Right angles are everywhere: walls meet floors, streets cross, edges of boxes meet. Wherever two perpendicular lengths are known, the Pythagorean theorem finds the distance across. When the answer is a square root, estimate it.
Why it works
A ladder against a wall, a diagonal across a screen, a shortcut across a field: each is the hypotenuse of a right triangle whose legs you know (or can measure). In a box, no single triangle reaches from a bottom corner to the opposite top corner, but two do: one across the floor, then one standing up on that floor diagonal.
Key idea
Sketch the right triangle, label the legs and the hypotenuse, then use $a^{2} + b^{2} = c^{2}$. For the space diagonal of a box: first the floor diagonal $d^{2} = l^{2} + w^{2}$, then $D^{2} = d^{2} + h^{2}$.
Worked examples
Worked example: A ladder
A 10-foot ladder leans against a wall with its foot 6 feet from the wall. How high does it reach?
- The ladder is the hypotenuse: $6^{2} + h^{2} = 10^{2}$.
- $h^{2} = 100 - 36 = 64$, so $h = 8$.
Answer: 8 feet
Worked example: A screen
A screen is 16 inches wide and 9 inches tall. How long is its diagonal, to the nearest tenth of an inch?
- $d^{2} = 16^{2} + 9^{2} = 256 + 81 = 337$.
- $18^{2} = 324$ and $19^{2} = 361$, so $d$ is between 18 and 19.
- $18.3^{2} = 334.89$ and $18.4^{2} = 338.56$; 337 is closer to 338.56.
Answer: about 18.4 inches
Worked example: A box
A box is 4 by 3 by 12 inches. How long is the longest rod that fits, from a bottom corner to the opposite top corner?
- Floor diagonal: $d^{2} = 4^{2} + 3^{2} = 25$, so $d = 5$.
- Space diagonal: $D^{2} = 5^{2} + 12^{2} = 25 + 144 = 169$.
- $D = 13$.
Answer: 13 inches
Common mistake
Using the theorem once in a box, or adding the three lengths. The space diagonal is not $4 + 3 + 12$; it takes two right triangles. (A shortcut: $D^{2} = l^{2} + w^{2} + h^{2}$.)