Year 10AC9M10M03

Pythagoras in Three Dimensions

Solve practical problems applying Pythagoras' theorem and trigonometry of right-angled triangles, including problems involving direction and angles of elevation and depression.

A 3D problem is two 2D problems. Find a right triangle on the base first, then stand a second one on it.

Method

Do it in two stages

Base diagonal first, then the space diagonal.

The base diagonal becomes a side of the second triangle

Keep the first answer exact

d^2 = a^2 + b^2

Carry $d^2$ forward rather than a rounded $d$.

The space diagonal of a box

d = \sqrt{a^2+b^2+c^2}

Both stages combined into one formula.

Worked example

e.g.

\text{A box is } 3 \times 4 \times 12. \text{ Find the length of its longest diagonal.}

Draw the base and find its diagonal. $d_{\text{base}}^2 = 3^2 + 4^2 = 25$
Keep it squared. Do not take the root yet. $d_{\text{base}}^2 = 25$
Stand the second triangle on that diagonal, with the height. $d^2 = 25 + 12^2$
Work it out. $d^2 = 25 + 144 = 169$
Take the root now. $d = 13$
Check with the combined formula. $\sqrt{9+16+144} = \sqrt{169} = 13 \quad\checkmark$

Practice

Base first, then height. Carry the squared value forward.

\text{Box } 2\times3\times6. \text{ Longest diagonal?}

Answer

7

\text{Cube of side } 5. \text{ Space diagonal?}

Answer

5\sqrt{3} \approx 8{,}66

\text{Base diagonal of a } 6\times8 \text{ rectangle?}

Answer

10

\text{Why keep the first answer squared?}

Answer

\text{Rounding early moves the final answer.}

\text{A } 10 \text{ m pole leans in a } 6\times8 \text{ room corner to corner. Does it fit flat?}

Answer

\text{Yes — the floor diagonal is exactly } 10.

\text{How many right triangles does a 3D problem need?}

Answer

\text{Two.}