CBSE Class 10 · Maths · board exam 2027
Chapter 12: Surface Areas and Volumes — notes
The chapter’s 4 parts in the book’s order — the first idea of each and the points Medha Ma’am writes on the board — and the summary at the end. The full lesson and a check question for each part are in the app.
1. Basic solids and their formulas
Every question of this chapter uses a few formulas again and again.
- • Cuboid: length l, breadth b, height h
- • Volume of cuboid = l × b × h
- • TSA of cuboid = 2(lb + bh + hl)
- • LSA of cuboid (four walls) = 2(l + b)h
- • Cube: each edge a
- • Volume of cube = a³
- • TSA of cube = 6a², LSA of cube = 4a²
- • Diagonal of cube = √(a² + a² + a²) = √(3a²) = a√3
- • Cylinder: radius r, height h
- • CSA of cylinder = 2πr × h = 2πrh
- • TSA of cylinder = 2πrh + 2πr² = 2πr(h + r)
- • Volume of cylinder = πr²h
- • Cone: radius r, height h, slant height l
- • l² = r² + h² ⇒ l = √(r² + h²)
- • CSA of cone = πrl
- • TSA of cone = πrl + πr² = πr(l + r)
- • Volume of cone = 1/3πr²h
- • Example: r = 7 cm, h = 24 cm, l = ?
- • r² = 7 × 7 = 49, h² = 24 × 24 = 576
- • r² + h² = 49 + 576 = 625
- • l = √625 = 25 cm (25 × 25 = 625)
- • Volume = 1/3 × 22/7 × 7 × 7 × 24
- • 24/3 = 8, 22/7 × 7 = 22
- • Volume = 22 × 7 × 8 = 154 × 8 = 1232 cm³
- • Sphere: radius r
- • Surface area of sphere = 4πr²
- • Volume of sphere = 4/3πr³
- • Hemisphere: radius r, height = r
- • CSA of hemisphere = 1/2 × 4πr² = 2πr²
- • TSA of solid hemisphere = 2πr² + πr² = 3πr²
- • Volume of hemisphere = 1/2 × 4/3πr³ = 2/3πr³
- • r = 7, 14, 3.5, 2.1, … → π = 22/7
- • Question says π = 3.14 → use 3.14
- • π on both sides → it cancels
2. Surface area of a combination of solids
The real topic of this chapter.
- • Oil tanker = hemisphere + cylinder + hemisphere
- • Test tube = cylinder + hemisphere
- • Surface area of a combination = sum of the surfaces we can see
- • TSA of tanker = CSA of hemisphere + CSA of cylinder + CSA of hemisphere
- • Rasheed's top: cone on a hemisphere
- • Total height = 5 cm, diameter = 3.5 cm
- • Find: the area to be coloured
- • r = 3.5/2 = 1.75 cm
- • Height of hemisphere = r = 1.75 cm
- • Height of cone h = 5 − 1.75 = 3.25 cm
- • l = √(r² + h²)
- • 1.75 × 1.75 = 3.0625, 3.25 × 3.25 = 10.5625
- • l = √(3.0625 + 10.5625) = √13.625
- • l ≈ 3.7 cm
- • Area to colour = CSA of hemisphere + CSA of cone
- • = 2πr² + πrl = πr(2r + l)
- • Area = 22/7 × 1.75 × (3.5 + 3.7)
- • 22/7 × 1.75 = 22 × 0.25 = 5.5
- • 3.5 + 3.7 = 7.2
- • Area ≈ 5.5 × 7.2 = 39.6 cm²
- • Joined flat faces are hidden: never count them
- • Hemisphere scooped out of a face
- • Area lost = flat circle = πr²
- • Area gained = inside of the bowl = 2πr²
- • Net change = 2πr² − πr² = +πr²
- • Example: cube of side 14 cm, hemisphere of diameter 14 cm scooped out
- • TSA of cube = 6 × 14 × 14 = 6 × 196 = 1176 cm²
- • r = 14/2 = 7 cm
- • CSA of hemisphere = 2 × 22/7 × 7 × 7 = 2 × 154 = 308 cm²
- • Circle removed = 22/7 × 7 × 7 = 154 cm²
- • TSA of remaining solid = 1176 + 308 − 154
- • = 1484 − 154 = 1330 cm²
3. Volume of a combination of solids
Volume.
- • Volume of a combination = sum of the volumes of the parts
- • Shanta's shed: cuboid with a half cylinder on top
- • Cuboid: 15 m × 7 m × 8 m
- • Machinery = 300 m³, 20 workers × 0.08 m³ each
- • Find: volume of air in the shed
- • Volume of cuboid = 15 × 7 × 8
- • = 105 × 8 = 840 m³
- • Half cylinder: r = 7/2 = 3.5 m, h = 15 m
- • Volume of half cylinder = 1/2 × πr²h
- • = 1/2 × 22/7 × 3.5 × 3.5 × 15
- • 22/7 × 3.5 = 11, 11 × 3.5 = 38.5
- • 38.5 × 15 = 577.5
- • Volume of half cylinder = 1/2 × 577.5 = 288.75 m³
- • Volume of shed = 840 + 288.75 = 1128.75 m³
- • Workers = 20 × 0.08 = 1.6 m³
- • Space taken = 300 + 1.6 = 301.6 m³
- • Air in the shed = 1128.75 − 301.6 = 827.15 m³
- • Toy: cone on a hemisphere
- • Volume of toy = 1/3πr²h + 2/3πr³
- • Something taken out → subtract its volume
- • Wood left = πr²h − 1/3πr²h = 2/3πr²h
- • Capacity of glass = volume of cylinder − volume of hemisphere
4. Converting one solid into another
Sometimes a solid is melted, or reshaped, into another shape.
- • Melted or reshaped: the volume stays the same
- • Volume before = volume after
- • Example: clay cone → sphere
- • Cone: h = 24 cm, r = 6 cm
- • Find: radius of the sphere
- • Volume of cone = 1/3πr²h = 1/3 × π × 6 × 6 × 24
- • 24/3 = 8, 6 × 6 = 36, 36 × 8 = 288
- • Volume of cone = 288π cm³
- • Let the radius of the sphere be r cm
- • 4/3πr³ = 288π
- • 4/3r³ = 288
- • r³ = 288 × 3/4 = 72 × 3 = 216
- • 216 = 6 × 6 × 6 = 6³
- • r = 6, so the radius of the sphere = 6 cm
- • Number of small pieces = big volume ÷ volume of one piece
- • Water flowing in a time t
- • Volume = area of cross-section × distance flowed in time t
Summary
Very good! You learnt a lot today, so let us revise it once together. First we revised the basic solids, the cuboid, cube, cylinder, cone, sphere and hemisphere, with their surface areas and volumes. Then we saw that many everyday objects are combinations of these solids. For the surface area of a combination, we add only the surfaces we can see, because the joined faces are hidden. For the volume, we simply add the volumes of the parts, and subtract a part that is taken out. And when a solid is melted or reshaped, the volume stays the same. Write all the formulas in your notebook once, without looking. Your homework is on your screen. In the next chapter we will start Statistics. If anything was not clear, press the Ask button and ask me. Done, you worked very well today!
AskMedha · Medha Ma’am — CBSE Class 10 Maths, Chapter 12. Written from the textbook, checked twice.
Questions parents and students ask
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- From the chapter’s own textbook text, part by part, in the book’s own terms; a second, independent model then checked every part against the book. The full lesson, a check question for each part and the practice are in the AskMedha app.
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