Chapter 12 · Kerala SSLC Class 10 Maths
Solids
Two ways to learn — a crisp formula reference, or a story that makes the maths feel obvious.
A story that builds each formula naturally.
How story mode works
Each scene is a real object. The formulas come from the shape of that object. Read the story, then tap "The Maths" to see the formula.
The Water Tank — Cylinder
The school needs a new overhead water tank. The plumber says the tank will be cylindrical — a circular cross-section all the way up. Radius 7 cm (scaled model), height 10 cm. How much water can it hold, and how much sheet metal is needed?
Volume is straightforward — it is just the area of the circle times the height. And for the sheet metal, I need the curved wall plus the top and bottom circles.
V = πr²h = π × 7² × 10 = 490π cm³
CSA (curved wall only) = 2πrh = 2π × 7 × 10 = 140π cm²
TSA (wall + both circles) = 2πr(r+h) = 2π × 7 × (7+10) = 238π cm²
The curved wall is the label on a tin can. The TSA is the full tin — label plus both lids.
The Ice Cream Cone — Cone and Hemisphere Combination
Riya buys an ice cream. The cone has r = 3 cm and h = 4 cm. On top sits a hemisphere of ice cream with the same radius r = 3 cm. She wants to know: total volume of ice cream, and the total surface area the waffle cone + dome occupy.
The cone is underneath, the scoop is a hemisphere on top. I can calculate each part separately and add.
First, find the slant height of the cone — needed for CSA.
l = √(r²+h²) = √(9+16) = √25 = 5 cm
V(cone) = ⅓πr²h = ⅓ × π × 9 × 4 = 12π cm³
V(hemisphere) = ⅔πr³ = ⅔ × π × 27 = 18π cm³
Total V = 12π + 18π = 30π cm³
The outer surface: the cone's slanted face (CSA) plus the dome of the hemisphere (CSA only — no flat base because that sits on the cone).
Cone CSA = πrl = π × 3 × 5 = 15π cm²
Hemisphere CSA (dome) = 2πr² = 2π × 9 = 18π cm²
Total outer surface = 15π + 18π = 33π cm²
Football Manufacturing — Sphere Surface Area and Volume Conservation
A sports factory makes footballs. Each ball is a sphere of radius 11 cm. How much leather (surface area) is needed per ball? Later, the factory melts down defective balls (spheres, r = 3 cm) and recasts them as small cylindrical medals (r = 1 cm, h = 1 cm). How many medals per ball?
Leather needed = surface area of the sphere. Medals from one ball = volume of sphere divided by volume of one medal. Volume is conserved when you melt and recast.
SA(sphere, r=11) = 4πr² = 4π × 121 = 484π cm²
V(sphere, r=3) = 4/3 πr³ = 4/3 × π × 27 = 36π cm³
V(cylinder medal, r=1, h=1) = πr²h = π × 1 × 1 = π cm³
Number of medals = 36π / π = 36
36 medals from one defective ball. The key insight: volume is conserved when melting and recasting. Set the two volumes equal and divide.
The big picture
Identify the shape, pick the right formula, and never mix up r vs diameter, h vs l, or CSA vs TSA.
Cylinder volume?
V = πr²h
Cone CSA? (use slant height!)
l = √(r²+h²) → CSA = πrl
Sphere surface area?
SA = 4πr²
Melted and recast?
V₁ = V₂ (volume conserved)
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