Chapter 12, Surface Areas and Volumes, is one of the highest-weightage chapters in the Class 10 Maths board exam. It combines mensuration formulas from earlier classes (cubes, cuboids, cylinders, cones, spheres, hemispheres) into problems on combinations of solids — a favourite source of 3-5 mark board questions. Getting comfortable with visualising which surfaces are “hidden” when two solids are joined is the single most important skill this chapter teaches, and it carries directly into similar problems in competitive exams like NTSE and Olympiads.
As per the current 2026-27 rationalised NCERT syllabus, this chapter has two exercises — 12.1 (Surface Area of a Combination of Solids) and 12.2 (Volume of a Combination of Solids) — 17 questions in total. Some older solution sites online still show a third exercise, 12.3, on converting one solid into another; that exercise was removed in the 2023 rationalisation, so ignore any “12.3” content you find elsewhere — it does not exist in the current book.
Exercise 12.1 — Surface Areas of Combination of Solids (use π = 22/7 unless stated)
Q1. Two cubes, each of volume 64 cm³, are joined end to end. Find the surface area of the resulting cuboid.
Solution: Side of each cube, a³ = 64 ⟹ a = 4 cm. Joining two cubes end to end gives a cuboid of dimensions 8 cm × 4 cm × 4 cm.
Surface area = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32+16+32) = 2×80 = 160 cm².
Q2. A vessel is a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution: r = 7 cm. Height of cylindrical part = 13 − 7 = 6 cm.
Inner surface area = 2πrh + 2πr² = 2πr(h + r) = 2×(22/7)×7×(6+7) = 2×22×13 = 572 cm².
Q3. A toy is a cone of radius 3.5 cm mounted on a hemisphere of the same radius. Total height of the toy is 15.5 cm. Find its total surface area.
Solution: r = 3.5 cm, height of cone h = 15.5 − 3.5 = 12 cm. Slant height l = √(r²+h²) = √(12.25+144) = √156.25 = 12.5 cm.
TSA = πrl + 2πr² = πr(l + 2r) = (22/7)×3.5×(12.5+7) = 11×19.5 = 214.5 cm².
Q4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have, and what is the surface area of the solid?
Solution: Greatest possible diameter = side of cube = 7 cm ⟹ r = 3.5 cm.
Surface area = (surface area of cube) − (base circle of hemisphere) + (curved surface of hemisphere) = 6×7² + πr² (since −πr²+2πr² = πr²) = 294 + (22/7×3.5×3.5) = 294 + 38.5 = 332.5 cm².
Q5. A hemispherical depression is cut from one face of a cubical wooden block of edge l, such that the diameter of the hemisphere equals l. Find the surface area of the remaining solid.
Solution: r = l/2. Surface area = 6l² − πr² + 2πr² = 6l² + πr² = 6l² + π(l/2)² = (6 + π/4)l², i.e. [(24+π)/4]l².
Q6. A medicine capsule is a cylinder with two hemispheres stuck to its ends. Total length 14 mm, diameter 5 mm. Find its surface area.
Solution: r = 2.5 mm. Cylinder length = 14 − 2×2.5 = 9 mm.
Surface area = 2πrh + 2(2πr²) = 2πr(h+2r) = 2×(22/7)×2.5×(9+5) = 2×(22/7)×2.5×14 = 220 mm².
Q7. A tent is a cylinder surmounted by a cone. Cylinder height 2.1 m, diameter 4 m, slant height of the cone 2.8 m. Find the area of canvas used and its cost at ₹500/m².
Solution: r = 2 m. Canvas area = 2πrh + πrl = πr(2h+l) = (22/7)×2×(4.2+2.8) = (22/7)×2×7 = 44 m². Cost = 44 × 500 = ₹22,000.
Q8. From a solid cylinder of height 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and diameter is hollowed out. Find the total surface area of the remaining solid (nearest cm²).
Solution: r = 0.7 cm. Slant height of cavity l = √(r²+h²) = √(0.49+5.76) = √6.25 = 2.5 cm.
TSA = CSA of cylinder + base circle + CSA of cone cavity = 2πrh + πr² + πrl = 10.56 + 1.54 + 5.5 = 17.6 ≈ 18 cm².
Q9. A wooden article is made by scooping out a hemisphere from each end of a solid cylinder of height 10 cm and base radius 3.5 cm. Find its total surface area.
Solution: TSA = CSA of cylinder + 2×(CSA of hemisphere) = 2πrh + 2(2πr²) = 2πr(h+2r) = 2×(22/7)×3.5×(10+7) = 22×17 = 374 cm².
Exercise 12.2 — Volumes of Combination of Solids
Q1. A solid cone stands on a hemisphere, both of radius 1 cm, and the cone’s height equals its radius. Find the volume in terms of π.
Solution: Volume = (1/3)πr²h + (2/3)πr³ = (1/3)π(1)(1) + (2/3)π(1) = π/3 + 2π/3 = π cm³.
Q2. Rachel’s model: a cylinder with two cones attached at its ends, diameter 3 cm, total length 12 cm, each cone of height 2 cm. Find the volume of air inside.
Solution: r = 1.5 cm. Cylinder height = 12 − 2×2 = 8 cm. Volume = πr²[h + (2/3)(2)] = π(2.25)(8 + 4/3) = π×2.25×(28/3) = 21π = 21×22/7 = 66 cm³.
Q3. A gulab jamun (cylinder with two hemispherical ends), length 5 cm, diameter 2.8 cm, holds sugar syrup up to 30% of its volume. Find the approximate syrup in 45 gulab jamuns.
Solution: r = 1.4 cm, cylinder height = 5 − 2.8 = 2.2 cm. Volume of one = πr²h + (4/3)πr³ ≈ 25.05 cm³. For 45 pieces: 45×25.05 ≈ 1127.25 cm³. Syrup (30%) ≈ 0.3×1127.25 ≈ 338 cm³.
Q4. A wooden pen stand, a cuboid 15 cm × 10 cm × 3.5 cm, has four conical depressions of radius 0.5 cm and depth 1.4 cm each. Find the volume of wood.
Solution: Cuboid volume = 15×10×3.5 = 525 cm³. One cone = (1/3)π(0.25)(1.4) ≈ 0.367 cm³; four cones ≈ 1.47 cm³. Wood volume = 525 − 1.47 ≈ 523.53 cm³.
Q5. An inverted conical vessel, height 8 cm, top radius 5 cm, is filled to the brim. Lead shots of radius 0.5 cm are dropped in, and one-fourth of the water flows out. Find the number of lead shots.
Solution: Cone volume = (1/3)π(25)(8) = 200π/3. Water displaced = (1/4)×200π/3 = 50π/3. One shot’s volume = (4/3)π(0.5)³ = π/6. Number of shots = (50π/3)/(π/6) = 100.
Q6. An iron pole is a cylinder of height 220 cm, base diameter 24 cm, surmounted by another cylinder of height 60 cm, radius 8 cm. Find its mass (1 cm³ iron ≈ 8 g, use π = 3.14).
Solution: Volume = π(144×220 + 64×60) = π×35520 ≈ 3.14×35520 = 111,532.8 cm³. Mass = 111,532.8×8 g = 892,262.4 g ≈ 892.26 kg.
Q7. A cone (h=120 cm, r=60 cm) on a hemisphere (r=60 cm) stands upright, touching the bottom, inside a cylinder full of water (r=60 cm, h=180 cm). Find the water left in the cylinder.
Solution: Cylinder volume = π(3600)(180) = 648000π. Cone + hemisphere = 144000π + 144000π = 288000π. Water left = 648000π − 288000π = 360000π = 360000×22/7 ≈ 1,131,428.57 cm³ (≈1.131 m³).
Q8. A spherical vessel with a cylindrical neck (8 cm long, 2 cm diameter) and spherical part diameter 8.5 cm is measured by a child to hold 345 cm³. Using π = 3.14, check her answer.
Solution: Sphere volume = (4/3)(3.14)(4.25)³ ≈ 321.39 cm³. Neck volume = 3.14×(1)²×8 = 25.12 cm³. Total ≈ 346.51 cm³ — very close to the child’s measured 345 cm³, so her measurement is essentially correct within experimental/rounding error.
Why This Chapter Matters for Boards
Surface Areas and Volumes typically contributes 6-8 marks to the Class 10 Maths board paper, almost always as one long-answer combination-of-solids question. Examiners particularly like the “convert one shape into another, keeping volume constant” style of reasoning, so make sure the Extra Questions/HOTS companion post below is not skipped — several of its questions are built exactly around that reasoning pattern.
FAQs
Q: Does the 2026-27 NCERT book still have Exercise 12.3?
A: No. The 2023 rationalisation removed the old “converting one solid into another” exercise (previously 13.3 in the pre-rationalised book); only 12.1 and 12.2 remain.
Q: Is π = 22/7 or 3.14 used in this chapter?
A: Both appear — use whichever value the specific question specifies; when unspecified, 22/7 is the NCERT default.
