Babylonian Mathematics/Examples/Sliding Ladder

Example of Babylonian Mathematics

A ladder of length $0; 30$ stands upright against a wall.

The upper end slides down a distance $0; 6$.

How far away will the lower end move out from the wall?

All lengths are expressed in Babylonian form.

Solution

The lower end of the ladder will move $0; 18$ away from the wall.

Proof

The ladder will be the hypotenuse of a right triangle whose legs are formed by the wall and the floor.

Let $x$ be the distance of the lower end from the wall after the ladder has finished moving.

We have:

\(\ds 0; 30^2\)

\(=\)

\(\ds \paren {0;30 - 0;6}^2 + x^2\)

Pythagoras's Theorem

\(\ds \leadsto \ \ \)

\(\ds x^2\)

\(=\)

\(\ds \paren {\dfrac {30} {60} }^2 - \paren {\dfrac {30} {60} - \dfrac 6 {60} }^2\)

expressing in conventional fraction form

\(\ds \)

\(=\)

\(\ds \dfrac 1 4 - \dfrac {16} {100}\)

arithmetic

\(\ds \)

\(=\)

\(\ds \dfrac {25 - 16} {100}\)

further arithmetic

\(\ds \)

\(=\)

\(\ds \dfrac 9 {100}\)

further arithmetic

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \dfrac 3 {10}\)

Positive Square Root only required

\(\ds \)

\(=\)

\(\ds \dfrac {18} {60}\)

multiplying top and bottom by $6$

\(\ds \)

\(=\)

\(\ds 0; 18\)

expressing in Babylonian form

$\blacksquare$

Sources

• 1976: Howard Eves: Introduction to the History of Mathematics (4th ed.)
• 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Pythagorean Triples: $14$