Chiu Chang Suann Jing/Examples/Example 6

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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.

$\blacksquare$


Sources