Pythagoras and trigonometry in 3D
Introduced| English | Meaning |
|---|---|
| space diagonal/speɪs daɪˈæɡənl/ | space diagonal |
| base diagonal/beɪs daɪˈæɡənl/ | base diagonal |
The diagonal through a room
- How long is the longest pole that fits inside a box? It's the space diagonal 空间对角线 — running from one corner to the far opposite corner.
- 3D Pythagoras and trig reduce this to a flat triangle hidden inside the solid.
The strategy: find a triangle inside
- Every 3D problem becomes a 2D problem: spot a right-angled triangle inside the solid, then use Pythagoras or trig on it.
- Often you find a base diagonal 底面对角线 first, then bring in the height to form a second triangle.
Pythagoras' theorem
In a right-angled triangle a² + b² = c². The same idea, applied twice, gives lengths inside 3-D solids.
A box has a base 6 cm by 8 cm. Find the length of the base diagonal (cm).
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√(6² + 8²) = √100 = 10 cm.
In a 3-D problem you look for a right-angled triangle inside the solid to work with.
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The whole 3-D method is to reduce it to a flat right-angled triangle.
In 3D problems, you often find the ______ diagonal first, then use it with the height.
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The base diagonal (found using 2D Pythagoras) becomes one side of a second right triangle with the height.
Worked example — box diagonal
- A box has base $6\text{ cm} \times 8\text{ cm}$, height $5\text{ cm}$.
- Step 1 — base diagonal: $\sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}$.
- Step 2 — space diagonal: $\sqrt{10^2 + 5^2} = \sqrt{125} = 11.2\text{ cm}$.
A box has dimensions 6, 8 and 5 cm. The space diagonal is √(6² + 8² + 5²). Find it (1 dp).
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√(36 + 64 + 25) = √125 ≈ 11.2 cm.
A cube has side 3 cm. Its space diagonal is √(3² + 3² + 3²) = √27. Find it (2 dp).
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√27 = 3√3 ≈ 5.20 cm.
Worked example — angle to the base
- The base diagonal ($10$) and height ($5$) form a right triangle.
- $\tan\theta = \dfrac{5}{10} = 0.5$, so $\theta = \tan^{-1}(0.5) = 26.6^{\circ}$.
3D Pythagoras shortcut. The space diagonal of a box $l \times w \times h$ is $\sqrt{l^2 + w^2 + h^2}$. For our box: $\sqrt{36 + 64 + 25} = \sqrt{125} = 11.2$ cm.
Two steps, not one. For the angle, you need the base diagonal AND the height to form the second triangle. Using the space diagonal as a side gives the wrong angle.
The base diagonal is 10 cm and the box height is 5 cm. The angle to the base has tan θ = 5/10. Find θ (degrees, 1 dp).
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θ = tan⁻¹(0.5) = 26.6°.
Where 3D trig appears
- Architecture: the angle of a roof rafter, the diagonal of a room.
- Engineering: the length of a support strut, the angle of a crane arm.
3D Pythagoras is just 2D Pythagoras applied twice — first on the base, then with the height.
You've got it
- in 3D, spot a right-angled triangle inside the solid, then use Pythagoras/trig
- often find a base diagonal first, then bring in the height
- box $6 \times 8 \times 5$: base diagonal $10$, space diagonal $\sqrt{125} \approx 11.2$, angle $26.6^{\circ}$