Pythagoras in three dimensions

In this cuboid, AD (labelled d) joins opposite corners through the interior: it is a space diagonal. Which two segments form the perpendicular sides of a triangle with AD as its hypotenuse? Diagram not to scale

These bars compare lengths for a cuboid with a fixed base diagonal of 15 cm. The last bar adds that diagonal to the vertical height. Start at height zero, then increase the height and compare the three bars

At positive heights, where does the space-diagonal bar lie?

Use two right triangles

In a cuboid, a base diagonal is perpendicular to a vertical edge. Find its square from the two base edges, then use that square in the second triangle. The space diagonal is the second hypotenuse

1. 1Work in the base Let a and b be the base edge lengths and f the base diagonal \(\displaystyle f^2=a^2+b^2\)
2. 2Include the height Let h be the vertical height and d the space diagonal. Keep the first square exact \(\displaystyle \begin{aligned}d^2&=f^2+h^2\\&=a^2+b^2+h^2\end{aligned}\)
3. 3Finish with a length Take the positive square root. Round only the requested final length \(\displaystyle d=\sqrt{a^2+b^2+h^2}\)

Calculate the space diagonal. Give the exact length. Diagram not to scale

Find the square of the base diagonal first. Use that square in the triangle containing the vertical edge

Find the space diagonal to one decimal place. Diagram not to scale

All three perpendicular edge lengths contribute a square. Do not round the base diagonal before continuing. Write the final length to one decimal place, including a zero if needed

Perpendicular edge lengths

\(\displaystyle \begin{gathered}a=4\text{ cm}\\b=6\text{ cm}\\h=11\text{ cm}\end{gathered}\)

A student is finding a cuboid's space diagonal. Which change fixes the method?

A pyramid needs the correct endpoint

In a right square pyramid the apex is vertically above the base centre. A slant edge ends at a base corner. A face slant height ends at a side midpoint. These use different horizontal lengths

1. 1Centre to corner Let s be the base side and f the distance from its centre to a corner. The two perpendicular half-sides form a right triangle \(\displaystyle f^2=\left(\frac{s}{2}\right)^2+\left(\frac{s}{2}\right)^2\)
2. 2Apex to corner Let h be the vertical height and x the slant edge. Use the centre-to-corner square without rounding it \(\displaystyle x^2=f^2+h^2\)
3. 3Centre to midpoint For a face slant height, use only half a base side as the horizontal leg \(\displaystyle m^2=h^2+\left(\frac{s}{2}\right)^2\)

The apex is 9 cm vertically above the square base centre. Find the slant edge to the labelled corner, to one decimal place. Diagram not to scale

The target ends at a corner. The horizontal leg is longer than half a base side

A cone uses its radius

Cut a right circular cone through its apex and the centre of its base. Half of this section is a right triangle. Its slant height is the hypotenuse

1. 1Choose the horizontal leg The leg runs from the centre to the circumference. If the diameter D is given, halve it to obtain the radius r \(\displaystyle r=\frac{D}{2}\)
2. 2Use the vertical height Let h be the perpendicular height and s the slant height \(\displaystyle s^2=r^2+h^2\)

Calculate the cone's slant height. Give the exact length. The axial section is not to scale

The right triangle uses the centre-to-rim distance rather than the full width

The longest cylinder rod spans its diameter

In a closed right circular cylinder, a thin straight rod has its greatest length between opposite bottom and top rim points. The axial section through those points is a rectangle

1. 1Use the whole width The rectangle spans the diameter D. Double a given radius r \(\displaystyle D=2r\)
2. 2Find the rectangle diagonal Let h be the perpendicular cylinder height and L the longest rod length \(\displaystyle L^2=D^2+h^2\)

This axial section contains the longest thin straight rod that fits inside a closed cylinder. Find its length to one decimal place. Diagram not to scale

The base centre divides the rectangle's width into two equal radii

Measure depth from the bottom

A hemispherical bowl is the lower half of a sphere. Its rim plane passes through the sphere centre. A horizontal water surface below the rim forms a circle. The gap below the centre is different from the water depth above the bottom

1. 1Identify the hypotenuse Let R be the sphere radius, r the water-surface radius and q the vertical gap below the centre \(\displaystyle q^2+r^2=R^2\)
2. 2Find the shorter side Subtract the water-radius square from both sides. Take the positive root for the gap \(\displaystyle \begin{aligned}q^2+r^2-r^2&=R^2-r^2\\q^2&=R^2-r^2\\q&=\sqrt{R^2-r^2}\end{aligned}\)
3. 3Find the water depth The centre-to-bottom distance is the sphere radius. Subtract the unrounded gap to obtain the depth h \(\displaystyle h=R-\sqrt{R^2-r^2}\)

The water surface is below the rim of this hemispherical bowl. Find the water depth from the bottom to one decimal place. Diagram not to scale

The sloping radius is the hypotenuse. The vertical gap is a shorter side, not the requested water depth