First turn coordinates into lengths: The exam often gives you a solid as a set of corner coordinates, not as ready-made lengths.
Your first job is always the same: use the 3D distance and midpoint formulas to find the lengths you need — an edge, a radius, or a height — before any volume or angle work.
Which formula gives which length: Edge / slant edge = 3D distance between its two end corners.
Centre of a base or sphere = midpoint of a diameter (or the point directly below the apex).
Radius = half a diameter, or the distance from the centre to any point on the surface.
IB-style question — an edge length from coordinates
A right pyramid has apex V(3, 2, 1) and base corner A(3, 14, 6).
Find the length of the edge VA.
Step by step
- The distance between two points in space is the square root of the sum of the squared coordinate gaps (the difference in x, in y and in z).
- Substitute V(3, 2, 1) and A(3, 14, 6): the gaps are 3−3, 14−2 and 6−1.
- Square, add, root.
Final answer
VA = 13.
The edge VA from the question: V(3, 2, 1) and A(3, 14, 6) are two points in space; the gaps 0, 12, 5 give VA = √(0² + 12² + 5²) = 13 (not to scale).
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IB-style question (a) — the centre of the base
The base of a cone is a circle with diameter [AB], where A(−1, 2, 2) and B(5, 2, 10).
Find the centre of the base.
Step by step
- The centre of a diameter is its midpoint: average the two endpoints coordinate by coordinate.
- Substitute A(−1, 2, 2) and B(5, 2, 10).
Final answer
Centre M(2, 2, 6).
IB-style question (b) — the radius
The same cone base has diameter [AB] with A(−1, 2, 2) and B(5, 2, 10).
Find the radius.
Step by step
- The radius is half the diameter, so first find the diameter AB with the 3D distance formula.
- Substitute A(−1, 2, 2) and B(5, 2, 10): the gaps are 6, 0 and 8.
- Halve it for the radius.
Final answer
Radius r = 5.
The cone's base circle on diameter [AB]: the centre is the midpoint M(2, 2, 6), and the radius is half of AB = 10, i.e. 5 (not to scale).
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Get the lengths, then use the usual formulas: Once the coordinates have given you the radius, edge or height, the volume and surface-area formulas are the ordinary ones. The only new step is reading those lengths off the coordinates first.
For a cone or pyramid, the height is the distance from the apex to the centre of the base — a vertical drop, not a slant edge.
Interactive: every solid with its volume and surface-area formulas — including the hemisphere (total surface area 3πr²).
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IB-style question — volume of a cone from coordinates
A cone has a base circle of centre (2, 2, 6) and radius 5 (from the previous section).
Its apex is V(2, 10, 6). Find the exact volume of the cone.
Step by step
- The height is the distance from the apex to the centre of the base, found with the 3D distance formula.
- Substitute V(2, 10, 6) and the base centre (2, 2, 6): the only gap is in y.
- The volume of a cone is one third of the base area times the height.
- Substitute r = 5 and h = 8.
- Simplify.
Final answer
V = 200π/3 ≈ 209.
IB-style question — surface area of a hemisphere from coordinates
A solid hemisphere has centre C(4, 1, −2), and the point P(4, 1, 7) lies on its curved surface.
Find (a) the radius and (b) the total surface area.
Step by step
- (a) The radius is the distance from the centre to any point on the surface, found with the 3D distance formula.
- Substitute C(4, 1, −2) and P(4, 1, 7): the only gap is in z, 7 − (−2) = 9.
- (b) A solid hemisphere has a curved dome (2πr²) and a flat circular base (πr²), so its total surface area adds both.
- Substitute r = 9.
Final answer
(a) r = 9. (b) A = 243π ≈ 764.
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Two angle types you will meet: Two kinds of 3D angle come up — both work by pulling a flat triangle out of the solid:
1. Between two edges (an angle at a corner): find the triangle's three side lengths with the distance formula, then use the cosine rule.
2. Between an edge and the flat base: drop the edge's top straight down to the base to make a right-angled triangle, then tan θ = height ÷ the flat distance across the base.
The idea behind both types: take the corners you care about and redraw them as one flat triangle. For an edge-to-base angle that triangle is right-angled — the height (straight up), the flat distance across the base, and the slant edge.
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IB-style question — angle at a vertex (cosine rule)
A pyramid has apex V(0, 0, 9) and base corners B(8, 0, 0) and C(0, 6, 0).
Find the size of angle BV̂C.
Step by step
- To find an angle at a vertex, first get the three side lengths of the triangle with the 3D distance formula.
- Apply it to each side of triangle BVC, using V(0, 0, 9), B(8, 0, 0) and C(0, 6, 0).
- The cosine rule, rearranged for the angle, gives the angle opposite a known side. The angle is at V, so the opposite side is BC.
- Substitute VB² = 145, VC² = 117 and BC² = 100.
- Evaluate and take the inverse cosine.
Final answer
Angle BV̂C ≈ 51.6°.
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Triangle BVC lifted out of the pyramid (not to scale). First get its three side lengths with the 3D distance formula; the angle we want is at V, opposite the known side BC = 10 — so the cosine rule gives cos(BV̂C) = (VB² + VC² − BC²) ÷ (2·VB·VC) ≈ 51.6°.
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IB-style question — angle between an edge and the base
A right pyramid has apex V(2, 2, 10) and base centre X(2, 2, 0). C is a base corner, and the horizontal distance XC = 6.
Find the angle the edge VC makes with the base.
Step by step
- The angle between the edge and the base lives in the right triangle VXC, right-angled at the base centre X. The tangent of an angle is the opposite side over the adjacent side.
- The angle to the base is at corner C: the opposite side is the vertical height VX, the adjacent side is the horizontal distance XC. From the coordinates VX = 10 and XC = 6.
- Take the inverse tangent.
Final answer
The edge makes about 59.0° with the base.
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The pyramid as given: the apex V(2, 2, 10) sits straight above the base centre X(2, 2, 0), so the vertical height VX = 10. C is a base corner; the flat step from the centre X out to that corner is XC = 6. The angle the edge VC makes with the base lives in the triangle VXC, redrawn flat below.
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The right triangle VXC, right-angled at the base centre X: the height VX = 10 is opposite θ, the horizontal distance XC = 6 is adjacent, and the slant edge VC is the hypotenuse. We have the two legs, so tan θ = 10 ÷ 6 → θ ≈ 59.0°.
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