Class 9 Mathematics Chapter 13: Perimeter and Area — Important Questions with Full Solutions

The CBSE Class 9 Mathematics syllabus (rationalized 2024–25) emphasizes conceptual understanding of perimeter and area over rote memorization. Examiners focus on three pillars: (1) Direct formula application (area of parallelogram = base × height; area of triangle = ½ × base × height; area of circle = πr²), (2) Problem-solving involving composite/irregular shapes—requiring decomposition into rectangles, triangles, and circular sectors, and (3) Real-life contexts like finding the cost of painting a wall, carpet area, or the radius of a circular plot. The board pattern includes 1-mark MCQs (definition and quick recall), 2-mark conceptual questions (formula justification or simple multi-step problems), 3-mark calculations (combining 2–3 shapes), and 5-mark integrated problems (word problems with diagram analysis). Case-study questions (introduced in recent years) test your ability to extract data and apply multiple formulas in sequence. These 18 questions mirror actual board papers, helping you recognize patterns and manage time during the exam.

1-mark MCQs test definition recall, formula identification, and quick numerical sense. These appear in Section A of the board paper and require only 1–2 minutes per question. **Q1:** If the area of a parallelogram is 240 cm² and its base is 15 cm, what is its height? (A) 16 cm (B) 15 cm (C) 14 cm (D) 20 cm **Answer:** (A) 16 cm. Using area = base × height, we get 240 = 15 × h, so h = 240 ÷ 15 = 16 cm. **Q2:** The circumference of a circle is 44 cm. What is its radius? (Use π = 22/7) (A) 7 cm (B) 14 cm (C) 21 cm (D) 28 cm **Answer:** (A) 7 cm. Circumference = 2πr, so 44 = 2 × (22/7) × r, giving r = (44 × 7) ÷ 44 = 7 cm. **Q3:** If the area of a triangle with base 12 cm and height 8 cm is A, then the area of a parallelogram with the same base and height is: (A) A (B) 2A (C) A/2 (D) 4A **Answer:** (B) 2A. Triangle area = ½ × 12 × 8 = 48 cm². Parallelogram area = 12 × 8 = 96 cm² = 2 × 48 = 2A. **Q4:** A rectangular plot has length 50 m and width 30 m. Its perimeter is: (A) 1500 m² (B) 160 m (C) 80 m (D) 240 m **Answer:** (B) 160 m. Perimeter = 2(l + w) = 2(50 + 30) = 160 m. **Q5:** If the radius of a circle is doubled, its area becomes: (A) 2 times (B) 4 times (C) √2 times (D) half **Answer:** (B) 4 times. Original area = πr². New area = π(2r)² = 4πr² = 4 × original area.

2-mark questions require brief calculation, formula justification, or a one-step word problem. These test conceptual clarity and are often used to check if you can link definitions to real contexts. **Q1:** A parallelogram has a base of 18 cm and a height of 10 cm. Find its area. Also, explain why we use height instead of slant side in the area formula. **Answer:** Area = base × height = 18 × 10 = 180 cm². We use perpendicular height (not slant side) because area measures space enclosed perpendicularly. The perpendicular distance from base to opposite side is the true 'width' of the parallelogram. **Q2:** Two triangles have the same base (20 cm) and height (12 cm). Are their areas equal? Justify. **Answer:** Yes, their areas are equal. Area = ½ × base × height = ½ × 20 × 12 = 120 cm² for both. Area depends only on base and perpendicular height, not on the shape or position of vertices. **Q3:** A circle has an area of 154 cm². Find its circumference. (Use π = 22/7) **Answer:** Area = πr² gives 154 = (22/7) × r². So r² = 154 × (7/22) = 49, thus r = 7 cm. Circumference = 2πr = 2 × (22/7) × 7 = 44 cm. **Q4:** A farmer has a rectangular field of length 200 m and width 150 m. He needs to fence it. If fencing costs ₹50 per meter, what is the total cost? **Answer:** Perimeter = 2(l + w) = 2(200 + 150) = 700 m. Cost = 700 × 50 = ₹35,000. **Q5:** A square plot has a perimeter of 80 m. What is its area? **Answer:** Perimeter = 4 × side, so 80 = 4 × side, giving side = 20 m. Area = side² = 20² = 400 m².

3-mark questions typically combine 2–3 formulas or require decomposition of irregular shapes into standard figures. These test problem-solving depth and are common in board exams. **Q1:** A rhombus has diagonals of lengths 16 cm and 12 cm. Find its area. Then, if a circle with the same area is drawn, find the radius of that circle. (Use π = 22/7) **Answer:** Area of rhombus = ½ × d₁ × d₂ = ½ × 16 × 12 = 96 cm². For the circle, πr² = 96, so (22/7) × r² = 96, giving r² = (96 × 7)/22 ≈ 30.55, thus r ≈ 5.5 cm. **Q2:** A composite figure consists of a rectangle (length 10 cm, width 6 cm) with a semicircle attached to one of its shorter sides. Find the total area. (Use π = 3.14) **Answer:** Area of rectangle = 10 × 6 = 60 cm². Semicircle has diameter = 6 cm, so radius = 3 cm. Area of semicircle = ½πr² = ½ × 3.14 × 3² = ½ × 3.14 × 9 = 14.13 cm². Total area = 60 + 14.13 = 74.13 cm². **Q3:** A trapezium has parallel sides of 12 cm and 18 cm, and a height of 8 cm. Find its area. If this trapezium is divided into two triangles by a diagonal, explain why both triangles do not have equal areas. **Answer:** Area of trapezium = ½(a + b) × h = ½(12 + 18) × 8 = ½ × 30 × 8 = 120 cm². A diagonal divides the trapezium into two triangles. Triangle 1 has base 12 cm and height 8 cm (area = 48 cm²); Triangle 2 has base 18 cm and height 8 cm (area = 72 cm²). They are unequal because their bases are different. **Q4:** A circular garden has a radius of 7 m. A path of width 2 m runs around the inside perimeter. Find the area of the path. (Use π = 22/7) **Answer:** Area of outer circle = πr₁² = (22/7) × 7² = 154 m². Inner circle radius = 7 − 2 = 5 m. Area of inner circle = (22/7) × 5² = (22/7) × 25 ≈ 78.57 m². Area of path = 154 − 78.57 ≈ 75.43 m².

5-mark questions are comprehensive, integrating real-world contexts with multiple calculations. These demand clear step-by-step solutions and are worth maximum marks. **Q1:** A plot of land is in the shape of a trapezium. Its parallel sides are 80 m and 60 m, and the perpendicular distance between them is 40 m. (i) Find the area of the plot. (ii) If the plot is to be divided into two equal parts by a line parallel to the parallel sides, at what distance from the 80 m side should this line be drawn? (iii) The plot is to be surrounded by a fence. If the non-parallel sides are each 50 m, find the total perimeter and the cost of fencing at ₹100 per meter. **Answer:** (i) Area = ½(a + b) × h = ½(80 + 60) × 40 = ½ × 140 × 40 = 2,800 m². (ii) For equal division, each part should have area 1,400 m². Let the dividing line be at distance x from the 80 m side. The upper trapezium has parallel sides 80 m and (80 − (80−60)x/40) = 60 + 20x/40. Area = ½(80 + 60 + 20x/40) × x = 1,400. Solving: 70x + 0.25x² = 1,400, so x ≈ 18.6 m. (iii) Perimeter = 80 + 60 + 50 + 50 = 240 m. Cost = 240 × 100 = ₹24,000. **Q2:** A circular park has a radius of 21 m. Inside, there is a square-shaped play area whose vertices touch the circle (inscribed square). (i) Find the area of the park. (ii) Find the side length of the inscribed square. (iii) Find the area of the park excluding the play area. (Use π = 22/7) **Answer:** (i) Area of park = πr² = (22/7) × 21² = (22/7) × 441 = 1,386 m². (ii) For a square inscribed in a circle, the diagonal of the square equals the diameter of the circle. Diagonal = 2 × 21 = 42 m. If side = a, then a√2 = 42, so a = 42/√2 = 21√2 ≈ 29.7 m. (iii) Area of square = a² = (21√2)² = 882 m². Remaining area = 1,386 − 882 = 504 m². **Q3:** A rectangular room is 12 m long and 8 m wide. (i) Calculate the area and perimeter of the room. (ii) Tiles are to be laid on the floor. If each tile is a square of side 0.5 m and costs ₹20, find the total cost. (iii) A border of width 1 m (without tiles) is left around the inner walls. What is the area of the border? (iv) If paint costs ₹15 per m², find the cost to paint the walls (excluding the border) of height 3 m on all four sides. **Answer:** (i) Area = 12 × 8 = 96 m². Perimeter = 2(12 + 8) = 40 m. (ii) Area of one tile = 0.5² = 0.25 m². Number of tiles = 96 ÷ 0.25 = 384 tiles. Total cost = 384 × 20 = ₹7,680. (iii) Inner room area (after 1 m border) = (12 − 2) × (8 − 2) = 10 × 6 = 60 m². Border area = 96 − 60 = 36 m². (iv) Wall area (excluding border) = Perimeter × height = 40 × 3 = 120 m². Paint cost = 120 × 15 = ₹1,800.

**Case-Study Question:** A food-delivery company designs pizza boxes in the shape of circular bases with varying sizes. The marketing team is analyzing two pizza sizes: **Pizza A:** Diameter = 28 cm **Pizza B:** Diameter = 35 cm The company charges based on area. Pizza A costs ₹300 and Pizza B costs ₹500. (i) Calculate the area of each pizza. (Use π = 22/7) (ii) Find the cost per cm² for each pizza. Which represents better value? (iii) A square box is being designed for a new rectangular pizza of dimensions 20 cm × 15 cm. What should be the side of the smallest square box to fit this pizza? (iv) If the company wants to add a decorative circular sticker of diameter 5 cm on the Pizza B box, what area remains uncovered on the pizza? **Full Solution:** (i) Pizza A: Area = πr² = (22/7) × 14² = (22/7) × 196 = 616 cm². Pizza B: Area = (22/7) × 17.5² = (22/7) × 306.25 ≈ 961.63 cm². (ii) Pizza A cost per cm² = 300 ÷ 616 ≈ ₹0.487/cm². Pizza B cost per cm² = 500 ÷ 961.63 ≈ ₹0.520/cm². Pizza A offers better value. (iii) Diagonal of rectangle = √(20² + 15²) = √(400 + 225) = √625 = 25 cm. Side of square box = 25 cm. (iv) Sticker area = π × 2.5² = (22/7) × 6.25 ≈ 19.64 cm². Uncovered area = 961.63 − 19.64 ≈ 941.99 cm². This case-study integrates area calculations, cost analysis, and geometric reasoning—mirroring the applied mathematics focus in CBSE's rationalized curriculum.

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