Know the booklet formulas: Volume and surface-area formulas for the standard solids are in the formula booklet — but you must know which to use and read the right radius/height.
Interactive — pick a solid, then tap “Volume” or “Surface area”: the diagram highlights what that formula actually uses (and shades the curved surface so you can see it). Includes the prism.
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What's given vs what you memorise: ✓ in the booklet — you can look it up in the exam.
★ not given — memorise it, or build it from the given parts (add a base circle, halve a sphere, sum the faces…).
| Solid | Volume | Surface area |
|---|---|---|
| Cuboid | ✓ | ★ |
| Cylinder | ✓ | curved ✓; closed adds ★ |
| Cone | ✓ | curved ✓; total adds base ★ |
| Sphere | ✓ | ✓ |
| Hemisphere | ★ | ★ (½ sphere + flat base) |
| Pyramid | ✓ (A = base area) | ★ |
| Prism | ✓ | sum of all faces ★ |
The given 2D areas, drawn — so you can see what b, h, a and r mean. These shapes are the bases / cross-sections that feed the 3D formulas above.
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| Shape | Area — also given | Where you use it |
|---|---|---|
| Parallelogram | cross-section of a slanted prism | |
| Triangle | the base A of a triangular prism / a pyramid | |
| Trapezoid | cross-section of a trough / trapezoidal prism | |
| Circle | a cylinder or cone base, or a hemisphere's flat face | |
| Circle — rim | the distance around the edge (e.g. how far a wheel rolls in one turn) |
Where these 2D areas get USED: a prism is its cross-section pushed along a length. Tap a cross-section — the shaded front face is exactly that 2D shape (its area A), and the volume is V = A × length. A slanted prism has a parallelogram cross-section, a trough a trapezoid, and so on.
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Pick the formula, substitute, evaluate: Identify the solid, read off the radius and height, then substitute. Keep π exact unless told to round.
IB-style question — volume of a cone
Find the volume of a cone with base radius 3 and height 4.
Step by step
- Write the general volume formula for a cone, where r is the base radius and h is the perpendicular height.
- Substitute the question's numbers: r = 3 and h = 4.
- Evaluate (only r is squared).
Final answer
V = 12π ≈ 37.7.
The cone from the question (r = 3, h = 4) — volume = ⅓πr²h = 12π.
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Square the radius, not the height: In V = ⅓πr²h, only r is squared. A common slip is squaring h too.
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Add up every face: Total surface area adds all the surfaces. A cone's slant uses l = √(r² + h²); a closed cylinder is 2 circles + the curved wrap (2πr² + 2πrh).
IB-style question — cone surface area
A cone has base radius 6 and height 8. Find its total surface area.
Step by step
- The slant height l is the sloping distance from the rim to the apex. Find it with the general formula.
- Substitute r = 6 and h = 8.
- Write the general total surface area of a cone: the circular base plus the curved (lateral) surface.
- Substitute r = 6 and l = 10.
- Evaluate.
Final answer
A = 96π ≈ 302.
The cone from the question (r = 6, h = 8) — the slant l = √(r²+h²) = 10 is found first, then A = πr² + πrl.
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IB-style question — surface area of a solid hemisphere
A solid hemisphere has radius 6. Find its total surface area.
Step by step
- Build it from what the booklet gives. The booklet has the whole sphere's surface (4πr²) and the area of a circle (πr²) — a solid hemisphere is the curved HALF of the sphere plus one flat circular lid.
- Add the two given pieces.
- Substitute r = 6.
Final answer
A = 108π ≈ 339 (curved ½-sphere 2πr², given, + flat circle πr², given).
The solid hemisphere from the question (r = 6) — its total surface area is the curved dome 2πr² plus the flat base πr², giving 3πr².
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Open or closed?: Read whether a surface is included — an open cylinder (no lid) drops one circle; a hemisphere's flat face is one extra circle.
Add the pieces (or subtract a hole): For a solid made of parts (e.g. a cylinder topped by a hemisphere), add the volumes. For surface area, add only the exposed faces (a shared join is not counted twice).
A cylinder topped by a hemisphere. Switch between Volume (add the two pieces) and Surface area (count only the exposed faces — the dashed join circle is internal, so leave it out).
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IB-style question — cylinder + hemisphere
A solid is a cylinder (r = 3, h = 10) with a hemisphere (r = 3) on top. Find its volume.
Step by step
- The total volume is the cylinder plus the hemisphere. State both general formulas (a hemisphere is half a sphere).
- Substitute the cylinder's r = 3, h = 10 and the hemisphere's r = 3.
- Evaluate each piece and add.
Final answer
V = 90π + 18π = 108π ≈ 339.
IB-style question — surface area of a capsule (cylinder + 2 hemispheres)
A capsule is a cylinder of radius r and height h with a hemisphere on each end.
Find a formula for its total surface area S.
Step by step
- The curved side of the cylinder is 2πrh. The two hemispheres are domes only — their flat faces sit against the cylinder, so they aren't exposed.
- Two hemisphere domes = one whole sphere's surface.
- Add the exposed surfaces (no flat circles — they're internal joins).
Final answer
S = 2πrh + 4πr².
The capsule from the question: the curved cylinder side (2πrh) plus the two domes, which together make one whole sphere (4πr²). The flat circles where the domes meet the cylinder are internal joins — not part of the outside surface.
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Don't double-count the join: When adding surface areas, the circle where two pieces meet is internal — leave it out.
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Hollow cylinder (a pipe): A pipe is a cylinder with a smaller cylinder removed (the hole). Picture the ring-shaped end: outer radius R, inner radius r.
Volume of metal = big cylinder − hole = .
Surface area = — the outer curved wall, the inner curved wall, and the two ring-shaped ends.
The pipe end is a ring (an annulus): its area is the big circle minus the hole, πR² − πr². That ring runs the full length h, so the volume of metal is π(R² − r²)h.
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IB-style question — volume of metal in a pipe
A metal pipe is a hollow cylinder with outer radius 5 cm, inner radius 3 cm and length 20 cm.
Find the volume of metal in the pipe.
Step by step
- The metal is the outer cylinder with the inner cylinder (the hole) taken out — so subtract the two volumes. Both share the same length h, so factorise.
- Substitute R = 5, r = 3 and h = 20.
- Evaluate.
Final answer
V = 320π ≈ 1005 cm³ of metal.
Factorise before you substitute: π(R² − r²)h with the numbers put in last is much quicker than working out πR²h and πr²h separately. The two cylinders share the same length h, so factor it out — and remember it's R² − r², not (R − r)².
Given the volume, solve for r or h: Set the formula equal to the given volume (or area) and solve for the unknown dimension — often a cube root or a square root.
Quick reference: tap each solid to recall the volume formula you're rearranging — then set it equal to the given value and solve.
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IB-style question — find the radius
A sphere has volume 36π. Find its radius.
Step by step
- State the general volume formula for a sphere.
- Set it equal to the given volume 36π.
- Divide both sides by π (it cancels), then multiply by ¾ to isolate r³.
- Cube-root both sides.
Final answer
r = 3.
Cancel the π: If π appears on both sides, divide it out first — the numbers get much friendlier.