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(2, 1, 8) and base corner A(2, 7, 0).
Find the length of the edge VA.
Step by step
- Use the 3D distance formula on V and A — the coordinate gaps are 0, 6 and 8.
- Square, add, root.
Final answer
VA = 10.
[Diagram: math-distance-formula] - Available in full study mode
IB-style question — centre and radius from a diameter
The base of a cone is a circle with diameter [AB], where A(−1, 2, 2) and B(5, 2, 10).
Find (a) the centre of the base and (b) the radius.
Step by step
- (a) The centre is the midpoint of the diameter — average each coordinate.
- (b) The radius is half the diameter. First find AB with the distance formula.
- Halve it.
Final answer
(a) Centre (2, 2, 6). (b) Radius 5.
[Diagram: math-distance-formula] - Available in full study mode
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.
[Diagram: math-solid-volume] - Available in full study mode
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 V to the centre of the base.
- Now use the cone volume formula with 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(3, −2, 6), and the point P(3, −2, 12) 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 a surface point.
- (b) A solid hemisphere has a curved dome AND a flat circular base, so add both.
- Substitute r = 6.
Final answer
(a) r = 6. (b) A = 108π ≈ 339.
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Two angle types you will meet: Angle at a vertex (between two edges): find the three side lengths of the triangle with the distance formula, then use the cosine rule.
Angle between an edge and the base: this is the line–plane angle — drop to the base centre to make a right-angled triangle, then tan θ = height ÷ (horizontal distance to that corner).
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
- Find the three side lengths of triangle BVC with the 3D distance formula.
- The angle is at V, so the side opposite is BC. Use the cosine rule rearranged for the angle.
- Evaluate and take the inverse cosine.
Final answer
Angle BV̂C ≈ 51.6°.
[Diagram: math-cuboid-diagonal] - Available in full study mode
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 height VX is vertical: from the coordinates, VX = 10. The angle sits in the right triangle VXC, with the right angle at X.
- The base is horizontal, so the angle to the base is at C: opposite = height VX, adjacent = XC.
- Inverse tan.
Final answer
The edge makes about 59.0° with the base.