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Complete NCERT Solutions for CBSE Class 10 Maths Chapter 12 — Surface Areas and Volumes: standard solids, combinations of solids, and melting/recasting problems.
Key Formulas (Quick Recap)
| Solid | Volume | Curved/Lateral SA | Total SA |
|---|---|---|---|
| Cube (side a) | a³ | 4a² | 6a² |
| Cuboid | lbh | 2h(l+b) | 2(lb+bh+hl) |
| Cylinder | πr²h | 2πrh | 2πr(r+h) |
| Cone (l=√(r²+h²)) | ⅓πr²h | πrl | πr(l+r) |
| Sphere | 4/3 πr³ | — | 4πr² |
| Hemisphere | 2/3 πr³ | 2πr² | 3πr² |
Exercise 12.1 — Surface Area of Combinations
Q1. Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid. Each cube: a³ = 64 → a = 4 cm. Cuboid = 8 × 4 × 4. TSA = 2(lb + bh + hl) = 2(8·4 + 4·4 + 8·4) = 2(32 + 16 + 32) = 160 cm².
Q2. A cone of radius 7 cm and height 24 cm. Find its slant height, curved surface area, and volume. l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm. CSA = πrl = (22/7)(7)(25) = 550 cm². Volume = ⅓πr²h = ⅓ × (22/7) × 49 × 24 = ⅓ × 22 × 7 × 24 = 1232 cm³.
Q3. A toy is in the shape of a cone mounted on a hemisphere, both of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the total surface area. (Use π = 22/7) Hemisphere radius r = 3.5 cm; cone height h = 15.5 − 3.5 = 12 cm. Cone slant l = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm. TSA = CSA(cone) + CSA(hemisphere) = πrl + 2πr² = (22/7)(3.5)(12.5) + 2(22/7)(3.5²) = 137.5 + 77 = 214.5 cm².
Exercise 12.2 — Volume of Combinations
Q1. A solid is a cylinder with a cone on top. Radius 3.5 cm, cylinder height 10 cm, cone height 6 cm. Find the total volume. (Use π = 22/7) Cylinder V = πr²h = (22/7)(3.5²)(10) = (22/7)(12.25)(10) = 385 cm³. Cone V = ⅓πr²h = ⅓(22/7)(12.25)(6) = 77 cm³. Total = 385 + 77 = 462 cm³.
Q2. Find the volume of a sphere of radius 7 cm. (Use π = 22/7) V = 4/3 πr³ = 4/3 × (22/7) × 343 = 4/3 × 22 × 49 = 4312/3 = 1437.33 cm³.
Melting / Recasting
Q1. A metallic sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder. Volume conserved: 4/3 π(6³) = π(4²)h → 4/3 × 216 = 16h → 288 = 16h → h = 18 cm.
Q2. How many spherical lead shots of radius 1 cm can be made from a solid cube of lead of edge 22 cm? (Use π = 22/7) Cube V = 22³ = 10648 cm³. Each shot V = 4/3 π(1³) = 4/3 × 22/7 = 88/21 cm³. Number = 10648 ÷ (88/21) = 10648 × 21/88 = 2541 shots.
Common Mistakes
- Adding total surface areas of joined solids instead of only the exposed curved areas.
- Forgetting l = √(r² + h²) for the cone slant height.
- In recasting, equating surface areas instead of volumes.
Exam Tips
- For combinations, sketch the solid and shade only the visible surfaces.
- Keep π = 22/7 when radii are multiples of 3.5 or 7.
- Recasting/melting always means equal volumes — set them equal and solve.