IGCSE Maths · Mensuration · Core & Extended

Surface Area, Volume and Compound Shapes

Six solids, six sets of formulas — most of them given to you on the exam formula sheet, so the real skill is knowing which formula to reach for and how to combine them. This note also covers compound shapes and solids (two shapes joined together) and, for Extended, frustums.

1. Formula Reference

Most — but not all — of these are given in the exam’s List of Formulas. Check the last column carefully: a few (marked “not given”) you’re expected to build yourself from simpler pieces you already know.

SolidVolumeCurved/total surface areaGiven?
CuboidV = lwhSA = 2(lw+lh+wh)Neither given
PrismV = (cross-section area) × lengthSA = 2×(cross-section) + (perimeter × length)Volume given; SA not given
CylinderV = πr²hCurved SA = 2πrh (add 2πr² for total)Both given (curved SA only — you add the circles yourself)
ConeV = ⅓πr²hCurved SA = πrl (add πr² for total)Both given (curved SA only)
SphereV = (4/3)πr³SA = 4πr²Both given, in full
PyramidV = ⅓ × base area × heightbase + triangular facesVolume given; SA not given
Memory tip — the “one third” connection Notice that both pyramids and cones use V = ⅓ × (base area) × height — because a cone is really just a pyramid with a circular base. Even better: a cone/pyramid always has exactly ⅓ the volume of the cylinder/prism that shares the same base and height. Picture pouring sand from a cone-shaped cup into a cylinder-shaped cup of the same base and height — it takes exactly 3 cone-fuls to fill the cylinder. That single image anchors all four volume formulas at once.
Memory tip — spotting what’s NOT given Look for the pattern: curved, single-surface shapes (cylinder, cone, sphere) have their formulas given in full or in curved-part form. Shapes built from flat polygon faces (cuboid, prism, pyramid) don’t — you’re expected to add up the flat faces yourself, the same way you would work out a net’s area. If a solid has straight edges and flat faces you could unfold into a net, assume you’ll need to build its surface area formula from scratch.

2. Cuboids and Prisms

Worked example — cuboid Find the surface area and volume of a cuboid 8 cm × 5 cm × 4 cm.
Surface area = 2(8×5 + 8×4 + 5×4) = 2(40+32+20) = 2(92) = 184 cm²
Volume = 8 × 5 × 4 = 160 cm³
Worked example — triangular prism A triangular prism has a right-angled triangular cross-section with base 8 cm, height 6 cm, and hypotenuse 10 cm. The prism is 10 cm long. Find its volume and surface area.
Cross-section area = ½ × 8 × 6 = 24 cm²
Volume = 24 × 10 = 240 cm³
Perimeter of cross-section = 8+6+10 = 24 cm
Surface area = 2(24) + (24 × 10) = 48 + 240 = 288 cm²
Try it — Cuboids and Prisms
  1. Find the surface area and volume of a cuboid 6 cm × 4 cm × 3 cm.
  2. A triangular prism has cross-section area 18 cm² and is 12 cm long. Find its volume.
  3. The prism in question 2 has a cross-section perimeter of 20 cm. Find its surface area.
Show answers
  1. SA=2(24+18+12)=108 cm². V=72 cm³
  2. 18×12 = 216 cm³
  3. 2(18)+(20×12) = 36+240 = 276 cm²

3. Cylinders

Worked example r=4 h=10 Find the curved surface area, total surface area, and volume of a cylinder with radius 4 cm and height 10 cm.
Curved SA = 2π(4)(10) = 251.3 cm²
Total SA = 251.3 + 2π(4)² = 251.3 + 100.5 = 351.9 cm²
Volume = π(4)²(10) = 502.7 cm³
Try it — Cylinders
  1. Find the volume of a cylinder with radius 5 cm and height 12 cm.
  2. Find the curved surface area of the same cylinder.
  3. Find its total surface area.
Show answers
  1. π(25)(12) = 942.5 cm³
  2. 2π(5)(12) = 377.0 cm²
  3. 377.0 + 2π(25) = 377.0+157.1 = 534.1 cm²

4. Cones

Worked example l=5 r=3 Find the curved surface area, total surface area, and volume of a cone with radius 3 cm, slant height 5 cm, and (perpendicular) height 4 cm.
Curved SA = π(3)(5) = 47.1 cm²
Total SA = 47.1 + π(3)² = 47.1+28.3 = 75.4 cm²
Volume = ⅓π(3)²(4) = 37.7 cm³
Worked example — finding slant height first A cone has radius 6 cm and perpendicular height 8 cm. Find its slant height, then its curved surface area.
Slant height (Pythagoras): l = √(6²+8²) = √100 = 10 cm
Curved SA = π(6)(10) = 188.5 cm²
Try it — Cones
  1. Find the volume of a cone with radius 5 cm and height 9 cm.
  2. A cone has radius 5 cm and height 12 cm. Find the slant height, then the curved surface area.
  3. Find the total surface area of the cone in question 2.
Show answers
  1. ⅓π(25)(9) = 235.6 cm³
  2. l=√(25+144)=13. Curved SA=π(5)(13)=204.2 cm²
  3. 204.2+π(25) = 204.2+78.5 = 282.7 cm²

5. Spheres

Memory tip — “skin vs stuffing” Surface area is the skin of the sphere — a 2D covering, so it only needs r squared: SA = 4πr². Volume is the stuffing inside — a 3D amount of space, so it needs r cubed: V = (4/3)πr³. If you’re ever unsure which formula is which, check the power on the r: squared means area, cubed means volume — this works for every 3D shape, not just spheres.
Worked example Find the surface area and volume of a sphere with radius 6 cm.
Surface area = 4π(6)² = 452.4 cm²
Volume = (4/3)π(6)³ = 904.8 cm³
Try it — Spheres
  1. Find the surface area of a sphere with radius 4 cm.
  2. Find the volume of a sphere with radius 10 cm.
  3. A sphere has volume 288π cm³. Find its radius.
Show answers
  1. 4π(16) = 201.1 cm²
  2. (4/3)π(1000) = 4188.8 cm³
  3. (4/3)πr³=288π → r³=216 → r=6 cm

6. Pyramids

Worked example 6 × 6 base h=8 A pyramid has a square base 6 cm × 6 cm and a perpendicular height of 8 cm. Find its volume.
Base area = 36 cm²
Volume = ⅓ × 36 × 8 = 96 cm³
Worked example — surface area (finding the slant height first) Using the same pyramid, find its total surface area (apex directly above the centre of the base).
Slant height of each triangular face: l = √(8² + 3²) = √73 ≈ 8.544 cm (3 is half the base edge)
Area of one triangular face = ½ × 6 × 8.544 = 25.63 cm²
Total surface area = base + 4 faces = 36 + 4(25.63) = 36 + 102.53 = 138.5 cm²
Try it — Pyramids
  1. Find the volume of a pyramid with a rectangular base 5 cm × 4 cm and height 9 cm.
  2. A square-based pyramid has base 10 cm × 10 cm and height 12 cm. Find its volume.
  3. Find the slant height of the triangular faces in question 2, then the total surface area.
Show answers
  1. ⅓(20)(9) = 60 cm³
  2. ⅓(100)(12) = 400 cm³
  3. l=√(144+25)=13. Each face=½(10)(13)=65. Total SA=100+4(65)=360 cm²

7. Compound Shapes and Solids

A compound shape or solid is made by joining two (or more) simpler ones together. Work out each part separately using the formulas above, then add (or subtract) as needed.

Worked example — a cylinder with a hemisphere on top A solid is made from a cylinder of radius 5 cm and height 12 cm, topped with a hemisphere of the same radius. Find the total volume and total surface area.
Volume = cylinder + hemisphere = π(5)²(12) + ½×(4/3)π(5)³ = 942.5 + 261.8 = 1204.3 cm³
Surface area (external surfaces only — no flat circle where they join, and no flat circle on top of the cylinder since it’s covered):
= curved SA of cylinder + bottom circle + curved SA of hemisphere
= 2π(5)(12) + π(5)² + 2π(5)² = 377.0 + 78.5 + 157.1 = 612.6 cm²
Try it — Compound Solids
  1. A solid is a cone (radius 4 cm, height 9 cm) sitting on a cylinder (same radius, height 6 cm). Find the total volume.
  2. Find the volume of a hemisphere with radius 9 cm.
  3. A shape is a rectangle 10 cm × 6 cm with a semicircle (diameter 6 cm) attached to one short end. Find the total area.
Show answers
  1. Cone: ⅓π(16)(9)=150.8. Cylinder: π(16)(6)=301.6. Total=452.4 cm³
  2. (2/3)π(729) = 1526.8 cm³
  3. Rectangle=60. Semicircle radius3: ½π(9)=14.14. Total=74.1 cm²

8. Parts of Solids

Worked example Find the volume of half of a sphere with radius 9 cm.
Full sphere volume = (4/3)π(9)³ = 3053.6 cm³
Half: 3053.6 ÷ 2 = 1526.8 cm³
Try it — Parts of Solids
  1. Find the volume of a quarter cylinder (radius 6 cm, height 10 cm) — i.e. a quarter of the full circular cross-section.
  2. Find the curved surface area of a hemisphere with radius 7 cm (curved part only, not including the flat circular face).
Show answers
  1. Full volume=π(36)(10)=1131.0. Quarter=282.7 cm³
  2. Full sphere SA=4π(49)=615.8. Half (curved only)=307.9 cm²

9. Frustums Extended only

A frustum is a cone (or pyramid) with the top sliced off by a cut parallel to the base. Find it by calculating the full solid and the small removed solid separately, then subtracting.

Diagram r=3 R=6 The dashed lines show the removed top of the original cone.
Worked example A cone with base radius 6 cm and height 12 cm has its top cut off, leaving a smaller similar cone with radius 3 cm (and height 6 cm, by similarity — half the linear scale). Find the volume and total surface area of the remaining frustum.
Volume:
Full cone: ⅓π(6)²(12) = 452.4 cm³
Small cone removed: ⅓π(3)²(6) = 56.5 cm³
Frustum volume = 452.4 − 56.5 = 395.8 cm³

Surface area (curved part + both circular ends):
Full cone slant height: √(6²+12²) = 13.42 cm. Small cone slant height: √(3²+6²) = 6.71 cm
Curved SA of frustum = π(6)(13.42) − π(3)(6.71) = 252.9 − 63.2 = 189.7 cm²
Add both circles: + π(6)² + π(3)² = 189.7 + 113.1 + 28.3 = 331.0 cm²
Try it — Frustums Extended
  1. A cone (radius 8 cm, height 16 cm) has its top cut off, leaving a small similar cone with radius 4 cm and height 8 cm. Find the frustum’s volume.
  2. Using the same frustum, find the slant heights of the full cone and the small cone.
Show answers
  1. Full: ⅓π(64)(16)=1072.3. Small: ⅓π(16)(8)=134.0. Frustum=938.3 cm³
  2. Full slant=√(64+256)=17.89 cm. Small slant=√(16+64)=8.94 cm
Common mistakes to watch for
  • Using the slant height in the volume formula instead of the perpendicular height (or vice versa) — volume always needs the perpendicular height.
  • Forgetting to add the flat circle(s) when a question asks for total surface area of a cylinder or cone, not just the curved part.
  • In compound solids, including a flat face that’s actually hidden inside the joined shape (e.g. the circle where a hemisphere meets a cylinder).
  • For frustums, forgetting the small removed cone is similar to the full cone — its dimensions scale by the same ratio, not some other value.
  • Not converting units consistently before combining volumes or areas from different parts of a compound solid.

10. Quick-Fire Challenge Round

True or False?
  1. The volume formula for a cone uses the perpendicular height, not the slant height.
  2. The curved surface area formula for a cylinder already includes the two circular ends.
  3. A frustum’s volume can be found by subtracting a small cone’s volume from a full cone’s volume.
  4. In a compound solid, every flat face from both original solids should be included in the total surface area.
Show answers
  1. True.
  2. False — “curved” surface area is only the curved part; the two circles must be added separately for total surface area.
  3. True. Ext.
  4. False — any face that becomes hidden where the two solids join should be excluded.
Spot the Error
  1. “Volume of a cone, radius 4, slant height 10: V = ⅓π(4)²(10).”
  2. “Total surface area of a cylinder, radius 5, height 8: SA = 2π(5)(8).”
  3. “A hemisphere’s curved surface area is half of 4πr², so it’s 2πr².”
Show answers
  1. The volume formula needs the perpendicular height, not the slant height — these are only the same if the cone is unusually shaped; normally you’d need to find the perpendicular height first (using Pythagoras if the slant height and radius are given).
  2. This only gives the curved surface area — total surface area needs the two circular ends added: 2π(5)(8) + 2π(5)² = 251.3+157.1 = 408.4 cm².
  3. This one is actually correct — the curved part of a hemisphere is exactly half a full sphere’s surface area, giving 2πr².

11. Mixed Practice — Bringing It All Together

  1. Find the volume of a cuboid 7 cm × 5 cm × 3 cm.
  2. Find the volume and curved surface area of a cylinder, radius 6 cm, height 9 cm.
  3. Find the volume of a cone, radius 5 cm, height 12 cm.
  4. Find the volume and surface area of a sphere with radius 3 cm.
  5. Find the volume of half a cylinder (radius 4 cm, height 10 cm).
Show answers
  1. 105 cm³
  2. V=π(36)(9)=1017.9 cm³. Curved SA=2π(6)(9)=339.3 cm²
  3. ⅓π(25)(12)=314.2 cm³
  4. V=(4/3)π(27)=113.1 cm³. SA=4π(9)=113.1 cm² (a fun coincidence at r=3 — check both separately, they aren’t equal in general!)
  5. Full volume=π(16)(10)=502.7. Half=251.3 cm³

Practice at Home

A large set to close out Mensuration — take it steadily.

Part A — Cuboids and prisms
  1. Find the surface area and volume of a cuboid 9 cm × 6 cm × 4 cm.
  2. A triangular prism has cross-section area 15 cm² and length 14 cm. Find its volume.
Part B — Cylinders and cones
  1. Find the volume and total surface area of a cylinder, radius 7 cm, height 15 cm.
  2. Find the curved surface area of a cone with radius 6 cm and slant height 14 cm.
  3. A cone has radius 9 cm and height 12 cm. Find its slant height, then its volume.
Part C — Spheres and pyramids
  1. Find the surface area and volume of a sphere with radius 8 cm.
  2. Find the volume of a pyramid with base area 40 cm² and height 9 cm.
  3. A square-based pyramid has base 8 cm × 8 cm and height 15 cm. Find its volume.
Part D — Compound shapes and parts of solids
  1. A solid is a hemisphere (radius 6 cm) sitting on a cylinder (same radius, height 10 cm). Find its total volume.
  2. Find the volume of a third of a cylinder (radius 9 cm, height 20 cm).
  3. A shape is a semicircle (radius 5 cm) attached to a rectangle 10 cm × 8 cm along its 10 cm edge. Find the total area.
Part E — Frustums Extended
  1. A cone (radius 10 cm, height 20 cm) has its top removed, leaving a small similar cone with radius 5 cm and height 10 cm. Find the frustum’s volume.
  2. Using the same frustum, find the total surface area (curved part plus both circles).
Part F — Applied / challenge
  1. A water tank is a cylinder with radius 0.5 m and height 1.2 m. Find its capacity in litres (1 m³ = 1000 litres).
  2. An ice cream cone has radius 3 cm and height 10 cm, topped with a hemisphere of ice cream (same radius). Find the total volume of ice cream and cone combined.
  3. A metal sphere of radius 6 cm is melted down and recast into a cylinder of radius 4 cm. Find the height of the cylinder (the volume stays the same).

Key Vocabulary Recap

Curved surface area — the area of the curved part only, excluding flat faces
Slant height — the distance from the apex to the base edge, along the surface
Compound solid — a solid made by joining two or more simpler solids
Frustum — a cone or pyramid with the top sliced off Ext.

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) sections C5.4 and C5.5 (Core subject content) and sections E5.4 and E5.5 (Extended subject content) — Surface area and volume, and Compound shapes and parts of shapes. Nearly all content is shared between tiers; frustums are explicitly an Extended-only worked example in the syllabus, marked above. This closes out the Mensuration unit.

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