Key Concepts & Formulae at a Glance
- Surface Area of a Combination:Add up only the exposed curvedflat surfaces of each part — any face where two solids are joined disappears from the total and must not be counted twice.
- Volume of a Combination:Always additive: total volume = sum of the volumes of the individual solids. Nothing is "hidden" the way surface area is.
- Common Radius Rule:When two solids are joined at a flat circular face (e.g. a cone on a cylinder), that face's radius must match in both solids for a smooth join — this is almost always given or implied.
- CSA / TSA formulas:Cylinder: CSA=2πrh, TSA=2πr(h+r). Cone: CSA=πrl, l=√(r²+h²). Sphere: TSA=4πr². Hemisphere: CSA=2πr², TSA=3πr². Cuboid: TSA=2(lb+bh+hl).
- Volume formulas:Cuboid=lbh. Cylinder=πr²h. Cone=⅓πr²h. Sphere=⁴⁄₃πr³. Hemisphere=₂⁄₃πr³.
- Hollowed-Out Solids:When a shape is scooped out of another (a cavity), subtract the removed solid's volume; for surface area, drop the flat face that disappeared and add the newly exposed curved surface instead.
Exercise 12.1
9 Questions
12 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Two 4×4×4 cubes joined end to end — the resulting cuboid is 8×4×4.
Volume of each cube = 64 cm³ → edge³ = 64 → edge = 4 cm.
(joining two cubes end to end doubles only the length) Resulting cuboid: l=8 cm, b=4 cm, h=4 cm.
TSA of cuboid = 2(lb+bh+hl) = 2(8×4 + 4×4 + 4×8) = 2(32+16+32) = 2×80
(joining two cubes end to end doubles only the length) Resulting cuboid: l=8 cm, b=4 cm, h=4 cm.
TSA of cuboid = 2(lb+bh+hl) = 2(8×4 + 4×4 + 4×8) = 2(32+16+32) = 2×80
Surface area = 160 cm².
2A vessel is in the form of 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.
Cylindrical mouth (open) mounted on a hollow hemispherical bowl of the same radius.
Radius r = 142 = 7 cm. Height of cylindrical part = total height − radius of hemisphere = 13 − 7 = 6 cm.
Inner surface area = CSA of cylinder + CSA of hemisphere (the flat circular face where they join is internal, not part of the inner surface, but it's also where the cylinder opens into the hemisphere, so only the two curved surfaces are exposed to the inside)
= 2πrh + 2πr² = 2×(227)×7×6 + 2×(227)×7² = 264 + 308
Inner surface area = CSA of cylinder + CSA of hemisphere (the flat circular face where they join is internal, not part of the inner surface, but it's also where the cylinder opens into the hemisphere, so only the two curved surfaces are exposed to the inside)
= 2πrh + 2πr² = 2×(227)×7×6 + 2×(227)×7² = 264 + 308
Inner surface area = 572 cm².
3A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Cone mounted on a hemisphere of the same radius (like the Example 1 top).
r = 3.5 cm. Height of cone = total height − radius = 15.5 − 3.5 = 12 cm.
Slant height l = √(r²+h²) = √(3.5²+12²) = √156.25 = 12.5 cm.
TSA of toy = CSA of cone + CSA of hemisphere = πrl + 2πr² = πr(l+2r) = (227)×3.5×(12.5+7) = 11×19.5
Slant height l = √(r²+h²) = √(3.5²+12²) = √156.25 = 12.5 cm.
TSA of toy = CSA of cone + CSA of hemisphere = πrl + 2πr² = πr(l+2r) = (227)×3.5×(12.5+7) = 11×19.5
Total surface area = 214.5 cm².
4A 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.
The hemisphere's diameter can't exceed the cube's edge, or it would overhang the top face.
The hemisphere sits on the cube's top face, so its diameter can be at most the length of that face's edge.
Greatest diameter = 7 cm → r = 3.5 cm.
Surface area of solid = TSA of cube − base area of hemisphere + CSA of hemisphere (base area cancels against part of curved area added)
= 6(side)² + πr² = 6×49 + (227)×3.5² = 294 + 38.5
Greatest diameter = 7 cm → r = 3.5 cm.
Surface area of solid = TSA of cube − base area of hemisphere + CSA of hemisphere (base area cancels against part of curved area added)
= 6(side)² + πr² = 6×49 + (227)×3.5² = 294 + 38.5
Greatest diameter = 7 cm. Surface area = 332.5 cm².
5A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
A hemispherical dent (dashed) is scooped into the top face; its diameter equals the cube's edge.
Let edge of cube = l, so hemisphere radius r = l2.
Surface area = TSA of cube − base area of hemisphere + CSA of hemisphere
(the flat circular patch is removed, and the curved inner bowl is exposed in its place)
= 6l² − πr² + 2πr² = 6l² + πr² = 6l² + π(l2)² = 6l² + πl²4
Surface area = TSA of cube − base area of hemisphere + CSA of hemisphere
(the flat circular patch is removed, and the curved inner bowl is exposed in its place)
= 6l² − πr² + 2πr² = 6l² + πr² = 6l² + π(l2)² = 6l² + πl²4
Surface area of remaining solid = (6 + π4)l² = (24+π)l²4.
6A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Cylindrical middle section with a hemispherical cap fused onto each end.
r = 52 = 2.5 mm. Length of cylindrical part = 14 − 2r = 14 − 5 = 9 mm.
Surface area = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 4πr² = 2πr(h+2r)
= 2×(227)×2.5×(9+5) = 2×(227)×2.5×14
Surface area = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 4πr² = 2πr(h+2r)
= 2×(227)×2.5×(9+5) = 2×(227)×2.5×14
Surface area = 220 mm².
7A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹500 per m². (Note that the base of the tent will not be covered with canvas.)
Conical roof of slant height 2.8 m sitting on a cylindrical wall of height 2.1 m.
r = 42 = 2 m, cylinder height h = 2.1 m, cone slant height l = 2.8 m.
Canvas area = CSA of cylinder + CSA of cone = 2πrh + πrl = πr(2h+l)
= (227)×2×(2×2.1+2.8) = (227)×2×7 = 44 m².
Cost = 44 × ₹500 = ₹22,000.
Canvas area = CSA of cylinder + CSA of cone = 2πrh + πrl = πr(2h+l)
= (227)×2×(2×2.1+2.8) = (227)×2×7 = 44 m².
Cost = 44 × ₹500 = ₹22,000.
Canvas area = 44 m². Cost = ₹22,000.
8From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
A conical cavity (dashed) of the same height and diameter is bored down through the top of the cylinder.
r = 1.42 = 0.7 cm, h = 2.4 cm.
Slant height of cavity l = √(r²+h²) = √(0.49+5.76) = √6.25 = 2.5 cm.
TSA of remaining solid = CSA of cylinder + area of bottom base + CSA of conical cavity
(the flat top disappears, replaced by the cavity's curved inner surface)
= 2πrh + πr² + πrl = πr(2h+r+l) = (227)×0.7×(4.8+0.7+2.5) = (227)×0.7×8 = 17.6 cm²
Slant height of cavity l = √(r²+h²) = √(0.49+5.76) = √6.25 = 2.5 cm.
TSA of remaining solid = CSA of cylinder + area of bottom base + CSA of conical cavity
(the flat top disappears, replaced by the cavity's curved inner surface)
= 2πrh + πr² + πrl = πr(2h+r+l) = (227)×0.7×(4.8+0.7+2.5) = (227)×0.7×8 = 17.6 cm²
Total surface area ≈ 18 cm² (to the nearest cm²).
9A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Both flat ends of the cylinder are scooped into hemispherical dents of the same radius.
Cylinder: r = 3.5 cm, h = 10 cm. Since the scooped hemisphere radius equals the cylinder's radius, each flat circular face is entirely replaced — nothing left over.
TSA = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 4πr² = 2πr(h+2r)
= 2×(227)×3.5×(10+7) = 2×(227)×3.5×17
TSA = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 4πr² = 2πr(h+2r)
= 2×(227)×3.5×(10+7) = 2×(227)×3.5×17
Total surface area = 374 cm².
Exercise 12.2
8 Questions
1A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.
A squat toy — cone height equals its own radius, which also equals the hemisphere's radius.
r = 1 cm, h (cone) = r = 1 cm.
Volume = volume of hemisphere + volume of cone = ₂⁄₃πr³ + ⅓πr²h = ₂⁄₃π(1)³ + ⅓π(1)²(1) = ₂⁄₃π + ⅓π
Volume = volume of hemisphere + volume of cone = ₂⁄₃πr³ + ⅓πr²h = ₂⁄₃π(1)³ + ⅓π(1)²(1) = ₂⁄₃π + ⅓π
Volume = π cm³.
2Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
A cylinder with an identical cone capping each end, like a rugby-ball / capsule frame.
r = 32 = 1.5 cm. Length of cylindrical part = 12 − 2×2 = 8 cm.
Volume of air = volume of cylinder + 2 × volume of cone = πr²h + ₂⁄₃πr²h₁
= (227)×1.5²×8 + ₂⁄₃×(227)×1.5²×2 = 18π + 3π = 21π = 21×227
Volume of air = volume of cylinder + 2 × volume of cone = πr²h + ₂⁄₃πr²h₁
= (227)×1.5²×8 + ₂⁄₃×(227)×1.5²×2 = 18π + 3π = 21π = 21×227
Volume of air = 66 cm³.
3A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.
Plumper than the capsule of Q6 — a shorter cylindrical middle relative to its radius.
r = 2.82 = 1.4 cm. Length of cylindrical part = 5 − 2×1.4 = 2.2 cm.
Volume of one jamun = πr²h + ⁴⁄₃πr³ = (227)×1.4²×2.2 + ⁴⁄₃×(227)×1.4³ ≈ 13.552 + 11.499 ≈ 25.05 cm³.
Volume of 45 jamuns ≈ 45 × 25.05 ≈ 1127.28 cm³.
Syrup (30% of volume) ≈ 0.30 × 1127.28
Volume of one jamun = πr²h + ⁴⁄₃πr³ = (227)×1.4²×2.2 + ⁴⁄₃×(227)×1.4³ ≈ 13.552 + 11.499 ≈ 25.05 cm³.
Volume of 45 jamuns ≈ 45 × 25.05 ≈ 1127.28 cm³.
Syrup (30% of volume) ≈ 0.30 × 1127.28
Approximate syrup content ≈ 338.18 cm³.
4A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Four identical conical depressions (radius 0.5 cm, depth 1.4 cm) bored into the top face.
Volume of cuboid = 15×10×3.5 = 525 cm³.
Volume of one conical depression = ⅓πr²h = ⅓×(227)×0.5²×1.4 = 1130 cm³.
Volume of 4 depressions = 4×1130 = 4430 = 2215 ≈ 1.467 cm³.
Volume of wood = 525 − 2215
Volume of one conical depression = ⅓πr²h = ⅓×(227)×0.5²×1.4 = 1130 cm³.
Volume of 4 depressions = 4×1130 = 4430 = 2215 ≈ 1.467 cm³.
Volume of wood = 525 − 2215
Volume of wood ≈ 523.53 cm³.
5A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
An inverted-cone vessel, full of water, with lead shots dropped in from the top.
Volume of water (= volume of cone) = ⅓πr²h = ⅓×(227)×5²×8 = 440021 cm³.
Water that overflows = ¼ of this = 110021 cm³ — this equals the total volume of all lead shots.
Volume of one lead shot = ⁴⁄₃πr³ = ⁴⁄₃×(227)×0.5³ = 1121 cm³.
Number of shots = (110021) ÷ (1121) = 110011
Water that overflows = ¼ of this = 110021 cm³ — this equals the total volume of all lead shots.
Volume of one lead shot = ⁴⁄₃πr³ = ⁴⁄₃×(227)×0.5³ = 1121 cm³.
Number of shots = (110021) ÷ (1121) = 110011
Number of lead shots = 100.
6A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)
A tall base cylinder topped by a narrower cylinder (dimensions not to true relative scale).
Volume = πr₁²h₁ + πr₂²h₂ = 3.14×12²×220 + 3.14×8²×60 = 3.14×31680 + 3.14×3840 = 3.14×35520 = 111532.8 cm³.
Mass = volume × 8 gcm³ = 111532.8 × 8 = 892262.4 g
Mass = volume × 8 gcm³ = 111532.8 × 8 = 892262.4 g
Mass of the pole ≈ 892.26 kg.
7A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
The cone-on-hemisphere solid sits inside the water-filled cylinder, displacing water around it.
Volume of cylinder = πR²H = π×60²×180.
Volume of solid (cone+hemisphere) = ⅓πr²h + ₂⁄₃πr³ = ⅓π×60²×120 + ₂⁄₃π×60³ = 144000π + 144000π = 288000π.
Volume of water left = 648000π − 288000π = 360000π = 360000×227
Volume of solid (cone+hemisphere) = ⅓πr²h + ₂⁄₃πr³ = ⅓π×60²×120 + ₂⁄₃π×60³ = 144000π + 144000π = 288000π.
Volume of water left = 648000π − 288000π = 360000π = 360000×227
Volume of water left = 1131428.57 cm³ (≈ 1.131 m³).
8A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
A flask — spherical body with a narrow cylindrical neck on top.
Sphere radius = 8.52 = 4.25 cm. Neck radius = 22 = 1 cm, neck length = 8 cm.
Volume = ⁴⁄₃πR³ + πr²h = ⁴⁄₃×3.14×4.25³ + 3.14×1²×8 ≈ 321.39 + 25.12
Volume = ⁴⁄₃πR³ + πr²h = ⁴⁄₃×3.14×4.25³ + 3.14×1²×8 ≈ 321.39 + 25.12
Actual volume ≈ 346.51 cm³, so the child's figure of 345 cm³ is very close (off by about 1.5 cm³) — essentially correct, allowing for rounding.
Frequently Asked Questions
Because wherever two solids are joined, a flat face on each of them gets covered up and is no longer part of the outer surface. Only the surfaces you can actually see and touch from outside count, so those hidden joining faces must be left out of the total.
Ask whether that face is still visible on the finished solid. If a flat circular face is completely covered by another solid sitting on it (like a cone's base sitting on a hemisphere of the same radius), drop it entirely. If only part of it remains exposed (like a smaller cylinder's base sitting on a larger cone's base), keep the exposed ring and subtract only the covered part.
Volume measures the space a solid occupies, and joining two solids together doesn't make either of them occupy less space — it just glues the spaces together. Surface area only measures the outer boundary, and that boundary genuinely shrinks wherever the two solids meet.
For volume, subtract the volume of the removed shape from the volume of the original solid. For surface area, remove the flat face that the cavity opened up through, and add the newly exposed curved surface of the cavity in its place — the rest of the original solid's surface is unaffected.
Yes — a printable Formula Card covering all 7 units, including Surface Areas and Volumes, is available for ₹99.
Continue to the Next Chapter
Chapter 13 (Statistics) moves from geometry back into data analysis.