An ancient geometry problem

Let ${x}$ be the the length that the peacock has to travel. Now using pythagoras' theorem, the length of the distance from the hole (where the snake and peacock will meet) = $sqrt{x^2 - 81}$ . This gives : $x = 27 - \sqrt{x^2 - 81}\Rightarrow\sqrt{x^2 - 81}\ = 27 - x$ $x^2 -81 = 729 - 54x\Rightarrow\ 54x = 729 + 81 = 27(27+3) = 27\times\ 30$ $x= \frac{27\times\ 30}{54}\ = 15\therefore\sqrt{x^2 - 81}\ = \sqrt{225 - 81}\ = \sqrt{144}\ = \boxed{12}$

This problem is from the 12 century Indian math text Lilavati written by Bhaskaracharya or Bhaskara II. Here is an animated solution I posted long back.

$\tan \theta=\dfrac{9}{27}$ $\implies$ $\theta = \tan^{-1}\left(\dfrac{9}{27}\right) \approx 18.435$

$\beta =180-2(18.435) \approx 143.13$

By pythagorean theorem,

$x=\sqrt{27^2+9^2}=\sqrt{810}$

By cosine law,

$(\sqrt{810})^2=y^2+y^2-2(y)(y)(\cos 143.13)$ $\implies$ $y=15$

Finally,

$z=27-15=$ $\boxed{12}$