Slant ≠ vertical height. The “
h” in the volume formula is always the perpendicular (vertical) height. The “
l” in the cone SA formula is the slant. Mixing them is the single biggest source of marks lost on this topic.
Worked examples
WE 1Cylinder volume — water bottle capacity
A cylindrical water bottle has internal radius 5 cm and internal height 12 cm. Find its capacity in millilitres (1 cm³ = 1 mL).
Identify and substitute
V = πr²h
= π(5²)(12)
= π(25)(12)
= 300π cm³
Convert
300π ≈ 942.5 cm³ = 942.5 mL
V = 300π cm³ ≈ 942 mL (3 sf)
cm³ and mL are interchangeable. Keep 300π as the exact form, then evaluate only at the end — avoids accumulating round-off errors.
WE 2Cuboid surface area — gift box wrapping
A rectangular gift box measures 20 cm × 15 cm × 8 cm. Find the total surface area of paper needed to wrap it exactly (no overlap).
Cuboid has 3 pairs of identical rectangular faces
SA = 2(lw + lh + wh)
Substitute l = 20, w = 15, h = 8
SA = 2(20·15 + 20·8 + 15·8)
= 2(300 + 160 + 120)
= 2(580)
= 1160 cm²
SA = 1160 cm²
remember the “×2” — every face has a parallel twin on the opposite side of the box. Forgetting it halves your answer.
WE 3Cone — volume and surface area
A solid chocolate is in the shape of a cone with radius 3 cm and vertical height 4 cm. Find (a) the volume of chocolate; (b) the total surface area (including the flat base).
(a) Volume
V = (1/3)πr²h = (1/3)π(9)(4)
= 12π
≈ 37.7 cm³ (3 sf)
(b) Need slant l first — Pythagoras inside the cone
l = √(r² + h²) = √(9 + 16) = √25 = 5
Total SA = base + curved
SA = πr² + πrl
= π(9) + π(3)(5)
= 9π + 15π = 24π
≈ 75.4 cm² (3 sf)
(a) V = 12π ≈ 37.7 cm³ · (b) SA = 24π ≈ 75.4 cm²
r = 3, h = 4 → l = 5 is the classic 3-4-5 triangle inside a cone. Use vertical h for volume, slant l for the curved surface area.
WE 4Sphere — hot-air balloon
A spherical hot-air balloon has radius 5 m. Find (a) the volume of air it holds; (b) the surface area of fabric used to make it.
(a) Volume of sphere
V = (4/3)πr³
= (4/3)π(5³)
= (4/3)π(125)
= 500π/3
≈ 524 m³ (3 sf)
(b) Surface area of sphere
A = 4πr² = 4π(25)
= 100π
≈ 314 m² (3 sf)
(a) V = 500π/3 ≈ 524 m³ · (b) A = 100π ≈ 314 m²
both sphere formulae use r alone — no h, no slant. The volume has r³ (cubed), the surface has r² (squared).
WE 5Square-based pyramid — total surface area
A glass paperweight is a right pyramid with a square base of side 6 cm and vertical height 4 cm. Find the total surface area of the paperweight.
Step 1 — slant of a triangular face
half-base = 6/2 = 3
slant of face = √(3² + 4²) = √25 = 5
Step 2 — area of one triangular face
A_tri = (1/2)(base)(slant)
= (1/2)(6)(5) = 15
Step 3 — total surface area
SA = (square base) + 4 × (triangle face)
= 6² + 4(15)
= 36 + 60 = 96 cm²
SA = 96 cm²
“slant of a face” uses HALF the base side (3, not 6) as one leg of the right triangle. The other leg is the vertical height of the pyramid.
WE 6Combined shape — grain silo
A grain silo consists of a cylinder of radius 3 m and height 8 m, with a cone of the same radius and height 4 m sitting on top. Find the total volume of the silo.
Split into parts: cylinder + cone
Step 1 — cylinder volume
V_cyl = πr²h = π(9)(8) = 72π
Step 2 — cone volume (same r, h = 4)
V_cone = (1/3)πr²h
= (1/3)π(9)(4) = 12π
Step 3 — add
V_total = 72π + 12π = 84π
≈ 263.9 m³ (4 sf)
V = 84π ≈ 264 m³
combined-shape rule: split → compute each → add (or subtract for “removed” pieces). Keep π exact through the addition; convert at the end.
💡 Top tips
- Always sketch and label: mark r, h, l on the diagram before substituting. Avoids the “which letter is which” trap.
- Keep π exact during working; convert to decimal only at the end. The exact form is often worth a mark.
- The “1/3” rule: pyramid and cone are exactly 1/3 of the prism / cylinder with the same base and height. Useful for ratio questions.
- Slant via Pythagoras: cone → l = √(r² + h²). Pyramid face → half-base and vertical height as the legs.
- Combined shapes: write the two formulae, then add or subtract. For SA, exclude any face where two solids meet (it’s internal).
⚠ Common mistakes
- Using slant instead of vertical height in the volume formula: V = (1/3)πr²h uses the perpendicular height, NOT the slant l.
- Forgetting the ×2 in the cuboid surface area: 3 faces × 2 (for the opposite faces) = 6. SA = 2(lw + lh + wh).
- Confusing diameter and radius: if a sphere has diameter 10, then r = 5. Using 10 directly inflates the answer enormously.
- Including hidden faces in combined shapes: if a cone sits on a cylinder, the circular face where they meet is internal — count it once for the cylinder and not for the cone (or vice versa). Often for VOLUME problems this is fine (just add); for SA problems, subtract the shared face.
- Wrong half-base in pyramids: for a square pyramid of side s, the half-base used in Pythagoras for the slant is s/2, not s.
That’s the end of the Geometry of 3D Shapes chapter. With these formulae and the 3D coordinate distance from the previous note, you can handle any AI SL question involving solids: capacity, packaging, scaling, and combined shapes in modelling problems. Next chapter: Right-Angled Trigonometry — SOH-CAH-TOA, then angles of elevation and depression.
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