E7.6 - Pythagoras and trigonometry in 3D (Extended)

Plane section and perpendicular height

A plane is a flat surface extending in every direction. A triangle whose three vertices lie in one plane can be treated as an ordinary two-dimensional triangle. A line perpendicular to a plane meets every line in that plane through its foot at a right angle.

Worked example 1: choosing two connected triangles

Question. A cuboid has $AB=8\,\mathrm{cm}$, $BC=6\,\mathrm{cm}$ and $CG=5\,\mathrm{cm}$, labelled as above. To find $AG$, which base length should you find first, and which right triangle should you then use?

Plan. Join the lower endpoint $A$ to the foot $C$ directly below the upper endpoint $G$.

First use base triangle $ABC$, right-angled at $B$, to find $AC$. Then use triangle $ACG$, right-angled at $C$, because $CG$ is perpendicular to the base plane containing $AC$.

Answer. Find $AC$ first; then use right triangle $ACG$.

Check. $AC$ and $CG$ meet at $C$, and the side joining their other endpoints is exactly the required $AG$. This gives one connected triangle rather than mixing edges from different faces.

Worked example 2: a pyramid's sloping edge

Question. A square-based pyramid has base $ABCD$ of side $6\,\mathrm{cm}$. The apex $V$ is directly above the centre $O$, and $VO=8\,\mathrm{cm}$. Which right triangle would you use to find sloping edge $VB$, and how would you obtain its horizontal leg?

Plan. Use the perpendicular height $VO$ and join its foot to $B$.

Triangle $VOB$ is right-angled at $O$. Its horizontal leg $OB$ is half the base diagonal $BD$. Find $BD$ from a right triangle with legs 6 and 6, then halve it. Thus $OB=\dfrac{\sqrt{6^2+6^2}}2=3\sqrt2\,\mathrm{cm}$.

Answer. Use triangle $VOB$, with legs $VO=8$ and $OB=3\sqrt2$, and hypotenuse $VB$.

Check. Every corner is the same distance from the square's centre, so this triangle would give the same sloping-edge length for all four corners. Using half an edge, 3 cm, would instead reach a side midpoint, not $B$.

Watch out

The perpendicular height of a pyramid is not its sloping edge. Also, centre-to-corner distance is half a diagonal; centre-to-side-midpoint distance is half an edge. Choose the one that reaches your target endpoint.

You already know

Pythagoras gives $h^2=p^2+q^2$ for a right triangle with hypotenuse $h$. To find a leg, subtract the other leg's square from the hypotenuse's square. Pythagoras is not separately printed in the supplied formula list: learn it.

Worked example 3: an interior diagonal

Question. A cuboid has base sides $AB=8\,\mathrm{cm}$, $BC=6\,\mathrm{cm}$ and height $CG=5\,\mathrm{cm}$. Calculate its interior diagonal $AG$.

Plan. Find base diagonal $AC$ in $ABC$, then use $AC$ and $CG$ in right triangle $ACG$.

Answer. $AG=11.2\,\mathrm{cm}$, to 3 significant figures.

Check. Use a different pair of faces: $BG^2=6^2+5^2=61$, then $AG^2=AB^2+BG^2=64+61=125$. Both routes give the same squared diagonal.