 1.5 m runs around the inside edge of the entire plot. Calculate: (i) Area of the garden. (ii) Total length of fencing needed around the playground perimeter. (iii) Area available for walking track. (iv) Whether the walking track consumes more than 15% of total land area. **Solution:** **Step 1 (i): Area of garden.** Total plot area = 60 × 40 = 2,400 m² Playground area = 35 × 25 = 875 m² Garden area = 2,400 − 875 = **1,525 m²** **Step 2 (ii): Fencing length around playground.** Perimeter of playground = 2(35 + 25) = 2 × 60 = 120 m Fencing cost = 120 × 100 = **₹12,000** (Length = 120 m) **Step 3 (iii): Area of walking track.** Inner rectangle (after removing 1.5 m track on each side): Length = 60 − 2(1.5) = 60 − 3 = 57 m Breadth = 40 − 2(1.5) = 40 − 3 = 37 m Inner area = 57 × 37 = 2,109 m² Walking track area = 2,400 − 2,109 = **291 m²** **Step 4 (iv): Percentage check.** Percentage = (291 ÷ 2,400) × 100 = 12.125% **Answer:** The walking track consumes 12.125%, which is **less than 15%**. ✓

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