G20 - Pythagoras and trigonometry in three dimensions

G20 - Pythagoras and trigonometry in three dimensions

A solid can hide a familiar triangle. You will learn to choose a flat section through a cuboid or pyramid, calculate spatial lengths in stages, and identify the angle between a line and a plane or between two lines. The triangle methods stay the same; choosing the correct triangle is the new skill.

Finding the correct triangle inside a solid (Higher)

Will a straight 11 cm11\,\mathrm{cm} rod fit completely inside a cuboid measuring 6 cm6\,\mathrm{cm} by 8 cm8\,\mathrm{cm} by 5 cm5\,\mathrm{cm}? No edge is long enough, so look for a triangle containing the longest corner-to-corner diagonal.

You already know

Pythagoras applies only to a right triangle. The sine and cosine rules work in general triangles. Any three points that are not on one line determine a flat triangle, even if the points belong to a solid.

Let A,CA,C be opposite base corners of the cuboid, and GG the upper corner directly above CC. The base triangle gives AC=62+82=10 cmAC=\sqrt{6^2+8^2}=10\,\mathrm{cm}. Triangle ACGACG contains the base diagonal, the vertical height and the body diagonal. Since CGCG is perpendicular to the base plane, it is perpendicular to ACAC, which passes through its foot CC. The right angle is at CC.

[DIAGRAM: asset_name: A right triangle inside a cuboid; asset_slug: 1ma1_m61_cuboid; caption: The body diagonal is the hypotenuse of a triangle with base diagonal ten and vertical height five.; description: Cuboid with base dimensions 6 cm by 8 cm and height 5 cm. Base corners A and C are opposite, G lies directly above C, and the highlighted section ACG has AC 10 cm, CG 5 cm and body diagonal AG.]
The body diagonal is the hypotenuse of a triangle with base diagonal ten and vertical height five.

A perspective drawing can make right angles look slanted, and unrelated lines look perpendicular. Use stated geometry, not a measurement from the picture. It often helps to redraw your chosen triangle on its own.

A right square-based pyramid has its apex TT directly above the base centre OO. A face slant height ends at a base-side midpoint MM; a sloping edge ends at a corner VV. These use different horizontal distances.

[DIAGRAM: asset_name: Two pyramid sections; asset_slug: 1ma1_m61_pyramid_sections; caption: The face slant height ends at M, but the sloping edge ends at V; both sections use the vertical height TO.; description: Oblique projection of a right square-based pyramid with base side 6 cm and perpendicular height TO 4 cm. M is the midpoint of a base side and V is its corner. Distinct highlighted segments show face slant height TM, sloping edge TV, perpendicular height TO and the different horizontal distances OM and OV. A projected right-angle mark at O identifies triangle TOM as right angled.]
The face slant height ends at M, but the sloping edge ends at V

Worked example 1: selecting the face-height section

Question. A right square-based pyramid has base side 6 cm6\,\mathrm{cm} and perpendicular height TO=4 cmTO=4\,\mathrm{cm}. Which triangle gives its face slant height TMTM? Find its two perpendicular sides.

Plan. Join the apex to the base centre and the midpoint of a base side.

OM=6/2=3 cmcentre to side midpointTO=4 cmgiven perpendicular height∠TOM=90∘height perpendicular to base\begin{aligned} OM&=6/2=3\,\mathrm{cm} && \text{centre to side midpoint} \\ TO&=4\,\mathrm{cm} && \text{given perpendicular height} \\ \angle TOM&=90^\circ && \text{height perpendicular to base} \end{aligned}

Why end at MM? The perpendicular height of the triangular face meets the midpoint of its base edge, because that face is isosceles.

Answer. Use right triangle TOMTOM, with legs 44 and 3 cm3\,\mathrm{cm}, and hypotenuse TMTM.

Check. The vertical height ends at OO; the face height ends at MM. They are different segments.

Change only the requested segment: find the sloping edge to a corner instead of the face height. The solid and its dimensions stay fixed.

Worked example 2: selecting the corner-edge section

Question. For the same 6 cm6\,\mathrm{cm}-base pyramid with height 4 cm4\,\mathrm{cm}, which triangle gives the sloping edge TVTV? Find its two perpendicular sides.

Plan. Use the apex, base centre and corner; calculate the centre-to-corner distance in the square base.

OV2=32+32=18half-width and half-lengthOV=32 cmpositive square rootTO=4 cmsame vertical height\begin{aligned} OV^2&=3^2+3^2=18 && \text{half-width and half-length} \\ OV&=3\sqrt2\,\mathrm{cm} && \text{positive square root} \\ TO&=4\,\mathrm{cm} && \text{same vertical height} \end{aligned}

Why use two half-sides? The route from the square's centre to a corner changes both base coordinates.

Answer. Use right triangle TOVTOV, with legs 44 and 32 cm3\sqrt2\,\mathrm{cm}, and hypotenuse TVTV.

Check. 32>33\sqrt2>3, so the corner is farther from the centre than the side midpoint is.

Choosing the endpoint correctly prevents a calculation that is accurate but answers the wrong question.

Watch out

Wrong: use OV=3 cmOV=3\,\mathrm{cm} for the corner-edge section. This is the centre-to-side distance, so it gives the face height instead. Correct: OV=32+32=32 cmOV=\sqrt{3^2+3^2}=3\sqrt2\,\mathrm{cm}, then use triangle TOVTOV.

Try the corner section with only the base-side length changed.

Calculating spatial lengths in stages (Higher)

A length missing from one triangle can be found first in another triangle. Start in the base, then move into the vertical section.

Pythagoras' theorem

In a right triangle with perpendicular sides a,ba,b and hypotenuse cc,

a2+b2=c2a^2+b^2=c^2

On the formula sheet (2026 version; the 2027 sheet will have the same content).

This is a two-dimensional rule. Using it twice explains why three squared edge lengths appear in a cuboid calculation.

Worked example 3: the longest cuboid diagonal

Question. Find the body diagonal of the opening 6 cm6\,\mathrm{cm} by 8 cm8\,\mathrm{cm} by 5 cm5\,\mathrm{cm} cuboid. Give an exact answer and decide whether the 11 cm11\,\mathrm{cm} rod fits.

Plan. Find the base diagonal ACAC, then use right triangle ACGACG.

AC2=62+82=100base triangleAC=10positive rootAG2=AC2+CG2=100+25=125vertical sectionAG=55 cmsimplify the root\begin{aligned} AC^2&=6^2+8^2=100 && \text{base triangle} \\ AC&=10 && \text{positive root} \\ AG^2&=AC^2+CG^2=100+25=125 && \text{vertical section} \\ AG&=5\sqrt5\,\mathrm{cm} && \text{simplify the root} \end{aligned}

Why use ACAC, rather than 66 or 88? The body diagonal projects onto the whole base diagonal.

Answer. AG=55 cm≈11.18 cmAG=5\sqrt5\,\mathrm{cm}\approx11.18\,\mathrm{cm}. The straight 11 cm11\,\mathrm{cm} rod fits along it.

Check. 112=121<125=AG211^2=121<125=AG^2, so the unrounded comparison confirms the fit.

Change only the cuboid height to 9 cm9\,\mathrm{cm}. The base diagonal is still 10 cm10\,\mathrm{cm}.

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