CBSE • Class 9 • Mathematics • Chapter 11 (Surface Areas and Volumes) • Exercise 12.1

Exercise 12.1: Surface Areas and Volumes — NCERT Solutions

Surface areas of cuboid, cube, cylinder and combinations.

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What this exercise covers

CSA vs TSACylinder wrapComposite solids

Step-by-step solutions — Exercise 12.1

5 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 12.1 Q1 • 2 marks

A closed cuboidal box has length 1212 cm, breadth 99 cm and height 55 cm. Find its total surface area.
Hint (Socratic — try this first)▾
Which formula adds up all six rectangular faces of a cuboid?
Step-by-step solution▾

Given: l=12l = 12 cm, b=9b = 9 cm, h=5h = 5 cm.

Formula: Total surface area of a cuboid =2(lb+bh+hl)= 2(lb + bh + hl).

Substitute: =2(12×9+9×5+5×12)= 2(12\times 9 + 9\times 5 + 5\times 12) =2(108+45+60)= 2(108 + 45 + 60) =2×213=426 cm2= 2 \times 213 = 426 \text{ cm}^2

Total surface area =426= 426 cm².

Common mistake:
Forgetting to multiply the bracket by 2 (which would give the area of only three faces).
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Exercise 12.1 Q2 • 2 marks

The side of a cube is 77 cm. Find (i) its lateral surface area and (ii) its total surface area.
Hint (Socratic — try this first)▾
How many faces make up the lateral surface, and how many make the total surface?
Step-by-step solution▾

Given: edge a=7a = 7 cm.

(i) Lateral surface area (four side faces) =4a2= 4a^2: =4×72=4×49=196 cm2= 4 \times 7^2 = 4 \times 49 = 196 \text{ cm}^2

(ii) Total surface area (all six faces) =6a2= 6a^2: =6×49=294 cm2= 6 \times 49 = 294 \text{ cm}^2

LSA =196= 196 cm², TSA =294= 294 cm².

Common mistake:
Confusing lateral surface area (4a24a^2) with total surface area (6a26a^2).
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Exercise 12.1 Q3 • 3 marks

A cylindrical pillar has diameter 5050 cm and height 3.53.5 m. Find the cost of painting its curved surface at ₹1212 per m². (Use π=227\pi = \tfrac{22}{7}.)
Hint (Socratic — try this first)▾
What is the radius in metres, and which surface area do you paint on a pillar?
Step-by-step solution▾

Given: diameter =50= 50 cm ⇒r=25\Rightarrow r = 25 cm =0.25= 0.25 m; height h=3.5h = 3.5 m.

Curved surface area =2πrh= 2\pi r h: =2×227×0.25×3.5= 2 \times \frac{22}{7} \times 0.25 \times 3.5 =2×227×0.875=38.57×...= 2 \times \frac{22}{7} \times 0.875 = \frac{38.5}{7} \times ... Let us compute step by step: =447×0.25×3.5=44×0.8757=38.57=5.5 m2= \frac{44}{7} \times 0.25 \times 3.5 = \frac{44 \times 0.875}{7} = \frac{38.5}{7} = 5.5 \text{ m}^2

Cost =5.5×12=₹66= 5.5 \times 12 = ₹66.

Cost of painting =₹66= ₹66.

Common mistake:
Using diameter instead of radius, or forgetting to convert centimetres to metres before computing.
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Exercise 12.1 Q4 • 3 marks

A cylindrical vessel open at the top has base radius 77 cm and height 1010 cm. Find the total area of metal sheet needed to make it. (Use π=227\pi = \tfrac{22}{7}.)
Hint (Socratic — try this first)▾
An open cylinder has a curved surface plus how many circular ends?
Step-by-step solution▾

Given: r=7r = 7 cm, h=10h = 10 cm. The vessel is open at the top, so it has curved surface + one base.

Metal sheet =2πrh+πr2= 2\pi r h + \pi r^2.

Curved surface =2×227×7×10=2×22×10=440= 2 \times \frac{22}{7} \times 7 \times 10 = 2 \times 22 \times 10 = 440 cm².

Base =227×72=22×7=154= \frac{22}{7} \times 7^2 = 22 \times 7 = 154 cm².

Total =440+154=594= 440 + 154 = 594 cm².

Metal sheet required =594= 594 cm².

Common mistake:
Including both circular ends (as in a closed cylinder) instead of only one, since the top is open.
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Exercise 12.1 Q5 • 2 marks

A cuboidal room is 88 m long, 66 m wide and 44 m high. Find the area of the four walls that must be painted.
Hint (Socratic — try this first)▾
Which surface area of a cuboid covers only the walls and not the floor or ceiling?
Step-by-step solution▾

Given: l=8l = 8 m, b=6b = 6 m, h=4h = 4 m.

The four walls form the lateral surface area =2(l+b)h= 2(l + b)h.

Substitute: =2(8+6)×4=2×14×4=112 m2= 2(8 + 6)\times 4 = 2 \times 14 \times 4 = 112 \text{ m}^2

Area of four walls =112= 112 m².

Common mistake:
Using the total surface area formula, thereby wrongly including the floor and ceiling.
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How to approach Exercise 12.1

  1. Re-read the chapter summary first. Open Surface Areas and Volumes and refresh the key concepts: Cuboid, Cube, Cylinder, Cone.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Surface Areas and Volumes

  1. Exercise 12.1 — Surface areas of cuboid, cube, cylinder and combinations.
  2. Exercise 12.2 — Volumes of solids — cones, spheres, hemispheres and capacity conversions.

Exercise 12.2 →

Volumes of solids — cones, spheres, hemispheres and capacity conversions.

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