Space Diagonals: The Pythagorean Theorem in 3D | Grade 8 Math | LevelKite

A box in this lesson is a rectangular prism: a solid whose six faces are all rectangles, like a shipping carton. Two corners are opposite when they do not lie on any common face. A segment joining two opposite corners passes through the inside of the box and is called a space diagonal.

A space diagonal does not lie on any face, so no flat triangle drawn on a face contains it. Its length still matters. For example, it tells you whether a long, thin rod can fit in the box from one corner to the opposite corner. In this lesson you find that length by using the Pythagorean theorem twice.

Label the bottom corners $A$ (front left), $B$ (front right), $C$ (back right) and $D$ (back left). Label the top corners $E$ , $F$ , $G$ and $H$ , directly above them. The box's length is $AB$ , its width is $BC$ and its height is $CG$ , all measured in the same unit. The space diagonal you want is $AG$ .

Two right triangles connect these lengths.

Triangle $ABC$ lies flat on the bottom face. Its right angle is at $B$ , because the bottom face is a rectangle. The side opposite the right angle is called the hypotenuse. Here it is $AC$ , the base diagonal of the bottom face.
Triangle $ACG$ stands upright inside the box. Its right angle is at $C$ . Edge $CG$ rises straight up from the bottom face, so it is perpendicular to (at a right angle with) every segment on the bottom face that starts at $C$ , including $AC$ . Its hypotenuse is $AG$ .

So $AC$ does two jobs. It is the hypotenuse of the flat triangle. It is also a leg (a side that forms the right angle) of the upright one.

Key concept

The space diagonal in two right-triangle steps

The Pythagorean theorem, $a^2 + b^2 = c^2$ , holds in a right triangle with legs $a$ and $b$ and hypotenuse $c$ , the side opposite the right angle. For a box, use it twice:

In triangle $ABC$ , with its right angle at $B$ , find the base diagonal $AC$ from the length $AB$ and the width $BC$ .
In triangle $ACG$ , with its right angle at $C$ , find the space diagonal $AG$ from $AC$ and the height $CG$ .

Side lengths are positive, so take the nonnegative square root each time. Keep a square root such as $\sqrt{13}$ exact until the last step. Do not replace it with a rounded decimal.

AC^2 = AB^2 + BC^2 \qquad AG^2 = AC^2 + CG^2

This box has

AB = 12

cm,

BC = 9

cm and

CG = 8

cm, and the right-angle marks at

B

and

C

show the two triangles used in the example below.

Worked example

The box in the figure above is $12$ cm long, $9$ cm wide and $8$ cm tall, so $AB = 12$ , $BC = 9$ and $CG = 8$ . How long is its space diagonal $AG$ ?

Start with the flat triangle $ABC$ . Its right angle is at $B$ , so its hypotenuse is $AC$ and its legs are $AB = 12$ and $BC = 9$ . Substitute them into the theorem and work out each square.
$AC^2 = AB^2 + BC^2 = 12^2 + 9^2 = 144 + 81 = 225$
$AC$ is a side length, so it is positive. The positive number whose square is $225$ is $15$ . The other square root, $-15$ , cannot be a length.
$AC = \sqrt{225} = 15 \text{ cm}$
Now use the upright triangle $ACG$ . Its right angle is at $C$ , so its hypotenuse is $AG$ . Its legs are the base diagonal $AC = 15$ and the height $CG = 8$ .
$AG^2 = AC^2 + CG^2 = 15^2 + 8^2 = 225 + 64 = 289$
$AG$ is also a side length, so take the nonnegative square root.
$AG = \sqrt{289} = 17 \text{ cm}$
Check the sizes. In a right triangle, the hypotenuse (the side opposite the right angle) is longer than each leg. Here $15$ is longer than $12$ and $9$ , and $17$ is longer than $15$ and $8$ , so both results are reasonable.

Answer

The space diagonal $AG$ of this box is $17$ cm long.

Here

AB = 3

inches,

BC = 2

inches and

CG = 6

inches, so the base diagonal

AC

is not a whole number, but

AG

turns out to be one.

Worked example

The box in the figure above is $3$ inches long, $2$ inches wide and $6$ inches tall. Find the exact length of its space diagonal $AG$ .

In triangle $ABC$ the right angle is at $B$ , so $AC$ is the hypotenuse. Substitute the legs $AB = 3$ and $BC = 2$ .
$AC^2 = AB^2 + BC^2 = 3^2 + 2^2 = 9 + 4 = 13$
No whole number squares to $13$ , so write the nonnegative root exactly as $\sqrt{13}$ . Keep it that way. $(\sqrt{13})^2$ is exactly $13$ , but a rounded value such as $3.6$ squares to $12.96$ . That error would carry into the next step.
$AC = \sqrt{13} \text{ in}$
In triangle $ACG$ the right angle is at $C$ , so $AG$ is the hypotenuse, with legs $AC = \sqrt{13}$ and $CG = 6$ . Squaring $\sqrt{13}$ gives back exactly $13$ .
$AG^2 = AC^2 + CG^2 = (\sqrt{13})^2 + 6^2 = 13 + 36 = 49$
Take the nonnegative square root, since $AG$ is a side length.
$AG = \sqrt{49} = 7 \text{ in}$

Answer

The space diagonal $AG$ is exactly $7$ inches long, even though the base diagonal $AC$ is $\sqrt{13}$ inches.

The shipping box in the next problem has

AB = 16

inches,

BC = 12

inches and

CG = 10

inches, and a rod from corner

A

to corner

G

would lie along the space diagonal.

Worked example

A thin, straight rod is $22$ inches long; treat it as a segment with no thickness. The shipping box in the figure above is $16$ inches long, $12$ inches wide and $10$ inches tall. Can the rod lie flat along the base diagonal $AC$ ? Can it fit along the space diagonal $AG$ ?

Triangle $ABC$ has its right angle at $B$ , so $AC$ is its hypotenuse; substitute the legs $16$ and $12$ . Take the nonnegative root, because $AC$ is a side length. Since $20 < 22$ , the rod is too long to lie flat along $AC$ .
$AC^2 = 16^2 + 12^2 = 256 + 144 = 400, \quad AC = \sqrt{400} = 20 \text{ in}$
Triangle $ACG$ has its right angle at $C$ , so $AG$ is its hypotenuse; substitute the legs $AC = 20$ and $CG = 10$ .
$AG^2 = AC^2 + CG^2 = 20^2 + 10^2 = 400 + 100 = 500$
No whole number squares to $500$ , so the exact length is the nonnegative root $\sqrt{500}$ inches. The calculation is finished, so only now should you round. A calculator gives about $22.36$ inches, to the nearest hundredth.
$AG = \sqrt{500} \approx 22.36 \text{ in}$
Compare the rod with $AG$ . The rod's $22$ inches is less than about $22.36$ inches. So it fits along the space diagonal with roughly $0.36$ inch to spare.
$22 < 22.36$

Answer

The rod is too long to lie flat along the $20$ -inch base diagonal. It fits along the space diagonal, which is exactly $\sqrt{500}$ inches, or about $22.36$ inches, long.

Takeaway: to find a box's space diagonal, use two right triangles that share the base diagonal. First use $AC^2 = AB^2 + BC^2$ , with the right angle at $B$ . Then use $AG^2 = AC^2 + CG^2$ , with the right angle at $C$ . In each triangle the hypotenuse is the side opposite the right angle, and each length is the nonnegative square root. Keep square roots exact until the final step.