Pythagoras' Theorem in Solids

Find the right-angled triangles inside a cuboid, a cube and a cylinder, and use Pythagoras' theorem twice to find the longest straight line inside a box.

Before you start

A cuboid has six rectangular faces, and its edges meet at right angles. In a cuboid $ABCDEFGH$ , $ABCD$ is the base and $EFGH$ the top, with $E$ above $A$ , $F$ above $B$ , $G$ above $C$ and $H$ above $D$ .
An upright edge, such as $CG$ , is at right angles to every line drawn on the base through its foot $C$ .
The diagonal of a rectangle is the hypotenuse of a right-angled triangle made by two of its sides (lesson 9.6).

Face Diagonals and Space Diagonals

A face diagonal joins opposite corners of one face, such as $AC$ across the base. A space diagonal joins opposite corners of the whole cuboid and runs through its inside, such as $AG$ from a bottom corner to the top corner furthest from it.

The space diagonal is the hypotenuse of a right-angled triangle standing inside the cuboid: $\triangle ACG$ , shaded below. Its right angle is at $C$ , between the base diagonal $AC$ and the upright edge $CG$ .

The space diagonal in two steps

1. On the base: $AC^2 = AB^2 + BC^2$ .

2. In $\triangle ACG$ , with $\angle ACG = 90^{\circ}$ : $AG^2 = AC^2 + CG^2$ .

Carry $AC^2$ from step 1 into step 2 as it is: there is no need to find $AC$ itself, or to round it.

Example 1: A room

A room is a cuboid $ABCDEFGH$ with floor $ABCD$ , $AB = 5$ m, $BC = 4$ m and height $CG = 3$ m. Find (a) the length of the diagonal $AC$ of the floor, (b) the length of a straight wire from the floor corner $A$ to the ceiling corner $G$ .

Solution

(a) The floor is a rectangle, so $\angle ABC = 90^{\circ}$ : $AC^2 = 5^2 + 4^2 = 25 + 16 = 41\text{,}$ and $AC = \sqrt{41} = 6.403\ldots = 6.40$ m (3 s.f.).

(b) $CG$ is upright, so $\angle ACG = 90^{\circ}$ and $AG$ is the hypotenuse of $\triangle ACG$ .

$AG^2 = AC^2 + CG^2 = 41 + 3^2 = 41 + 9 = 50$

$AG = \sqrt{50} = 7.071\ldots = 7.07$ m (3 s.f.)

(a) $AC = 6.40$ m (b) the wire is $7.07$ m long.

Example 2: Naming the triangle

In the cuboid $ABCDEFGH$ , $AB = 8$ cm, $BC = 6$ cm and $CG = 5$ cm. (a) The space diagonal $BH$ is the hypotenuse of a right-angled triangle whose other sides are a diagonal of the base and an edge. Name the triangle and its right angle. (b) Find $BH$ .

Solution

(a) $H$ is above $D$ , so the edge $DH$ is upright. The base diagonal from $B$ that ends at $D$ is $BD$ . The triangle is $\triangle BDH$ , with its right angle at $D$ : $\angle BDH = 90^{\circ}$ .

(b) $AD = BC = 6$ cm, and $\angle BAD = 90^{\circ}$ : $BD^2 = 8^2 + 6^2 = 64 + 36 = 100\text{,}$ so $BD = 10$ cm.

$DH = CG = 5$ cm. $BH^2 = BD^2 + DH^2 = 100 + 25 = 125\text{,}$ so $BH = \sqrt{125} = 11.18\ldots = 11.2$ cm (3 s.f.).

(a) $\triangle BDH$ , with $\angle BDH = 90^{\circ}$ (b) $BH = 11.2$ cm

One Formula for the Space Diagonal

Put the two steps together. For a cuboid of length $l$ , width $w$ and height $h$ , step 1 gives $AC^2 = l^2 + w^2$ , and step 2 gives $AG^2 = AC^2 + h^2$ .

The space diagonal of a cuboid

If $d$ is the length of a space diagonal of a cuboid with edges $l$ , $w$ and $h$ , then $d^2 = l^2 + w^2 + h^2$ .

A cuboid has four space diagonals, all of this length. It is the longest straight line that fits inside the cuboid.

Example 3: A box

Find the length of the longest straight stick that fits inside a box $30$ cm long, $20$ cm wide and $15$ cm high, correct to 1 decimal place.

Solution

The longest stick lies along a space diagonal.

$d^2 = 30^2 + 20^2 + 15^2 = 900 + 400 + 225 = 1525$

$d = \sqrt{1525} = 39.051\ldots = 39.1$ (1 d.p.)

Round once, from the calculator's full value. Writing $39.05$ and then rounding that again is rounding twice, which can give a wrong answer.

The longest stick is $39.1$ cm.

Example 4: Decimals and a cube

(a) A shipping container is a cuboid $12$ m long, $2.4$ m wide and $2.6$ m high. Find the length of a space diagonal. (b) Find the length of a space diagonal of a cube of edge $10$ cm.

Solution

(a) $d^2 = 12^2 + 2.4^2 + 2.6^2 = 144 + 5.76 + 6.76 = 156.52\text{,}$ so $d = \sqrt{156.52} = 12.51\ldots = 12.5$ m (3 s.f.).

(b) All three edges are $10$ cm: $d^2 = 10^2 + 10^2 + 10^2 = 300\text{,}$ so $d = \sqrt{300} = 17.32\ldots = 17.3$ cm (3 s.f.).

(a) $12.5$ m (b) $17.3$ cm

Example 5: A missing edge

A box has a base $12$ cm by $4$ cm, and its space diagonal is $13$ cm. How high is the box?

Solution

Let the height be $h$ cm. $13^2 = 12^2 + 4^2 + h^2$

$169 = 144 + 16 + h^2$ , so $h^2 = 169 - 160 = 9$

$h = \sqrt{9} = 3$

The box is $3$ cm high.

The Longest Rod in a Cylinder

The longest straight rod inside a closed cylinder runs from a point on the rim of the base to the opposite point on the rim of the top, through the axis. It is the diagonal of the rectangle made by a diameter and the height.

Example 6: A rod in a tin

A closed cylindrical tin has a radius of $3$ cm and a height of $8$ cm. Find the length of the longest straight rod that fits inside it.

Solution

The rectangle through the axis has the diameter $PR$ as its base: $PR = 2 \times 3 = 6$ cm. Its height is $RQ = 8$ cm, and $\angle PRQ = 90^{\circ}$ .

$PQ^2 = 6^2 + 8^2 = 36 + 64 = 100\text{,}$ so $PQ = 10$ .

The longest rod is $10$ cm.

Common Mistakes

Stopping at the face diagonal. For a $3$ cm by $4$ cm by $12$ cm box, the base diagonal is $5$ cm, but the space diagonal is $\sqrt{25 + 144} = 13$ cm.
Adding the edges before squaring. $(3 + 4 + 12)^2 = 361$ gives $19$ cm, which is the three edges added. Square each edge first: $9 + 16 + 144 = 169$ , so $13$ cm.
Using the radius in a cylinder. The rod crosses the whole base: use the diameter.
Rounding the face diagonal. Carry $AC^2$ into the second step; a rounded $AC$ can change the final answer.

Practice Questions

Try each question before you open its solution. Name the right-angled triangle you use. Give lengths correct to 3 significant figures where they are not exact.

Question 1

A cuboid is $12$ cm long, $9$ cm wide and $8$ cm high. Find (a) the diagonal of its base, (b) its space diagonal.

ShowHide solution

(a) $\sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15$ cm

(b) $\sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$ cm

(a) $15$ cm (b) $17$ cm

Question 2

In the cuboid $ABCDEFGH$ , $AB = 4$ cm, $BC = 3$ cm and $CG = 12$ cm. (a) Name the right-angled triangle that has $AG$ as its hypotenuse and $AC$ as a side, and its right angle. (b) Find $AG$ .

ShowHide solution

(a) $\triangle ACG$ , with the right angle at $C$ : $\angle ACG = 90^{\circ}$ , since $CG$ is upright.

(b) $AC^2 = 4^2 + 3^2 = 25$ ; $AG^2 = 25 + 12^2 = 25 + 144 = 169\text{,}$ so $AG = 13$ cm.

(a) $\triangle ACG$ , $\angle ACG = 90^{\circ}$ (b) $13$ cm

Question 3

Find the length of a space diagonal of a box $25$ cm by $15$ cm by $10$ cm.

ShowHide solution

$d^2 = 25^2 + 15^2 + 10^2 = 625 + 225 + 100 = 950$

$d = \sqrt{950} = 30.82\ldots = 30.8$ (3 s.f.)

$30.8$ cm

Question 4

A cube has edges of $7$ cm. Find the length of (a) a diagonal of one face, (b) a space diagonal.

ShowHide solution

(a) $\sqrt{7^2 + 7^2} = \sqrt{98} = 9.899\ldots = 9.90$ cm (3 s.f.)

(b) $\sqrt{7^2 + 7^2 + 7^2} = \sqrt{147} = 12.12\ldots = 12.1$ cm (3 s.f.)

(a) $9.90$ cm (b) $12.1$ cm

Question 5

A classroom is $9$ m long, $7$ m wide and $3.5$ m high. Find the length of the longest straight pole that would fit inside it.

ShowHide solution

The longest pole lies along a space diagonal.

$d^2 = 9^2 + 7^2 + 3.5^2 = 81 + 49 + 12.25 = 142.25$

$d = \sqrt{142.25} = 11.92\ldots = 11.9$ (3 s.f.)

$11.9$ m

Question 6

A cuboid has a base $6$ cm by $2$ cm, and its space diagonal is $7$ cm. Find its height.

ShowHide solution

$7^2 = 6^2 + 2^2 + h^2$ , so $49 = 36 + 4 + h^2$ and $h^2 = 9$ .

$h = 3$

$3$ cm

Question 7

For a box $2$ cm long, $6$ cm wide and $9$ cm high, Arjun gives the space diagonal as $6.32$ cm and Lina gives it as $17$ cm. Explain each mistake and find the correct length.

ShowHide solution

Arjun found $\sqrt{2^2 + 6^2} = \sqrt{40} = 6.324\ldots = 6.32$ cm (3 s.f.), the diagonal of the base, and stopped after the first step.

Lina found $2 + 6 + 9 = 17$ : she added the edges instead of their squares.

$d = \sqrt{2^2 + 6^2 + 9^2} = \sqrt{4 + 36 + 81} = \sqrt{121} = 11$ cm.

Arjun stopped at the face diagonal; Lina added the edges. The space diagonal is $11$ cm.

Question 8

A closed cylindrical tin has a radius of $4$ cm and a height of $15$ cm. Can a straight chopstick $16$ cm long fit inside it with the lid on?

ShowHide solution

The longest rod inside runs across a diameter, $2 \times 4 = 8$ cm, and up the height, $15$ cm.

$\sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$ cm

$16$ cm is less than $17$ cm.

Yes: the longest rod that fits is $17$ cm.

Question 9

A cylindrical water tank has a radius of $0.5$ m and a height of $1.2$ m. Find the length of the longest straight pole that can be placed inside it with the lid closed.

ShowHide solution

Diameter $= 2 \times 0.5 = 1$ m.

$d^2 = 1^2 + 1.2^2 = 1 + 1.44 = 2.44\text{,}$ so $d = \sqrt{2.44} = 1.562\ldots = 1.56$ (3 s.f.).

$1.56$ m

Question 10

In the cuboid $ABCDEFGH$ , $AB = 9$ cm, $BC = 7$ cm and $BF = 6$ cm. The space diagonal $DF$ is the hypotenuse of a right-angled triangle whose other sides are a diagonal of the base and an edge. Name the triangle and its right angle, and find $DF$ .

ShowHide solution

$F$ is above $B$ , so $BF$ is upright, and the base diagonal ending at $B$ is $BD$ . The triangle is $\triangle BDF$ , with $\angle DBF = 90^{\circ}$ .

$BD^2 = AB^2 + AD^2 = 9^2 + 7^2 = 81 + 49 = 130$ ( $AD = BC = 7$ cm). Carry $BD^2$ into the next step as it is.

$DF^2 = BD^2 + BF^2 = 130 + 6^2 = 130 + 36 = 166\text{,}$ so $DF = \sqrt{166} = 12.88\ldots = 12.9$ cm (3 s.f.).

$\triangle BDF$ , with $\angle DBF = 90^{\circ}$ ; $DF = 12.9$ cm.

Question 11

A pencil case is a closed box $20$ cm long, $6$ cm wide and $4$ cm high. (a) Can a pencil $21$ cm long lie flat on the bottom of the case? (b) Can it fit inside the closed case at all?

ShowHide solution

(a) The longest line on the bottom is its diagonal: $\sqrt{20^2 + 6^2} = \sqrt{436} = 20.88\ldots$ cm, less than $21$ cm. It cannot lie flat.

(b) The longest line inside is a space diagonal: $\sqrt{436 + 4^2} = \sqrt{452} = 21.26\ldots$ cm, more than $21$ cm.

(a) No ( $20.9$ cm) (b) Yes, slanting along a space diagonal ( $21.3$ cm).

Key Points

A face diagonal crosses one face; a space diagonal runs through the inside, from a corner to the opposite corner.
The space diagonal $AG$ of cuboid $ABCDEFGH$ is the hypotenuse of $\triangle ACG$ , with the right angle at $C$ between the base diagonal $AC$ and the edge $CG$ .
Two steps: find $AC^2$ on the base, then $AG^2 = AC^2 + CG^2$ . In one formula: $d^2 = l^2 + w^2 + h^2$ .
The space diagonal is the longest straight line inside a cuboid; in a cylinder, the longest is $\sqrt{(\text{diameter})^2 + (\text{height})^2}\text{.}$
Square the edges before adding, and round only at the end.