Right-triangle sections in solids

A three-dimensional problem often needs two right triangles. Draw the relevant flat section: first find a diagonal in a horizontal face, then use it with a perpendicular height. A sloping edge and a perpendicular height are different lengths.

Example

A cuboid has length 12 cm, width 9 cm and height 8 cm. Its base diagonal is \(\sqrt{12^2+9^2}=15\text{ cm}\). The space diagonal is \(\sqrt{15^2+8^2}=17\text{ cm}\). The vertical edge is perpendicular to every line in the horizontal base through its foot, including that diagonal.

The diagram labels the lengths; it is not drawn to scale. If a shortest path must lie on the surface, unfold the relevant faces into a net before using Pythagoras. The straight space diagonal passes through the solid and is not a surface path.