Surface Areas And Volumes: Class 9 Maths NCERT Guide

Key Formulae at a Glance

Cuboid (length l, breadth b, height h)

• Lateral Surface Area (LSA) = 2h(l + b) • Total Surface Area (TSA) = 2(lb + bh + hl) • Volume = l × b × h

Cube (side a)

• Lateral Surface Area (LSA) = 4a² • Total Surface Area (TSA) = 6a² • Volume = a³

Right Circular Cylinder (radius r, height h)

• Curved Surface Area (CSA) = 2πrh • Total Surface Area (TSA) = 2πr(r + h) • Volume = πr²h

Right Circular Cone (radius r, height h, slant height l)

• Slant Height (l) = √(h² + r²) • Curved Surface Area (CSA) = πrl • Total Surface Area (TSA) = πr(l + r) • Volume = (1/3)πr²h

Sphere (radius r)

• Surface Area = 4πr² • Volume = (4/3)πr³

Hemisphere (radius r)

• Curved Surface Area (CSA) = 2πr² • Total Surface Area (TSA) = 3πr² • Volume = (2/3)πr³

Beyond Formulas: What Are Surface Area and Volume?

It's easy to get lost in the formulas, but let's take a step back and understand the core concepts. Imagine you have a gift box (a cuboid). The Surface Area is the total area of all the faces of the box. It's the amount of wrapping paper you'd need to cover it completely, with no overlaps. We often split this into two types:

1. Lateral or Curved Surface Area (LSA/CSA): This is the area of the 'sides' only, excluding the top and bottom. For your room, the LSA is the area of the four walls you might want to paint.
2. Total Surface Area (TSA): This is the LSA plus the area of the top and bottom. For your room, it would be the four walls plus the floor and the ceiling.

Now, imagine filling that same gift box with sand. The amount of sand the box can hold is its Volume. Volume measures the three-dimensional space an object occupies. It's about capacity. So, if you're painting a water tank, you need its surface area. If you're filling it with water, you need its volume. Remembering this simple distinction—covering vs. filling—will help you choose the right formula for any problem.

Worked Example: Calculating Cost of Painting a Well

1. Step 1: Understand the Problem and Identify the Shape — Problem: A cylindrical well has an inner diameter of 3.5 m and is 10 m deep. Find the cost of plastering its inner curved surface at the rate of ₹40 per m². • The shape is a cylinder. • We need to find the inner curved surface area (CSA) because we are plastering the inside wall, not the bottom or top. • Given: Diameter = 3.5 m, Height (depth) = 10 m, Rate = ₹40/m².
2. Step 2: Calculate the Radius — The formula for CSA uses the radius (r), not the diameter. Radius (r) = Diameter / 2 r = 3.5 m / 2 = 1.75 m
3. Step 3: Apply the CSA Formula for a Cylinder — The formula for the Curved Surface Area of a cylinder is CSA = 2πrh. Let's substitute the values we have (use π = 22/7 as 3.5 is a multiple of 7): CSA = 2 × (22/7) × 1.75 × 10 CSA = 2 × (22/7) × (175/100) × 10 CSA = 2 × 22 × (25/100) × 10 (since 175 / 7 = 25) CSA = 44 × (1/4) × 10 CSA = 11 × 10 CSA = 110 m²
4. Step 4: Calculate the Total Cost — The rate of plastering is ₹40 per square meter. We have 110 square meters to plaster. Total Cost = Area × Rate Total Cost = 110 m² × ₹40/m² Total Cost = ₹4400
5. Final Answer — The inner curved surface area of the well is 110 m², and the total cost of plastering it is ₹4400.

Common Mistakes to Avoid in Exams

Pay close attention to these points to maximize your marks!

• CSA vs. TSA: Always read the question carefully. 'Area of four walls' or 'canvas for a tent' usually implies LSA/CSA. 'Painting the entire box' or 'material for a closed tank' means TSA. Don't use TSA when only the curved surface is needed.
• Units: This is a major source of errors. Ensure all dimensions (length, radius, height) are in the same unit before you start calculating. If a question gives radius in cm and height in m, convert one of them first! Remember 1 m = 100 cm, 1 m² = 10000 cm², and 1 m³ = 1000000 cm³. Also, 1000 cm³ = 1 Litre.
• Radius vs. Diameter: Many questions give the diameter. It's a common mistake to use the diameter value directly in the formula instead of the radius. Always halve the diameter first to find 'r'.
• Combined Solids: For shapes like a tent (cone on a cylinder), be careful. The TSA is NOT the sum of the two individual TSAs. You need to add the CSA of the cone and the CSA of the cylinder. The base of the cone and the top of the cylinder are not part of the external surface.

Practice Questions with Solutions

• Q: The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 per m². A: Step 1: Identify the surfaces to be whitewashed. This includes the four walls (Lateral Surface Area) and the ceiling. Step 2: Calculate the area of the four walls (LSA of a cuboid). LSA = 2h(l + b) = 2 × 3(5 + 4) = 6 × 9 = 54 m². Step 3: Calculate the area of the ceiling. Area of ceiling = l × b = 5 × 4 = 20 m². Step 4: Calculate the total area to be whitewashed. Total Area = Area of walls + Area of ceiling = 54 + 20 = 74 m². Step 5: Calculate the total cost. Cost = Total Area × Rate = 74 × 7.50 = ₹555. Final answer: The total cost of whitewashing is ₹555.
• Q: A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent and the cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹70. A: Step 1: Find the slant height (l) using the formula l = √(r² + h²). Given h = 10 m, r = 24 m. l = √(24² + 10²) = √(576 + 100) = √676 = 26 m. Step 2: The canvas required is the Curved Surface Area (CSA) of the cone. CSA = πrl. Step 3: Calculate the CSA. CSA = (22/7) × 24 × 26 = 13728/7 ≈ 1961.14 m². Step 4: Calculate the total cost. Cost = CSA × Rate = (13728/7) × 70 = 13728 × 10 = ₹137280. Final answer: The slant height is 26 m and the cost of the canvas is ₹137,280.
• Q: A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it contains? A: Step 1: Identify the shape and the required formula. The shape is a hemisphere and we need to find its volume. Step 2: The formula for the volume of a hemisphere is V = (2/3)πr³. Given r = 3.5 cm. Step 3: Substitute the values and calculate. V = (2/3) × (22/7) × (3.5)³ = (2/3) × (22/7) × 3.5 × 3.5 × 3.5 V = (2/3) × 22 × 0.5 × 3.5 × 3.5 = (44/3) × 0.5 × 12.25 V = (22/3) × 12.25 = 269.5 / 3 ≈ 89.83 cm³. Final answer: The volume of water the bowl contains is approximately 89.83 cm³.
• Q: The diameter of a sphere is 21 cm. Calculate its surface area and volume. A: Step 1: Find the radius from the diameter. Radius (r) = Diameter / 2 = 21 / 2 = 10.5 cm. Step 2: Calculate the surface area using the formula SA = 4πr². SA = 4 × (22/7) × (10.5)² = 4 × (22/7) × 10.5 × 10.5 = 4 × 22 × 1.5 × 10.5 = 88 × 15.75 = 1386 cm². Step 3: Calculate the volume using the formula V = (4/3)πr³. V = (4/3) × (22/7) × (10.5)³ = (4/3) × (22/7) × 10.5 × 10.5 × 10.5 V = 4 × 22 × 0.5 × 10.5 × 10.5 = 44 × 110.25 = 4851 cm³. Final answer: The surface area of the sphere is 1386 cm² and its volume is 4851 cm³.

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