Exercise 12.1 — Surface Areas and Volumes

1. 12 marksNCERT Cl-10 Maths, Ex 12.1, Q1

   Two cubes, each of volume 64 cm³, are joined end to end. Find the surface area of the resulting cuboid.

   Hint. First find the cube's side from its volume. The join hides two square faces — one from each cube.

   Step 1 — Find the side of each cube. Volume = side³ = 64, so side = ∛64 = 4 cm

   Step 2 — Describe the resulting cuboid. Joining two 4 cm cubes end to end gives a cuboid of length 8 cm, breadth 4 cm, height 4 cm.

   Step 3 — Apply the cuboid surface-area formula. TSA = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 2(80)

   ✦ Answer: 160 cm²

   Where students slip. Adding the two cubes' individual surface areas (6×16 = 96 each, so 192 total) without removing the two faces that are now glued together and hidden inside the cuboid.

   Another way. Count faces directly: the combined shape has 2 square end-faces (4×4 = 16 each) and 4 rectangular side-faces (4×8 = 32 each): 2(16) + 4(32) = 32 + 128 = 160 — the same total, reached by counting rather than the formula.

2. 23 marksNCERT Cl-10 Maths, Ex 12.1, 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.

   Hint. Since it's hollow and open, only curved surfaces are exposed on the inside — no flat circles anywhere.

   Step 1 — Find the radius and the cylinder's height. radius = 14/2 = 7 cm The hemisphere itself takes up 7 cm of the total 13 cm height, so the cylinder's height is 13 − 7 = 6 cm.

   Step 2 — Identify what is exposed on the inside. Being hollow, the inner surface consists of the cylinder's curved surface plus the hemisphere's curved surface — there is no flat base to include, since the vessel is open at the top and the hemisphere's curved bowl is what you'd feel reaching inside.

   Step 3 — Apply the formula. inner surface = 2πrh + 2πr² = 2πr(h + r) = 2 × (22/7) × 7 × (6 + 7)

   Step 4 — Compute. = 2 × 22 × 13 = 572

   ✦ Answer: 572 cm²

   Where students slip. Adding a flat circular base to the total. A vessel is open — there is no solid bottom to include; the hemisphere's curved bowl already forms the bottom of the inside.

   Another way. Factor r out early: 2πr(h + r) with r = 7 and h + r = 13 gives 2 × (22/7) × 7 × 13 = 2 × 22 × 13 in one line, since the 7s cancel before any multiplication.

3. 34 marksNCERT Cl-10 Maths, Ex 12.1, Q3

   A toy is a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

   Hint. Subtract the hemisphere's radius from the total height to get the cone's own height, then find the slant height before anything else.

   Step 1 — Find the cone's height. The hemisphere contributes 3.5 cm to the total height, so the cone's own height is 15.5 − 3.5 = 12 cm

   Step 2 — Find the slant height. l = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm

   Step 3 — Identify the exposed surfaces. Where the cone sits on the hemisphere, the flat circular face is hidden inside — only the cone's curved surface and the hemisphere's curved surface are visible. TSA = πrl + 2πr²

   Step 4 — Substitute and compute. = π × 3.5 × 12.5 + 2π × 3.5² = π(43.75 + 24.5) = π × 68.25 = (22/7) × 68.25 = 1501.5/7

   ✦ Answer: 214.5 cm²

   Where students slip. Using 15.5 cm as the cone's own height. The total height includes the hemisphere's radius, which must be subtracted first to get the cone's height for the slant-height formula.

   Another way. Factor π × 3.5 out of both terms: π × 3.5 × (12.5 + 2×3.5) = π × 3.5 × 19.5 = (22/7)(3.5)(19.5), reaching the same 214.5 without expanding each term separately.

4. 44 marksNCERT Cl-10 Maths, Ex 12.1, Q4

   A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

   Hint. The hemisphere sits on the cube's top face, so its diameter cannot exceed that face's side.

   Step 1 — Find the greatest possible diameter. The hemisphere's flat circular base must fit entirely within the cube's 7 cm × 7 cm top face. The largest circle that fits inside a square has diameter equal to the square's side.

   ✦ Greatest diameter = 7 cm (so radius = 3.5 cm)

   Step 2 — Identify the exposed surfaces. The cube contributes all six faces, except that a circular patch on the top face is covered by the hemisphere's base. In its place, the hemisphere's curved surface is exposed. TSA = (TSA of cube) − (base circle of hemisphere) + (CSA of hemisphere) = 6a² − πr² + 2πr² = 6a² + πr²

   Step 3 — Substitute a = 7, r = 3.5. = 6 × 49 + (22/7) × 3.5² = 294 + (22/7) × 12.25

   Step 4 — Compute the second term. 12.25/7 = 1.75, so (22)(1.75) = 38.5. = 294 + 38.5

   ✦ Answer: 332.5 cm²

   Where students slip. Forgetting to subtract the circular patch that the hemisphere covers, and simply adding 6a² + 2πr² — that double-counts nothing removed, when in fact the base circle's area must first come out of the cube's flat top.

   Another way. Notice the −πr² and +2πr² combine to a net +πr², so the whole calculation reduces to 'cube's TSA, plus one extra base-circle's worth of area' — a pattern worth remembering, since it recurs in Q5.

5. 54 marksNCERT Cl-10 Maths, Ex 12.1, Q5

   A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter of the hemisphere equals the edge of the cube (call the edge l). Determine the surface area of the remaining solid.

   Hint. This time the hemisphere is a hole, not a bump — but the surface-area bookkeeping works out the same way as Q4.

   Step 1 — Set up in terms of the edge l. The hemisphere's diameter equals l, so its radius is l/2.

   Step 2 — Identify the exposed surfaces. Removing a hemispherical scoop from one face takes away a circular patch of that face (area π(l/2)²) but exposes the hemisphere's curved bowl (area 2π(l/2)²) in its place. Surface area = 6l² − π(l/2)² + 2π(l/2)² = 6l² + π(l/2)²

   Step 3 — Simplify (l/2)² = l²/4. = 6l² + πl²/4

   ✦ Answer: 6l² + πl²/4 (equivalently, l²(24 + π)/4)

   This is exactly the same net effect as Q4 — a hemisphere joined outward or scooped inward both remove one flat circle and add one curved bowl of the same size, so both give '+πr²' on top of the flat shape's own area.

   Where students slip. Reasoning that a *hole* should reduce the total surface area compared with a plain cube. It does not — carving in a curved bowl actually adds a little more surface than the flat circle it replaced, since a hemisphere's curved area (2πr²) is twice the flat circle it covers (πr²).

   Another way. Compare directly with Q4: both problems remove a flat circle of area πr² and expose a curved area of 2πr² in its place, so both reduce to (shape's own area) + πr² regardless of whether the hemisphere bulges out or is scooped in.

6. 63 marksNCERT Cl-10 Maths, Ex 12.1, Q6 (Fig. 12.10)

   A medicine capsule is a cylinder with two hemispheres stuck to each end. The length of the entire capsule is 14 mm and the diameter is 5 mm. Find its surface area.

   Hint. Both ends are hemispheres, so the cylinder's own length is the total length minus twice the radius.

   Step 1 — Find the radius and the cylinder's length. radius = 5/2 = 2.5 mm Each hemispherical end contributes its radius to the total length, so the cylindrical part's length is 14 − 2(2.5) = 14 − 5 = 9 mm

   Step 2 — Identify the exposed surfaces. Only curved surfaces are exposed: the cylinder's curved surface, plus the two hemispherical ends' curved surfaces — which together make one full sphere's worth of curved area. surface area = 2πrh + 2(2πr²) = 2πr(h + 2r)

   Step 3 — Substitute r = 2.5, h = 9. = 2 × (22/7) × 2.5 × (9 + 5) = 2 × (22/7) × 2.5 × 14

   Step 4 — The 14 and 7 simplify neatly: 14/7 = 2. = 2 × 22 × 2.5 × 2 = 220

   ✦ Answer: 220 mm²

   Where students slip. Treating the diameter (5 mm) as the radius directly in the formula, which doubles every term that depends on r. Halve the diameter first.

   Another way. Recognise that two hemispherical ends make a full sphere, so the total is simply (cylinder's CSA) + (one sphere's surface area) = 2πrh + 4πr² = 2πr(h + 2r) — the same expression, seen as 'cylinder plus a whole sphere' rather than 'cylinder plus two half-spheres'.

7. 74 marksNCERT Cl-10 Maths, Ex 12.1, Q7

   A tent is a cylinder surmounted by a conical top. The cylindrical part has height 2.1 m and diameter 4 m, and the slant height of the conical top is 2.8 m. Find the area of canvas used (the base is not covered), and the cost at ₹500 per m².

   Hint. No base circle is needed at all — a tent has no floor of canvas. Just the two curved surfaces.

   Step 1 — Find the radius. radius = 4/2 = 2 m

   Step 2 — Identify what needs canvas. A tent is open at the bottom, so there is no base circle to cover — only the cylinder's curved wall and the cone's curved roof need canvas. canvas area = 2πrh + πrl = πr(2h + l)

   Step 3 — Substitute r = 2, h = 2.1, l = 2.8. = π × 2 × (2×2.1 + 2.8) = π × 2 × (4.2 + 2.8) = π × 2 × 7

   Step 4 — Compute. = 14π = 14 × (22/7) = 44

   ✦ Canvas area = 44 m²

   Step 5 — Find the cost. cost = 44 × 500 = 22,000

   ✦ Cost = ₹22,000

   Where students slip. Adding a base circle (πr²) to the canvas total. The question explicitly says the base is not covered, and physically a tent needs no floor — you stand on the ground, not on canvas.

   Another way. Factor πr out at the start: πr(2h + l) with r = 2, so the whole thing is 2π(2h + l) = 2π(7) = 14π — the 2h + l = 7 simplification is what makes the final multiplication by 22/7 come out to a whole number.

8. 84 marksNCERT Cl-10 Maths, Ex 12.1, 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, to the nearest cm².

   Hint. The remaining solid keeps the cylinder's curved side and one flat base — the other end has a conical dent instead of a flat circle.

   Step 1 — Find the radius and the slant height of the cavity. radius = 1.4/2 = 0.7 cm, height = 2.4 cm l = √(r² + h²) = √(0.7² + 2.4²) = √(0.49 + 5.76) = √6.25 = 2.5 cm

   Step 2 — Identify the three surfaces that remain. (a) The cylinder's full curved side — unaffected by the cavity, since the hollowing is only at one end. (b) The flat circular base at the other end — also unaffected. (c) At the hollowed end, instead of a flat circle, the cone-shaped dent exposes its slanted inner surface.

   TSA = 2πrh + πr² + πrl

   Step 3 — Substitute r = 0.7, h = 2.4, l = 2.5. 2πrh = 2 × (22/7) × 0.7 × 2.4 = 10.56 πr² = (22/7) × 0.49 = 1.54 πrl = (22/7) × 0.7 × 2.5 = 5.5

   Step 4 — Add them. 10.56 + 1.54 + 5.5 = 17.6

   Step 5 — Round to the nearest cm², as asked. 17.6 → 18

   ✦ Answer: 18 cm²

   Where students slip. Subtracting the cone's curved surface instead of adding it. Hollowing out material does not remove surface — it replaces a flat circle with a larger slanted surface, which is *added* to the total, not taken away.

   Another way. Sanity-check the size: the flat circle removed had area πr² = 1.54 cm², while the slanted surface that replaces it (πrl = 5.5 cm²) is more than three times larger — makes sense, since a slant is always longer than the straight radius it stands in for.

9. 94 marksNCERT Cl-10 Maths, Ex 12.1, Q9 (Fig. 12.11)

   A wooden article is made by scooping out a hemisphere from each end of a solid cylinder. The cylinder has height 10 cm and radius 3.5 cm. Find the total surface area of the article.

   Hint. Both flat ends are replaced by hemispherical bowls — no flat circles remain anywhere on the article.

   Step 1 — Identify the surfaces. The cylinder's curved side is untouched. But both flat circular ends have had a hemisphere scooped out, so both flat circles are replaced by curved hemispherical bowls, each larger than the circle it replaced.

   TSA = (CSA of cylinder) + 2 × (CSA of hemisphere) = 2πrh + 2(2πr²) = 2πr(h + 2r)

   Step 2 — Substitute r = 3.5, h = 10. = 2 × (22/7) × 3.5 × (10 + 7)

   Step 3 — Compute. 22/7 × 3.5 = 11. = 2 × 11 × 17 = 374

   ✦ Answer: 374 cm²

   Where students slip. Including flat circular area anywhere in the total. After scooping from *both* ends, there is no flat circular surface left on the whole object — every part of its surface is either the cylinder's curved side or one of the two hemispherical bowls.

   Another way. This is structurally identical to Q6 (the capsule), just inverted — two hemispherical scoops from a cylinder give the same formula, 2πr(h + 2r), as two hemispherical bumps on a cylinder. The sign of the bulge doesn't change which surfaces are exposed.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (this chapter now has two exercises, 12.1 with 9 questions and 12.2 with 8; the frustum of a cone and conversion of solids, with the old third, fourth and fifth exercises, are no longer part of this chapter). Questions are referenced from the NCERT textbook for identification.