Right triangles inside solids
Choose a triangle by its endpoints and perpendicular sides. The cuboid formula uses three mutually perpendicular edges. Other solids need their own section
- Cuboid Use base edges a and b, vertical height h, base diagonal f and space diagonal d. \(\displaystyle \begin{aligned}f^2&=a^2+b^2\\d^2&=f^2+h^2\\d&=\sqrt{a^2+b^2+h^2}\end{aligned}\)
- Right square pyramid For base side s and vertical height h, the slant edge x reaches a corner. A face slant height m reaches a side midpoint. Here f is the distance from the base centre to a corner \(\displaystyle \begin{aligned}f^2&=2\left(\frac{s}{2}\right)^2\\x^2&=h^2+f^2\\m^2&=h^2+\left(\frac{s}{2}\right)^2\end{aligned}\)
- Exact intermediate values Carry a diagonal's square into the next triangle. Apply the requested rounding only to the final length
Worked example
Find the space diagonal of this cuboid to one decimal place. The three given edges are mutually perpendicular; f is the base diagonal and d joins opposite corners through the solid
\[\begin{gathered}a=3\text{ cm}\\b=7\text{ cm}\\h=10\text{ cm}\end{gathered}\]
- \(\displaystyle f^2=3^2+7^2\)The base diagonal is the hypotenuse in the base rectangle
- \(\displaystyle f^2=9+49\)Square both base lengths
- \(\displaystyle f^2=58\)Keep the exact squared length
- \(\displaystyle d^2=58+10^2\)The base diagonal and the vertical edge are perpendicular
- \(\displaystyle d^2=58+100\)Square the vertical height
- \(\displaystyle d^2=158\)Add the two squared lengths
- \(\displaystyle d=\sqrt{158}\text{ cm}\)Take the positive square root
- \(\displaystyle d\approx12.6\text{ cm}\)Round the final length to one decimal place
Common mistakes
Stopping at the base diagonal
✕\(\displaystyle d=\sqrt{3^2+7^2}\)✓\(\displaystyle d=\sqrt{3^2+7^2+10^2}\)
A cuboid has base edges 3 cm and 7 cm and vertical height 10 cm. Find its space diagonal d. The first expression stops at the base diagonal; the space diagonal also uses the vertical heightSquaring a rounded diagonal
✕\(\displaystyle \begin{aligned}d&=\sqrt{7.2^2+11^2}\\d&\approx13.1\end{aligned}\)✓\(\displaystyle \begin{aligned}d&=\sqrt{52+11^2}\\d&\approx13.2\end{aligned}\)
A cuboid has base edges 4 cm and 6 cm and vertical height 11 cm. Find its space diagonal d in centimetres to one decimal place. Retain the exact squared base diagonal, 52, because squaring its rounded length 7.2 changes the final answerConfusing a pyramid corner with a midpoint
✕\(\displaystyle x^2=h^2+\left(\frac{s}{2}\right)^2\)✓\(\displaystyle \begin{gathered}f^2=2\left(\frac{s}{2}\right)^2\\x^2=h^2+f^2\end{gathered}\)
A right square pyramid has base side s and apex vertically h above the base centre. Find the slant edge x from the apex to a base corner. The horizontal triangle uses two half-sides; one half-side instead reaches a side midpoint for a face slant height. Here f is the centre-to-corner distanceDifficulty levels
introFind a cube's space diagonal with equal edges
standardFind a cuboid's space diagonal using different edge lengths
challengeFind a right square pyramid's slant edge using its centre-to-corner distance
examRecognise the required length from calculations with the same spatial demand
Questions about this topic
Why is the base diagonal perpendicular to the height?+
A cuboid's vertical edge is perpendicular to its base plane, including the base diagonal meeting it
Must I find a decimal for the first diagonal?+
No. Its exact square is enough for the second Pythagoras calculation
Can I add three squares for every solid?+
No. That formula applies to a cuboid's three mutually perpendicular edges. First identify the right triangle in another solid
Can I find a missing height instead?+
Yes. Subtract the two base-edge squares from the space-diagonal square, then take the positive root
Does fitting the diagonal guarantee an object fits?+
No. Internal dimensions, object thickness, clearance and the insertion path can matter
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 10 Mathematics, chapter 9: Pythagoras and trigonometry
In this chapter
9A Finding the hypotenuse: Find a side of a right-angled triangleDetermine whether three sides form a right-angled triangle9B Finding a shorter side: Find a side of a right-angled triangle9C Pythagoras' theorem in 2D and 3D: Find a side of a right-angled trianglePythagoras in three dimensionsName sides and state trigonometric ratios9D Trigonometric ratios: Evaluate a trigonometric ratio9D Trigonometric ratios: Choose a trigonometric ratio9E Unknown side in the numerator: Find an unknown side of a right-angled triangle9E Unknown side in the numerator: Solve a practical right-angled-triangle problem9E Unknown side in the numerator: Evaluate a trigonometric ratio9E Unknown side in the numerator: Choose a trigonometric ratio9F Unknown side in the denominator: Find an unknown side of a right-angled triangle9F Unknown side in the denominator: Solve a practical right-angled-triangle problem9F Unknown side in the denominator: Choose a trigonometric ratioFind an angle of a right-angled triangle from two sides9G Finding angles with trigonometry: Solve a practical right-angled-triangle problem9G Finding angles with trigonometry: Evaluate a trigonometric ratio, and find an angle from its valueAngles of elevation and depressionRead and state three-figure bearingsSolve a navigation problem with bearingsTrigonometry in three dimensions