3D Pythagoras and Trigonometry

Find a right triangle inside the solid

A 3D sketch is drawn in perspective, so a right angle may not look like 90° on the page. Use the stated geometry. A cuboid has three mutually perpendicular edge directions. A diagonal across its base is perpendicular to a vertical edge, giving a second right triangle.

Space diagonal of a cuboid

Worked example — use Pythagoras twice

A cuboid has length 8 cm, width 6 cm and height 12 cm. The base diagonal d satisfies d²=8²+6²=100, so d=10 cm. The space diagonal s then satisfies s²=d²+12²=100+144=244. Therefore s=√244=2√61 cm≈15.6 cm (3 significant figures).

Combining the two steps gives s=√(length²+width²+height²). This shortcut applies to mutually perpendicular dimensions, such as cuboid edges; it is not a rule for any three labelled lengths.

Angle between a line and a plane

Project the line straight down onto the plane. The required angle is between the line and its projection. In the diagram, the angle AG makes with the base is ∠GAC, in right triangle ACG. It is not the angle between AG and an arbitrary base edge.

Worked example

For the 8 by 6 by 12 cm cuboid, AC=10 cm and CG=12 cm. Thus tanθ=12/10, giving θ≈50.2° to 1 decimal place. Use the full base diagonal as adjacent; using 8 would calculate a different angle.

Pyramids and cones

For a right square-based pyramid, the apex is vertically above the centre of the base. Distinguish its vertical height, the slant height down a triangular face, and the sloping edge to a base corner. They belong to different right triangles.

Worked example — two different sloping lengths

A right square-based pyramid has base side 10 cm and vertical height 12 cm. The distance from the base centre to a side midpoint is 5 cm. Face slant height=√(12²+5²)=13 cm. But the centre-to-corner distance is half the base diagonal: √(5²+5²)=√50 cm. The apex-to-corner edge is √(144+50)=√194 cm≈13.9 cm.

For a right circular cone, a cross-section through the axis gives slant height l, vertical height h and radius r, with l²=h²+r². A cone with r=5 cm and h=12 cm has l=13 cm. This connects the volume dimensions to curved surface area πrl.

Reverse and multi-step problems

Worked example — recover the height

A cuboid has base 3 cm by 4 cm and space diagonal 13 cm. The base diagonal is 5 cm. Therefore h²+5²=13², so h²=144 and h=12 cm. Its volume is 3×4×12=144 cm³. Keep the geometric steps separate from the final volume calculation.

Think Like a Mathematician

Redraw the right triangle you need on its own. Label which lengths come from edges, diagonals or projections. This turns a complicated solid into a familiar 2D calculation.

Common Mistakes

Do not trust the apparent angle in a perspective sketch. Do not use a base edge in place of a base diagonal. Do not confuse a pyramid’s face slant height with its corner edge. Keep intermediate square roots unrounded.

Practice — 3D triangles

1. Find the space diagonal of a cuboid measuring 2 cm by 3 cm by 6 cm.

2. A cuboid has base 9 m by 12 m and height 8 m. Find its base and space diagonals.

3. Find the angle the space diagonal in question 2 makes with the base, to 1 decimal place.

4. A right square pyramid has base side 12 cm and vertical height 8 cm. Find a face slant height.

5. A cuboid has base 6 cm by 8 cm and space diagonal 26 cm. Find its height.

Frequently Asked Questions

Must I always use Pythagoras twice?

Only if an intermediate side is unknown. If the required right triangle already has enough information, use it directly.

Why is this Higher?

The 3D applications require identifying appropriate planes, projections and right triangles; the underlying Pythagoras and trig rules are unchanged.

Continue Learning

Previous: Right-Angled Trigonometry and Exact Values

Next: Sine Rule, Cosine Rule and Triangle Area

Geometry and Measures topic overview