# Chiu Chang Suann Jing/Examples/Example 6

## Example of Problem from Chiu Chang Suann Jing

There is a pool $10$ feet square, with a reed growing vertically in the centre,
its roots at the bottom of the pool, which rises $1$ foot above the surface;
when drawn towards the shore it reaches exactly to the brink of the pool;
what is the depth of the water?

## Solution

The water is $12$ feet deep.

## Proof

Let the depth of the water be $d$.

The length of the reed is $d + 1$.

When drawn to the edge of the pool, the reed forms the hypotenuse of a right triangle.

One of the legs of that right triangle is the depth of the pool, which is $d$.

The other leg is the distance from the centre of the pool, which is $5$ feet.

Hence:

 $\ds \paren {d + 1}^2$ $=$ $\ds d^2 + 5^2$ Pythagoras's Theorem $\ds \leadsto \ \$ $\ds 2 d + 1$ $=$ $\ds 25$ simplification $\ds \leadsto \ \$ $\ds d$ $=$ $\ds 12$ simplification

The right triangle in question here is the $\text{5-12-13}$ triangle.

$\blacksquare$