Class 9 Mathematics Chapter 9 Mensuration Important Questions with Solutions

Mensuration is a permanent pillar of Class 9 Mathematics, consistently weighted at 8–12% of the total marks across all board exam papers nationwide. The 2024-25 rationalized CBSE syllabus restructured this chapter to emphasize conceptual depth over rote memorization: examiners now expect students to derive area formulas (trapezium, regular polygons), connect 2D areas to 3D surface calculations, and solve multi-step volume problems involving real contexts (water tanks, paint coverage, material wastage). The four sub-topics carry distinct weightage: **Area of Trapezium & Polygons (3–4 marks):** 1-mark MCQs and 2-mark direct-formula questions dominate. Expect one question requiring decomposition of irregular polygons into triangles or trapeziums. **Surface Area of Solids (3–4 marks):** Mix of formula recall (1–2 marks) and application (2–3 marks). Cylinders appear frequently because their curved surface area formula (2πrh) is often confused with total surface area (2πrh + 2πr²). **Volume of Solids (2–3 marks):** Usually one comparative question (which tank holds more?) or a scenario-based problem (cost to fill a cylindrical well). **HOTS / Case Studies (1–2 marks in Section D):** Real-world contexts—a cylindrical paint tin, a trapezoidal field, a composite solid made of cube and hemisphere—require integrated formula knowledge and logical reasoning. Mastering these 18 questions ensures you cover ~90% of likely exam scenarios and build the speed needed to score full marks under exam pressure.

**Q1. The area of a trapezium with parallel sides 6 cm and 8 cm, and height 5 cm, is:** (A) 30 cm² (B) 35 cm² (C) 40 cm² (D) 70 cm² **Answer: (B) 35 cm²** *Working:* Area of trapezium = ½ × (sum of parallel sides) × height = ½ × (6 + 8) × 5 = ½ × 14 × 5 = 35 cm² --- **Q2. A cube has edge length 4 cm. Its total surface area is:** (A) 16 cm² (B) 48 cm² (C) 64 cm² (D) 96 cm² **Answer: (D) 96 cm²** *Working:* Total surface area of cube = 6a² = 6 × 4² = 6 × 16 = 96 cm² --- **Q3. The curved surface area of a cylinder with radius 3 cm and height 10 cm is:** (A) 30π cm² (B) 60π cm² (C) 90π cm² (D) 180π cm² **Answer: (B) 60π cm²** *Working:* Curved surface area = 2πrh = 2 × π × 3 × 10 = 60π cm² --- **Q4. The volume of a cuboid with length 5 cm, width 4 cm, and height 3 cm is:** (A) 12 cm³ (B) 30 cm³ (C) 60 cm³ (D) 120 cm³ **Answer: (C) 60 cm³** *Working:* Volume = l × w × h = 5 × 4 × 3 = 60 cm³ --- **Q5. A regular polygon has 6 sides. Each interior angle measures:** (A) 90° (B) 108° (C) 120° (D) 135° **Answer: (C) 120°** *Working:* Interior angle of regular n-gon = [(n − 2) × 180°] ÷ n = [(6 − 2) × 180°] ÷ 6 = (4 × 180°) ÷ 6 = 720° ÷ 6 = 120°

**Q1. A trapezium has parallel sides of length 10 cm and 6 cm. If its area is 48 cm², find its height.** **Solution:** Area of trapezium = ½ × (sum of parallel sides) × height 48 = ½ × (10 + 6) × h 48 = ½ × 16 × h 48 = 8h h = 6 cm --- **Q2. A cuboid has dimensions 8 cm × 6 cm × 4 cm. Calculate its total surface area.** **Solution:** Total surface area of cuboid = 2(lw + bh + hl) = 2[(8 × 6) + (6 × 4) + (4 × 8)] = 2[48 + 24 + 32] = 2 × 104 = 208 cm² --- **Q3. A cylinder has radius 7 cm and height 15 cm. Find its total surface area (use π = 22/7).** **Solution:** Total surface area = 2πr(h + r) = 2 × (22/7) × 7 × (15 + 7) = 2 × 22 × 22 = 968 cm² --- **Q4. The volume of a cube is 512 cm³. Find the length of its edge.** **Solution:** Volume of cube = a³ 512 = a³ a = ∛512 = 8 cm --- **Q5. A regular hexagon has side length 5 cm. Express the formula for its area and calculate (use √3 ≈ 1.73).** **Solution:** Area of regular hexagon = (3√3/2) × side² = (3 × 1.73/2) × 5² = (5.19/2) × 25 = 2.595 × 25 = 64.875 cm² ≈ 64.88 cm²

**Q1. A field is in the shape of a trapezium with parallel sides 20 m and 30 m, and the perpendicular distance between them is 15 m. A path of width 2 m runs all around the inside of the field. Find the area of the path.** **Solution:** Area of trapezium field = ½ × (20 + 30) × 15 = ½ × 50 × 15 = 375 m² Inner trapezium (after 2 m inset on all sides): parallel sides become (20 − 2 − 2) = 16 m and (30 − 2 − 2) = 26 m, height becomes 15 − 2 − 2 = 11 m Area of inner trapezium = ½ × (16 + 26) × 11 = ½ × 42 × 11 = 231 m² Area of path = 375 − 231 = 144 m² --- **Q2. A cuboid-shaped water tank has dimensions 2 m × 1.5 m × 1 m. It is filled to 80% of its capacity. How many litres of water does it contain? (1 m³ = 1000 litres)** **Solution:** Volume of tank = 2 × 1.5 × 1 = 3 m³ Volume filled to 80% = 0.8 × 3 = 2.4 m³ In litres = 2.4 × 1000 = 2400 litres --- **Q3. A cylinder with radius 5 cm and height 12 cm is melted and recast into a sphere. Find the radius of the sphere (use π = 3.14).** **Solution:** Volume of cylinder = πr²h = 3.14 × 5² × 12 = 3.14 × 25 × 12 = 942 cm³ Volume of sphere = (4/3)πr³ 942 = (4/3) × 3.14 × r³ 942 = 4.1867 × r³ r³ = 942 ÷ 4.1867 ≈ 225 r ≈ ∛225 ≈ 6.08 cm --- **Q4. A regular pentagon has side 6 cm. The area of a regular pentagon with side a is (a²/4) × √(25 + 10√5). Calculate its approximate area (use √5 ≈ 2.236).** **Solution:** First, 25 + 10√5 = 25 + 10(2.236) = 25 + 22.36 = 47.36 √47.36 ≈ 6.88 Area = (36/4) × 6.88 = 9 × 6.88 = 61.92 cm²

**Q1. A composite solid consists of a cylinder of radius 5 cm and height 10 cm topped with a hemisphere of radius 5 cm. Find (a) the total surface area, and (b) the total volume. (Use π = 3.14)** **Solution:** **Part (a) – Total Surface Area:** The total surface area includes: - Curved surface of cylinder = 2πrh = 2 × 3.14 × 5 × 10 = 314 cm² - Curved surface of hemisphere = 2πr² = 2 × 3.14 × 5² = 2 × 3.14 × 25 = 157 cm² - Base of cylinder = πr² = 3.14 × 25 = 78.5 cm² Total surface area = 314 + 157 + 78.5 = 549.5 cm² **Part (b) – Total Volume:** Volume of cylinder = πr²h = 3.14 × 25 × 10 = 785 cm³ Volume of hemisphere = (2/3)πr³ = (2/3) × 3.14 × 125 = 261.67 cm³ Total volume = 785 + 261.67 = 1046.67 cm³ ≈ 1047 cm³ --- **Q2. An irregular hexagon ABCDEF can be divided into a rectangle ABCD and a trapezium CDEF. Rectangle ABCD has length 10 cm and width 6 cm. Trapezium CDEF has parallel sides CD = 10 cm, EF = 8 cm, and height 4 cm. Calculate (a) the total area, (b) the perimeter if AB = 10 cm, BC = 6 cm, CD = 10 cm, DE = 5 cm, EF = 8 cm, FA = 5 cm.** **Solution:** **Part (a) – Total Area:** Area of rectangle ABCD = 10 × 6 = 60 cm² Area of trapezium CDEF = ½ × (10 + 8) × 4 = ½ × 18 × 4 = 36 cm² Total area = 60 + 36 = 96 cm² **Part (b) – Perimeter:** Perimeter = AB + BC + CD + DE + EF + FA = 10 + 6 + 10 + 5 + 8 + 5 = 44 cm --- **Q3. A cylindrical pipe has an outer radius of 8 cm and an inner radius of 6 cm. Its length is 25 cm. Find (a) the volume of the material (metal), (b) the total outer surface area, and (c) the total inner surface area. (Use π = 22/7)** **Solution:** **Part (a) – Volume of Material:** Volume = π × (R² − r²) × h, where R = 8 cm, r = 6 cm, h = 25 cm = (22/7) × (64 − 36) × 25 = (22/7) × 28 × 25 = (22/7) × 700 = 22 × 100 = 2200 cm³ **Part (b) – Outer Surface Area:** Outer curved surface = 2πRh = 2 × (22/7) × 8 × 25 = (44/7) × 200 = 1257.14 cm² Outer annular (ring) ends = 2 × π(R² − r²) = 2 × (22/7) × 28 = 176 cm² Total outer surface area = 1257.14 + 176 = 1433.14 cm² ≈ 1433 cm² **Part (c) – Inner Surface Area:** Inner curved surface = 2πrh = 2 × (22/7) × 6 × 25 = (44/7) × 150 = 942.86 cm² Inner annular ends = 176 cm² (same as outer) Total inner surface area = 942.86 + 176 = 1118.86 cm² ≈ 1119 cm²

**Q. An engineer is designing a water conservation system. A cylindrical tank (radius 3 m, height 4 m) is connected to a cuboid-shaped reservoir (length 6 m, width 4 m, depth 2.5 m). Both are to be painted on their outer surfaces (excluding the base) with waterproof coating. The cost is ₹50 per m².** **Part (i): Calculate the curved surface area of the cylinder.** **Part (ii): Calculate the lateral surface area of the cuboid (all four sides, excluding top and base).** **Part (iii): If the paint covers 10 m² per litre, how many litres are required for both structures?** **Part (iv): What is the total cost of painting?** **Solution:** **Part (i) – Curved Surface Area of Cylinder:** CSA = 2πrh = 2 × (22/7) × 3 × 4 = 2 × (22/7) × 12 = (44/7) × 12 = 528/7 = 75.43 m² (use π ≈ 22/7 or 3.14 for exact value) **Part (ii) – Lateral Surface Area of Cuboid:** Lateral area = 2(l + b) × h = 2(6 + 4) × 2.5 = 2 × 10 × 2.5 = 50 m² **Part (iii) – Paint Required:** Total area to paint = 75.43 + 50 = 125.43 m² Litres needed = 125.43 ÷ 10 = 12.543 litres ≈ 12.55 litres (or round to 13 litres if buying in whole units) **Part (iv) – Total Cost:** If we use 12.55 litres: Cost = 12.55 × 50 = ₹627.50 If rounding to 13 litres: Cost = 13 × 50 = ₹650 **Key Skill Tested:** Integration of multiple solids, real-world application, unit conversion, and cost estimation.

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