E6.6 - Lengths and angles in three dimensions | Cambridge IGCSE Mathematics

A space diagonal belongs to a flat triangle even though the surrounding object is three-dimensional. You will locate useful right triangles in solids, calculate lengths in stages, and identify the angle between a line and a plane using a perpendicular projection. This lesson is for Extended.

Right triangles inside solids

A cuboid has edges $8\,\mathrm{cm}$ , $6\,\mathrm{cm}$ , and $12\,\mathrm{cm}$ . Its longest diagonal goes through the interior. Which two lengths meet at a right angle and can reveal that diagonal?

Label its rectangular base $ABCD$ , in order, with $AB=8$ and $BC=6$ . Put $E,F,G,H$ directly above $A,B,C,D$ , respectively, with vertical edges $12$ . A face diagonal joins opposite corners of one face; a space diagonal, such as $AG$ , joins opposite vertices through the interior.

[DIAGRAM: asset_name: Face and space diagonals in a cuboid; asset_slug: cuboid_triangles; caption: Find a base diagonal first, then use it as one leg of a vertical triangle.; description: A labelled cuboid ABCD EFGH has base AB 8 centimetres, BC 6 centimetres and height CG 12 centimetres. Face diagonal AF and space diagonal AG are distinguished. Alongside, the flat right triangles ABF and ACG are redrawn, with the right angles at B and C respectively. AC is the base diagonal, length 10 centimetres.]
Find a base diagonal first, then use it as one leg of a vertical triangle.

In a cuboid, a vertical edge is perpendicular to the whole horizontal base, so it is perpendicular to a base diagonal meeting its foot. This justifies the right angle at $C$ in triangle $ACG$ . Perspective on a page may distort that angle.

Worked example 1: a face diagonal

Question. In this cuboid, calculate the exact length of face diagonal $AF$ .

Plan. Use triangle $ABF$ , which lies on the front rectangular face and is right-angled at $B$ .

\begin{aligned} AF^2&=AB^2+BF^2 && \text{Pythagoras in the front face}\\ &=8^2+12^2=208\\ AF&=\sqrt{208}=4\sqrt{13} && \text{positive length} \end{aligned}

Answer. $4\sqrt{13}\,\mathrm{cm}$ .

Check. The value is about $14.4\,\mathrm{cm}$ , greater than both $8$ and $12$ , but less than their sum $20$ . Its square is $16(13)=208$ , agreeing with the sum of the two edge squares.

Now change the endpoint from $F$ to the opposite corner $G$ . The new diagonal leaves the front face, so a second triangle is needed.

Worked example 2: a space diagonal

Question. In the same cuboid, calculate the exact length of $AG$ .

Plan. Find the base diagonal $AC$ , then use the vertical triangle $ACG$ .

\begin{aligned} AC^2&=8^2+6^2=100 && \text{right angle in the base}\\ AC&=10\\ AG^2&=AC^2+CG^2 && \text{vertical is perpendicular to the base}\\ &=100+12^2=244\\ AG&=\sqrt{244}=2\sqrt{61} \end{aligned}

Answer. $2\sqrt{61}\,\mathrm{cm}$ , approximately $15.6\,\mathrm{cm}$ .

Check. A different route uses the front diagonal: $AG^2=AF^2+FG^2=208+6^2=244$ . Also, the space diagonal is longer than the face diagonal.

Watch out

The line $AG$ is not the hypotenuse of a triangle with just sides $8$ and $12$ : those lengths describe the front face and lead to $AF$ . Name the three vertices of your triangle before substituting.

Lengths through several planes

A pyramid's sloping edge, face height and vertical height are different lengths. A useful cross-section passes through the apex and the point directly below it; that point establishes the right angle.

Take a pyramid with rectangular base $ABCD$ . Its apex $V$ is directly above the base's centre $O$ . With base $12\,\mathrm{cm}$ by $8\,\mathrm{cm}$ , the centre-to-corner distance is half the base diagonal, not half either base edge:

OA^2=6^2+4^2=52.

This first triangle lies in the base plane. The next, $VOA$ , lies in a vertical plane and is right-angled at $O$ .

[DIAGRAM: asset_name: A pyramid and its perpendicular projection; asset_slug: pyramid_projection; caption: The sloping edge projects onto a centre-to-corner base segment.; description: A rectangular-base pyramid has apex V directly over centre O. Base sides are 12 and 8 centimetres, with half lengths 6 and 4 shown on a base triangle leading from O to corner A. The vertical height VO is perpendicular to the base. Sloping edge VA and the shorter face height VM to the midpoint M of AB are distinguished. A flat section VOA marks the line-plane angle at A between AV and AO.]
The sloping edge projects onto a centre-to-corner base segment.

Worked example 3: vertical height to sloping edge

Question. This pyramid has vertical height $VO=9\,\mathrm{cm}$ . Find the exact length of a sloping edge $VA$ .

Plan. Find the centre-to-corner distance in the base, then use the vertical cross-section.

\begin{aligned} OA^2&=6^2+4^2=52 && \text{half the base diagonal}\\ VA^2&=VO^2+OA^2 && \text{right angle at }O\\ &=9^2+52=133\\ VA&=\sqrt{133} \end{aligned}

Answer. $\sqrt{133}\,\mathrm{cm}$ , approximately $11.5\,\mathrm{cm}$ .

Check. The base diagonal is $\sqrt{12^2+8^2}=\sqrt{208}$ ; half of it is $\sqrt{52}$ , confirming $OA$ . The sloping edge must exceed the vertical height $9\,\mathrm{cm}$ .

Now the sloping edge is known instead of the vertical height. Keep the same base triangle, but subtract its square from the hypotenuse square.

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E6.6 - Lengths and angles in three dimensions

Right triangles inside solids

Lengths through several planes

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